George G. Roussas Introduction to Probability, 2nd Edition (Instructor's Solution Manual)

INSTRUCTORS SOLUTION MANUAL

Introduction to Probability George G. Roussas 2-nd edition, 2014 ANSWERS TO CHAPTER EXERCISES

Chapter 2

C.2.1 4 × 6 × 3 = 72.

C.2.2

× 31 × 42 × 21 × 22 × 11 = (15 × 3) × (6 × 2) × (1 × 1) = 45 × 12 × 1 = 540.

(ii) 63 × 32 × 11 = 20 × 3 × 1 = 60. (i)

6 2

6!

C.2.3 In 1!1!2!1!1! = 6!

2! = 3 × 4 × 5 × 6 = 360 ways.

C.2.4 One selects n people out of 2n in 2n places them to the right side of the chairperson, n ways,

2n and rearranges them in n! ways to get n ×n!. Then one selects n people out of the renaming

n, places them to the left of the chairperson, and rearranges them in n! ways to get nn × n!.

(2n)!

Then the answer is: 2n

n × n! × n! = n!n! × n!n! = (2n)!.

C.2.5

(i) Write {1, 2, . . . , n − 1, n} = {1, 2, . . . , n1 } ∪ {n}. Then, clearly, the nk choices can be

arrived at by choosing k from {1, 2, . . . , n − 1} (in n−1 ways), then choosing k − 1 from k

n−1 the same set {1, 2, . . . , n − 1} (in k−1 ways), and then attaching to these choices of k − 1 numbers the number n to get (k − 1) + 1 = k numbers. (ii) Indeed,

n−1 n−1 (n − 1)! (n − 1)! + = + k k−1 k!(n − k − 1)! (k − 1)!(n − k)! n(n − 1)! k n(n − 1)! n−k = × + × n k!(n − k − 1)!(n − k) n (k − 1)!k(n − k)! n−k n! k n! = × + × n k!(n − k)! n k!(n − k)!

n−k n k n n = + = . n k n k k

C.2.6 Refer to Exercise C.2.5, and at the left-hand side of the expression there replace n successively by i = k, . . . , n to obtain:

i i−1 i−1 = + . k k k−1 1

2

Chapter 2

Summing up over i, as i goes from k to n, we get:

k k+1 n−1 n + + ... + + k k k k

k−1 k k+1 n−1 = + + + ... + + k k k k

k−1 k n−1 + + ... + . k−1 k−1 k−1 After the cancellation of same terms on the left-hand side and right-hard side in the last relation above, we are left with:

n k−1 k n−1 = + + ... + , k k−1 k−1 k−1 (since

k−1 k

= 0), as was to be seen.

Chapter 3

C.3.1 Since the order does not matter, there are 16 ways of selecting 3 numbers from the total of 3

6 10 6 16, and out of these 3 + 2 × 1 produce negative products. The required probability is: × 6 (63)+(10 71 2 ) (1) = 20+45

16 560 = 560 ' 0.127.

(3)

C.3.2 (i) The number of selecting 3 balls is 63 . Next, remove balls numbered 5 and 6, select 2 balls out of the remaining 4, and finally, add to these ball numbered 5. The required (4)×1 6 probability is: 2 6 = 20 = 0.3.

(3) (ii) Here there are 6 × 5 × 4 ways of selecting three balls. Next, remove balls numbered 5 and 6, and select 2 balls from the remaining in 4 × 3 ways. Finally, for each one of these 4 × 3 selections, place ball numbered 5 in 3 possible ways (first, second, third). The required 4×3×3 3 probability is: 6×5×4 = 10 = 0.3, same as in part (i).

C.3.3 Recall that in a triangle, each side is smaller than the sum of the other two. This happens for the following choices of the line segments: 3,5,7; 3,7,9; 5,7,9. Therefore the required 3 probability is 35 = 10 = 0.3.

(3) C.3.4

4 5 3 2 (i) P (C1 ) = 15 ' 0.267, P (C2 ) = 15 ' 0.333, P (C3 ) = 15 = 0.2, P (C4 ) = 15 ' 0.133, 1 P (C5 ) = 15 ' 0.067.

10 9 8 4 ' 0.111, P (F2 ) = 36 ' 0.278, P (F3 ) = 36 = 0.25, P (F4 ) = 36 ' 0.222, (ii) P (F1 ) = 36 5 P (F5 ) = 36 ' 0.139.

20!

C.3.5 (i) 20 6 = 6!×14! = 8 × 15 × 17 × 19 = 38, 760.

(ii) 86 12 0 = 4 × 7 = 28.

8!

12! (iii) 82 12 4 = 2!×6! × 4!×12! = 28 × (5 × 9 × 11) = 28 × 495 = 13, 860.

8 12 (iv) 86 12 0 + 0 6 = 28 + 924 = 952.

(v) 13,860 38,760 ' 0.358.

C.3.6

(i) 0 4 5 4 1 P (of at least one 6) = 1 − P (no 6’s) = 1 − 0 6 4 4 5 =1− ' 1 − 0.48225 = 0.51775.

6 3

4

Chapter 3

(ii) P (of at least one double 6) = 1 − P (no double 6’s) 24 0 24 35 35 24 1 =1− =1− 36 36 36 0 ' 1 − 0.50860 = 0.49140. (iii) The probability in part (i) is larger than that in part (ii).

C.3.7 The number of possible choices for the 3 passengers are: 3 × 3 × 3 = 27. (i) Here the number of outcomes is: 3 × 1 × 1 = 3, so that the required probability is: 3 1 27 = 9 ' 0.1111. (ii) Here the number of outcomes is: 3 × 2 × 1 = 6, so that the required probability is: 6 2 27 = 9 = 0.2222.

Alternatively, if order is taken into consideration, the above three numbers become:

3 2 1 4 1 × 3 × 1 × 3 × 1 × 3 = 3 × 3 × 2 × 3 × 3 = 3 × 2 = 162,

3 2 1 2 1 × 3 × 1 × 1 × 1 × 1 = 3 × 3 × 2 = 3 × 2 = 18,

1 2 3 2 2 1 × 3 × 1 × 2 × 1 × 1 = 3 × 3 × 2 × 2 = 3 × 2 = 36, 18 36 so that the probabilities in part (i) and (ii) are respectively: 162 = 19 and 162 = 92 , the same as before.

C.3.8 (i) 84 = 70.

(ii) One from each class, so that: 21 × 21 × 21 × 21 = 24 = 16.

(iii) The classes to be represented may be: (Fr., So.), (Fr., Jr.), (Fr., Sr.), (So., Jr.), (So., Sr.), (Jr., Sr.).

Then it is necessary that 2 students from each one of the two classes, so

are selected

4 2 2 that the required number is: 2 × 2 × 2 = 6 × 1 × 1 = 6.

8 (iv) For part (ii): 16 70 = 35 ' 0.22857, 6 3 For part(iii): 70 = 35 ' 0.08571.

C.3.9 Without regard to order: Three balls can be selected in 63 = 20 ways, and if the largest number is to be 5, then the other two are to be selected from among the ones numbered 1, 2, 3, 4 in 42 = 6 ways. Thus, the required probability is 6/20 = 3/10 = 0.3.

When order matters: The denominators is 6 × 5 × 4 = 120, and the numerator is 42 × 2! × 3 = 36, so that the probability is 36/120 = 3/10 = 0.3.

R2 3 2 2 2 C.3.10 (i) From 0 2c(2x − x2 )dx = 4c × x2 0 − 2c × x3 0 = 1, we get c = 3/8.

Chapter 3

5

2

1/2

(ii) P (X < 1/2) = 4 × 38 × x2 0

3

1/2

− 2 × 83 × x3 0 x

5 = 32 = 0.15625.

x

2

2c x 2 3 (iii) For 0 < x < 2, F (x) = 4c 2 × y 0 − 3 × y 0 = 4 × (3 − x), so that   for x ≤ 0 0, 2 x F (x) = 4 (3 − x), for 0 < x < 2   1, for x ≥ 2.

C.3.11 Indeed, for y ≥ 0, FY (y) = P (Y ≤ y) = P (|X| ≤ y) = P (−y ≤ X ≤ y) = P (X ≤ y) − P (X ≤ −y) = FX (y) − FX (−y).

Therefore: fY (y) =

d FY (y) = fX (y) − fX (−y)(−1) = fX (y) + fX (−y); dy

i.e., fY (y) = fX (y) + fX (−y).

C.3.12 Indeed, P(

n \

Aci ) = P

i=1

!c #

n [

"

Ai

(by DeMorgan’s laws)

i=1

=1−P

n [

!

Ai

.

i=1

However, by the finite-subadditive property, ! n n [ X P Ai ≤ P (Ai ). i=1

Therefore P(

n \ i=1

Aci ) ≥ 1 −

i=1

n X

P (Ai ).

i=1

T S c c] = 1 − C.3.13 P (A ) = 1 is equivalent toSP (Aci ) = 0, i ≥ 1. Then P ( ∞ [( ∞ i=1 Ai ) = P S i=1 Ai ) S Si∞ c P ( i=1 Aci ). But, clearly, { ni=1 Aci }, n ≥ 1 is nondecreasing and limn→∞ ni=1 Aci S = ∞ i=1 Ai .

S∞ n c c Then it P follows by the theorem cited Ai ) = limn→∞ P ( i=1 Ai ) ≤ S∞ in cthe Hint that P ( Ti=1 n ∞ c limn→∞ i=1 P (Ai ) = 0; i.e., P ( i=1 Ai ) = 0, so that P ( i=1 Ai ) = 1.

C.3.14 We conclude of this problem as follows. On a straight line, consider r + w points to be occupied by the r + w balls. Then the number of selecting the r + w balls is the number of (distinct) arrangements on the line of the r + w balls. This number is (r+w)! r!w! .

6

Chapter 3

Next, that all r red balls are drawn before the w white balls are drawn means that the (r +w)th point is occupied by a white ball, whereas the previous r + w − 1 points are occupied by the r red balls and the remaining w − 1 balls. This can happen in [r + (w − 1)]!/r!(w − 1)! ways.

Therefore the required probability is: (r+w−1)! r!w!(r + w − 1)! w r!(w−1)! = = , (r+w)! r!(w − 1)!(r + w)! r+w r!w!

as to be seen.

C.3.15

(i) The number of ways that k balls are drawn from 2n is: 2n(2n − 1) . . . (2n − k + 1) (since 2n − (k − 1) = 2n − k − 1). In order to determine the number of choices for which no matches occur, we can think as follows: The 1st ball can be chosen in 2n ways. After its selection, we are left with 2n − 1 balls out of which, we remove the ball which would form a match with the ball chosen. So, we are left with 2n − 2 = 2(n − 1) balls to choose the 2-nd one. After the choice of the 2-nd ball and the removal of the one which would form a match with it, we are left with 2n − 2 − 2 = 2(n − 2) balls for the selection of the 3-rd ball. Continuing like this, the number of balls from which the k-th ball is to be chosen is 2n − 2(k − 1) = 2(n − k − 1).

Therefore the number of ways that k balls are selected is: 2n×2(n−1)×. . .×2(n−k+1) = 2k (n−1)...(n−k+1) 2k n(n − 1) . . . (n − k + 1) and the required probability is 2n(2n−1)...(2n−k+1) . k

2 (n−1)...(n−k+1) . (ii) 1 − 2n(2n−1)...(2n−k+1)

(iii) The numerical values for parts (i) and (ii) are 8/95 ' 0.084 and 87/95 ' 0.916, respectively.

Chapter 4

C.4.1

(i) Denoting by A and B the respective events, we have: P [(A ∩ B c ) ∪ (Ac ∩ B)] = P [(A ∪ B) − (A ∩ B)] = P (A ∪ B) − P (A ∩ B) = P (A) + P (B) − P (A ∩ B) = P (A) + P (B) − 2P (A)P (B) = p1 + p2 − 2p1 p2 . (ii) 0.1 + 0.09 − 2 × 0.1 × 0.09 = 0.172.

C.4.2 Consider the following array: 1 • H T T T .. .

2 • H H T T .. .

3 • H H H T .. .

··· ··· ··· ··· ··· ··· .. .

x−2 • H H H H .. .

x−1 • H H H H .. .

x • T T T T .. .

T

T

T

···

T

H

T

(i) That is, when starting with a T , there is only one desirable outcome, which occurs when T is followed by an H. Sliding T by one position at a time, we get x − 2 desirable outcomes. Together with the outcome starting with an H, there are x − 1 outcomes for which X = x, and each of the outcomes has probability 21x . Thus, P (X = x) =

x−1 , 2x

x = 2, 3 . . .

(ii) ∞ X

f (x) =

x=1

=

∞ X x−1 x=2 ∞ X x=2

2x x×

∞

∞

∞

x=2

x=1

x=1

X 1 X 1 1 1 X 1 1 − = x× x − − + x x x 2 2 2 2 2 2

1/2 1/2 = − = 2 − 1 = 1, 2 (1 − 1/2) 1 − 1/2 P P∞ 1 1/2 1/2 1 since ∞ x=1 x × 2x = (1−1/2)2 and x=1 2x = 1−1/2 . Thus, f (x) is, indeed a p.d.f.

7

8

Chapter 4

C.4.3 Indeed, P (B|A) = C.4.4

P (A ∩ B) P (A|B)P (B) P (A)P (B) = > = P (B).

P (A) P (A) P (A)

(i) With the notation introduced: P (Gi ) = P (Gi |G)P (G) + P (Gi |Gc )P (Gc ) = (0.8 × 0.7) + (0.1 × 0.3) = 0.56 + 0.03 = 0.59,

i = 1, 2, 3.

(G) 56 = 0.56 (ii) From part (i), P (G|Gi ) = P (GPi |G)P 0.59 = 59 ' 0.949, i = 1, 2, 3.

(Gi )

(iii) If A is the required event, then: P (A) = P (G1 ∩ G2 ∩ Gc3 ) + P (G1 ∩ Gc2 ∩ G3 ) + P (Gc1 ∩ G2 ∩ G3 ) + P (G1 ∩ G2 ∩ G3 ) = 3 × (0.59)2 × 0.41 + (0.59)3 = 0.428163 + 0.205379 = 0.633542.

C.4.5 If the events A, B, C are independent, then so are the events: Ac , B, C; Ac , B c , C; Ac , B c , C c . Therefore: P (Ac ∩ B ∩ C) = P (Ac )P (B)P (C) = 0.7 × 0.2 × 0.6 = 0.084, P (Ac ∩ B c ∩ C) = P (Ac )P (B c )P (C) = 0.7 × 0.8 × 0.6 = 0.336, P (Ac ∩ B c ∩ C c ) = P (Ac )P (B c )P (C c ) = 0.7 × 0.8 × 0.4 = 0.224.

C.4.6

(i) P (B) = P (B ∩ A1 ) + P (B ∩ A2 ) + P (B ∩ A3 ) = P (B|A1 )P (A1 ) + P (B|A2 )P (A2 ) + P (B|A3 )P (A3 ) = 0.3 × 0.4 + 0.6 × 0.35 + 0.5 × 0.25 = 0.12 + 0.21 + 0.125 = 0.455.

0.12 1 ∩B) (ii) P (A1 |B) = P (A = 120/455 ' 0.264, = 0.455 P (B) 0.21 2 ∩B) P (A2 |B) = P (A = 0.455 = 210/455 ' 0.462, P (B) 3 ∩B) P (A3 |B) = P (A = 0.125 0.455 = 125/455 ' 0.275.

P (B)

C.4.7

(i) P (V ) = 0.10, P (V c ) = 0.90. (ii) (a) P (G|V c ) = 0.08 (b) P (I|V ) = 0.15. (iii) P (G) = P (G ∩ V ) + P (G ∩ V c ) = P (G|V )P (V ) + P (G|V c )P (V c ) = (0.85 × 0.10) + (0.08 × 0.90) = 0.085 + 0.072 = 0.157. )P (V ) 85 (iv) P (V |G) = P (G|V = 0.085 0.157 = 157 ' 0.541.

P (G)

C.4.8

(i) P (R3 |R1 ) = P (R1 ∩ R3 )/P (R1 ), and P (R1 ∩ R3 ) = P (R1 ∩ R2 ∩ R3 ) + P (R1 ∩ R2c ∩ R3 ) = P (R3 |R1 ∩ R2 )P (R2 |R1 )P (R1 ) + P (R3 |R1 ∩ R2c )P (R2c |R1 )P (R1 ) = P (R3 |R2 )P (R2 |R1 )P (R1 ) + +P (R3 |R2c )P (R2c |R1 )P (R1 ) = p21 P (R1 ) + p2 (1 − p1 )P (R1 ) = [p21 + (1 − p1 )p2 ]P (R1 ), so that P (R3 |R1 ) = p21 + (1 − p1 )p2 .

Chapter 4

9

(ii) P (R3 |R1 ) = (0.4)2 + (0.6) × (0.2) = 0.28.

C.4.9

(i) P (Ac ∩ B) P [B − (A ∩ B)] P (B − P (A ∩ B)) = = P (B) P (B) P (B) P (A ∩ B) =1− = 1 − P (A|B).

P (B)

P (Ac |B) =

(ii) Let S = {1, 2, 3, 4, 5} with the uniform probability, and let A = {1, 2}, B = {2, 3, 4}, so that B c = {1, 5}. Then: 1/5 {2} P (A|B c ) = PP (B) = 2/5 = 21 , whereas (2) 1/5 P (A|B) = PP(B) = 3/5 = 1/3, so that 1 − P (A|B) = 2/3 6= 1/2 = P (A|B c ).

C.4.10

(i) P [(A ∪ B) ∩ C] P [(A ∩ C) ∪ (B ∩ C)] = P (C) P (C) P (A ∩ C) + P (B ∩ C) − P (A ∩ B ∩ C) = P (C) P (A ∩ C) P (B ∩ C) P [(A ∩ B) ∪ C] + − = P (C) P (C) P (C) = P (A|C) + P (B|C) − P (A ∩ B|C).

P (A ∪ B|C) =

2/5 = 4/5 = 1/2, P (C|A) = (ii) Here A ∪ B = {1, 2, 3, 4}, so that P (C|A ∪ B) = P [C∩(A∩B)] P (A∪B) P (C∩A) 1/5 P (C∩B) 1/5 1 P (A) = 2/5 = 1/2, P (C|B) = P (B) = 2/5 = 1/2, and therefore P (C|A ∪ B) = 2 6= 1 1

2 + 2 = P (C|A) + P (C|B).

C.4.11

(i) Denoting by S0 , Sa , and M the events that a dinner had soup, salad, and a meat dish, respectively, we have: P (So ) = 0.40, P (Sa ) = 0.60, P (M c |S0 ) = 0.25, P (M c |Sa ) = 0.35.

Then: P (M |S0 )P (S0 ) 0.75 × 0.40 0.30 = = , whereas P (M ) P (M ) P (M ) P (M ) = P (M |S0 )P (S0 ) + P (M |Sa )P (Sa ) = 0.75 × 0.4 + 0.65 × 0.60 = 0.30 + 0.39 = 0.69,

P (S0 |M ) =

30 so that P (S0 |M ) = 0.30 0.69 = 69 ' 0.43478. a )P (Sa ) 39 (ii) P (Sa |M ) = P (MP|S(M = 0.39 0.69 = 69 ' 0.56522. )

C.4.12 Denote by Ci , Di , Hi and Si the events that the i-th card drawn is club, diamond, heart, and spade, respectively, i = 1, 2. Then:

10

Chapter 4

1 (i) P (C2 ) = 13 52 = 4 = 0.25 = (P (D2 ) = P (H2 ) = P (S2 )).

(ii) Here P (C2 ) = P [(C1 ∩ C2 ) ∪ (D1 ∩ C2 ) ∪ (H1 ∩ C2 ) ∪ (S1 ∩ C2 )] = P (C2 |C1 )P (C1 ) + P (C2 |D1 )P (D1 ) + P (C2 |H1 )P (H1 ) + P (C2 |S1 )P (S1 ) 12 13 13 13 13 13 13 13 = × + × + × + × 51 52 51 52 51 52 51 52 13 × 51 13 1 3 × 132 + 12 × 13 = = = = 0.25. = 51 × 52 51 × 52 52 4 (The above would be also true for P (D2 ), P (H2 ) and P (S2 )). (iii) The probabilities are equal. This should not be surprising, because the second drawing takes place without any information regrading the first drawing. Things would 12 be different, if some information was provided. For example: P (C2 |C1 ) = 51 (< 13 52 ), 13 13 P (C2 |D1 ) = 51 (= P (C2 |H1 ) = P (C2 |S1 )) (> 52 ).

C.4.13

(i) D = (A ∩ B ∩ C c ) ∪ (A ∩ B c ∩ C) ∪ (Ac ∩ B ∩ C). (ii) P (D) = P (A∩B ∩C c )+P (A∩B c ∩C)+P (Ac ∩B ∩C). But A∩B ∩C c = (A∩B)∩C c = (A ∩ B) − (A ∩ B ∩ C), so that P (A ∩ B ∩ C c ) = P (A ∩ B) − P (A ∩ B ∩ C), and likewise, P (A ∩ B c ∩ C) = P (A ∩ C) − P (A ∩ B ∩ C), P (Ac ∩ B ∩ C) = P (B ∩ C) − P (A ∩ B ∩ C).

Then: P (D) = P (A ∩ B) + P (A ∩ C) + P (B ∩ C) − 3P (A ∩ B ∩ C). (iii) P (D) = P (A)P (B) + P (A)P (C) + P (B)P (C) − 3P (A)P (B)P (C) = (0.2 × 0.7) + (0.2 × 0.4) + (0.7 × 0.4) − 3 × (0.2 × 0.7 × 0.4) = 0.332.

C.4.14 Denoting by D, “+” and “–” the events of an individual having the disease, and that the test results in a positive and negative diagnosis, respectively, we have: (i) P (+) = P (+|D)P (D) + P (+|Dc )P (Dc ) = (0.95)(0.005) + (0.65)(0.995) = 0.00475 + 0.04975 = 0.0545.

(D) (ii) P (D|+) = P (+|D)P = 0.00475 0.0545 ' 0.0872.

P (+)

(iii) On the average, 1 individual in 200 has the disease, so that (on the average) in 200 individuals, 1 has the disease and 199 do not The test will confirm the disease: Correctly in 1 × 0.95 = 0.95 individuals and Incorrectly in 199 × 0.05 = 9.95 individuals.

Chapter 4

11

Thus, the proportion of correct confirmations is: 0.95 0.95 = ' 0.0872, 0.95 + 9.95 10.9 and that is exactly the probability in part (ii).

C.4.15

(A) (i) P (A|A ∪ B) = P [A∩(A∪B)] = P [A∪(A∩B)] = P P(A∪B) , since A ∪ (A ∩ B) = A. P (A∪B) P (A)

(ii) Immediate from part (i) since P (A ∪ B) = P (A) + P (B). (iii) P (A) (by part (i)) P (A ∪ B) P (A ∩ B) (since A ⊇ A ∩ B) = P (A ∪ B) P (A ∩ B) − (since P (A ∪ B) ⊇ P (B)) P (B) = P (A|B).

P (A|A ∪ B) =

C.4.16

(i) Let A6 and A36 be the events of the numbers 6 and 36 winning, respectively. Then 1 and P (A6 ∩ A36 ) = 0 since A6 ∩ A36 = ∅. Therefore: P (A6 ) = P (A36 ) = 38 (a) P (Two wins) = P (A6 ∩ A36 ) = 0; 2 1 (b) P (At least one win) = P (A6 ∪ A36 ) = P (A6 ) + P (A36 ) − P (A6 ∩ A36 ) = 38 = 19 ' 0.05263; 2 1 (c) P (Exactly one win) = P [(A6 ∩ Ac36 ) ∪ (Ac6 ∩ A36 )] = P (A6 ) + P (A36 ) = 38 = 19 ' 0.05263. (ii) Let B36 be the event that number 36 wins the second time. Then: 1 (a) P (A6 ∩ B36 ) = P (A6 )P (B36 ) (by independence) = 3812 = 1,444 ' 0.00069; 2 25 (b) P (A6 ∪ B36 ) = P (A6 ) + P (B36 ) − P (A6 )P (B36 ) = 38 − 3812 = 1,444 ' 0.05194; c c c c (c) P [(A ∩ B36 ) ∪ (A6 ∩ B36 )] = P (A6 ∩B36 )+P (A6 ∩B36 ) = (1/38)(37/38)+(37/38)(1/38) = 37/722 ' 0.05125.

C.4.17 Denote by Ri , Bi , and Wi the events that the i-th ball is red, blue, and white, respectively.

Then the required probability is given by: def

P [(R1 ∩ R2 ) ∪ (B1 ∩ B2 ) ∪ (W1 ∩ Wc )|W3c ] = p.

We have: P (R1 ∩ R2 ∩ W3c ) = P (W3c |R1 ∩ R2 )P (R2 |R1 )P (R1 ) = [1 − P (W3 |R1 ∩ R2 )]P (R2 |R1 )P (R1 )

2 5 1 4 = 1− × × = , 8 9 10 6

12

Chapter 4

P (B1 ∩ B2 ∩ W3c ) = P (W3c |B1 ∩ B2 )P (B2 |B1 )P (B1 ) = [1 − P (W3 |B1 ∩ B2 )]P (B2 |B1 )P (B1 )

2 3 1 2 × × = , = 1− 8 9 10 20 P (W1 ∩ W2 ∩ W3c ) = P (W3c |W1 ∩ W2 )P (W2 |W1 )P (W1 ) = [1 − P (W3 |W1 ∩ W2 )]P (W2 |W1 )P (W1 ) 1 2 1 = (1 − 0) × × = .

9 10 45 and P (W3c ) = 1 − P (W3 ), where P (W3 ) = P (W1c ∩ W2c ∩ W3 ) + P (W1 ∩ W2c ∩ W3 ) + P (W1c ∩ W2 ∩ W3 ) 7 2 2 8 1 8 2 1 8 × × + × × + × × = 10 9 8 10 9 8 10 9 8 7 1 1 9 1 = + + = = = 0.2.

5×9 5×9 5×9 5×9 5 Thus, the required probability is p=

43 1/6 + 1/20 + 1/45 = ' 0.2986.

1 − 0.2 144

C.4.18 Let W be the event that the ball drawn was white. Then, in obvious notation, we have: P (U1 or U2 |W ) = P (U1 |W ) + P (U2 |W ) 1 [P (W |U1 )P (U1 ) + P (W |U2 )P (U2 )] = P (W ) 1 = [P (W |U1 ) + P (W |U2 )] .

3P (W ) 1 2 However, P (W |U1 ) + P (W |U2 ) = m1m+n + m2m+n , 1 2

P (W ) = P (W |U1 )P (U1 ) + P (W |U2 )P (U2 ) + P (W |U3 )P (U3 )

1 m1 m2 m3 = + + , 3 m1 + n1 m2 + n2 m3 + n3 and

P (W ) =

m1 m2 + m1 + n + 1 m2 + n2

.

C.4.19 The total number of outcomes is 63 . Then:

m1 m2 m3 + + m1 + n1 m2 + n2 m3 + n3

.

Chapter 4

13

= 65 ' 0.833. (i) P (R 6= B) = 6×5×6 63 (ii) P (R 6= B) = P (R < B) + P (R > B) = P (R < B) + P (B < R) = 2 × P (R < B), so that 5 P (R < B) = 12 ' 0.417. = 95 ' 0.054. (iii) P (R 6= B 6= Y ) = 6×5×4 63 (iv) P (R 6= B 6= Y ) = P (R < B < Y ) + P (R < Y < B) + P (B < R < Y ) + P (B < Y < R) + P (Y < R < B) + P (Y < B < R) = 6 × P (R < B < Y ), since, by symmetry, the probabilities on the right-hand side are equal. So, 6P (R < B < Y ) = 59 , and 5 P (R < B < Y ) = 54 ' 0.093. (v) P (R < B < Y |R 6= B) = P [(R < B < Y ) ∩ (R 6= B)] /P (R 6= B) = {P [(R < B < Y ) ∩ (R < B)] + P [(R < B < Y ) ∩ (R > B)]} /P (R 6= B) = P [(R < B < Y ) ∩ (R < B)] /P (R 6= B) = P (R < B < Y )/P (R 6= B) 5 .5 1 = = ' 0.111 (by parts (i) and (iv)).

54 6 9 C.4.20

(i) The required (conditional) probability is, in obvious notation: def

p1 = P [(R1 ∩ R2 )|(R1 ∩ R2 ) ∪ (B1 ∩ B2 )] P (R1 ∩ R2 ) . = P (R1 ∩ R2 ) + P (B1 ∩ B2 ) r(r−1) r r−1 But P (R1 ∩ R2 ) = P (R2 |R1 )P (R1 ) = r+b × r+b−1 = (r+b)(r+b−1) , P (B1 ∩ B2 ) = b(b−1) r(r−1) b−1 b × r+b−1 = (r+b)(r+b−1) P (B2 |B1 )P (B1 ) = r+b . Therefore p1 = r(r−1)+b(b−1) .

(ii) Here the required probability is: def

p2 = P [(R1 ∩ B2 ∩ R3 ) ∪ (B1 ∩ R2 ∩ R3 )] = P (R1 ∩ B2 ∩ R3 ) + P (B1 ∩ R2 ∩ R3 ) = P (R3 |R1 ∩ B2 )P (B2 |R1 )P (R1 ) + P (R3 |B1 ∩ R2 )P (R2 |B1 )P (B1 ) b r−1 b r r−1 r × × + × × , = r+b r+b−1 r+b−2 r+b r+b−1 r+b−2 2r(r−1)b so that p2 = (r+b)(r+b−1)(r+b−2) .

4×3 1 (iii) p1 = (4×3)+(6×5) = 27 ' 0.2857, p2 = 2×4×3×6 10×9×8 = 5 = 0.2.

C.4.21 Denote by B the event that the ball drawn from urn is black, and, for j = 1, . . . , k − 1, let Bj,j+1 , Wj,j+1 stand for the events: Bj,j+1 = “a black ball is transfered from urn Uj to urn Uj+1 ”, Wj,j+1 = “a white ball is transfered from urn Uj to urn Uj+1 ”,

14

Chapter 4

m m n and let pj = P (Wj,j+1 ), qj = 1 − pj = P (Bj,j+1 ). Then: p1 = m+n , q1 = 1 − m+n = m+n .

Next

p2 = P (W2,3 ) = P (W2,3 |W1,2 )P (W1,2 ) + P (W2,3 |B1,2 )P (B1,2 )

m m+1 = × p1 + × q1 m+n+1 m+n+1

m+1 m m m n = + = × × , m+n+1 m+n m+n+1 m+n m+n m n m n so that q2 = 1 − m+n = m+n ; i.e., p2 = p1 = m+n , q2 = q1 = m+n . m m Now, assume that pj = m+n and show that pj+1 = m+n , j = 1, . . . , k − 1. Indeed,

pj+1 = P (Wj,j+1 ) = P (Wj,j+1 |Wj−1,j )P (Wj−1,j ) + P (Wj,j+1 |Bj−1,j )P (Bj−1,j )

m m n m m+1 × + × = , = m+n+1 m+n m+n+1 m+n+1 m+n n , j = 1, . . . , k − 1. Therefore and then qj+1 = m+n

P (B) = P (B|Wk−1,k )P (Wk−1,k ) + P (B|Bk−1,k )P (Bk−1,k ) n n+1 = × pk + × qk m+n+1 m+

n + 1

n m n+1 n = × + × m+n+1 m+n m+n+1 m+n+1 n n = ; i.e., P (B) = . m+n m+n C.4.22 Use the notation: Wj,3 = “a white ball was transfered from urn Uj to urn U3 ”, Bj,3 = “a black ball was transferred from urn Uj to urn U3 ”, j = 1, 2, B = “a black ball was drawn from urn U3 ”. (i) Then: P (B) =P [B ∩ (W1,3 ∩ W2,3 )] + P [B ∩ (W1,3 ∩ B2,3 )] + P [B ∩ (B1,3 ∩ W2,3 )] + P [B ∩ (B1,3 ∩ B2,3 )] =P [B|(W1,3 ∩ W2,3 )] P (W1,3 ∩ W2,3 ) + P [B|W1,3 ∩ B2,3 ] P (W1,3 ∩ B2,3 ) + P [B|B1,3 ∩ W2,3 ] P (B1,3 ∩ W2,3 ) + P [B|B1,3 ∩ B2,3 ] P (B1,3 ∩ B2,3 )

0 m1 m2 1 m1 n2 = × × + × × 2 m1 + n1 m2 + n2 2 m1 + n1 m2 + n2

1 n1 m2 2 n1 n2 + × × + × × 2 m1 + n1 m2 + n2 2 m1 + n1 m2 + n2 m1 n2 + n1 m2 + 2n1 n2 = , 2(m1 + n1 )(m2 + n2 )

Chapter 4

15

since by independence, P (W1,3 ∩W2,3 ) = P (W1,3 )P (W2,3 ), P (W1,3 ∩W2,3 ) = P (W1,3 )P (W2,3 ), P (B1,3 ∩ W2,3 ) = P (B1,3 )P (W2,3 ), P (B1,3 ∩ B2,3 ) = P (B1,3 )P (B2,3 ). (ii) For the numerical values given: P (B) = 33 65 ' 0.50769.

C.4.23

(i) (a) Since R3 = (R1 ∩ R2 ∩ R3 ) ∪ (R1 ∩ W2 ∩ R3 ) ∪ (W1 ∩ R2 ∩ R3 ) ∪ (W1 ∩ W2 ∩ R3 ), we have P (R3 ) = P (R3 |R1 ∩ R2 )P (R2 |R1 )P (R1 ) + P (R3 |R1 ∩ W2 )P (W2 |R1 )P (R1 ) + P (R3 |W1 ∩ R2 )P (R2 |W1 )P (W1 ) + P (R3 |W1 ∩ W2 )P (W2 |W1 )P (W1 ) r3 + 1 r2 + 1 r1 = + × × r3 + w3 + 1 r2 + w2 + 1 r1 + w1 r3 w2 r1 × × + r3 + w3 + 1 r2 + w2 + 1 r1 + w1 r3 + 1 r2 w1 × × + r3 + w3 + 1 r2 + w2 + 1 r1 + w1 r3 w2 + 1 w1 × × r3 + w3 + 1 r2 + w2 + 1 r1 + w1 r1 (r2 + 1)(r3 + 1) + r1 w2 r3 + w1 r2 (r3 + 1) + w1 (w2 + 1)r3 = . (r1 + w1 )(r2 + w2 + 1)(r3 + w3 + 1) 3 ∩R1 ) (b) P (R3 |R1 ) = P (R and P (R1 )

P (R3 ∩ R1 ) = P (R1 ∩ R2 ∩ R3 ) + P (R1 ∩ W2 ∩ R3 ) = P (R3 |R1 ∩ R2 )P (R2 |R1 )P (R1 ) + P (R3 |R1 ∩ W2 )P (W2 |R1 )P (R1 )

r2 + 1 r3 w2 r3 + 1 × + × P (R1 ) = r3 + w3 + 1 r2 + w2 + 1 r3 + w3 + 1 r2 + w2 + 1 (r2 + 1)(r3 + 1) + w2 r3 = P (R1 ), (r2 + w2 + 1)(r3 + w3 + 1) 2 +1)(r3 +1)+w2 r3 so that P (R3 |R1 ) = (r(r2 +w .

2 +1)(r3 +w3 +1) 3 ∩W1 ) (c) P (R3 |W1 ) = P (R and P (W1 )

P (R3 ∩ W1 ) = P (W1 ∩ R2 ∩ R3 ) + P (W1 ∩ W2 ∩ R3 ) = P (R3 |W1 ∩ R2 )P (R2 |W1 )P (W1 ) + P (R3 |W1 ∩ W2 )P (W2 |W1 )P (W1 )

r3 + 1 r2 r3 w2 + 1 = × + × P (W1 ) r3 + w3 + 1 r2 + w2 + 1 r3 + w3 + 1 r2 + w2 + 1 r2 (r3 + 1) + (w2 + 1)r3 = P (W1 ), (r2 + w2 + 1)(r3 + w3 + 1) (r3 +1)+(w2 +1)r3 . so that P (R3 |W1 ) = (rr22+w 2 +1)(r3 +w3 +1)

16

Chapter 4

(ii) Comparing the expressions for P (R3 |W1 ) and P (R3 ), after cancellation, the inequality P (R3 |W1 ) < P (R3 ) is equivalent to 0 < r1 . Likewise, the inequality P (R3 ) < P (R3 |R1 ) is equivalent to 0 < w1 .

C.4.24 Introduce the following notation: For i = 1, 2, let: Ei =“even number appears in the i-th rolling”, Oi =“odd number appears in the i-th rolling”, i =“a white ball was transferred from U to U the i-th time”, W12 1 2 i =“a white ball was transferred from U to U the i-th time”, W21 2 1 i =“a black ball was transferred from U to U the i-th time”, B12 1 2 i =“a black ball was transferred from U to U the i-th time”, B21 2 1 W =“the number of white balls in U2 remains the same”. (i) Then, clearly:

1 2 1 2 1 2 W = E1 ∩ W12 ∩ O2 ∩ W21 ∪ E1 ∩ B12 ∩ E2 ∩ B21 ∪ E1 ∩ B12 ∩ O2 ∩ B21

1 2 1 2 1 2 ∪ O1 ∩ W21 ∩ E2 ∩ W12 ∪ O1 ∩ B12 ∩ E2 ∩ B12 ∪ O1 ∩ B21 ∩ O2 ∩ B21 , and the P (W ) is the sum of the probabilities of the events on the right-hand side above.

However, 1 2 2 1 1 1 P (E1 ∩ W12 ∩ O2 ∩ W21 ) = P (W21 |E1 ∩ W12 ∩ O2 )P (O2 |E1 ∩ W12 )P (W12 |E1 )P (E1 ) 1 m1 m2 + 1 1 = × × × 2 m1 + n1 2 m2 + n2 + 1 (evaluating the probabilities starting from the right to the left),

1 2 2 1 1 1 P (E1 ∩ B12 ∩ E2 ∩ B12 ) = P (B12 |E1 ∩ B12 ∩ E2 )P (E2 |E1 ∩ B12 )P (B12 |E1 )P (E1 ) 1 n1 1 n1 − 1 = × × × , 2 m1 + n1 2 m1 + n1 − 1 1 2 2 1 1 1 P (E1 ∩ B12 ∩ O2 ∩ B21 ) = P (B21 |E1 ∩ B12 ∩ O2 )P (O2 |E1 ∩ B12 )P (B12 |E1 )P (E1 ) n1 1 n2 + 1 1 × × , = × 2 m1 + n1 2 m2 + n2 + 1 1 2 2 1 1 1 P (O1 ∩ W21 ∩ E2 ∩ W12 ) = P (W12 |O1 ∩ W21 ∩ E2 )P (E2 |O1 ∩ W21 )P (W12 |O1 )P (O1 ) m2 1 m1 + 1 1 = × × × , 2 m2 + n2 2 m1 + n1 + 1 1 2 2 1 1 1 P (O1 ∩ B21 ∩ E2 ∩ B12 ) = P (B12 |O1 ∩ B21 ∩ E2 )P (E2 |O1 ∩ B21 )P (B21 |O1 )P (O1 ) 1 n1 1 n1 + 1 = × × × , 2 m1 + n1 2 m1 + n1 + 1

Chapter 4

17

1 2 2 1 1 1 P (O1 ∩ B21 ∩ O2 ∩ B21 ) = P (B21 |O1 ∩ B21 ∩ O2 )P (O2 |O1 ∩ B21 )P (B21 |O1 )P (O1 ) 1 1 n1 n2 − 1 = × × × , and 2 m1 + n1 2 m2 + n2 − 1

m1 (m2 + 1) 1 n1 (n1 − 1) P (W ) = + + 4 (m1 + n1 )(m2 + n2 + 1) (m1 + n1 )(m1 + n1 − 1) n1 (n2 + 1) m2 (m1 + 1) + + (m1 + n1 )(m2 + n2 + 1) (m2 + n2 )(m1 + n1 + 1)

n1 (n1 + 1) n1 (n2 − 1) . + (m1 + n1 )(m1 + n1 + 1) (m1 + n1 )(m2 + n2 − 1) (ii) For the numerical values given, we have: P (W ) =

1 1, 773 1, 773 × = ' 0.336.

4 1320 5, 280

Chapter 5

C.5.1 P (|X − EX| ≤ c) = P (|X| < c) = P (−c ≤ X ≤ c) = P (X = −c) + P (X = c) = 12 + 12 = 1.

The upper bound of Tchebichev’s inequality is: σ 2 (X)/c2 = c2 /c2 = 1.

C.5.2

(i) EY = −7P (X < c) + 4P (X ≥ c) = −7[1 − P (X ≥ c)] + 4P (X ≥ c) = −7 + 11P (X ≥ c) = −7 + 11e−λc = 11e−λc − 7. (ii) EY = 0 or 11e−λc − 7 = 0 gives c = λ1 log(7/11) ' 0.45/λ.

1 3 36 9 4 = 13 , P (face) = 12 C.5.3 Here P (ace) = 52 52 = 13 , and P (anything else) = 52 = 13 . Therefore, 1 3 9 if X is a r.v. taking the values 10, 5, and 0 with respective probabilities 13 , 13 and 13 , its 1 3 9 25 expected value is: EX = 10 × 13 + 5 × 13 + 0 × 13 = 13 ' 1.92. So, you should expected to lose approximately $2 − $1.92 = $0.08.

Rc n+1 c n+1 C.5.4 (i) 0 xn dx = xn+1 0 = cn+1 = 1, or cn+1 = n + 1, or c = (n + 1)1/(n+1) .

Rx n+1 x xn+1 p = p, or xp = [(n + 1)p]1/(n+1) . (ii) 0 p xn dx = p, or xn+1 0 p = p, or n+1

(iii) c = 40.25 ' 1.4142, x0.50 = (4 × 0.50)0.25 = 20.25 ' 1.1892.

C.5.5

(i) Let A=“person afflicted with disease”. Then: P (Dp ) = P (Dp ∩A)+P (Dp ∩Ac ) = P (Dp ∩A) = P (Dp |A)P (A) = 0.6×0.0001 = 0.00006, P (Dc ) = P (Dc |A)P (A) = 0.3 × 0.0001 = 0.00003, P (Dn ) = P (Dn |A)P (A) + P (Dn |Ac )P (Ac ) = 0.1 × 0.0001 + 1 × (1 − 0.0001) = 0.00001 + 0.9999 = 0.99991. (ii) Let X be the r.v. denoting the payments made by the company. Then: x P (X = x)

0 0.99991

20,000 0.00006

50,000 0.00003

C.5.6

Then EX = 20, 000 × 0.00006 + 50, 000 × 0.00003 = 2.70, and the required premium is $2.70.

R1 R1 a b a 2 (i) 0 (ax + b)dx = + b = 1 or a + 2b = 2; 2 0 (ax + bx)dx = 3 + 2 = 0.6 or 2a + 3b = 3.6. ( a + 2b = 2 That is, , so that a = 1.2 and b = 0.4.

2a + 3b = 3.6

C.5.7

(i)

x f (x)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.01 0.04 0.03 0.01 0.01 0.01 0.01 0.06 0.18 0.16 0.20 0.13 0.09 0.03 0.03

The p.d.f. takes on the integer values 1 through 15 with respective probabilities as indicated above, and all other values with probability 0.

18

Chapter 5

19

(ii) Median = 10, since P (X < 10) = 0.36 (< 0.50) and P (X ≤ 10) = 0.52 (> 0.50); mode= 11. (iii) EX = (1 × 0.01) + (2 × 0.04) + (3 × 0.03) + (22 × 0.01) + (8 × 0.06) + (9 × 0.18) + (10 × 0.16) + (11 × 0.2) + (12 × 0.13) + (13 × 0.09) + (29 × 0.03) = 0.01 + 0.08 + 0.09 + 0.22 + 0.48 + 1.62 + 1.6 + 2.2 + 1.56 + 1.17 + 0.87 = 9.9, EX 2 = (1 × 0.01) + (4 × 0.04) + (9 × 0.03) + (126 × 0.01) + (64 × 0.06) + (81 × 0.18) + (100 × 0.16) + (121 × 0.20) + (144 × 0.13) + (169 × 0.09) + (421 × 0.03) = 0.01 + 0.16 + 0.27 + 1.26 + 3.84 + 14.58 + 16 + 24.2 + 18.72 + 15.21 + 12.63 = 106.88, √ so that σ 2 (X) = 106.88 − (9.9)2 = 106.88 − 98.01 = 8.87 and σ(X) = 8.87 ' 2.978.

So, EX = 9.9, σ 2 (X) = 8.87, σ(X) ' 2.978.

C.5.8

(i) σ 2 (Y ) = σ 2 (aX + b) = a2 σ 2 (X) = a2 σ 2 , so that σ(Y ) = σY = |a|σ. (ii) Here σY = 4 × 3 = 12.

C.5.9

R1 3 1 1 (i) From c 0 (1 − x2 )dx = c x 0 − x3 0 = c 1 − 31 = 2c 3 = 1 and c = 3/2.

3 1 1 7 7 (ii) P (X > 1/2) = c x 1/2 − x3 1/2 = c(1 − 1/2 − 1/3 + 1/8) = 32 × 3×8 = 16 ' 0.4375.

3 3 3 3 p p 11 = 3p−p = 16 , or 3p − p3 = (iii) If we call it p, then: c x 0 − x3 0 = c p − p3 = 32 × 3p−p 3 2 11/8. We observe that for p = 1/2, equality obtains. So x11/16 = 1/2.

C.5.10 Let X be the r.v. denoting the number of bananas bought. Then X takes on the values 2n, probabilities P (X = 2n) = (0, 9)n , n = 1, 2, . . .. So, EX = P∞n = 1, 2, . . .n with0.9respective 0.9

n=1 n × (0.9) = (0.1)2 = 0.01 = 90; i.e., EX = 90.

C.5.11 Let {A1 , A2 , A3 } be a partition of an underlying sample space S with P (A1 ) = P (A2 ) = P (A3 ) = 1/3, and let X = −1IA1 + 0IA2 + 3IA3 , Y = −4IA1 + 0IA2 + 2IA3 . Then X ≥ Y , EX = 2/3, EY = −2/3, EX 2 = 10/3, EY 2 = 20/3, so that σ 2 (X) = 10/3 − 4/9 = 26/9, σ 2 (Y ) = 20/3 − 4/9 = 56/9. Thus σ 2 (X) = 26/9 < 56/9 = σ 2 (Y ).

0

0

0

C.5.12 For |x| ≥ 1, |x|r ≤ |x|r ; for |x| < 1, |x|r ≤ 1, so, |x|r ≤ 1 + |x|r for all x. Replacing x by the 0 0 r.v. X, we have: |X|r ≤ 1 + |X|r and therefore E|X|r ≤ 1 + E|X|r < ∞.

C.5.13 Indeed, EX = 1 × P (A) = P (A); EX 2 = 12 × P (A), so that σ 2 (X) = P (A) − [P (A)]2 = P (A) [1 − P (A)] = P (A)P (Ac ).

C.5.14

(i) EX = 0, EX 2 = 1, so that σ 2 (X) = 1 and σ(X) = 1. (ii) MX (t) = EetX = (e−t × 21 ) + (et × 21 ) = 12 (et + e−t ), t ∈ <.

2

d d 1 t −t ) (iii) dt MX (t) t=0 = 12 (et − e−t ) t=0 = 0 = EX, dt = 1 = EX 2 , 2 MX (t) t=0 = 2 (e + e t=0

so that σ 2 (X) = 1 and σ(X) = 1.

20

Chapter 5

C.5.15 Indeed, α EX = c

Z ∞

c α+1

Z ∞

x−α dx x× dx = αc x c c Z ∞ αcα =− dx−α+1 (α 6= 1) α−1 c ∞ αcα αcα =− × x1−α = × c1−α (α > 1) α−1 α−1 c αc = . α−1 α

R∞ ∞ For α < 1, x1−α c = ∞, so that EX = ∞. Also, for α = 1, f (x) = xc2 , and EX = c c dx x = ∞ c log x c = ∞.

C.5.16

Rb Rb Rb Rb (i) EX = a xf (x)dx ≤ a bf (x)dx = b a f (x)dx = b, and likewise, EX ≥ a af (x)dx = a, so that a ≤ EX ≤ b. In the discrete case, replace the integral by a summation. (ii) have seen that σ 2 (X) ≤ E(X − c)2 for any constant c. Take c = a+b 2 to obtain:

a+b 2 a + b 2 (b − a)2 σ (X) ≤ E X − ≤E b− = , 2 2 4 2

a+b because clearly, |x − a+b 2 | ≤ |b − 2 |.

C.5.17

(i) Expanding h(x) around x0 , according to Taylor’s formula, we get: h(x) = h(x0 ) + (x − x0 )h0 (x0 ) +

(x − x0 )2 00 ∗ h (x ), 2!

some x∗ between x and x0 . Hence h(x) ≥ h(x0 ) + (x − x0 )h0 (x0 ),

x ∈ <.

Replace x by X and x0 by EX to get h(X) ≥ h(EX) + (X − EX)h0 (EX).

Now, take expectations on both sides to obtain Eh(X) ≥ h(EX) + (EX − EX)h0 (EX) = h(EX). (ii) For h(x) = xn , we have h0 (x) = nxn−1 and h00 (x) = n(n − 1)xn−2 ≥ 0, x ∈ <. Therefore Eh(X) = EX n ≥ (EX)n . n−1

n

n (iii) For h(x) = x1n , we have h0 (x) = − nxx2n = − xn+1 and h00 (x) = n(n+1)x = n(n+1) ≥ 0, x2n+2 xn+2 n n x ∈ <. Therefore Eh(X) = E (1/X) ≥ (1/EX) = h(EX).

Chapter 5

C.5.18

21

d d (i) We have: dt G(t) = dt

P

∞ j j=0 pj t

=

P∞

d j j=0 dt (pj t ) (where the interchange of the differ-

entiation and the summation sign is allowed here) = P∞ j=0 jpj = EX.

P∞

j−1 , so that d G(t) j=0 jpj t dt

t=1

=

(ii) ∞

X dk

dk G(t) = pj tj k k dt dt =

j=0 ∞ X

j(j − 1) . . . (j − k + 1)pj tj−k ,

j=0

P∞ P∞ dk so that dt k G(t) t=1 = j=k j(j − 1) . . . (j − k + 1)pj = j=0 j(j − 1) . . . (j − k + 1)pj = E [X(X − 1) . . . (X − k + 1)].

C.5.19 EX = EX 2 =

n X k=1 n X k=1

n

k×

1 1 n(n + 1) 1X n+1 k= × = = , n n n 2 2

k2 ×

k=1 n X

1 1 = n n

k=1

k2 =

1 n(n + 1)(2n + 1) (n + 1)(2n + 1) × = , n 6 6 2

(n+1)(n−1) so that σ 2 (X) = EX 2 − (EX)2 = (n+1)(2n+1) − (n+1) = n+1 .

6 4 12 (4n + 2 − 3n − 3) = 12

C.5.20 X is the r.v. denoting the number of tossings required, then X has the Negative Binomial distribution with p = 1/2, r = 10, and its p.d.f. is

x x−1 1 f (x) = , x = 10, 11, . . .

9 2 Then: √ √ 10 (i) EX = pr = 1/2 = 20, σ 2 (X) = 10×1/2 20 = 2 5 ' 4.47.

1/4 = 20, and σ(X) =

P P15 x−1 x−1 x x (ii) P (X > 15) = 1 − P (X ≤ 15) = 1 − 15 x=10 10 9 (0.5) , where 9 (0.5) = 10 11 12 13 14 15 (0.5) +10×(0.5) +55×(0.5) +220×(0.5) +715×(0.5) +2, 003×(0.5) ' 0.151, and therefore P (X > 15) ' 1 − 0.151 = 0.849.

C.5.21

(i) EX = (0 × 12 ) + (3 × 13 ) + (6 × 61 ) = 2, EX 2 = (0 × 12 ) + (32 × 31 ) + (62 × 16 ) = 9, so that √ V ar(X) = EX 2 − (EX)2 = 9 − 4 = 5, and s.d. of X = 5 ' 2.236. (ii) P (|X − EX| < 3) = P (|X − 2| < 3) = P (−3 < X − 2 < 3) = P (−1 < X < 5) = P (X = 0 or X = 3) = 21 + 13 = 56 ' 0.834. (iii) P (|X − EX| < 3) ≥ 1 − V ar(X) = 1 − 59 = 49 ' 0.445.

32

22

C.5.22

Chapter 5

(i)

1/2 1/2 1 x = 1−1/2 = 1/2 = 1.

2

P∞

x=1

(ii) EX =

P∞ 1/2 1/2 1 x 1 x = (1−1/2) = 2 = 1/4 = 2, and E[X(X − 1)] = x=1 x × 2 x=2 x(x − 1) × 2

P∞

2(1/2)2 = 4.

(1−1/2)3

(iii) Since E[X(X − 1)] = EX 2 − EX, we have: EX 2 = 4 + 2 = 6, and therefore: σ 2 (X) = EX 2 − (EX)2 = 6 − 4 = 2.

C.5.23

(i) Indeed, E[X(X − 1)] = E(X 2 − X) = EX 2 − EX, so that EX 2 = E[X(X − 1)] + EX. (ii) Likewise, E[X(X − 1)(X − 2)] = E(X 3 − 3X 2 + 2X) = EX 3 − 3EX 2 + 2EX, so that EX 3 = E[X(X − 1)(X − 2)] + 3EX 2 − 2EX. (iii) Finally, E[X(X − 1)(X − 2)(X − 3)] = E(X 4 − 6X 3 + 11X 2 − 6X) = EX 4 − 6EX 3 + 11EX 2 − 6EX, so that EX 4 = E[X(X − 1)(X − 2)(X − 3)] + 6EX 3 − 11EX 2 + 6EX.

C.5.24

d (i) f (x) = dx F (x) = 3x2 − 2x + 1 for 0 < x < 1, and 0 for x < 0 and x ≥ 1.

(ii) Z 1 EX =

x × (3x2 − 2x + 1)dx = 3 ×

0

EX 2 =

Z 1

x4 1 x3 1 x2 1 7 −2× + = , 4 0 3 0 2 0 12

x2 × (3x2 − 2x + 1)dx = 3 ×

0

x5 1 x4 1 x3 1 13 −2× + = , 5 0 4 0 3 0 30

13 49 67 so that σ 2 (X) = EX 2 − (EX)2 = 30 − 144 = 720 .

1/2

R 1/2 2 3 (iii) P X > 12 = 1−P (X ≤ 12 ) = 1− 0 (3x2 −2x+1)dx = 1− 3 × x3 − 2 × x2 + x = 0

1 − 18 − 41 + 12 = 58 = 0.625.

C.5.25 Indeed, for the continuous case (the discrete case is treated similarly),

Z ∞ X −µ 3 x−µ 3 = f (x)dx γ1 = E σ σ −∞

Z 0 Z ∞ x−µ 3 x−µ 3 = f (x)dx + f (x)dx. σ σ −∞ 0

Next, look at the first integral on the right-hand side above and set: x−µ σ = y, hence x = µ − σy, dx = −σdy, y : ∞, −µ/σ to get: Z 0

x−µ σ

−∞ Z −µ/σ

=σ ∞

3

Z −µ/σ f (x)dx =

(−y)3 f (µ − σy)(−σdy)

∞

y 3 f (µ − σy)dy = −σ

Z ∞ −µ/σ

y 3 f (µ − σy)dy = −σ

Z ∞ −µ/σ

y 3 f (µ + σy)dy

Chapter 5

23

(since f (µ − σy) = f (µ + σy), y ∈ <). Furthermore, setting x−µ σ = y, hence x = µ + σy, dx = σdy, y : −µ/σ, ∞, in the second integral, we get:

Z ∞ Z ∞ Z ∞ x−µ 3 y 3 f (µ + σy)dy, y 3 f (x + σy)σdy = σ f (x)dx = σ −µ/σ −µ/σ 0 so that γ1 = 0.

C.5.26 We have Z Z f (x)dx = nk−δ f (x)dx nk−δ P (|X| ≥ n) = nk−δ (|x|≥n) (|x|≥n) Z Z |x|k f (x)dx (since |x| ≥ 1) |x|k−δ f (x)dx ≤ ≤ (|x|≥n) (|x|≥n) Z ∞ Z Z k k k = |x| f (x)dx − |x| f (x)dx = E|X| − |x|k f (x)dx; (∗) −∞

However,

R

(|x|<n)

k −→ (|x|<n) |x| f (x)dx n→∞

(|x|<n)

R∞

k k −∞ |x| f (x)dx = E|X| (as is intuitively clear; its rigorous

justification requires what is known as “Dominated Convergence Theorem”). Then, the expression on the right-hand side in (∗) −→ 0, hence nk−δ P (|X| ≥ n) −→ 0. n→∞

C.5.27

n→∞

≤ y) = P (1 − X 2 ≤ y) = P (X 2 ≥ 1 − y) = P (X ≥

(i) For 0 < y < 1, FY (y) = P (Y √ √ √ 1 − y) = 1 − P (X ≤ 1 − y) = 1 − FY ( 1 − y). so that p p 1 1 d FY (y) = −f ( 1 − y) × √ (−1) = √ f ( 1 − y) fY (y) = dy 2 1−y 2 1−y 1 3p 3p = √ × 3(1 − y) = 1 − y; i.e., fY (y) = 1 − y, 0 < y < 1.

2 2 2 1−y Rb √ R √1−a √ (ii) P (a < Y < b) = a 23 1 − ydy = 32 √1−b x × 2xdx (by setting x = 1 − y ) = √

x3

√

1−a 1−b

= (1−a)3/2 −(1−b)3/2 , and therefore P (1/3 < Y < 2/3) = (2/3)3/2 −(1/3)3/2 '

0.352.

R1 √ R1 R1 (iii) EY = 32 0 y 1 − ydy = 32 0 (1−x2 )2 x×2xdx = 3 0 x2 (1−x2 )dx = 3(1/3−1/5) = 2/5, R1 √ R1 R1 R1 EY 2 = 23 0 y 2 1 − ydy = 23 0 (1 − x2 )2 x × 2xdx = 3 0 x2 (1 − x2 )2 dx = 3 0 (x2 − 2x4 + 8 8 4 12 x6 )dx = 3(1/3 − 2 × 1/5 + 1/2) = 35 , so that σ 2 (Y ) = 35 − 25 = 175 ; i.e., EY = 25 , 12 σ 2 (Y ) = 175 .

C.5.28 We have: (i) Z ∞ 2

Z ∞ f (x)dx =

−∞

−|x−µ|

e

Z µ dx =

−∞

e −∞ Z µ

= −∞

−|x−µ|

Z ∞

dx + e−|x−µ| dx µ Z ∞ x−µ e dx + e−(x−µ) dx. µ

24

Chapter 5

In the first integral, set x − µ = −y, and in the second integral, set x − µ = y. Then, we continue thus: Z 0

−y

−

e

Z ∞ e

dy +

−∞

−y

Z ∞ dy = 2

e−y dy = 2 × 1 = 2,

0

0

so that f (x) is a p.d.f. (ii) As in part (i), Z ∞ xe

2EX =

−|x−µ|

Z ∞ (µ − y)e

dx =

−∞

−y

Z ∞ dy +

Z ∞ = 2µ

(µ + y)e−y dy

0

0

e−y dy + 0 = (2µ) × 1 = 2µ,

0

so that EX = µ. (iii) Again as in part (i), 2

Z ∞

Z ∞

(µ − y) e dy + (µ + y)2 e−y dy 0 Z ∞ = 2µ2 + 2 y 2 e−y dy + 0 Z0 ∞ 1 3−1 −y 2 = 2µ + 4 y e dy = 2µ2 + 4 × 1 = 2µ2 + 4, Γ(3) 0

2EX =

2 −y

0

so that EX 2 = µ2 + 2. Then σ 2 (X) = 2.

C.5.29 σ E|X| = π

Z ∞

|x|dx

−∞

σ 2 + (x − µ)2

Z |x|dx σ ∞ |x|dx + 2 2 2 π 0 σ + (x − µ)2 −∞ σ + (x − µ) Z 0 Z σ xdx σ ∞ xdx =− + 2 2 2 π −∞ σ + (x − µ) π 0 σ + (x − µ)2 Z 0 Z ∞ 1 xdx 1 xdx =− 2 + . x−µ σπ −∞ 1 + σπ 0 1 + x−µ 2 σ σ σ = π

Z 0

In the first integral on the right-hand side, set x−µ σ = −y, so that x = µ − σy, dx = −σdy, y : ∞, −µ/σ; and in the second integral on the same side, set x−µ σ = y, so that x = µ + σy,

Chapter 5

25

dx = σdy, y : −µ/σ, ∞. Then we continue thus: Z −µ/σ Z ∞ 1 (µ − σy)(−σ)dy 1 (µ + σy)σdy − + 2 σπ ∞ 1+y σπ −µ/σ 1 + y2 Z Z Z Z σ −µ/σ ydy µ ∞ dy σ ∞ ddy µ −µ/σ dy − + + = π ∞ 1 + y2 π ∞ 1 + y 2 π −µ/σ 1 + y 2 π −µ/σ 1 + y 2 Z Z Z µ ∞ dy dy ydy µ ∞ 2σ ∞ =− + + 2 2 π −µ/σ 1 + y π −µ/σ 1 + y π −µ/σ 1 + y 2 Z ∞ Z ∞ ∞ ydy d(1 + y 2 ) σ σ 2σ = ∞, = = log(1 + y 2 ) = 2 2 π −µ/σ 1 + y π −µ/σ 1 + y π −µ/σ R∞

dy −µ/σ 1+y 2

is finite, since, for y = tan θ, we have: 1 + y 2 = cos12 θ , dy = cosdθ2 θ with R∞ R b 1/ cos2 θ b dy a ≤ θ ≤ b, say. Then −µ/σ 1+y

2 = a 1/ cos2 θ dθ = θ a = b − a. Thus, E|X| = ∞.

and

C.5.30

(i) EX =

∞ X

jP (X = j) = lim

n→∞

j=1

n X

jP (X = j) = lim

n→∞

j=1

n X

jpj

j=1

= lim {(1 × p1 ) + (2 × p2 ) + (3 × p3 ) + (4 × p4 ) + . . . + [(n − 1) × pn−1 ] + (n × pn )} n→∞

= lim p1 + p2 + p3 + p4 + . . . + pn−1 + pn + n→∞

p2 + p3 + p4 + . . . + pn−1 + pn + p3 + p4 + . . . + pn−1 + pn + + p4 + . . . + pn−1 + pn + ...

...

... + pn−1 + pn + pn

= lim [P (X ≥ 1) + P (X ≥ 2) + P (X ≥ 3) + P (X ≥ 4) + . . . + P (X ≥ n − 1) + P (X ≥ n)] n→∞

= lim

n→∞

n X j=1

P (X ≥ j) =

∞ X

P (X ≥ j).

j=1

(ii) EX ≥ P (X ≥ 1) + P (X ≥ 2) = P (X ≥ 1) + P (X ≥ 1) − P (X = 1) = 2P (X ≥ 1) − P (X = 1) = 2 − P (X = 1) (since P (X ≥ 1) = 1). (iii) EX ≥ P (X ≥ 1) + P (X ≥ 2) + P (X = 3) = P (X ≥ 1) + [P (X ≥ 1) − P (X = 1)] + [P (X ≥ 1) − P (X = 1) − P (X = 2)] = 3P (X ≥ 1) − 2P (X = 1) − P (X = 2) = 3 − 2P (X = 1) − P (X = 2).

26

C.5.31

Chapter 5

(i) Working as in part (i) and (ii) of Exercise 87, we have:

3

Z ∞

Z ∞

Z ∞

dy + (µ + y)3 e−y dy 0 Z ∞ = (µ3 − 3µ2 y + 3µy 2 − y 3 )e−y dy + (µ3 + 3µ2 y + 3µy 2 + y 3 )e−y dy 0 0 Z ∞ Z ∞ y 2 e−y dy + 0 e−y dy + 6µ = 2µ3

2EX =

3 −|x−µ|

x e

3 −y

(µ − y) e

dx =

Z−∞ ∞

0

0

0 3

3

= 2µ + (6µ) × 2 = 2µ + 12µ, so that EX 3 = µ3 + 6µ. (ii) Z ∞

4

2EX =

4 −|x−µ|

x e

Z ∞ dx =

4 −y

(µ − y) e

Z ∞ dy +

(µ + y)4 e−y dy

0 0 Z−∞ ∞ = (µ4 − 4µ3 y + 6µ2 y 2 − 4µy 3 + y 4 )e−y dy+ 0 Z ∞ (µ4 + 4µ3 y + 6µ2 y 2 + 4µy 3 + y 4 )e−y dy 0 Z ∞ Z ∞ 2 −y 4 2 y e dy + 2 y 4 e−y dy = 2µ + 12µ 0 Z ∞0 1 5−1 = 2µ4 + (12µ2 ) × 2 + 2Γ(5) y dy Γ(5) 0 = 2µ4 + 24µ2 + 2 × 24,

so that EX 4 = µ4 + 12µ2 + 24. (iii) Since σ 2 = 2 (so that σ 4 = 4), we have:

4E

X −µ σ

4

= E(X − µ)4 = E(X 4 − 4µEX 3 + 6µ2 EX 2 − 4µ3 EX + µ4 ) = µ4 + 12µ2 + 24 − 4µ(6µ + µ3 ) + 6µ2 (µ2 + 2) − 4µ3 × µ + µ4 = 24,

so that γ2 = 24 4 − 3 = 3 > 0.

Chapter 5

C.5.32

27

(i) For c > 0, Z c

Z 0 xdF (x) =

Z c xdF (x) +

−c

−c

0 =xF (x) −c −

Z 0

−c Z 0

=cF (−c) −

xdF (x) 0

c F (x)dx + xF (x) 0 −

F (x)dx 0

Z c

F (x)dx + cF (c) − −c Z 0

F (x)dx 0

Z c F (x)dx − c + c

F (x)dx + cF (c) −

=cF (−c) −

Z c

−c Z 0

0

Z c F (x)dx − c[1 − F (c)] +

=cF (−c) − −c

[1 − F (x)]dx.

(∗)

0

However, R∞ R∞ R∞ c[1 − F (c)] = c c dF (x) ≤ c xdF (x) → 0 as c → ∞ since 0 xF (x) < ∞; also R −c R −c R −c R −c cF (−c) = c −∞ dF (x) = −∞ cdF (x) ≤ −∞ |x|dF (x) → 0 as c → ∞, since −∞ |x|dF (x) = R −c − −∞ xdF (x) ∈ <.

Taking the limit in (∗), as c → ∞, we get: Z ∞

Z c lim

n→∞ −c Z ∞

xdF (x) = EX = 0 −

xdF (x) = −∞ Z 0

[1 − F (x)]dx −

=

Z ∞

Z 0 F (x)dx − 0 + −∞

[1 − F (x)]dx 0

F (x)dx. −∞

0

R0 (ii) Observe that −∞ F (x)dx represents the area under the curve y = F (x), x ≤ 0, and R∞ 0 [1 − F (x)]dx represents the area above the curve y = 1 − F (x), x > 0.

F (x) 1 − F (x)

F (x) 0

x

C.5.33

(i) EX =

R∞

−∞ xf (x)dx =

R∞ −∞ xf (x)dx + c xf (x)dx.

Rc

28

Chapter 5

Set x − c = y, so that x = c + y, dx = dy, y : −∞, 0 for −∞ < x ≤ c and y : 0, ∞ for c ≤ x < ∞. Therefore, we continue thus: Z ∞ Z 0 (c + y)f (c + y)dy (c + y)f (c + y)dy + −∞

0

Z ∞

Z 0 −∞ Z ∞

−∞ Z 0

0 Z ∞

yf (c + y)dy

yf (c + y)dy +

f (c + y)dy +

f (c + y)dy + c

=c

Z ∞

Z 0

0

f (c + y)dy + yf (c + y)dy + (−t)f (c − t)(−dt) 0 ∞ Z ∞ Z ∞ yf (c − y)dy yf (c + y)dy − f (y)dy + =c 0 0 −∞ Z ∞ Z ∞ =c + yf (c + y)dy − yf (c − y)dy = c. =c

(by setting y = −t)

Z−∞ ∞

0

0

(ii) 2n+1

Z ∞

2n+1

Z 0

2n+1

EX = x f (x)dx = x f (x)dx + −∞ −∞ Z ∞ Z ∞ =− y 2n+1 f (−y)dy + x2n+1 f (x)dx Z ∞ 0 Z ∞0 2n+1 x f (x)dx − x2n+1 f (−x)dx = Z0 ∞ Z0 ∞ = x2n+1 f (x)dx − x2n+1 f (x)dx = 0.

0

0

Z ∞ 0

x2n+1 f (x)dx

Chapter 6

C.6.1

X>n) (i) P (X > n + k|X > n) = P (X>n+k, = PP(X>n+k) P (X>n) (X>n) =

P∞ x−1 x=n+k+1 pq P ∞ x−1 pq x=n+1

n+k

=

p× q1−q n

q p× 1−q

=

q n+k k qn = q .

1 5 1 = 32 = 0.03125.

2

(ii) P (X > n + 5|X > n) = C.6.2

(i) From P (X < x) = P

X−µ x−20 σ < 4

= P Z < x−20 4

= Φ

x−20 4

= 0.95, so that

x−20 = 1.645 (by linear interpolation), and hence x = 26.58; i.e., x = 26.58 minutes = 4

26 minutes and 34.8 seconds. (ii) Depart on 9 hours 33 minutes and 25.2 seconds, since (33 minutes and 25.2 seconds) + (26 minutes and 34.8 seconds) = 59 minutes and 60 seconds = 60 minutes after 9:00 am which is 10:00 am.

C.6.3

0

−λy . (i) For P (Y > y) = P (0 events occurred in the time interval (0, y]) = e−λy (λy) 0! = e

(ii) By differentiation, −fY (y) = −λe−λy , or fY (y) = λe−λy , y > 0. (iii) fY is the Negative Exponential p.d.f. with λ = 1.

C.6.4 Both roots x1 and x2 of 2x2 + 5Y x + (Y + 0.1) = 0 are real, if and only if (5Y )2 − 4 × 2 × (Y + 0.1) ≥ 0 or 25Y 2 − 8Y − 0.8 ≥ 0. The roots of 25Y 2 − 8Y − 0.8 = 0 are given by: ( √ Y1 = −0.08 8 ± 64 + 4 × 25 × 0.8 8 ± 12 = = 50 50 Y2 = 0.4.

Then 25Y 2 − 8Y − 0.8 = 25(Y + 0.08)(Y − 0.4) ≥ 0 if and only if Y ≤ −0.08 or Y ≥ 0.4.

Thus, both roots of 2x2 + 5Y x + (Y + 0.1) = 0 are real if and only if Y ≤ −0.08 or Y ≥ 0.4, and the probability of this happening is: P (Y ≤ −0.08) + P (Y ≥ 0.4) = P (Y ≥ 0.4) = P (0.4 ≤ Y ≤ 1) = 0.6.

C.6.5

−10 ≤ = P (Z ≤ −2) = 1 − P (Z ≤ 2) = 1 − Φ(2) = (i) P (X < 0) = P (X ≤ 0) = P X−µ σ 5 1 − 0.977250 = 0.02275.

5−10 (ii) P (X > 5) = P X−µ > = P (Z > −1) = 1 − P (Z ≤ −1) = 1 − Φ(−1) = Φ(1) = σ 5 0.841345.

X−µ 5−10 (iii) P (−5 < X < 5) = P −5−10 < < = P (−3 < Z < −1) = P (Z ≤ −1) − 5 σ 5 P (Z ≤ −3) = Φ(−1) − Φ(−3) = Φ(3) − Φ(1) = 0.998650 − 0.841345 = 0.157305.

29

30

Chapter 6

15−10 (iv) P (X > 15) = P X−µ > = P (Z > 1) = 1−P (Z ≤ 1) = 1−0.841345 = 0.158655. σ 5

Rc Rc c C.6.6 (i) P (X ≤ c) = 0 λe−λx dx = − 0 de−λx = −e−λx 0 = −e−λc + 1 = p, or e−λc = 1 − p, or λ = − log(1 − p)/c. (ii) EX = 1/λ = −c/ log(1 − p), σ 2 (X) = 1/λ2 = c2 /[log(1 − p)]2 , σ(X) = −c/ log(1 − p), −2,000λ = e2,000 log(1−p)/c = (1 − p)2,000/c .

P (X > 2, 000) = −e−λx |∞ 2,000 = e (iii) λ = − log(0.25)/1, 000 ' 1.386/1, 000 = 0.001386, EX = −1, 000/ log(0.25) ' 1, 000/1.386 ' 721.5, σ 2 (X) = (1, 000)2 /[log(0.25)]2 ' 1, 000, 000/1.92 ' 520, 833.3, σ(X) = −1, 000/ log(0.25) ' 721.5, P (X > 2, 000) = (0.25)2 = 0.0625.

C.6.7

C.6.8

80 2 (i) EX = 40, σ 2 (X) = 2 × 40 = 80, E(X/n) = 40 40 = 1, σ (X/n) = 1,600 = 0.05.

X

X 0.05 (ii) P X n − 1 ≤ 0.5 = P n − E( n ) ≤ 0.5 ≥ 1 − 0.0025 = 0.80.

(iii) P X n − 1 ≤ 0.5 = P (20 ≤ X ≤ 60) = P (X ≤ 60) − P (X ≤ 20) ' 0.977269 − 0.002574 (by linear interpolation) = 0.974695 (as opposed to 0.80).

(i) By the Hint, n

Z ∞

−λx

1 dx = n λ

Z ∞

x × λe y n e−y dy 0 Z ∞ Γ(n + 1) 1 n! y (n+1)−1 e−y dy = = n = n. λ 0 λn λ

EX =

n

0

(ii) EX = λ1 , EX 2 = λ22 , EX 3 = λ63 . (iii) As always, E(X − µ)3 = EX 3 − 3µEX 2 + 3µ2 EX − µ3 = λ63 − λ3 × λ22 + λ32 × λ1 − λ13 = λ23 , 3

2/λ and since σ 2 = λ12 , which implies σ 3 = λ13 , we have γ1 = 1/λ 3 = 2.

C.6.9

(i) The p.d.f. of X is the Hypergeometric as follows: m w P (X = x) =

x

r−x

m+w r

,

x = 0, 1, . . . , r,

wr and EX = m+w (see relations (6.10) and (6.11) in Chapter 6.)

(ii) Here EX = 4×3 10 = 1.2, and P (X = x) =

4 x

6 3−x

10 3

,

x = 0, 1, 2, 3.

6 6 4 Since 10 3 = 120, 3 = 20, 2 = 15, 2 = 6, the above probabilities are: 20 60 36 P (X = 0) = 120 = 1/6 ' 0.16667, P (X = 1) = 120 = 1/2 = 0.5, P (X = 2) = 120 = 3 4 1 ' 0.3, P (X = 3) = = ' 0.03333.

10 120 30

Chapter 6

C.6.10

31

k (i) Here P [X > k(n − X)] = P (X > kn − kX) = P [(k + 1)X > kn] = P (X > k+1 n) = k 1 − P (X ≤ k+1 n).

(ii) Let p = 0.5, n = 10. Then for k = 1, the probability is: 1 − P (X ≤ 5) = 1 − 0.6230 = 0.377; and for k = 1/4, the probability is: 1 − P (X ≤ 2) = 1 − 0.0547 = 0.9453. For p = 0.25, n = 25, and k = 1/4, the probability is 1 − P (X ≤ 5) = 1 − 0.3783 = 0.6217.

C.6.11

R8 R8 1 2 8 1 3 8 2 (i) EX = 18 0 xdx = 16 x 0 = 4, EX 2 = 18 0 x2 dx = 24 x 0 = 64 3 , so that σ (X) = 64 16 3 − 16 = 3 . (ii) P (|X − 4| > 3) = 1 − P (|X − 4| ≤ 3) = 1 − P (−3 ≤ X − 4 ≤ 3) = 1 − P (1 < X < 7) = 1 1 − 7−1 8 = 4 = 0.25.

16 (iii) P (|X − 4| > 3) = P (|X − EX| > 3) ≤ 16/3 9 = 27 ' 0.59.

C.6.12

(i) E

Y n

= n1 EY = n1 × nλ = λ, σ 2

Y n

= n12 σ 2 (Y ) = n12 × nλ = nλ .

σ2 ( Y ) (ii) P Yn − λ ≤ c = P Yn − E Yn ≤ c ≥ 1− c2n = 1− ncλ2 . Then suffices 1− ncλ2 ≥ p, λ 2 or n ≥ (1−p)c 2 . The required n is the smallest n which is ≥ λ/(1 − p)c .

200,000 2 2 λ (iii) (1−p)c 2 = 0.1×(0.25)2 = 0.1×0.0625 = 625 = 320. Thus, n = 320 pages.

C.6.13

2

(β−α) 8 16 16 64 2 2 (i) EX = β−α = 64 2 = 2 = 4, σ (X) = 12 12 = 3 , so that EX = 3 + 16 = 3 .

(ii) P (|X − 4| > 3) = 1 − P (|X − 4| ≤ 3) = 1 − P (1 ≤ X ≤ 7) = 1 − 68 = 14 = 0.25.

2

16 (iii) P (|X − 4| > 3) = P (|X − EX| > 3) ≤ σ (X) = 16/3 9 9 = 27 ' 0.59.

C.6.14

(i)

P (X > Y ) =

=

=

n X y=0 n X y=0 n X y=0

P [(X > Y ) ∩ (Y = y)] =

n X

P (X > y, Y = y)

y=0

P (X > y)P (Y = y) =

n X

[1 − P (X ≤ y)]P (Y = y)

y=0

P (Y = y) −

n X y=0

P (X ≤ y)P (Y = y) = 1 −

n X y=0

P (X ≤ y)P (Y = y).

32

Chapter 6

(ii) P (X > Y ) = 1 −

5 X

P (X ≤ y)P (Y = y)

y=0

= 1 − [P (X = 0)P (Y = 0) + P (X ≤ 1)P (Y = 1) + P (X ≤ 2)P (Y = 2) + P (X ≤ 3)P (Y = 3) + P (X ≤ 4)P (Y = 4) + P (X ≤ 5)P (Y = 5)] = 1 − [(0.0312 × 0.0954) + (0.1875 × 0.2861) + (0.5000 × 0.3433) + (0.8125 × 0.2060) + (0.9687 × 0.0618) + (1 × 0.016)] ' 1 − 0.4715 = 0.5285.

C.6.15

(i) In maximizing (with respect to r) the V ar(X), it suffices to maximize g(r) = r(m+n−r). d d2 From dr g(r) = 0, we get r = m+n 2 , which is the maximizer, as dr2 g(r) = −2(< 0) for all r. m mn (ii) For r = m+n

2 , we get: EX = 2 , V ar(X) = 4(m+n−1) .

C.6.16 Set g(p) = np(1 − p) = np − np2 , so that dg(p) dp = n − 2np = 0 has the unique root p = 1/2.

2

d Furthermore, dp 2 g(p) = −2n(< 0) for all p, so that p = 1/2 maximizes g(p) and its maximum value is g(1/2) = n/4. Hence σ 2 (X) ≤ n/4. That σ 2 (X) = n/4 only for p = 1/2 follows from the fact that there is only one maximizer of g(p).

3 3 = 0. Then E X−µ C.6.17 (i) If Z = X−µ , then Z ∼ N (0, 1) and EZ = 0, or E(X − µ)3 = 0, σ σ

or EX 3 − 3µEX 2 + 3µ2 EX − µ3 = 0, or EX 3 − 3µ(σ 2 + µ2 ) + 3µ3 − µ3 = 0, so that EX 3 = −2µ3 + 3µ(σ 2 + µ2 ) = −2µ3 + 3µσ 2 + 3µ3 = µ3 + 3µσ 2 ; i.e., EX 3 = µ3 + 3µσ 2 . (ii) EX 3 = 1 + 12 = 13.

C.6.18

(i) Here EX = p < 1 = EX, and the answer is no! (ii) Here EX = λ > 0 and possibly 1, and the answer is yes!

P P∞ q q x x (iii) Here ∞ x=0 x × pq = p x=1 xq = p × (1−q)2 = p and this = 1 if and only if p = 1/2.

So, the answer is yes!

2 (iv) Here EX = p1 +2p2 = 1, EX 2 = p1 +4p2 = 13 +1 = 10 9 . So, we have: p0 +p1 +p2 = 1, 10 1 1 p1 + 2p2 = 1, p1 + 4p2 = 9 , from which we obtain: p0 = 18 , p1 = 16

18 , p2 = 18 .

C.6.19 E[X(X − 1) . . . (X − k + 1)] =

∞ X

x(x − 1) . . . (x − k + 1)e−λ

x=k

=e−λ =e

−λ

∞ X x=k ∞ X x=k

x(x − 1) . . . (x − k + 1)

λx x!

λx x(x − 1) . . . (x − k + 1) × (x − k)!

∞

∞

x=k

y=0

X λx−k X λy λx = λk e−λ = λk e−λ = λk e−λ eλ = λk . (x − k)! (x − k)! y!

Chapter 6

33

C.6.20 We have: ∞ X

P (X > k) = P (X ≥ k + 1) =

pq

x−1

x=k+1

=p

∞ X

qy = p ×

y=k

qk qk =p× = qk ; 1−q p

i.e., P (X > k) = q k (k ≥ 1, integer). Next, P (X > x + y|X > x) = as was to be seen.

C.6.21

P (X > x + y, X > x) P (X > x + y) q x+y = = x = q y = P (X > y), P (X > x) P (X > x) q

1 xα−1 e−x , x > 0, so that (i) Here f (x) = Γ(α) Z ∞ Z ∞ 1 1 EX 3 = x3 × xα−1 e−x dx = x(α+3)−1 e−x dx Γ(α) 0 Γ(α) 0 Z ∞ 1 Γ(α + 3) = × x(α+3)−1 e−x dx Γ(α) Γ(α + 3) 0 Γ(α + 3) ×1 = Γ(α) Γ(α + 3) (α + 2)(α + 1)αΓ(α) = = = α(α + 1)(α + 2). Γ(α) Γ(α)

(ii) E(X −α)3 = EX 3 −3αEX 2 +3α2 EX −α3 , where EX 3 = α(α+1)(α+2) = α3 +3α2 +2α (by part (i)), EX 2 = σ 2 (X) + (EX)2 = α + α2 , EX = α, so that E(X − α)3 = α3 + 3α2 + 2α − 3α(α + α2 ) + 3α3 − α3 = 3α2 + 2α − 3α2 − 3α3 + 3α3 = 2α, as was to be seen.

C.6.22 Indeed, E [X(X − 1) . . . (X − k + 1)] =

n X x=0

= = = =

n X x=k n X x=k n X x=k n X x=k

n x n−x x(x − 1) . . . (x − k + 1) p q x

n x n−x x(x − 1) . . . (x − k + 1) p q x x(x − 1) . . . (x − k + 1)

n! px q n−x x!(n − x)!

x(x − 1) . . . (x − k + 1)

n! px q n−x x(x − 1) . . . (x − k + 1) × (x − k) . . . 2 × 1 × (n − x)! n

X (n − k)!(n − k + 1) . . . (n − 1)n n! px q n−x = px q n−x ; (x − k)!(n − x)! (x − k)!(n − x)! x=k

34

Chapter 6

Set n − x = y, so that x = n − y, x − k = (n − k) − y, y: 0, n − k. Then, we continue thus: n−k X

(n − 1) . . . (n − k + 1)pk

y=0 k

=n(n − 1) . . . (n − k + 1)p

(n − k)! q y p(n−k)−y y![(n − k) − y]!

n−k X y=0

n − k y (n−k)−y q p y

k

=n(n − 1) . . . (n − k + 1)p × 1 = n(n − 1) . . . (n − k + 1)pk .

C.6.23

(i) First, E[X(X − 1)(X − 2)] =

∞ X

x(x − 1)(x − 2) × e−λ

x=3

= e−λ

∞ X

λx x! ∞

x(x − 1)(x − 2) ×

x=3 ∞ X λy 3 −λ

=λ e

y=0

y!

X λx−3 λ3 × λx−3 = λ3 e−λ (x − 3)!(x − 2)(x − 1)x (x − 3)! x=3

= λ3 e−λ eλ = λ3 ;

i.e., E[X(X − 1)(X − 2)] = λ3 . Hence E(X 3 − 3X 2 + 2X) = λ3 , so that EX 3 = λ3 + 3EX 2 − 2EX = λ3 + 3(λ + λ2 ) − 2λ = λ3 + 3λ2 + λ; i.e., EX 3 = λ3 + 3λ2 + λ. Then: E(X − λ)3 = EX 3 − 3λEX 2 + 3λ2 EX − λ3 = λ3 + 3λ2 + λ − 3λ(λ + λ2 ) + 3λ3 − λ3 = λ; i.e., E(X − λ)3 = λ. λ (ii) γ1 = σλ3 = λ√ = λ

√

λ λ .

C.6.24 We have:

γ2 = E

X −µ σ

4 −3=

1 E(X − µ)4 − 3, where µ = EX, and σ 2 = σ 2 (X). σ4

But E(X − µ)4 = EX 4 − 4µEX 3 + 6µ2 EX 2 − 4µ3 EX + µ4 = EX 4 − 4µEX 3 + 6µ2 EX 2 − 3µ4 , where µ=

β 3 − α3 β 4 − α4 β 5 − α5 (α − β)2 α+β , EX 2 = , EX 3 = , EX 4 = , and σ 2 = .

2 3(β − α) 4(β − α) 5(β − α) 12

Therefore: β 5 − α5 (α + β)(β 4 − α4 ) (α + β)2 (β 3 − α3 ) 3(α + β)4 − + − 5 2 2 16

1 5 5 4 4 2 3 3 4 = 16(β − α ) − 40(α + β)(β − α ) + 40(α + β) (β − α ) − 15(α + β) (β − α) 80 (β − α)E(X − µ)4 =

Chapter 6

35

or

80(β − α)E(X − µ)4 =16β 5 − 16α5 − 40αβ 4 + 40α5 − 40β 5 + 40α4 β + 40α2 β 3 − 40α5 + 80αβ 4 − 80α4 β + 40β 5 − 40α3 β 2 − 15α4 β − 15α5 − 60α3 β 2 + 60α4 β − 90α2 β 3 + 90α3 β 2 − 60αβ 4 + 60α2 β 3 − 15β 5 + 15αβ 4 =β 5 − 5β 4 α + 10β 3 α2 − 10β 2 α3 + 5βα4 − α5 = (β − α)5 .

4

5

4

4

(α−β) (β−α) 144 = (β−α) = (α−β) and σ14 E(X − µ)4 = (α−β) = 144 Hence E(X − µ)4 = 80(β−α) 4 × 80 80 80 80 = 1.8, so that γ2 = 1.8 − 3 = −1.2 < 0, as was to be seen.

C.6.25

(i) For α = β = 1: Γ(α)Γ(β) = Γ(1)Γ(1) = 1, Γ(1 + 1) = Γ(2) = 1 × Γ(1) = 1, so that f (x) = 1 for 0 < x < 1 (and 0 otherwise) which is the p.d.f. of U (0, 1). (ii) For α = β = 2: Γ(α)Γ(β) = Γ(2)Γ(2) = 1, Γ(α + β) = Γ(4) = 3! = 6, and f (x) = d 6x(1 − x) for 0 < x < 1. Then log f (x) = log 6 + log x + log(1 − x) and dx log f (x) = 1 1 1−2x d2 1 1 x − 1−x = x(1−x) = 0 if and only if x = 1/2, and dx2 f (x) = − x2 − (1−x)2 < 0 for all x and therefore for x = 1/2. It follows that x = 1/2 is the unique maximizing value of f (x); i.e., x = 1/2 is the unique mode of f (x) = 6x(1 − x), 0 < x < 1.

(iii) For α, β > 1, we have log xα−1 (1 − x)β−1 = (α − 1) log x + (β − 1) log(1 − x), so that its derivative is: α−1 β−1 (α − 1) − (α + β − 2)x − = , x 1−x x(1 − x)

α−1 . Since the second derivative of the above expression which is = 0 if and only if x = α+β−2 is:

−

α−1 β−1 − < 0 for all x, 2 x (1 − x)2

α−1 it follows that α+β−2 is the unique mode of the p.d.f.

36

Chapter 6

C.6.26 Indeed, Z ∞ Z ∞ e−λ 1 n−1 −x x e dx = xn−1 e−(x−λ) dx P (X > λ) = Γ(n) λ Γ(n) λ Z ∞ e−λ = [(x − λ) + λ]n−1 e−(x−λ) dx Γ(n) λ Z ∞ e−λ = (λ + y)n−1 e−y dy (by setting x − λ = y) Γ(n) 0 # Z ∞ "n−1 X n − 1 e−λ = λx y n−x−1 e−y dy Γ(n) 0 x x=0

Z ∞ n−1 e−λ X x n − 1 = λ y (n−x)−1 e−y dy Γ(n) x 0 x=0

Z ∞ n−1 1 e−λ X x n − 1 λ × Γ(n − x) × y (n−x)−1 e−y dy = Γ(n) x Γ(n − x) 0 x=0

n−1 e−λ X x n − 1 λ Γ(n − x) × 1 = x Γ(n) =

=

i.e., P (X > λ) =

x=0 n−1 −λ X e

Γ(n)

λx

(n − 1)! × (n − x − 1)! x!(n − x − 1)!

x=0 n−1 X

(n − 1)! Γ(n)

n−1

e−λ

x=0

λx (n − 1)! X −λ λx = e ; x! (n − 1)! x! x=0

Pn−1 −λ λk x=0 e x! .

Remark: If X ∼ Γ(α, β), then computations of probabilities P (X > λ) involves the so-called incomplete Gamma Tables. However, if X ∼ Γ(n, 1), then such probabilities can be read out of the Poisson Tables (provided, of course, λ is included in the tables!).

C.6.27

P Pn n j n−j (i) Here pj = nj pj q n−j ≥ 0 and ∞ = 1, since j=0 pj = j=0 j p q Therefore: G(t) =

=

∞ X

pj t =

j=0 n X j=0

i.e., G(t) = (pt + q)n .

j

n X n j=0

j

j n−j

p q

tj

n (pt)j q n−j = (pt + q)n ; j

n j

= 0 for j ≥ n+1.

Chapter 6

37

j

In this case, pj = e−λ λj! > 0 and G(t) =

∞ X j=0

= e−λ

−λ λj = 1. Therefore: j=0 e j!

P∞

∞ j X −λ λ tj pj t = e j! j

j=0

∞ X j=0

(λt)j = e−λ × eλt = eλ(t−1) ; j!

i.e., G(t) = eλ(t−1) . (ii) For X ∼ B(n, p), E[X(X − 1) . . . (X − k + 1)] dk (pt + q)n = n(n − 1) . . . (n − k + 1)pk (pt + q)n−k dtk t=1 t=1 k = n(n − 1) . . . (n − k + 1)p . =

k

d λ(t−1) For X ∼ P (λ), E[X(X − 1) . . . (X − k + 1)] = dt ke

C.6.28

(i) We have: P (Y = M ) = P (X ≥ M ) = M −1 p × q p = q M −1 . For y < M , the r.v. y) = pq y−1 . Alternatively,

P∞

x=M pq

t=1

x−1 = p

= λk eλ(t−1)

P∞

x=M q

t=1

= λk .

x−1 = p × q M −1 1−q

=P (Y = y|X < M )P (X < M ) + P (Y = y|X ≥ M )P (X ≥ M ) =P (Y = y|X < M )P (X < M ) + 0 =P (X = y|X < M )P (X < M ) =P (X = y, X < M ) = P (X = y) = pq y−1 .

∞ X y=1

∞ X y=M

yq

y−1

yq

PM −1

y−1

y=1

y × pq y−1 = M q M −1 + p

∞ X d

=

Y is equal to X, so that P (Y = y) = P (X =

P (Y = y) = P (Y = y, X < M ) + P (Y = y, X ≥ M )

(ii) EY = M × q M −1 +

PM −1 y=1

yq y−1 . Next,

∞

d X y = q = q dq dq y=1 y=1

d q 1 − q − q(−1) 1 = = = , dq 1 − q (1 − q)2 (1 − q)2 y

whereas,

∞ ∞ X d y d X y = q = q dq dq y=M y=M M q M q M −1 (1 − q) + q M d M q M −1 (1 − q) − q M (−1) = = = , dq 1 − q (1 − q)2 (1 − q)2

38

Chapter 6

so that

PM −1 y=1

Mq 1 yq y−1 = (1−q) 2 −

M −1 (1−q)+q M

(1−q)2

M −1

(1−q)−q = 1−M q (1−q) 2

M

and

1 − M q M −1 (1 − q) − q M (1 − q)2

1 − M q M −1 (1 − q) − q M 1 − qM 1 − qM = M q M −1 + = . = 1−q 1−q p

EY = M × q M −1 + (1 − q) ×

C.6.29

(i) Here eλ = e−λ =

∞ X λk k=0 ∞ X

k!

k!

X λk k∈E

k∈E

(−λ)k

k=0

=

X λk

=

k!

−

=

k!

X λk def. = SE + SO , k!

k∈O

X (−λ)k

k∈E X λk k∈O

+

k!

k!

+

X (−λ)k k∈O

k!

= SE − SO ;

i.e.; SE + SO = eλ , SE − SO = e−λ , so that SE = 12 eλ + e−λ , SO = 12 eλ − e−λ .

Therefore: P (X ∈ E) =

X

e−λ

λk e−λ (eλ + e−λ ) 1 + e−2λ = e−λ SE = = , k!

2 2

e−λ

e−λ (eλ − e−λ ) 1 − e−2λ λk = e−λ SO = = . k!

2 2

k∈E

P (X ∈ O) =

X k∈O

(ii) For λ = 5, e2λ = e10 ' 25, and P (E) = 0.52 and P (O) = 0.48.

C.6.30 Indeed, 1 EX = β−α n

C.6.31

Z β α

xn dx =

β 1 β n+1 − αn+1 × xn+1 = . (n + 1)(β − α) (n + 1)(β − α) α

(i) So, we have got to compute E(X−np)3 (= E(X−µ)3 ). But E(X−µ)3 = EX 3 −3µEX 2 + 3µ2 EX − µ3 . But, E[X(X − 1)] = E(X 2 − X) = EX 2 − EX = EX 2 − np = n(n − 1)p2 and hence EX 2 = n2 p2 + np(1 − p) = n2 p2 − np2 + np.

Chapter 6

39

Next, E[X(X − 1)(X − 2)] =

n X x=3

=

n X

x(x − 1)(x − 2)

x=3 n X

n x n−x x(x − 1)(x − 2) p q x

n! px q n−x x!(n − x)!

(n − 3)!(n − 2)(n − 1)n × p3 px−3 q (n−3)−(x−3) (x − 3)!(x − 2)(x − 1)x × [(n − 3) − (x − 3)]! x=3

n X n − 3 x−3 (n−3)−(x−3) 3 = n(n − 1)(n − 2)p p q x−3 =

x(x − 1)(x − 2)

x=3

= n(n − 1)(n − 2)p3 × 1 = n(n − 1)(n − 2)p3 ; i.e., E[X(X −1)(X −2)] = n(n−1)(n−2)p3 . Hence E(X 3 −3X 2 +2X) = n(n−1)(n−2)p3 and EX 3 = n(n − 1)(n − 2)p3 + 3EX 2 − 2EX = n(n − 1)(n − 2)p3 + 3n2 p2 − 3np2 + 3np02np = (n3 − 3n2 + 2n)p3 + (3n2 − 3n)p2 + np.

Therefore: E(X − µ)3 = EX 3 − 3µEX 2 + 3µ2 EX − µ3 = EX 3 − 3µEX 2 + 2µ3 = EX 3 − 3np(n2 p2 − np2 + np) + 2n3 p3 = EX 3 − (n3 − 3n2 )p3 − 3n2 p2 = (n3 − 3n2 + 2n)p3 − (n3 − 3n2 )p3 + (3n2 − 3n − 3n2 )p2 + np = 2np3 − 3np2 + np = np(2p2 − 3p + 1) = np × 2(p − 1)(p − 1/2) = np(p − 1)(2p − 1) = −np(1 − p)(2p − 1) = −npq(2p − 1).

√ √ √ Since σ 3 = σ 2 × σ = npq npq, then or γ1 = −npq(2p − 1) npq npq = −(2p − 1)/ npq, √ or γ1 = −(2p − 1) npq/npq. (ii) From the expression of γ1 , we have p = 1/2 implies γ1 = 0.

Remark: For p < 1/2, we have γ1 > 0, so that the distribution is skewed to the right, and it is skewed to the left for p > 1/2.

C.6.32

(i) We have: EX =

r X x=0

m n x r−x

m+n r

x×

=

r X x=1

m n x r−x

m+n r

x×

.

40

Chapter 6

But

m x × m! x × m! m! x× = = = x x!(m − x)! x(x − 1)!(m − x)! (x − 1)!(m − x)!

m−x m(m − 1)! =m , = (x − 1)![(m − 1) − (x − 1)]! x−1 and

m+n (m + n)(m + n − 1)! m+n m+n−1 = , = r r(r − 1)![(m + n − 1) − (r − 1)]! r r−1

so that r

mr X EX = m+n x=1

r−1 mr X = m+n y=0

m−1 x−1

n (r−1)−(x−1)

m+n−1 r−1

n m−1 mr mr (r−1)−y y

= ×1= . m+n−1 m+n m+n r−1

(ii) Working as in part (i), we have: E[X(X − 1)] =

r X x=0

m n x r−x

m+n r

x(x − 1) ×

=

r X

m n x r−x

m+n r

x(x − 1) ×

x=2

,

whereas

m x(x − 1) × m(m − 1)(m − 2)! m−2 x(x − 1) × = = m(m − 1) , and x x(x − 1)(x − 2)![(m − 2) − (x − 2)]! x−2

m+n (m + n)(m + n − 1)(m + n − 2)! = r r(r − 1)(r − 2)![(m + n − 2) − (r − 2)]!

(m + n)(m + n − 1) m+n−2 × , = r(r − 1) r−2 so that r

m(m − 1)r(r − 1) X E[X(X − 1)] = (m + n)(m + n − 1) x=2

r−2 m(m − 1)r(r − 1) X = (m + n)(m + n − 1) y=0

=

m−2 x−2

n (r−2)−(x−2)

m+n−2 r−2

m−2 n y (r−2)−y

m+n−2 r−2

m(m − 1)r(r − 1) m(m − 1)r(r − 1) ×1= . (m + n)(m + n − 1) (m + n)(m + n − 1)

Chapter 6

41

Next, from E[X(X − 1)] = E(X 2 − X) = EX 2 − EX, we get m(m − 1)r(r − 1) mr + (m + n)(m + n − 1) m + n m(m − 1)r(r − 1) + mr(m + n − 1) mr(mr + n − r) = = . (m + n)(m + n − 1) (m + n)(m + n − 1)

EX 2 = E[X(X − 1)] + EX =

(iii) From part (i) and (ii), we get: 2

mr(mr + n − r) mr − (m + n)(m + n − 1) m+n 2 2 mr(mr + n − r)(m + n) − m r (m + n − 1) = (m + n)2 (m + n − 1) mnr(m + n − r) mr(mn + n2 − nr) = . = 2 (m + n) (m + n − 1) (m + n)2 (m + n − 1)

σ 2 (X) = EX 2 − (EX)2 =

Chapter 7

C.7.1 We have: E(Y |X = x) = pq = 1−p = x2 , so that p = p(x) = x21+1 . Then: p(0) = 1, q p(1) = 1/2 = 0.5, p(2) = 1/5 = 0.2.

C.7.2 Indeed, for the continuous case (the discrete case is treated similarly), we have: Z ∞ g(x)h(y)fY |X (y|x)dy E[g(X)h(Y )|X = x] = −∞ Z ∞ =g(x) h(y)fY |X (y|x)dy = g(x)E [g(Y )|X = x] . −∞

C.7.3 y

x

2

From the above picture and symmetry, it is obvious that the required ares is πr4 = π4 . One arrives at the same result by√ considering the equation of the circle x2 + y 2 = 1 and integrating with respect to y from 0 to 1 − x2 and with respect to x from 0 to 1.

C.7.4

42

(i) If X = x, let Yx be the number of the children born to the x-th child, x = 0, 1, . . ..

Then the (conditional) expected number of grandchildren is E(Y0 + Y1 + . . . + Yx−1 ) = EY0 + EY1 + . . . + EYx−1 = xλ2 ; i.e., E(Y |X = x) = xλ2 . Therefore the (unconditional) expected number of grandchildren is E[E(Y |X)] = E(Xλ2 ) = λ2 EX = λ1 × λ2 .

Chapter 7

43

(ii) λ1 × λ2 = 3 × 4 = 12.

C.7.5

(i) fX (x) =

2

R1

y x 8xydy = 8x × 2

1 x

= 4x(1 − x)2 , 0 < x < 1.

8xy 2y (ii) fY |X (y|x) = 4x(1−x) 2 = 1−x2 , 0 < x < y < 1.

(iii) E(Y |X = x) =

1 2(1−x3 ) 2y y3 2 x y × 1−x2 dy = 1−x2 × 3 x = 3(1−x2 ) , 0 < x < 1.

R1

2(1−1/8) 7 (iv) E(Y |X = 1/2) = 3(1−1/4) = 7/4 9/4 = 9 .

C.7.6

R∞ R∞ (i) fX (x) = 0 cxe−x(y+c) dy = ce−cx 0 xe−xy dy = ce−cx × 1 = ce−cx , since xe−xy , y > 0 is the Negative Exponential p.d.f. with parameter x(> 0). Thus fX is Negative Exponential with parameter c, any positive constant. −x(y+c)

(ii) fY |X (y|x) = cxece−cx x(> 0).

= xe−xy , y > 0, which is Negative Exponential with parameter

(iii) From part (ii): E(Y |Y = X) = x1 , σ 2 (Y |X = x) = x12 .

R∞ R∞ c c −(y+c)x dx = c × 1 = (iv) fY (y) = 0 cxe−x(y+c) dx = y+c , since y+c y+c 0 x(y + c)e (y+c)2 y+c (y + c)e x, x > 0 is Negative Exponential with parameter y + c(> 0). hR iR R1R1 R1 1 1 c C.7.7 (i) 0 0 cx(1 − x)y 2 dydx = 0 cx(1 − x)dx 0 y 2 dy = 3c 0 x(1 − x)dx = 3c × 16 = 18 , so that c = 18. (ii) From the first integration in part (i), it follows that fX (x) = 3c x(1 − x) = 6x(1 − x), R1 0 < x < 1; and fY (y) = cy 2 0 x(1 − x)dx = 18y 2 × 61 = 3y 2 , 0 < y < 1.

R1 R1 1 (iii) EX = 6 0 x2 (1 − x)dx = 6 × 12 = 12 , and EY = 3 0 y 3 dy = 43 .

R1 R1R1 4 (iv) P (X < Y ) = 18 0 x x(1 − x)y 2 dydx = 6 0 x(1 − x)(1 − x3 )dx = 6 × 30 = 45 = 0.8; i.e., P (X < Y ) = 0.8.

C.7.8

R1 2 1 2 y cy dxdy = 0 cy (1 − y)dy = c × 12 = 1, so that c = 12.

Rx (ii) fX (x) = c 0 y 2 dy = 3c x3 = 4x3 , 0 < x < 1; and from the first integration in part (i), fY (y) = cy 2 (1 − y) = 12y 2 (1 − y), 0 < y < 1. (i)

R1R1 0

2

2

2

cy 3y y 1 (iii) fY |X (y|x) = 4x 3 = x3 , fX|Y (x|y) = 12y 2 (1−y) = 1−y , both for 0 < y < x < 1.

Rx 3 4 dy = x33 × x4 = 3x (iv) E(Y |X = x) = 0 3y 4 , 0 < x < 1, and E(Y |X = 0.7) = 0.525, x3 E(Y |X = 0.8) = 0.6, E(Y |X = 0.9) = 0.675.

C.7.9

(i) The required probability is: Z Z

R 1 2 2 P X +Y ≤ = dxdy.

2 2 {x2 +y 2 ≤R} πR

44

Chapter 7

∂y ∂x Set x = r cos θ, y = r sin θ, so that x2 + y 2 = r2 , ∂x ∂r = cos θ, ∂θ = −r sin θ; ∂r = sin θ, cos θ −r sin θ ∂y = r, and 0 ≤ r ≤ R2 , 0 ≤ θ < 2π. ∂θ = r cos θ. The Jacobian is: J = sin θ r cos θ Therefore the above integral equals:

1 πR2

Z 2π Z R/2 0

0

1 R 1 drdθ = × × 2π = ; i.e., P πR2 2 R

1 R 2 2 = .

X +Y ≤ 2 R

(ii) Here

R R R R R R = P − ≤ X ≤ ,− ≤ Y ≤ P |X| ≤ , |Y | ≤ 2 2 2 2 2 2 Z R/2 Z R/2 1 1 1 = dxdy = × R × R = ; i.e., 2 2 πR π −R/2 −R/2 πR

R 1 R = .

P |X| ≤ , |Y | ≤ 2 2 π √ R √R2 −x2 2 R2 −x2 1 √ (iii) Here fX (x) = πR , −R ≤ x ≤ R; and likewise, fY (y) = 2 − R2 −x2 dy = πR2 √ 2 2 2 R −y , −R ≤ y ≤ R. πR2

C.7.10

(i) fX (x) =

y2 1 0 (x + y)dy = x + 2 0 = x + 1/2, 0 < x < 1, and similarly, for fY ;

R1

fX|Y (x|y) =

x+y , y + 1/2

0 < x < 1, 0 < y < 1.

x3 1 x2 1 1 7 1 1 0 x(x + 1/2)dx = 3 0 + 2 × 2 0 = 3 + 4 = 12 , and similarly for EY ; R1 4 1 3 1 5 , and similarly for EY 2 ; hence EX 2 = 0 x2 (x + 1/2)dx = x4 0 + 21 × x3 0 = 41 + 16 = 13 5 49 11 σ 2 (X) = 12 − 144 = 144 , and similarly for σ 2 (Y ).

(ii) EX =

R1

3 1

2 1

x 1 R1 +y× x2 + y2 x+y 3 0 1 0 3 (iii) E(X|Y = y) = 0 x × y+1/2 dx = y+1/2 = y+1/2 = 3y+2 6y+3 , 0 x(x + y)dx = y+1/2 4y+3 R 1 4y+3 4y+3 1 2 2 12 E(X 2 |Y = y) = y+1/2 0 x (x + y)dx = 2y+1 = 12y+6 , so that σ (X|Y = y) = 12y+6 − 2

2 2 +6y+1 3y+2 . = 6y 6y+3 18(2y+1)2

R1

C.7.11

R1Ry 2 R 1 3 y R1 3 1 (i) c 0 0 xy dxdy = c 0 y1 x3 0 dy = 3c 0 y 2 dy = 3c × y3 0 = 9c = 1, so that c = 9.

R1 2 R1 R 2 1 d log y = 9x2 (log y) 1 = 9x2 (− log x) = −9x2 log x; (ii) fX (x) = x 9xy dy = 9x2 x dy = 9x y x x i.e., fX (x) = −9x2 log x, 0 < x < 1; Ry 2 3 y fY (y) = 0 9xy dx = y9 × x3 0 = 3y 2 , 0 < y < 1.

Chapter 7

45

2

2

(iii) fX|Y (x|y) = 9x3y/y = 3x , 0 < x < 1, and 2 y3 2

9x /y −1 fY |X (y|x) = −9x 2 log x = y log x , 0 < x < y < 1.

C.7.12

(i) Indeed,

=

0 x2 1

2 0

0

×y

1 0

+

ydxdy

xdxdy +

(x + y)dxdy = 0

Z 1Z 1

Z 1Z 1

Z 1Z 1

0

0

0

1 y2 1 1 1 × x = + = 1.

2 0 2 2 0

R1 R1 2 1 (ii) fX (x) = 0 xdy + 0 ydy = x+ y2 = x+ 21 , 0 < x < 1, and by symmetry, fY (y) = y + 12 , 0 0 < y < 1.

R1 3 1 2 1 7 7 (iii) EX = 0 (x2 + x2 )dx = x3 0 + 21 × x2 0 = 12 , and by symmetry, EY = 12 . Next, R 2 4 3 1 1 1 3 x x x 1 5 1 1 2 2 2 EX = 0 (x + 2 )dx = 4 0 + 2 × 3 0 = 4 + 6 = 12 , so that σ (X) = EX − (EX)2 = 49 11 11 5 2 12 − 144 = 144 , and by symmetry, σ (Y ) = 144 .

R R α 1 x β−α β−α (iv) P (X β < Y < X α ) = 0 xβ (x + y)dydx = (α+2)(β+2) + (2α+1)(2β+1) , by straightforward β−α β−α integration. By symmetry, P (Y β < X < Y α ) = (α+2)(β+2) + (2α+1)(2β+1) .

(v) For α = 1, β = 1, the probability is 3/20; for α = 0 and β = 1/2, the probability is 7/20 = 0.35.

Rc 2 c 3 c C.7.13 (i) EX = c22 0 x(c − x)dx = c22 c × x2 0 − x3 0 = 3c ,

Rc 3 c 2 4 c EX 2 = c22 0 x2 (c − x)dx = c22 c × xr 0 − x4 0 = c6 , 2

2

2

2

c c ; i.e., EX = 3c , σ2 (X) = 18 . so that σ 2 (X) = c6 − c9 = 18 R R 3 c c c 2 y4 c 2 2 c2 2 (ii) EY = c22 0 y × ydy = c22 × y3 0 = 2c 3 , EY = c2 0 y × ydy = c2 × 4 0 = 2 , so that 2 2 c2 c2 ; i.e., EY = 2c σ 2 (Y ) = c2 − 4c9 = 18 3 , σ2 (Y ) = 18 .

Ry 2 y (iii) E(X|Y = y) = 0 x × y1 dx = y1 × x2 0 = y2 , 0 < y < c, Rc 2 c dy 1 E(Y |X = x) = x y × c−x = c−x × y2 x = c+x 2 , 0 < x < c; i.e.;

y c+x , 0 < y < c; E(Y |X = x) = , 0 < x < c.

2 2

c c+X (iv) Then E[E(X|Y )] = E Y2 = 21 × 2c = 3 = 3 = EX, and E[E(Y |X)] = E 2

1 c 2c c + = = EY .

2 3 3

c2 c2 (v) V ar[E(X|Y )] = σ 2 Y2 = 14 × 18 < 18 = σ 2 (X), and V ar[E(Y |X)] = σ 2 c+X = 2 1 c2 c2 2

4 × 18 < 18 = σ (Y ).

E(X|Y = y) =

C.7.14

1 (i) For 0 < y < 1 − x, 0 < x < 1, we have: fX,Y (x, y) = fY |X (y|x)fX (x) = 1−x × 20x3 (1 − 3 3 x) = 20x ; i.e., fX,Y (x, y) = 20x , 0 < y < 1 − x, 0 < x < 1.

46

Chapter 7

(ii) Then fY (y) =

R 1−y 0

4

1−y

20x3 dx = 20 × x4 0

= 5(1 − y)4 ; i.e., fY (y) = 5(1 − y)4 , 0 < y < 1.

(iii) For 0 < y < 1 − x, 0 < x < 1, fX|Y (x|y) =

fX,Y (x,y) 4x3 20x3 fY (y) = 5(1−y)4 = (1−y)4 .

2

(1−x) 2 (iv) E(Y |X = x) = 1−x by the fact that Y |X ∼ U (0, 1 − x); 2 and σ (Y |X = x) = 12 R 1−y 4 x5 1−y 4 x × x3 dx = (1−y) E(X|Y = y) = (1−y) = 54 (1 − y), 4 0 4 × 5 0 R 1−y 2 4 x6 1−y 4 x × x3 dx = (1−y) = 23 (1 − y)2 , E(X 2 |Y = y) = (1−y) 4 0 4 × 6 0 2 so that σ 2 (X|Y = y) = E(X 2 |Y = y) − [E(X|Y = y)]2 = 75 (1 − y)2 , 0 < y < 1.

C.7.15

fX,Y

1 2 x2 y 2 2 0 0 cxy dxdy = c 0 y 2 (x, y) = 10xy , 0 < x < y < 1.

(i) Here

R1Ry

R

0

5

1

c dy = 2c × y5 0 = 10 = 1, so that c = 10. Thus

R1 3 1 3 (ii) fX (x) = 10x x y 2 dy = 10x × y3 x = 10 3 x(1 − x ), 0 < x < 1, Ry 2 y fY (y) = 10y 2 0 xdx = 10y 2 × x2 0 = 5y 4 , 0 < y < 1. (iii) fX|Y (x|y) =

fX,Y (x,y) 10xy 2 2x fY (y) = 5y 4 = y 2 , 0 < x < y < 1, 2

2

3y fY |X (y|x) = 10 10xy = 1−x 3 , 0 < x < y < 1. x(1−x2 ) 3

2 (iv) fX|Y (x|y = 1/2) = 8x, 0 < x < 1/2, fY |X (y|x = 1/3) = 81 26 y , 1/3 < y < 1.

C.7.16

R1

3

x6 1 x 1 3 0− 6 0

5 1 1 = 10 3 3 − 6 = 9,

R1 3 10 x4 1 x7 1 1 1 5 3 2 = 10 EX 2 = 10 3 0 x (1 − x )dx = 3 4 0− 7 0 3 4 − 7 = 14 , so that σ (X) =

(i) EX = 10 3

10 2 3 0 x (1 − x )dx = 3

5 25 55 14 − 81 = 1,134 .

R1 R1 6 1 7 1 (ii) EY = 5 0 y 5 dy = 5 × y6 0 = 56 , EY 2 = 5 0 y 6 dy = 5 × y7 0 = 57 , so that σ 2 (Y ) = 5 25 5 7 − 36 = 252 .

2y 2x 2 x3 y 0 x × y 2 dx = y 2 × 3 0 = 3 , 0 < y < 1,

(iii) E (X|Y = y) =

Ry

E (Y |X = x) =

R1

3(1−x4 ) 3y 2 y4 1 3 x y × 1−x3 dy = 1−x3 × 4 x = 4(1−x3 ) , 0 < x < 1,

and E(X|Y = 1/3) = 2/9, E(Y |X = 1/2) = 45/56.

R1 y6 1 5 = 23 0 y × 5y 4 dy = 10 3 × 6 0 = 9 = EX, h i

R R 4) 3 1 1−x4 10 3 )dx = 5 1 x(1−x4 )dx = 5 x2 1 − x6 1 = E[E(Y |X)] = E 3(1−X = × x(1−x 3 3 4 0 1−x 3 2 0 2 2 0 6 0 4(1−X ) 5 5 1 1 − = = EY .

2 2 6 6

(iv) E[E(X|Y )] = E

2Y 3

Chapter 7

47

C.7.17 Indeed, P (X = x) =

∞ X

P (X = x, Y = y) =

y=0

∞ X

P (X = x|Y = y)P (Y = y)

y=0

∞ ∞ y X X y x y−x −λ λy y x y−x −λ λ = p q e = p q ×e y! y! x x y=x y=0

= e−λ px

∞ X

y! λy q y−x x!(y − x)! y! y=x ∞

∞

e−λ (λp)x X q y−x λy−x e−λ (λp)x X (λq)y−x = = x! (y − x)! x! (y − x)! y=x y=x ∞

=

e−λ (λp)x X (λq)k x! k!

(by setting y − x = k)

k=0

= as was to be seen.

C.7.18

(λp)x −λ(1−q) (λp)x e−λ (λp)x × eλq = e = e−λp , x! x! x!

(i) Observe that, for 0 < x < 1, 0 < y < 1 − x. Then Z 1 Z 1−x

Z 1

2

cx(1 − x)y dydx = c 0

0

Z 1−x x(1 − x)

0

0

c y dydx = 3 2

Z 1

x(1 − x)4 dx

0

Z 1 Z 1 1 c c c 5 5 =− xd(1 − x) = − x(1 − x) + (1 − x)5 dx 15 0 15 15 0 0 Z 1 Z 1 1 c c c c = (1 − x)5 dx = − d(1 − x)6 = − (1 − x)6 = ; 15 0 90 0 90 90 0 so c/90 = 1 gives c = 90. (ii) Integration with respect to y in part (i) produces 3c x(1−x)4 , so that fX (x) = 30x(1−x)4 , 0 < x < 1. Next, Z 1−y

2 x 1−y x3 1−y fY (y) = cy x(1 − x)dx = cy − 2 0 3 0 0

2 3 (1 − y) (1 − y) = cy 2 − = 15y 2 (1 + 2y)(1 − y)2 , 2 3 2

2

or fY (y) = 15y 2 (1 − 3y 2 + 2y 3 ), 0 < y < 1.

2

6x(1−x) (iii) fX|Y (x|y) = 15y90x(1−x)y 2 (1−3y 2 +2y 3 ) = 1−3y 2 +2y 3 , x > 0, y > 0, (0 <)x + y < 1; 2

2

3y fY |X (y|x) = 90x(1−x)y = (1−x) 3 , x > 0, y > 0, (0 <)x + y < 1.

30x(1−x)4

48

Chapter 7

(iv) Z 1

Z 1

Z 1

1

x(1 − x)5 dx x d(1 − x) = −6x (1 − x) + 12 0 0 Z 1 Z 1 Z 1 2 1 (1 − x)6 dx = − = −2 xd(1 − x)6 = −2x(1 − x)6 + 2 d(1 − x)7 7 0 0 0 0 1 2 2 7 = − (1 − x) = ; i.e., 7 7 0 2

4

x (1 − x) dx = −6

EX = 30

2

5

2

5

0

0

EX = 2/7. Next, Z 1

4

y6 1 y7 1 y 1 EY = 15 y (1 − 3y + 2y )dy = 15 −3× +2× 4 0 6 0 7 0

0 15 1 1 2 − + = ; = 15 4 2 7 28 3

2

3

i.e., EY = 15/28. (v) Z 1−y 6 E(X|Y = y) = x2 (1 − x)dx 1 − 3y 2 + 2y 3 0 3

6 x 1−y x4 1−y = − 1 − 3y 2 + 2y 3 3 0 4 0

3 (1 − y) (1 − y)4 6 − = 1 − 3y 2 + 2y 3 3 4 3 3 (1 − y) (1 + 3y) (1 − y) (1 + 3y) (1 − y)(1 + 3y) = = = ; i.e., 2 3 2 2(1 − 3y + 2y ) (1 − y) (1 + 2y) 1 + 2y , x, y > 0, (0 <) < x + y < 1; therefore E(X|Y = 1/2) = 5/8.

E(X|Y = y) = (1−y)(1+3y) 1+2y

C.7.19

(i) From Z 30 Z 30 c 20

20

(x2 + y 2 )dxdy = c

Z 30 Z 30 20

20

x2 dxdy + c

Z 30 Z 30 20

30 30 y 3 30 x3 30 =c × ×y +c×x × 3 20 3 20 20 20 c 380, 000c = (27, 000 − 8, 000) × (30 − 20) × 2 = = 1, 3 3 3 we get c = 380,000 .

20

y 2 dxdy

Chapter 7

49

R 30 3 30 30 30x2 (ii) Here fX (x) = c 20 (x2 + y 2 )dy = cx2 × y 20 + c × y3 20 = 10cx2 + 19,000c = 380,000 + 3 19,000 3 3x2 1 3 1 × 380,000 = 380,000 + 20 ; i.e., fX (x) = 38,000 x2 + 20 , 20 ≤ x ≤ 30; 3 2 2 2 10(3x +1,900) 3x +1,900 3x 1 observe that fX (x) = 38,000 + 20 = 38,000 = , so that 380,000

fY |X (y|x) =

3(x2 + y 2 ) 3(x2 + y 2 )/380, 000 = , 20 ≤ x ≤ 30, 20 ≤ y ≤ 30.

10(3x2 + 1, 900)/380, 000 10(3x2 + 1, 900)

R 30 3 2 2 (iii) E(Y |X = x) = 10(3x2 +1,900) 20 y(x + y )dy.

R 30 2 30 4 30 2 But 20 y(x2 + y 2 )dy = x2 × y2 + y4 = x2 (900 − 400) + 14 (810, 000 − 160, 000) = 20

20

250x2 + 162, 500 = 250(x2 + 650), so that

E(Y |X = x) =

750(x2 + 650) 75(x2 + 650) = , 20 ≤ x ≤ 30.

10(3x2 + 1, 900) 3x2 + 1, 900

1,575 (iv) E(Y |X = 20) = 78,750 3,100 = 62 ' 25, 403, 3,825 E(Y |X = 25) = 95,625 3,775 = 151 ' 25.331, 2,325 E(Y |X = 30) = 116,250 4,600 = 92 ' 25, 272.

C.7.20

R∞ R∞ R∞Ry (i) Here c2 0 0 e−cy dxdy = c2 0 ye−cy dy = c 0 (y × ce−cy )dy = c × 1c = 1, since ce−cy , y > 0 is the Negative Exponential p.d.f. with parameter c, so that its expectation is 1c .

It follows that c can be any positive constant.

R∞ R∞ ∞ (ii) fX (x) = c2 x e−cy dy = −c x de−cy = −ce−cy x = ce−cx , x > 0, so that fX is Negative Exponential with parameter c.

Ry fY (y) = c2 0 e−cy dx = c2 ye−cy , y > 0, and this is the Gamma p.d.f. with α = 2 and β = 1/c. (iii) EX = 1/c, σ 2 (X) = 1/c2 ; EY = 2/c, σ 2 (Y ) = 2/c2 .

2 −cy

e (iv) fX|Y (x|y) = cc2 ye −cy = 1/y, 0 < x < y, which is the U (0, y) p.d.f. Therefore E(X|Y = y 2 y) = 2 and σ (Y ) = y 2 /12.

2 −cy

(v) fY |X (y|x) = ccee−cx = ce−c(y−x) , 0 < x < y. (vi)

50

Chapter 7

y y = 2x

y=x

0 0

x

Z ∞ Z 2x e−cy dy dx = c ce−cy dy dx 0 x 0 x

Z ∞ Z 2x Z ∞ 2x −cy −cy = −c de dx = − e dx x 0 x 0 Z ∞ Z ∞ −2cx = −c e dx + c e−cx dx 0 0 Z ∞ Z 1 1 ∞ 1 −(2c)x (2c)e dx + ce−cx dx = − + 1 = ; =− 2 0 2 2 0

P (Y < 2X) = c2

i.e., P (Y < 2X) = 12 .

C.7.21

Z ∞ Z 2x

(i) n

X 2 2 n(n + 1)(2n + 1) 2n + 1 EX = x2 = × = , (from Table #6) n(n + 1) n(n + 1) 6 3 x=1   n n n n X X X X 2 2 n EY = y(n − y + 1) = y− y2 + y n(n + 1) n(n + 1) y=1 y=1 y=1 y=1

2 n(n + 1) n(n + 1)(2n + 1) n(n + 1) n× − + = n(n + 1) 2 6 2

2n(n + 1) 2n + 1 n+2 = n− +1 = , 2n(n + 1) 3 3

Chapter 7

51

n

EX 2 =

X 2 2 n2 (n + 1)2 n(n + 1) x3 = × = , n(n + 1) n(n + 1) 4 2 x=1

EY 2 =

2 n(n + 1)

n X

y 2 (n − y + 1)

y=1

2 n(n + 1)(2n + 1) n2 (n + 1)2 n(n + 1)(2n + 1) = n× − + n(n + 1) 6 4 6

2 n + 3n + 2 2n(n + 1) n(2n + 1) n(n + 1) 2n + 1 = = − + .

2n(n + 1) 3 2 3 6

(ii)

n2 + n − 2 n(n + 1) (2n + 1)2 − = , 2 9 18 n2 + 3n + 2 (n + 2)2 n2 + n − 2 σ 2 (Y ) = − = .

6 9 18 σ 2 (X) =

C.7.22

(i) The required probability is:

Z 26 Z 26

2

Z 26 Z 26

2

(x + y )dxdy = c

c 20

=c ×

20 3 x 26

3 20

×y

26

+c×x

26

×

2

Z 26 Z 26

x dxdy + c 20 20 3 26 y

20

y 2 dxdy

20

3 20 114, 912 c = 0.3024; = (17, 576 − 8, 000) × 6 × 2 = 38, 304c = 3 380, 000 20

20

i.e., the required probability is 0.3024. (ii) The diagram below will facilitate specifying limits in several integrals. In the first place,

Z Z P (|Y − X| ≤ 2) = c {|y−x|≤2}

def

(x2 + y 2 )dxdy = I.

52

Chapter 7

(|y − x| ≤ 2) 30 y =x+2

y =x−2

22 20

20

22

28

30

Now, |y − x| ≤ 2 or −2 ≤ y − x ≤ 2 or x − 2 ≤ y ≤ x + 2. Therefore, Z 22 Z x+2 Z 28 Z x+2 Z 30 Z 30 2 2 2 2 I=c (x + y )dydx + c (x + y )dydx + c (x2 + y 2 )dydx 20

20

22

x−2

28

x−2

def

= I1 + I2 + I3 .

But Z 22 Z x+2

Z 22 Z x+2

y 2 dydx 20 20 20 20

Z 22 Z 22 x+2 c 2 3 =c x (x + 2 − 20)dx + y dx 3 20 20 20 Z 22 Z

c 22 2 (x + 2)3 − 203 dx =c x (x − 18)dx + 3 20 20 22 22 22 c 18c c c = × x4 − × x3 + × (x + 2)4 − × 8, 000 × 2 4 3 12 3 20 20 20 c = × (234, 256 − 160, 000) − 6c × (10, 648 − 8, 000) 4 c 16, 000c + × (331, 776 − 234, 256) − 12 3

24, 380 − 16, 000 c = 18, 564 − 15, 888 + 3

8, 380 16, 408c 16, 408 = 2, 676 + c= = ; 3 3 380, 000

I1 = c

16,408 i.e., I1 = 380,000 .

2

x dydx + c

Chapter 7

53

Next, Z 28 Z x+2

(x2 + y 2 )dydx

Z x+2 c 28 3 x+2 dx =c x ×y dx + y 3 22 x−2 x−2 22 Z 28 Z c 28 x2 dx + = 4c (12x2 + 16)dx (by expanding (x + 2)3 and (x − 2)3 ) 3 22 22 28 16c 4c 4c = (21, 952 − 10, 648) + × x3 + ×6 3 3 3 22 45, 216c 45, 216c 96c 90, 528 = + + = ; 3 3 3 380, 000

I2 = c

22 x−2 Z 28 2

90,528 i.e; I2 = 380,000 .

Finally, Z 30 Z 30 I3 = c

2

Z 30

2

x ×y

(x + y )dydx = c 28 Z 30

x−2

2

28

2

30 x−2

c dx + 3

Z 30 y 28

3

30 x−2

dx

Z

c 30 27, 000 − (x − 2)3 dx =c x (32 − x)dx + 3 28 28 30 30 30 32c c c c = × x3 − × x4 + × 54, 000 − × (x − 2)4 3 4 3 12 28 28 28 30 54, 000c c 161, 536c 4 − 48, 836c + − × (x − 2) = 3 3 12 28 69, 028c 39, 420c 29, 608c 29, 608 = − = = ; 3 3 3 380, 000 2

29,608 i.e.; I3 = 380,000 .

136,544 Therefore I = 16,408+90,528+29,608 = 380,000 , or 380,000

I=

C.7.23

4, 267 ' 0.359.

11, 875

1 1 1 1 (i) FX (x) = limz→∞ F (x, y, z) = limz→∞ 1 − 1+x 1 − 1+y 1 − 1+z 1 − 1+x+y+z = y→∞

y→∞

1 1 1 1 − 1+x ; i.e., FX (x) = 1 − 1+x , x > 0; and by symmetry, FY (y) = 1 − 1+y , y > 0, 1 FZ (z) = 1 − 1+z , z > 0.

d d 1 1 1 Then fX (x) = dx FX (x) = dx 1 − 1+x = (1+x) 2 ; i.e., fX (x) = (1+x)2 , x > 0, and by 1 1 symmetry, fY (y) = (1+y) 2 , y > 0; fZ (z) = (1+z)2 , z > 0.

54

Chapter 7

1 1 (ii) FX,Y (x, y) = limz→∞ F (x, y, z) = 1 − 1+x 1 − 1+y as above; i.e., FX,Y (x, y) =

1 1 1 1 1 − 1+x 1 − 1+y , x > 0, y > 0; and by symmetry, FX,Z (x, z) = 1 − 1+x 1 − 1+z ,

1 1 1 − 1+z , y > 0, z > 0. x > 0, z > 0; FY,Z (y, z) = 1 − 1+y

∂2 ∂2 1 1 Then fX,Y (x, y) = ∂x∂y FX,Y (x, y) = ∂x∂y 1 − 1+x 1 − 1+y = (1+x)21(1+y)2 ; i.e., fX,Y (x, y) = (1+x)21(1+y)2 , x > 0, y > 0; and by symmetry, fX,Z (x, z) = (1+x)21(1+z)2 , x > 0, z > 0; fY,Z (y, z) = (1+y)21(1+z)2 , y > 0, z > 0. (iii) Clear, for all values of the arguments involved, fX,Y (x, y) = fX (x)fY (x), fX,Z (x, z) = fX (x)fZ (z), fY,Z (y, z) = fY (y)fZ (z), so that the pairs (X, Y ), (X, Z) and (Y, Z) consist of independent components. On the other hand,

1 1 1 1 F (x, y, z) = 1 − 1− 1− 1− 1+x 1+y 1+z 1+x+y+z

1 1 1 = 1− 1− 1− 1+x 1+y 1+z 1 if and only if 1 − 1+x+y+z = 0 or x + y + z = ∞, which is not true. So, X, Y, Z are not independent.

R 1 dx 1 1 1 (iv) (a) P (X ≤ 1) = 0 (1+x) 2 = − 1+x 0 = 2 = 0.5;

(b) P (X ≤ 1, Y ≤ 1) = P (X ≤ 1)P (Y ≤ 1) (by independence) = 14 = 0.25;

(c) P (X ≤ 1, Y ≤ 1, Z ≤ 1) = F (1, 1, 1) = 1 − 21 1 − 12 1 − 12 1 − 14 = 12 × 21 × 1 3 3 2 × 4 = 32 = 0.09375; (d) P (X > 1) = 1 − P (X ≤ 1) = 0.5; (e) P (X ≤ 1, Y > 1) = P (X ≤ 1)P (Y > 1) = 12 × 12 = 14 = 0.25, P (X > 1, Y ≤ 1) = P (X > 1)P (Y ≤ 1) = 12 × 12 = 14 = 0.25; (f) P (X > 1, Y > 1) = P (X > 1)P (Y > 1) = 0.25; (g) From 14 = P (X ≤ 1, Y ≤ 1) = P (X ≤ 1, Y ≤ 1, Z ≤ 1) + P (X ≤ 1, Y ≤ 1, Z > 1) = 3 1 3 5 32 + P (X ≤ 1, Y ≤ 1, Z > 1), we get P (X ≤ 1, Y ≤ 1, Z > 1) = 4 − 32 = 32 = 0.15625; from 14 = FX,Z (1, 1) = P (X ≤ 1, Z ≤ 1) = P (X ≤ 1, Y ≤ 1, Z ≤ 1) + P (X ≤ 3 1, Y > 1, Z ≤ 1) = 32 + P (X ≤ 1, Y > 1, Z ≤ 1), we get P (X ≤ 1, Y > 1, Z ≤ 1) = 1 3 5 4 − 32 = 32 = 0.15625, P (X > 1, Y ≤ 1, Z ≤ 1) = 0.15625 on the basis of the immediately previous result and symmetry; (h) From P (X ≤ 1, Y > 1) = P (X ≤ 1, Y > 1, Z ≤ 1) + P (X ≤ 1, Y > 1, Z > 1), we get P (X ≤ 1, Y > 1, Z > 1) = P (X ≤ 1, Y > 1) − P (X ≤ 1, Y > 1, Z ≤ 1) = 0.25 − 0.15625 (by (e) and (g)) = 0.09375,

Chapter 7

55

P (X > 1, Y ≤ 1, Z > 1) = 0.09375 by the immediately previous case and symmetry; from P (Z ≤ 1) = P (X ≤ 1, Y ≤ 1, Z ≤ 1) + P (X ≤ 1, Y > 1, Z ≤ 1) + P (X > 1, Y ≤ 1, Z ≤ 1) + P (X > 1, Y > 1, Z ≤ 1), we get P (X > 1, Y > 1, Z ≤ 1) = P (Z ≤ 1) − P (X ≤ 1, Y ≤ 1, Z ≤ 1)− P (X ≤ 1, Y > 1, Z ≤ 1) − P (X > 1, Y ≤ 1, Z ≤ 1).

But 1 P (Z ≤ 1) = FZ (1) = , 2 P (X ≤ 1, Y ≤ 1, Z ≤ 1) = F (1, 1, 1) =

1 , 12

5 , 32 5 P (X > 1, Y ≤ 1, Z ≤ 1) = .

32 P (X ≤ 1, Y > 1, Z ≤ 1) =

1 5 5 5 Therefore P (X > 1, Y > 1, Z ≤ 1) = 12 − 12 − 32 − 32 = 48 ' 0.104; (i) Finally, from P (X > 1, Y > 1) = P (X > 1, Y > 1, Z > 1) + P (X > 1, Y > 1, Z ≤ 1), we get

P (X > 1, Y > 1, Z > 1) = P (X > 1, Y > 1) − P (X > 1, Y > 1, Z ≤ 1) 1 5 = − (by (f) and (h)), 4 48 7 ' 0.146. and this is equal to 48

Chapter 8

C.8.1 V ar(X + Y ) = σ 2 (X) + σ 2 (Y ) + 2Cov(X, Y ) = σ12 + σ22 + 2σ1 σ2 ρ ≤ σ12 + σ22 + 2σ1 σ2 (since ρ ≤ 1) = (σ1 + σ2 )2 .

C.8.2

(i) EX = −2 × 14 + −1 × 41 + (1 × 14 ) + 2 × 41 = 0 (ii) XY = X 3 . Thus: xy fXY (x, y)

-8 1/4

-1 1/4

1 1/4

8 1/4

1 1 (iii) Cov(X, Y ) = E(XY ) − (EX)(EY ) = E(XY ) (since EX = 0) = −8 × + −1 × 4 4 ×

1 × 14 + 8 × 14 = 0. (iv) Cov(X, Y ) = 0, means that X and Y are uncorrelated; i.e, they are completely not linearly related (whereas they are allowed to be related otherwise, as they are here).

C.8.3

(i) EX 2 = 4 × 41 + 1 × 14 + 1 × 14 = 25 = V ar(X). (ii) y P (Y = y)

1 1/2

4 1/2

EY = 1 × 12 + (4 × 12 ) = 5/2,

EY 2 = 1 × 21 + (16 × 12 ) = 17/2, so that 25 11 V ar(Y ) = EY 2 − (EY )2 = 17 2 − 4 = 4 .

(iii) V ar(2X + 6Y ) = 4V ar(X) + 36V ar(Y ) + 24Cov(X, Y ) = 4V ar(X) + 36V ar(Y ) 5 11 = 4 × + 36 × = 10 + 99 = 109.

2 4 56

Chapter 8

57

C.8.4 We have: E(aX + Y ) = aEX + bEY , so that Cov(X, aX + Y ) = E(aX 2 + XY ) − (EX)(aEX + EY ) = aEX 2 + E(XY ) − a(EX)2 − (EX)(EY ) = a[EX 2 − (EX)2 ] + [E(XY ) − (EX)(EY )] = aσ 2 (X) + Cov(X, Y ). ) .

Therefore Cov(X, aX + Y ) = 0 if and only if aσ 2 (X) + Cov(X, Y ) = 0, or a = − Cov(X,Y σ 2 (X) ) follows either by symmetry, or as in That Cov(Y, X + bY ) = 0 if and only if b = − Cov(X,Y σ 2 (Y ) the first case.

C.8.5

(i) fX (100) = fX (250) = 0.50; fY (0) = fY (100) = 0.25, fY (200) = 0.50. (ii) EX = (100 + 250) × 0.5 = 175, EX 2 = (10, 000 + 62, 500) × 0.5 = 36, 250, so that σ 2 (X) = 36, 250 − 1752 = 36, 250 − 30, 625 = 5, 625; i.e., EX = 175, σ 2 (X) = 5, 625. (iii) EY = (100 × 0.25) + (200 × 0.50) = 125, EY 2 = (10, 000 × 0.25) + (40, 000 × 0.5) = 22, 500, so that σ 2 (Y ) = 22, 500 − 1252 = 22, 500 − 15, 625 = 6, 875; i.e., EY = 125, σ 2 (Y ) = 6, 875. (iv) E(XY ) = (10, 000 × 0.1) + (25, 000 × 0.15) + (20, 000 × 0.2) + (50, 000 × 0.3) = 1, 000 + 3, 750 + 4, 000 + 15, 000 = 23, 750, Cov(X, Y ) = E(XY ) − (EX)(EY ) = 23, 750 − 175 × 125 = 23, 750 − 21, 875 = 1, 875, and hence √ 1, 875 11 √ ρ(X, Y ) = √ = ; 11 5, 625 6, 875 √

i.e.; ρ(X, Y ) = C.8.6

11 11 ' 0.3015.

(i) n X n n x X X X x(x + 1) 1 2 2 = xy = x× x2 (x + 1) n(n + 1) n(n + 1) 2 n(n + 1) x=1 y=1 x=1 x=1 2

2 1 n (n + 1) n(n + 1)(2n + 1) = + n(n + 1) 4 6 2

2 1 n + n 2n + 1 3n + 7n + 2 = + = .

2 2 3 12

E(XY ) =

(ii) 3n2 + 7n + 2 2n + 1 n + 2 n2 + n − 2 − × = , 12 3 3 36 n2 + n − 2 . n2 + n − 2 1 = . ρ(X, Y ) = 36 18 2

Cov(X, Y ) =

58

C.8.7

Chapter 8

(i) E(XY ) =

R R 2 dxdy = 10 1 y 3 x3 y dy = 10 1 y 6 dy = 10 . xy × 10xy 3 0 3 0 21 0 0

R1Ry 0

5 5 5 (ii) Cov(X, Y ) = E(XY ) − (EX)(EY ) = 10 21 − 9 × 6 = 378 . ) 5/378 q (iii) ρ(X, Y ) = Cov(X,Y σ(X)σ(Y ) = 55×5 1,134×352

(iv) E

X Y

R1Ry x

=

0 y

0

√

99,792 √ = 189 ' 0.504.

11

× 10xy 2 dxdy = 10

R1 0

3 y y × x3 0

dy = 32 ,

R1Ry R 1 3 y = 0 0 yx2 × 10xy 2 dxdy = 10 0 x3 0 dy = 56 , R1Ry y R1 y Y = 0 0 x × 10xy 2 dxdy = 10 0 y 3 × x 0 dy = 2.

E X

E

C.8.8

X Y2

(i) E(XY ) =

1 1 1 R1R1 2 y2 x3 x2 2 0 x ydxdy + 0 0 xy dxdy = 3 0 × 2 0 + 2 0 ×

R1R1

R1R1

0

0

0 xy(x + y)dxdy =

1

y3 1 1 7 2 49 1 1 1 = 13 − 144 = − 144 , 3 0 = 6 + 6 = 3 . Then Cov(X, Y ) = E(XY )−(EX)(EY ) = 3 − 12 1/144 1 and ρ(X, Y ) = − 11/144 = − 11 .

11 2ab (ii) σ 2 (Z) = σ 2 (aX + bY ) = a2 σ 2 (X) + b2 σ 2 (Y ) + 2abCov(X, Y ) = 144 (a2 + b2 ) − 144 =

11(a2 +b2 )−2ab 5 . Therefore, a = b = 1 gives σ 2 (X + Y ) = 36 , and a = −b = 1 gives 144 1 2 σ (X − Y ) = 6 .

C.8.9

(i) fX (1) = fX (2) = 0.5; fY (1) = 0.6, fY (2) = 0.4. f

(x,y)

(ii) fX|Y (x|y) = X,Y fY (y) , and specifically: 0.2 0.4 3 0.1 1 fX|Y (1, 1) = 0.6 = 31 , fX|Y (2|1) = 0.6 = 23 ; fX|Y (1|2) = 0.3 0.4 = 4 , fX|Y (2|2) = 0.4 = 4 . (iii) EX = (1 × 0.5) + (2 × 0.5) = 1.5, EX 2 = (1 × 0.5) + (4 × 0.5) = 2.5, so that σ 2 (X) = 2.5−(1.5)2 = 2.5−2.25 = 0.25 and σ(X) = 0.5; likewise, EY = (1×0.6)+(2×0.4) √ = 1.4, EY 2 = (1×0.6)+(4×0.4) = 2.2, so that σ 2 (Y ) = 2.2−1.96 = 0.24 and σ(Y ) = 0.24 ' 0.49. (iv) E(XY ) = (1 × 0.2) + (2 × 0.3) + (2 × 0.4) + (4 × 0.1) = 2.

0.1 √ (v) Cov(X, Y ) = 2 − (1.5 × 1.4) = 2 − 2.1 = −0.1, and ρ(X, Y ) = − 0.5×0.1 ' − 0.245 = 0.24 −0.408.

(vi) E(X|Y = 1) = 1 × 13 + 2 × 32 = 53 , E(X|Y = 2) = 1 × 43 + 2 × 41 = 54 .

(vii) E [E(X|Y )] = [E(X|Y = 1)] P (Y = 1)+[E(X|Y = 2)] P (Y = 2) = 35 × 0.6 + 54 × 0.4 = 1 + 0.5 = 1.5 = EX.

C.8.10

1 y2 2 x 8xydy = 8x × 2 x = 4x(1 − x ), 0 < x < 1; Ry 2 y fY (y) = 0 8xydx = 8y × x2 0 = 4y 3 , 0 < y < 1.

(i) fX (x) =

R1

R

R1 4 1 2 x5 1 1 1 8 x3 1 2 )dx = 4 x × 4x(1 − x x dx − 0 0 0 x dx = 4( 3 0 − 5 0 ) = 4 3 − 5 = 15 , R

R1 R1 4 1 6 1 1 EX 2 = 0 x2 × 4x(1 − x2 )dx = 4 0 x3 dx − 0 x5 dx = 4( x4 0 − x6 0 ) = 4 41 − 16 = 31 , R1 R1 1 1 EY = 0 y × 4y 3 dy = 45 × y 5 0 = 45 , EY 2 = 0 y 2 × 4y 3 dy = 46 × y 6 0 = 32 .

(ii) EX =

R1

Chapter 8

59

64 33 11 2 16 6 2 (iii) Then σ 2 (X) = 31 − 225 = 225−192 3×225 = 3×225 = 225 , σ (Y ) = 3 − 25 = 225 .

R1Ry R1 Ry R1 6 1 (iv) E(XY ) = 0 0 xy × 8xydxdy = 8 0 y 2 0 x2 dx dy = 85 0 y 2 × y 3 dy = 38 × y6 0 = 8 4 18 = 9 .

4 8 8 = 225 (v) Cov(X, Y ) = 49 − 15 × 45 = 34 13 − 25 .

√ (vi) ρ(X, Y ) = √ 4/225 11

6/225

√

= √466 = 2 3366 ' 0.492.

Chapter 9

C.9.1 The r.v. L − S is of the from c1 L + c2 S with c1 = 1, c2 = −1, and therefore by the Hint, c1 L + c2 S has Normal distribution with mean: c1 µ1 + c2 µ2 = 70.0 − 66.8 = 3.2 and variance c1 σ12 + 2c1 c2 σ1 σ2 + c22 σ22 = 4 − 2 × 0.68 × 2 × 2 + 4 = 2.56; i.e., L − S ∼ N (3.2, 2.56). Therefore

(L − S) − E(L − S) −3.2 P (S > L) = P (L − S < 0) = P <√ σ(L − S) 2.56

3.2 = 1 − Φ(2) =1−Φ √ 2.56 = 1 − 0.977250 = 0.02275; i.e., P (S > L) = 0.02275.

C.9.2 Indeed, since E(X2 |X1 ) = µ2 + ρ σσ12 (X1 − µ1 ), we have:

σ2 σ2 V ar [E(X2 |X1 )] = V ar µ2 + ρ (X1 − µ1 ) = ρ2 22 × σ12 = ρ2 σ22 ≤ σ22 ; σ1 σ1

i.e., V ar [E(X2 |X1 )] ≤ σ22 = σ 2 (X2 ), and similarly, V ar [E(X1 |X2 )] ≤ σ12 = σ 2 (X1 ).

60

Chapter 10

C.10.1 They are not independent, because the relationship fX,Y (x, y) = fX (x)fY (y) does not hold, e.g., for x = 1/3 and y = 1/4. Indeed,

fX,Y (1/3, 1/4) = 7/24 6=

245 = fX (1/3)fY (1/4).

3, 888

C.10.2

f (x, y) 0 1 2 x 3 4 5 fY (y)

y 1 0.015 0.030 0.075 0.090 0.060 0.030 0.3

0 0.025 0.050 0.125 0.150 0.100 0.050 0.5

2 0.010 0.020 0.050 0.060 0.040 0.020 0.2

fX (x) 0.05 0.10 0.25 0.30 0.20 0.10 1

(i) fX (x), fY (y) see table. (ii) Independent, because f (x, y) = fX (x)fY (y) for all x and y. (iii) P (X = 1) = 0.1, P (Y = 1) = 0.3.

C.10.3

(i) 61

62

Chapter 10

y

−1

0

1

fX,Y (x, y) = π1 , x2 + y 2 ≤ 1. For −1 ≤ x ≤ 1, fX (x) = π1 √ 2 1−y 2 symmetry, fY (y) = , −1 ≤ y ≤ 1. π

x

R √1−x2

√ 2 1−x2 √ dy = , and by 2 π − 1−x

p (ii) The r.v.s’s X and Y are not independent, since fX (x)fY (y) = π42 (1 − x2 )(1 − y 2 ) 6= π1 .

X−EX 42−40 C.10.4 (i) P (39 ≤ X ≤ 42) = P 39−40 ≤ ' Φ(1.34) − Φ(−0.67) = Φ(1.34) + ≤ 1.5 1.5 σ(X) Φ(0.67) − 1 = 0.909877 + 0.748571 − 1 = 0.658448. (ii) P (at least one of X1 , X2 , X3 , X4 ≥ 42) = 1 − P (all are < 42) = 1 − [P (X1 < 42)]4 ' 1 − [Φ(1.34)]4 = 1 − [0.909877]4 ' 1 − 0.685379 = 0.314621.

C.10.5

(i) X + Y ∼ P (λ1 + λ2 ). (ii) E(X + Y ) = λ1 + λ2 , σ 2 )(X + Y ) = λ1 + λ2 , so that σ(X + Y ) = (λ1 + λ2 )1/2 . √ √ (iii) E(X + Y ) = 18, σ(X + Y ) = 18 = 3 2 ' 4.243.

C.10.6 Indeed,

fX,Y (x, y) = fX (x)fY (y) (by independence) 1 1 2 2 2 2 =√ e−(x−µ1 ) /2σ1 × √ e−(y−µ2 ) /2σ2 2πσ1 2πσ2 − 12 1 = e 2πσ1 σ2

x−µ1 σ1

2 2 y−µ + σ 2 2

,

which is the p.d.f. of the Bivariate Normal with means µ1 , µ2 , variances σ12 , σ22 , and correlation coefficient ρ = 0.

Chapter 10

63

C.10.7 E(Y |X = x) = x2 , x = 0, 1, 2. So, the r.v. E(Y |X) takes on the values: 0, 1, 4 with probabilities: 1/18, 16/18, 1/18, respectively. Then: 16 1 (i) EY = E[E(Y |X)] = (1 × 18 ) + (4 × 18 ) = 10 9 , and E(XY ) = E[E(XY |X)] = 16 1 4 2 3 E[XE(Y |X)] = E(X ×X ) = EX = (1× 18 )+(8× 18 ) = 34 ; i.e., EY = 10 9 , E(XY ) = 3 .

(ii) If X and Y were independent, then E(XY ) = (EX)(EY ). Here (EX)(EY ) = 1 × 10 9 = 10 4

9 6= 3 = E(XY ), so that X and Y are not independent.

C.10.8 fX (x)fY (y) = 16x(1 − x2 )y 3 needs not be = 8xy = fX,Y (x, y) for all x and y. For example, for x = y = 12 , the left-hand side = 34 and the right-hand side = 2. So, X and Y are not independent.

C.10.9

√

n(X̄n −µ) σ (i) We know that X̄n ∼ N (µ, ∼ N (0, 1). Therefore P (X̄25 > 80) = h √ n ) or i σ ' 1 − Φ(2.08) = 1 − 0.981237 = 0.018763.

1 − P (X̄25 ≤ 80) = 1 − Φ 25(80−75) 12 h√ i ' 1 − Φ(3.33) = 1 − 0.999566 = 0.000434. (ii) P (X̄64 > 80) ' 1 − Φ 64(80−75) 12 2

(iii) X̄25 ∼ N (75, 5.76), X̄64 ∼ N (75, 2.25), and by independence, X̄25 − X̄64 ∼ N (0, 8.01).

Therefore

3 P (X̄25 − X̄64 > 3) = 1 − Φ √8.01 ∼ 1 − Φ(1.06) = 1 − 0.855428 = 0.144572.

C.10.10 Indeed, Cov(Xi − X̄, X̄) = Cov(Xi , X̄) − Cov(X̄, X̄) n 1X = Cov(Xi , Xj ) − σ 2 (X̄) n j=1

=

n 1X

n

Cov(Xi , Xj ) −

j=1

σ2 n

1 2 σ2 σ (Xi ) − (by independence) n n σ2 σ2 = − = 0, so that ρ(Xi − X̄, X̄) = 0, i = 1, . . . , n n n

=

C.10.11

(i) EX(N ) = E {E [X(N )|N ]} = E[(N + 1)p] = (EN + 1)p = (λ + 1)p;

(since X(N )|N = n ∼ B(n + 1, p))

i.e.; EX(N ) = (λ + 1)p.

P∞ P∞ n+1 0 n+1 Next, P [X(N ) = 0] = × n=0 P [X(n) = 0|N = n] P (N = n) = 0 0 p q P∞ (λq)n λn −λ −λ λq −λ(1−q) −λp −λp −λ e n! = qe × e = qe = qe ; i.e.; P [X(N ) = 0] = qe . n=0 n! = qe (ii) Since EX = np, we have EX > EX(N ) is equivalent to np > (λ + 1)p, or n > λ + 1.

64

Chapter 10

(iii) Here P (X = 0) = q n , so that P (X = 0) > P [X(N ) = 0] is equivalent to q n > qe−λp , or (1 − p)n−1 > e−λp .

C.10.12 Set U = X − Y , V = X + Y . To show that MU,V (t1 , t2 ) = MU (t1 )MV (t2 ) for all t1 , t2 ∈ <.

MU (t1 ) = MX+(−Y ) (t1 ) = MX (t1 )M−Y (t1 ) = MX (t1 )MY (−t1 ) 2

2

2

= et1 /2 × e(−t1 ) /2 = et1 , so that U ∼ N (0, 2), and likewise, 2

2

2

MV (t2 ) = et2 /2 × et2 /2 = et2 , and V ∼ N (0, 2).

Next, MU,V (t1 , t2 ) = Eet1 U +t2 V = Eet1 (X−Y )+t2 (X+Y ) = Ee(t1 +t2 )X−(t1 −t2 )Y = Ee(t1 +t2 )X Ee−(t1 −t2 )Y 2

2

= MX (t1 + t2 ) × MY [−(t1 − t2 )] = e(t1 +t2 ) /2 × e[−(t1 −t2 )] /2 2

2

2

2

= e[(t1 +t2 ) +(t1 −t2 ) ]/2 = et1 +t2 2

2

= et1 × et2 = MU (t1 )MV (t2 ), as was to be seen.

C.10.13

1 (x, 1 − y) x+y =1 (1 − x, y) y

0 0

x

R 1−x (i) fX (x) = 24x 0 ydy = 12x(1 − x)2 , 0 ≤ x < 1; R 1−y fY (y) = 24y 0 xdx = 12y(1 − y)2 , 0 ≤ y < 1.

1

Chapter 10

65

(ii) For example, for x = y = 1/2, we get: fX (1/2) = 1.5 = fY (1/2), and fX,Y (1/2, 1/2) = 6, so that fX,Y (1/2, 1/2) 6= fX (1/2)fY (1/2), and X and Y are dependent.

24xy 2x (iii) fX|Y (x|y) = 12y(1−y) 2 = (1−y)2 , 0 ≤ x < 1, 0 ≤ y < 1, 0 ≤ x + y < 1.

(iv) E(X|Y = y) =

R 1−y 0

E(X|Y = 1/2) = 1/3.

3

2 x 2x x × (1−y) 2 dx = (1−y)2 × 3

1−y 0

= 23 (1 − y), 0 ≤ y < 1, and

Chapter 11

C.11.1 From Y =

√

X, we get y =

2

2ye−y , y > 0.

√

2

−y × 2y = x or y 2 = x, dx dy = 2y, y > 0. Therefore fY (y) = e

C.11.2 ( X = Z12 + Z22 Y = Z32 + Z42

,

U=

X .

X +Y

Y /2 X First, X ∼ χ22 , Y ∼ χ22 and X, Y are independent, so that X/2 ∼ F2,2 . Next, U = X+Y = Y /2 1 1 1 1 1−u Y /2 = 1+W , where W = X/2 ∼ F2,2 . So, U = 1+W implies u = 1+w or w = u , 1+ X/2 1 1 1 2 w > 0, and dw du = − u2 . Since fW (w) = (1+w)2 , w > 0, we get: fU (u) = u × u2 = 1 with 1 Y 1+ X

=

0 < u < 1. So U ∼ U (0, 1).

C.11.3

R R n−1 dy = n 1 y n dy = n × y n+1 1 = n , EY 2 = n 1 y n+1 dy = y × ny n n+1 n+1 0 0 0

02 1 n n n n n n n+2 2 = n+2 , so that σ (Yn ) = n+2 − n+1 , = (n+1)2 (n+2) ; i.e., EYn = n+2 n+2 × y 0

(i) EYn =

R1

σ 2 (Yn ) = (n+1)n2 (n+2) . n (ii) EYn = n+2 = 1+1 2 −→ 1 and σ 2 (Yn ) = (n+1)n2 (n+2) = (n+1)21 1+ 2 −→ 0. ( n ) n→∞ n n→∞

C.11.4

R β−1 e−αttβ dt = − x de−αtβ = 1 − e−αxβ , so 0 0 αβt

β n−1 β n−1 (y) = n[1−F (y)] f (y) = n e−αy αβy β−1 e−αy =

(i) In the first place, for x > 0, F (x) = β

that 1−F (x) = e−αt . Therefore fY

Rx

β

(nα)βy β−1 e−(nα)y , y > 0, which is the Weibull p.d.f. with parameter nα and β. (ii) On account on the exercise cited, we have Γ EY =

C.11.5

66

1 β +1

(nα)1/β

σ 2 (Y ) =

Γ

2 β +1

h i2 − Γ β1 + 1

(nα)2/β

(i) From the transformations u = x, v = xy, we get x = u/v and y = v, and the Jacobian 1/v −u/v 2 = 1/v. From u/v = x < 1, we get u < v. of the transformations is: 0 1 Therefore, u 1 u fU,V (u, v) = + v × = 2 + 1, 0 < u < v < 1. v v v

Chapter 11

67

R1 R1 +(1−u) = u −1 + u1 +(1−u) = −u+1+1−u = (ii) Here fU (u) = u vu2 + 1 dv = u u dv v2 2(1 − u); i.e.; fU (u) = 2(1 − u), 0 < u < 1.

R∞ α R R α ∞ x−α−1 dx = −β α ∞ dx−α = −β α × 1 ∞ = β α = 1.

C.11.6 (i) Indeed, β xαβ α+1 dx = αβ xα β βα β β α

αβ α (ii) From y = βx , we have x = βy, dx = βdy, y ∈ (1, ∞). Then fY (y) = β α+1 × β = yα+1 , y α+1 y > 1, as asserted.

R∞ R R α ∞ 1 α ∞ x−α dx = − αβ α ∞ dx−α+1 = − αβ α × = (iii) EX = β x × xαβ α+1 dx = αβ α−1 β α−1 β xα−1 β

αβ α αβ 1 α−1 × β α−1 = α−1 , provided α > 1.

R∞ α ∞ αβ α αβ α R ∞ 1 −α+2 = − αβ α × 1 = α−2 Next, EX 2 = β x2 × xαβ = × β α−2 α+1 dx = −α+2 β dx α−2 xα−2 β 2 αβ α−2 , provided α > 2.

2 αβ 2 (α−1)2 −α2 β 2 (α−2) α2 β 2 αβ 2 Therefore σ 2 (X) = α−2 − (α−1) = (α−1)αβ2 (α−2) . That is, EX = 2 = (α−1)2 (α−2) αβ αβ 2 2 2 α−1 (α > 1), σ (X) = (α−1)2 (α−2) (α > 2). The EY and σ (Y ) follow from the results

just established by setting β = 1.

C.11.7

(i) From u = x, v = xy, w = xyz, we get: x = u, y = v/u, z = w/v, and J = 1/uv.

Furthermore, from 0 < x < 1, 0 < y < 1, 0 < z < 1, we get 0 < u < 1, 0 < uv < 1 or 0 < v < u, 0 < wv < 1 or 0 < w < v. Combining these results, we have: 0 < w < v < u < 1.

Next, v w 1 8w fU,V,W (u, v, w) = 8 × u × × × = ; i.e., u v uv uv fU,V,W (u, v, w) = 8w uv , 0 < w < v < u < 1.

Rv w Ru v R1 2 2 1 v 1 (ii) 0 uv dw = uv × v2 = 2u , 0 2u dv = 2u × u2 = u4 , 0 u4 du = 14 × 21 = 18 , so that Z 1Z uZ v 8w dwdvdu = 1.

0 0 0 uv

C.11.8

(i) Except for the constant 120, we have:

Z y3 Z y2 Z y5 Z 1 1 y32 y3 × − y3 + , dy1 dy2 × dy4 dy5 = 2 2 2 0 0 y3 y3 so that y2 fY3 (y) = 120 × 2

1 y2 −y+ 2 2

= 30y 2 (1 − 2y + y 2 ), 0 < y < 1.

(ii) Here Z 2/3

1 2/3 1 4 2/3 1 5 2/3 − y + y y 2 (1 − 2y + y 2 )dy = y 3 3 1/3 2 1/3 5 1/3 1/3 1 7 1 15 1 31 = × − × + × 3 27 2 31 5 243 141 = ' 0.058.

2, 430

68

C.11.9

Chapter 11

(i) We have Z ∞

1 √

2

2

e−(log x−log α) /2β dx

xβ 2π Z ∞ 1 2 2 √ = e−(y−log α) /2β × ey dy y e β 2π −∞ Z ∞ 1 2 2 √ e−(y−log α) /2β dy = 1, = −∞ β 2π 0

(by setting log x = y y so that x = ey , dx dy = e , y ∈ <)

since the integrand is the p.d.f. of the N (log α, β 2 ). (ii) With Y = log X, and y ∈ <, we have: FY (y) = P (Y ≤ y) = P (log X ≤ y) = P (X ≤ ey ) = FX (ey ), so that d FX (ey ) = fX (ey ) × ey dy 1 y 2 2 √ e−(log e −log α) /2β × ey = ey β 2π 1 2 2 = √ e−(y−log α) /2β , β 2π

fY (y) =

which is the p.d.f. of N (log α, β 2 ).

C.11.10

(i) Z ∞ x×

EX = Z0 ∞

1 2 2 √ e−(log x−log α) /2β dx xβ 2π

1 2 2 √ e−(log x−log α) /2β dx β 2π Z0 ∞ 1 2 2 = ey × √ e−(y−log α) /2β dy β 2π −∞

=

(by setting log x = y, so that x = ey , dx = ey dy, y ∈ <).

If Y ∼ N (log α, β 2 ), then the last expression above is MY (1). However, MY (t) = β 2 t2

β2

2

2

e(log α)t+ 2 ; thus, MY (1) = elog α+ 2 = αeβ /2 ; i.e., EX = αeβ /2 . (ii) 2

Z ∞

x 2 2 √ e−(log x−log α) /2β dx β 2π Z0 ∞ 1 2 2 = e2y × √ e−(y−log α) /2β dy β 2π −∞

EX =

2

2

2

(as above) 2

= MY (2) = e2 log α+2β = elog α +2β = α2 e2β ; 2

i.e., EX 2 = α2 e2β .

Chapter 11

69

2

2

2

2

(iii) Therefore σ 2 (X) = EX 2 − (EX)2 = α2 e2β − α2 eβ = α2 eβ (eβ − 1); i.e., σ 2 (X) = 2 2 α2 eβ (eβ − 1).

C.11.11

−x

, so that Theorem 11 gives: (i) For 0 < x < c, F (X) = 1−e 1−e−c n−1 −y1 e−y1 ne−y1 (e−y1 − e−c )n−1 e − e−c × = , g1 (y1 ) = n 1 − e−c 1 − e−c (1 − e−c )n

gn (yn ) = n

1 − e−yn 1 − e−c

n−1 ×

e−yn ne−yn (1 − e−yn )n−1 = , 1 − e−c (1 − e−c )n

0 < y1 < c,

0 < yn < c.

(ii) By the exercise cited,

g1n (y1 , yn ) = n(n − 1) =

1 − e−yn 1 − e−y1 − −c 1−e 1 − e−c

n−2 ×

e−y1 −yn (1 − e−c )2

n(n − 1) −y1 −yn −y1 e (e − e−yn )n−2 , 0 < y1 < yn < c.

(1 − e−c )n

(iii) We have R = Yn − Y1 , and let S = Y1 , so r = yn − y1 , s = y1 , and y1 = s, yn = r + s 0 1 with 0 < r < c, 0 < s < c − r; also, J = = −1. So, 1 0 n−2 n(n − 1) −r−2s −s e e − e−r−s −c n−2 (1 − e ) n(n − 1) −r = e (1 − e−r )n−2 en−s , 0 < r < c, 0 < s < c − r.

(1 − e−c )n−2

fR,S (r, s) =

Hence, Z c−r

n(n − 1) −r −r n−2 e 1−e e−ns ds fR (r) = (1 − e−c )n−2 0 i

n(n − 1) −r 1h −r n−2 −n(c−r) = e 1 − e × 1 − e (1 − e−c )n−2 n n−2 (n − 1) = e−r 1 − e−r (1 − e−nc e−nr ), 0 < r < c. −c n−2 (1 − e ) C.11.12

(i) Z y5 Z y4 Z y3 g(y1 , y5 ) = 120 y1

e−(y1 +...+y5 ) dy2 dy3 dy4

y1

= 120e−y1 −y5

y Z y51Z y4 Z y3 y1

y1

y1

e−y2 e−y3 e−y4 dy2 dy3 dy4 .

70

Chapter 11

Next, Z y3

Z y4

e−y3

e−y2 dy2 = e−y1 − e−y3 , y1 Z y4 Z y4

−y3 −y1 −y3 −y1 e dy3 − e−2y3 dy3 e −e dy3 = e y1

y1

y1

1 1 = e−2y1 − e−y1 −y4 + e−2y4 , 2 2

Z y5 1 −2y1 1 −2y4 −y4 −y1 −y4 e dy4 e −e e + e 2 2 y1 Z y5 Z y5 Z 1 1 y5 −3y4 e−2y4 dy4 + = e−2y1 e−y4 dy4 − e−y1 e dy4 2 2 y1 y1 y1 1 1 1 1 = e−3y1 − e−3y5 − e−2y1 −y5 + e−y1 −2y5 .

6 6 2 2 Therefore,

1 −3y1 1 −3y5 1 −2y1 −y5 1 −y1 −2y5 e g(y1 , y5 ) = 120e − e − e + e 6 6 2 2

−(y1 +y5 ) −3y1 −2y1 −y5 −y1 −2y5 = 20e e − 3e e + 3e e − e−3y5 −y1 −y5

= 20e−(y1 +y5 ) (e−y1 − e−y5 )3 ; i.e., g(y1 , y5 ) = 20e−(y1 +y5 ) (e−y1 − e−y5 )3 , 0 < y1 < y5 . (ii) From R = Y5 − Y1 , S = Y1 , we have r = y5 − y1 , s = y1 , so that y1 = s, y5 = r + s, and 0 1 J= = −1; also, r > 0, s > 0. Therefore: 1 1 fR,S (r, s) = 20e−(r+2s) (e−s − e−r e−s )3 = 20e−(r+2s) e−3s (1 − e−r )3 = 20e−5s e−r (1 − e−r )3 = 20e−5s e−r (1 − 3e−r + 3e−2r − e−3r )

= 20 e−r − 3e−2r + 3e−3r − e−4r e−5s ; i.e., fR,S (r, s) = 20(e−r − 3e−2r + 3e−3r − e−4r )e−5s , r > 0, s > 0. (iii) From part (ii), fR (r) = 20(e−r − 3e−2r + 3e−3r − e−4r )

Z ∞ 0

= −4(e

−r

−2r

− 3e

+ 3e

−3r

−4r

−e

)e−5s

e−5s ds ∞ 0

= 4(e−r − 3e−2r + 3e−3r − e−4r ) × 1, or fR (r) = 4(e−r − 3e−2r + 3e−3r − e−4r ), r > 0.

Chapter 11

71

R∞ R∞ R∞ R∞ (iv) ER = 4 0 re−r dr − 3 0 re−2r dr + 3 0 re−3r dr − 0 re−4r dr .

Observing that the integrals are expectations of Negative Exponential distributions, we have: Z ∞ re−r dr = 1, Z0 ∞ Z ∞ re−2r dr = 1/2 × r(2e−2r )dr = 1/2 × 1/2 = 1/4, Z0 ∞ Z0 ∞ r(3e−3r )dr = 1/3 × 1/3 = 1/9, re−3r dr = 1/3 × 0 0 Z ∞ Z ∞ −4r re dr = 1/4 × r(4e−4r )dr = 1/4 × 1/4 = 1/16.

0

0

Therefore. ER = 4 (1 − 3/4 + 3/9 − 1/16) = 25/12; i.e., ER = 25/12. (y−x)2

C.11.13

(y−x)2

1 1 (i) First, fX,Y (x, y) = fY |X (y|x)fX (x) = √2πx e− 2x2 × 1 = √2πx e− 2x2 , 0 < x < 1, y ∈ <.

The transformation u = y/x, v = x give: y = uv, x = v, and the relevant Jacobian is v u = v, so that 0 1

fU,V (u, v) = √

(uv−v)2 (u−1)2 1 1 e− 2v2 × v = √ e− 2 , u ∈ <, 0 < v < 1.

2πv 2π

(ii) From part (ii), fU (u) = √12π e−

(u−1)2 2

R1

0 dv =

√1 e− 2π

(u−1)2 2

, which is N (1, 1).

(iii) The r.v.’s U and V are independent, since fU,V (u, v) = √12π e− fU (u) × fV (v), u ∈ <, 0 < v < 1.

(u−1)2 2

= √12π e−

(u−1)2 2

×1 =

Chapter 12

C.12.1 Indeed, as n → ∞: FXn ,Yn (x, y) = P (Xn ≤ x, Yn ≤ y) ≤ P (Yn ≤ y) → 0, y < c. d

P

This is so, because Yn −→ c implies Yn −→ c, and the d.f. of c is: Fc (y) = 0, y < c, and Fc (y) = 1, y ≥ c. For y ≥ c, FXn ,Yn (x, y) = P (Xn ≤ x, Yn ≤ y) = P (Xn ≤ x) = FXn (x) −→ FX (x) for continuity point x of FX .

C.12.2

(i) S100 ∼ B(100, 0.1), because if X1 , . . . , Xk are independent and Xi ∼ B(ni , p), then X1 + . . . + Xk ∼ B(n1 + . . . + nk , p). (ii) ESn = np, V ar(Sn ) = npq, so that here ES100 = 100 × 0.1 = 10, V ar(S100 ) = 100 × 0.1 × 0.9 = 9.

16 84 ' 0.019292. (iii) P (S100 = 16) = 100 16 (0.1) (0.9)

S100 √ −np ≤ 16−10 (iv) P (S100 = 16) = P (15 < S100 ≤ 16) = P 15−10 < ' P (1.67 < 3 npq 3 S100 √ −np ≤ 2) = Φ(2) − Φ(1.67) = 0.977250 − 0.952540 = 0.02471; i.e., P (S100 = 16) ' npq

0.02471.

C.12.3

(i) Here Sn ∼ B(n, p), so that ESn = np, σ 2 (Sn ) = npq. Since n is large, the CLT applies and gives P (Sn = k) = P (k − 0.5 < Sn ≤ k + 0.5)

k − 0.5 − np Sn − np k + 0.5 − np =P < √ < √ √ npq npq npq

k + 0.5 − np k − 0.5 − np 'Φ − Φ , √ √ npq npq or P (Sn = k) ' Φ

k+0.5−np √ npq

−Φ

k−0.5−np √ npq

.

√ √ (ii) For n = 50, np = 25, npq = 25 × 12 , npq = 2.5 2. Therefore: √ √ k = 10: Φ(3.1√2) − Φ(2.9√2) ' 0; k = 15: Φ(2.1√2) − Φ(1.9√2) ' 0.002084; k = 20: Φ(1.1√2) − Φ(0.9 √ 2) ' 0.096005; k = 25: Φ(0.1 2) − Φ(−0.1 2) ' 0.111340.

72

Chapter 12

C.12.4

73

(i) By assuming that n is sufficiently large for the CLT to apply, we have: P [Sn > (c + 1)nµ] = P (Sn > cnµ + nµ) = P (Sn − nµ > cnµ)

√ Sn − nµ Sn − nµ cnµ cµ n √ √ =P =P > √ > σ σ n σ n σ n √

√ cµ n Sn − nµ cµ n √ '1−Φ ; i.e., =1−P ≤ σ σ σ n P [Sn > (c + 1)nµ] ' 1 − Φ

√ cµ n . σ

√ √ √ n = 0.05 n, so that (ii) Here cµσ n = 0.01×500 100

√ √ P (Sn > 505 n) ' 1 − Φ(0.05 n). (iii) For n = 1, 000: P (S1,000 > 505, 000) ' 1 − Φ(1.58) = 1 − 0.942947 = 0.057063.

For n = 5, 000: P (S5,000 > 2, 525, 000) ' 1 − Φ(3.54) ' 0. √ Pn √ P d n X C.12.5 Observe that Yn = 1 Pni=1X 2i and, as n → ∞, n ni=1 Xi −→ Z ∼ N (0, 1) by the CLT, i=1 i n P P whereas n1 ni=1 Xi2 −→ EXi2 = σ 2 (Xi ) = 1 by the WLLN. Then their ratio, which is Yn converges in distribution to Z1 = Z ∼ N (0, 1) by the Skutsky theorem (Theorem 6(iii), √ Pn √ P d d n Xi Chapter 12). Similarly, Zn = q 1 Pi=1 −→ Z ∼ N (0, 1), since as n → ∞, n ni=1 Xi −→ n 2 n

1 Pn

Z ∼ N (0, 1), and n

i=1 Xi

P

2 i=1 Xi −→ 1 by Theorem 5(i), Chapter 12.

2 4 20 10 10 1 1 16 2 + 18 = 1, EX12 = 16 C.12.6 EX1 = 18 18 + 18 = 18 = 9 , so that σ (X1 ) = 9 − 1 = 9 and σ(X1 ) = 3 . √

Set Sn = X1 + . . . + Xn . Then ESn = n, σ(Sn ) =

n 3 , and Tn = 100Sn .

(i) P (Tn > 97n) = P (100Sn > 97n) = P (Sn > 0.97n)

Sn − ESn 0.97n − n =P > √ σ(Sn ) n/3

√ Sn − ESn =P > −0.09 n σ(Sn )

√ Sn − ESn =1−P ≤ −0.09 n σ(Sn ) √ √ ' 1 − Φ(−0.09 n) = Φ(0.09 n); i.e., √ P (Tn > 97n) ' Φ(0.09 n).

74

Chapter 12

(ii) P (995, 000 < Tn < 1, 005, 000) = P (995, 000 < 100Sn ≤ 1, 005, 000) = P (9, 950 < Sn ≤ 10, 050)

9, 950 − 10, 000 Sn − ESn 10, 050 − 10, 000 =P < ≤ 100/3 σ(Sn ) 100/3 ' Φ(1.5) − Φ(−1.5) = 2Φ(1.5) − 1 = 2 × 0.933193 − 1 = 0.866386; i.e., for n = 10, 000, P (995, 000 < Tn < 1, 005, 000) ' 0.866386.

C.12.7

(i) The Xi ’s are i.i.d. with p.d.f. f (x) = λe−λx , x > 0, so that EX1 = 1/λ, σ 2 (X1 ) = 1/λ2 , √ and σ(X1 ) = 1/λ. Then ESn = n/λ, σ 2 (Sn ) = n/λ2 , and σ(Sn ) = n/λ. Therefore:

√ kn/λ − n/λ Sn − n/λ Sn − n/λ √ √ √ > > (k − 1) n P (Sn > k × n/λ) = P =P n/λ n/λ n/λ

√ √ Sn − n/λ √ =1−P ≤ (k − 1) n ' 1 − Φ[(k − 1) n]. n/λ

(ii) n = 40,

C.12.8

k = 1.2 k = 1.4, k = 1.5,

1 − Φ(1.26) = 1 − 0.896165 = 0.103835 1 − Φ(2.53) = 1 − 0.994297 = 0.005703 1 − Φ(3.16) = 1 − 0.999211 = 0.000789.

(i) Xi ∼ B(1, 0.25), and assume that they are independent. Then X ∼ B(750, 0.25). Therefore

200 X 750 (0.25)x (0.75)750−x .

P (X > 200) = 1 − P (X ≤ 200) = 1 − x x=0

(ii) √ Since EX = np = 750 × 0.25 = 187.5, σ 2 (X) = npq = 750 × 0.25 × 0.75 = 140.625, 140.625 ' 11.858, we have, by the CLT: P (X > 200) = 1 − P (X ≤ 200) = 1 − P (−0.5 < X ≤ 200) ' 1 − [Φ(b∗ ) − Φ(a∗ )], where a∗ = −0.5−187.5 ' −15.85, b∗ = 200−187.5 11.858 11.858 ' 1.05, so that: P (X > 200) ' 1 − Φ(1.05) + Φ(−15.85) = 1 − Φ(1.05) = 1 − 0.853141 = 0.146859. √

C.12.9

(i) Here Sn ∼ B(n, p = 9/19), so that ESn = 9n/19 and σ 2 (Sn ) = 9×10n , σ(Sn ) = 3 1910n .

192

Chapter 12

75

Therefore: P (Sn > c) = 1 − P (Sn ≤ c) = 1 − P (−0.5 < Sn ≤ c) −0.5 − 9n 19

=1−P

√ 3 10n 19

c − 9n Sn − np < √ ≤ √ 19 3 10n npq

!

19

9.5 + 9n Sn − np 19c − 9n =1−P − √ < √ ≤ √ npq 3 10n 3 10n

19c − 9n 9.5 + 9n √ '1−Φ −Φ − √ 3 10n 3 10n

19c − 9n 9.5 + 9n √ √ −Φ ; =Φ 3 10n 3 10n

19c−9n √ √ i.e., P (Sn > c) ' Φ 9.5+9n − Φ .

3 10n 3 10n

(ii) For n = 2k, k ≥ 1 integer, and c = k − 1, by part (i), we have " # " # 9.5 + 9(2k) 19(k − 1) − 9(2k) p p P (S2k > k − 1) ' Φ −Φ 3 10(2k) 3 10(2k)

9.5 + 18k k − 19 √ √ −Φ . =Φ 6 5k 6 5k (iii) For k = 20, we have 2k = 40 and:

369.5 1 P (S40 > 19) = Φ −Φ 60 60 ' Φ(6.158) − Φ(0.017)

' 1 − Φ(0.017) ' 1 − 0.5067813 (by linear interpolation) = 0.493187.

Revised Answers Manual to Introduction to Probability

George G. Roussas University of California, Davis

Chapter 2 Section 2.2 2.1 (i)

S = {(3, 0, 0), (2, 1, 0), (2, 0, 1), (1, 2, 0), (1, 0, 2), (1, 1, 1), (0, 2, 1), (0, 1, 2), (0, 3, 0), (0, 0, 3)}.

A = {(1, 1, 1)},

(ii)

B = {(3, 0, 0)}, C = {(3, 0, 0), (0, 3, 0), (0, 0, 3)}, D = {(2, 1, 0), (2, 0, 1), (3, 0, 0)}, E = {(0, 1, 2), (0, 2, 1)}.

2.2 (i)

S = {(r , r , r ), (r , r , b), (r , r , g), (r , b, r ), (r , b, b), (r , b, g), (r , g, r ), (r , g, b), (r , g, g), (b, r , r ), (b, r , b), (b, r , g), (b, b, r ), (b, b, b), (b, b, g), (b, g, r ), (b, g, b), (b, g, g), (g, r , r ), (g, r , b), (g, r , g), (g, b, r ), (g, b, b), (g, b, g), (g, g, r ), (g, g, b), (g, g, g)}.

(ii)

A = {(r , b, g), (r , g, b), (b, r , g), (b, g, r ), (g, r , b), (g, b, r )}, B = {(r , r , b), (r , r , g), (r , b, r ), (r , b, b), (r , g, r ), (r , g, g), (b, r , r ), (b, r , b), (b, b, r ), (b, b, g), (b, g, b), (b, g, g), (g, r , r ), (g, r , g), (g, b, b), (g, b, g), (g, g, r ), (g, g, b)}, C = A ∪ B = S − {(r , r , r ), (b, b, b), (g, g, g)}.

2.3 (i)

S = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)}, or S = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}.

Introduction to Probability, Second Edition. http://dx.doi.org/10.1016/B978-0-12-800041-0.00023-7 Copyright © 2014 Elsevier Inc. All rights reserved.

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(ii) For the case of the sample space S: A = {(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}, B = {(4, 5), (5, 4)}, C = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}, D = A. For the case of the sample space S : A = {(1, 2), (1, 3), (2, 3)}, B = {(4, 5)}, C = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}, D = A. Note The sample space S is more suitable in describing this experiment, because it reflects more closely the randomness stipulated. Also, it makes it easier to assign probabilities later on.

2.4 (i) Denoting by (x1 , x2 ) the cars sold in the first and the second sale, we have: S = {(a1 , a1 ), (a1 , a2 ), (a1 , a3 ), (a2 , a1 ), (a2 , a2 ), (a2 , a3 ), (a3 , a1 ), (a3 , a2 ), (a3 , a3 ), (a1 , b1 ), (a1 , b2 ), (a2 , b1 ), (a2 , b2 ), (a3 , b1 ), (a3 , b2 ), (a1 , c), (a2 , c), (a3 , c), (b1 , a1 ), (b1 , a2 ), (b1 , a3 ), (b2 , a1 ), (b2 , a2 ), (b2 , a3 ), (b1 , b1 ), (b1 , b2 ), (b2 , b1 ), (b2 , b2 ), (b1 , c), (b2 , c), (c, a1 ), (c, a2 ), (c, a3 ), (c, b1 ), (c, b2 ), (c, c)}. (ii)

A = {(a1 , a1 ), (a1 , a2 ), (a1 , a3 ), (a2 , a1 ), (a2 , a2 ), (a2 , a3 ), (a3 , a1 ), (a3 , a2 ), (a3 , a3 )}, B = {(a1 , b1 ), (a1 , b2 ), (a2 , b1 ), (a2 , b2 ), (a3 , b1 ), (a3 , b2 )}, C = B ∪ {(b1 , a1 ), (b1 , a2 ), (b1 , a3 ), (b2 , a1 ), (b2 , a2 ), (b2 , a3 )}, D = {(c, b1 ), (c, b2 ), (b1 , c), (b2 , c)}.

2.5 (i) S = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}. (ii) A = {(3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}, B = {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}, C = {(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (1, 2), (1, 3), (1, 4), (1, 5),

Revised Answers Manual to Introduction to Probability

(1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)}, D = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (4, 0)}.

2.6 E = Ac , F = C − D = C ∩ D c , G = B − C = B ∩ C c , H = Ac − B = Ac ∩ B c = (A ∪ B)c , I = B c .

2.7 S = {(−3, 0), (−3, 1), (−3, 2), (−3, 3), (−3, 4), (−2, 0), (−2, 1), (−2, 2), (−2, 3), (−2, 4), (−1, 0), (−1, 1), (−1, 2), (−1, 3), (−1, 4), (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (3, 0), (3, 1), (3, 2), (3, 3), (3, 4)}, A = {(0, 0), (1, 1), (2, 2), (3, 3)}, B = {(−3, 3), (−2, 2), (−1, 1), (0, 0)}, C = {(−3, 3), (−2, 2), (−1, 1), (0, 0), (1, 1), (2, 2), (3, 3)}, D = {(−2, 0), (−2, 1), (−1, 0), (−1, 1), (−1, 2), (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1)}.

2.8 (i) (ii) (iii) (iv) (v) (vi)

B0 = Ac1 ∩ Ac2 ∩ Ac3 .

B1 = (A1 ∩ Ac2 ∩ Ac3 ) ∪ (Ac1 ∩ A2 ∩ Ac3 ) ∪ (Ac1 ∩ Ac2 ∩ A3 ).

B2 = (A1 ∩ A2 ∩ Ac3 ) ∪ (A1 ∩ Ac2 ∩ A3 ) ∪ (Ac1 ∩ A2 ∩ A3 ).

B3 = A1 ∩ A2 ∩ A3 .

C = B0 ∪ B1 ∪ B2 .

D = B1 ∪ B2 ∪ B3 = A1 ∪ A2 ∪ A3 .

2.9 If A ∪ B ∪ C = A, then B ∪ C ⊆ A, which also implies that B ⊆ A and C ⊆ A. If A ∩ B ∩ C = A, then A ⊆ B ∩ C, which also implies that A ⊆ B and A ⊆ C. Thus, A = B = C. 2.10 If A = ∅, then A∩ B c = ∅, Ac ∩B = S ∩ B = B, so that (A∩ B c )∪(Ac ∩ B) = B for every B. Next, let (A ∩ B c ) ∪ (Ac ∩ B) = B and take B = ∅ to obtain A ∩ B c = Ac ∩ B = ∅. But A ∩ B c = A ∩ S = A, so that A = ∅.

2.11 (i) (A − B) ∪ B = (A ∩ B c ) ∪ B is correct, but (A ∩ B c ) ∪ B = B is incorrect, unless A = ∅. Indeed, if s ∈ ((A ∩ B c ) ∪ B), then s ∈ (A ∩ B c ) or s ∈ B but not both, since (A ∩ B c ) ∩ B = ∅. So, if s ∈ (A ∩ B c ), then s ∈ B c and hence s ∈ / B. (ii) (A ∪ B) − A = (A ∪ B) ∩ Ac is correct, but (A ∪ B) ∩ Ac = B is incorrect, unless at least one of A or B c = ∅. Indeed, if s ∈ B, then s ∈ (A ∪ B).

Thus, if s ∈ A, then s ∈ / Ac unless at least one of A or B c is ∅, and hence s∈ / ((A ∪ B) ∩ Ac ). (iii) (A∩B)∩(A−B) = (A∩B)∩(A∩B c ) is correct, and (A∩B)∩(A∩B c ) = ∅ is also correct, since (A ∩ B) ∩ (A ∩ B c ) = A ∩ (B ∩ (A ∩ B c )) = A ∩ ((A ∩ B c ) ∩ B) = A ∩ (A ∩ (B c ∩ B)) = A ∩ ∅ = ∅.

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(iv) Correct. Indeed, apply the distributive law as needed, to obtain: (A ∪ B)∩(B ∪C)∩(C ∪ A) = [(A ∪ B)∩(B ∪C)]∩(C ∪ A) = [(A ∪ B)∩ B ∪(A∪ B)∩C]∩(C ∪ A) = [(A∩ B)∪ B ∪(A∩C)∪(B ∩C)]∩(C ∪ A) = [B ∪ (A ∩ C)] ∩ (C ∪ A) = [B ∪ (A ∩ C)] ∩ C ∪ [B ∪ (A ∩ C)] ∩ A = (B ∩ C) ∪ (A ∩ C) ∪ (A ∩ B) ∪ (A ∩ C) = (A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A).

Alternatively, one may show easily that the event on either side is contained in the event on the other side, so that equality follows.

2.12 A ⊆ B implies that, for every s ∈ A, we have s ∈ B, whereas B ⊆ C implies that, for every s ∈ B, we have s ∈ C. Thus, for every s ∈ A, we have s ∈ C, so that A ⊆ C. 2.13 For s ∈ A ∩ (∪ j A j ), we have s ∈ A and s ∈ ∪ j A j , so that s ∈ A j for at least one j, say, j0 . Then s ∈ (A ∩ A j0 ), so that s ∈ ∪ j (A ∩ A j ). Next, if s ∈ ∪ j (A ∩ A j ), then s ∈ (A ∩ A j ) for at least one j, say, j0 , so that s ∈ A and s ∈ A j0 . It follows that s ∈ A and s ∈ ∪ j A j , so that s ∈ A ∩ (∪ j A j ).

The first distributive law is established. The validity of the second follows similarly.

2.14 For s ∈ ∪ j A j , let j0 ≥ 1 be the first j for which s ∈ A j0 . Then, if j0 = 1, it follows that s ∈ A1 and therefore s belongs in the right-hand side of the / A j , j = 1, . . . , j0 − 1, but s ∈ A j0 , so that relation. If j0 > 1, then s ∈ s ∈ Ac1 ∩ · · · ∩ Acj0 −1 ∩ A j0 and hence s belongs to the right-hand side of the relation. Next, let s belong to the right-hand side event. Then, if s ∈ A1 , it / A j for j = 1, . . . , j0 − 1 but s ∈ A j0 , it follows follows that s ∈ ∪ j A j . If s ∈ that s ∈ ∪ j A j . The identity is established. / ∪ j A j , hence s ∈ / A j for any j, so that s ∈ Acj for all 2.15 If s ∈ (∪ j A j )c , then s ∈ c c / Aj j, and thus s ∈ ∩ j A j . Next, if s ∈ ∩ j A j , then s ∈ Acj for all j, so that s ∈ for any j, hence s ∈ / ∪ j A j , so that s ∈ (∪ j A j )c . The first law is established.

The second is shown similarly.

2.16 (i) Since −5 + 1 < −5 + 1 and 20 − 1 < 20 − 1 , it follows that n+1 n n n+1 1 1 (−5 + n1 , 20 − n1 ) ⊂ (−5 + n+1 , 20 − n+1 ), or An ⊂ An+1 , so that {An } is 3 3 increasing. Likewise, 7 + n+1 < 7 + n3 , so that (0, 7 + n+1 ) ⊂ (0, 7 + n3 ), or Bn+1 ⊂ Bn ; thus, {Bn } is decreasing.

1 1 ∞ ∞ ∞ (ii) ∪∞ n=1 An = ∪n=1 (−5 + n , 20 − n ) = (−5, 20), and ∩n=1 Bn = ∩n=1 3 (0, 7 + n ) = (0, 7].

Section 2.3 3.1 Each one of the r.v.’s X i , i = 1, 2, 3 takes on the values: 0, 1, 2, 3, and X 1 + X 2 + X 3 = 3.

3.2 Each one of X and Y takes on the values: 0, 1, 2, 3 and X + Y = 3.

3.3 X takes on the values: 0, 1, 2, 3, 4, 5 and Y takes on the values: 0, 1, 2, 3, 4, 5, 6; X + Y takes on the values 0 through 11.

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3.4 X takes on the values: −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7, (X ≤ 2) = {(−3, 0), (−3, 1), (−3, 2), (−3, 3), (−3, 4), (−2, 0), (−2, 1), (−2, 2), (−2, 3), (−2, 4), (−1, 0), (−1, 1), (−1, 2), (−1, 3), (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (2, 0)}, (3 < X ≤ 5) = (4 ≤ X ≤ 5) = (X = 4 or X = 5) = {(0, 4), (1, 3), (1, 4), (2, 2), (2, 3), (3, 1), (3, 2)}, (X > 6) = (X ≥ 7) = {(3, 4)}.

3.5 The values of X are 10 through 365.

3.6 (i) S = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}. (ii) The values of X are: 2, 3, 4, 5, 6, 7, 8.

(X ≤ 3) = (X = 2 or X = 3) = {(1, 1), (1, 2), (2, 1)}, (iii) (2 ≤ X < 5) = (2 ≤ X ≤ 4) = (X = 2 or X = 3 or X = 4) = {(1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1)}, (X > 8) = ∅.

3.7 Each one of the r.v.’s X i , i = 1, 2, 3, 4 takes on the values 0 through n, and X 1 + X 2 + X 3 + X 4 = n.

3.8 (i) S = [8:00, 8:15]. (ii) The values of X consist of the interval [8:00, 8:15]. (iii) The event described is the interval [8:10, 8:15].

Section 2.4 4.1 The total number of telephone numbers whose first 3 digits are specified (752 here) is: 10 ×10 ×10 ×10 = 104 = 10,000. So, there aren’t enough telephone numbers.

4.2 Denoting by T1 , T2 , T3 the 3 temperatures, by P1 , P2 , P3 , and P4 the 4 pressures, and by C1 , C2 , C3 , C4 , and C5 the 5 catalysts, the answers to parts (i)–(iii) are, respectively: 3 × 4 × 5 = 60; 1 × 2 × 5 = 10; 3 × 4 × 1 = 12.

4.3 (i) A zip code is formed in five stages (k = 5) and the choices in each stage are: n 1 = · · · = n 5 = 10. Therefore the required number is: 105 = 100,000. (ii) Clearly, the possible values of X are: 0, 1, 2, 3, 4, 5. (iii) X (01200) = 2, X (13057) = 4, X (95616) = 5.

4.4 The 5-digit numbers are formed by filling in the 5 positions below as prescribed.

• • • • • The total number of 5-digit numbers without any restrictions 1 2 3 4 5 is : 55 = 3,125. Next, the numbers resulting under the prescriptions given are:

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(i) 3 × 2 × 3 × 2 × 3 = 108. (ii) 3 × 2 × 2 × 1 × 1 = 12.

4.5 The answers are as follows: (i) 10 × 10 × 10 = 103 = 1,000. (ii) 10 × 9 × 8 = 720. (iii) 1 × 10 × 1 = 10 with repetition, and 1 × 8 × 1 = 8 without repetitions.

4.6 Since for each one of the n spots there are 2 choices, the required number is: 2n .

For the specified values of n, we have: 25 = 32, 210 = 1,024, 215 = 32,768, and 225 =

220 = 1,048,576,

n33,554,432. n n! n! = (n−m)!m! , so that equality holds. Alterna4.7 First, m = m!(n−m)! and n−m tively, to each choice of m out of n, there corresponds a choice of n − m out of n, thus equality follows again. n+1 n (n+1)! n!(n+1) m! / m = (m+1)!(n−m)! × m!(n−m)! = (n+1)! × 4.8 We have: m+1 n! n! × (m+1)! = n! (n+1) m! m!(m+1) = (m+1) .

4.9 Indeed,

M −1 M −1 (M − 1)!

(M − 1)! + = + m m−1 m!(M − m − 1)! (m − 1)!(M − m)!

(M − 1)!M M! M (M − 1)!

(M − m + m) = = = . = m!(M − m)! m!(M − m)! m!(M − m)! m

4.10 In the first place, mx = 0 for x = m + 1, . . . , r if r ≥ m + 1. So, we may

ways can be parceled assume that r ≤ m. Now, selecting r out of m +n in m+n r n m out as choosing x from m in x ways and the remaining r − x from n in r −x n ways, so that mx × r −x is the number of ways of choosing x from m and r − x from n. Summing up those terms, by letting x run from 0 through r , we

get all possible ways of apportioning r among m and n; i.e., we get m+n r .

4.11 (i) The binomial expansion of (x +y)n is the summation of all terms of the form x k y n−k , k = 0, 1, . . . , n. These terms may be thought of being formed as follows: Out of n positions, select k and place x’s; in the remaining n − k positions place y’s. Multiplying them out, we get x k y n−k . There are nk ways of selecting k places for the x’s, and all these selections produce the same product x k y n−k . Thus, there are nk x k y n−k such terms. Letting k go from 0 to n, and summing up these terms, we get the desired result. (ii) Set x = y = 1 and x = −y = 1, respectively.

4.12 The required number is n3 , which for n = 10 becomes 10 3 = 120.

4.13 (i) The required number here is: 9 27 15 = 9 × 27 × 15 = 3,645.

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(ii) The required number here is the number in part (i) multiplied by 3! So, 3,645 × 3! = 3,645 × 6 = 21,870.

Revised Answers Manual to Introduction to Probability

4.14 The required number is found by pairing out the n countries in n2 ways and then this by 2 to account for the 2 ambassadors involved. Thus, n multiplying n(n−1) × 2 = × 2 = n(n−1) × 2 = n(n − 1). For the given values of n, we 2!

2 2 have, respectively, 9 × 10 = 90, 49 × 50 = 2,450, 99 × 100 = 9,900.

4.15 (i) 10 × 263 × 103 = 175,760,000. (ii) 10 × (26 × 25 × 24) × 103 = 156,000,000. (iii) 10 × 263 × (9 × 8 × 7) = 88,583,040. (iv) 10 × (26 × 25 × 24) × (9 × 8 × 7) = 768,624,000.

Chapter 3 Section 3.2 2.1 (i) From P(A ∪ B) = P(A) + P(B) − P(A ∩ B), we have P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.4 + 0.6 − 0.7 = 0.3. (ii) Let S = {1, 2, 3, 4, 5, 6}, P({i}) = 1/6, i = 1, . . . , 6, A = {1, 2, 3}, B = {2, 4, 5}, so that A− B = {1, 3}. Then P(A) = P(B) = 36 , P(A− B) = 26 , and 26 = 0 = 36 − 36 = P(A) − P(B).

2.2 Since A ∪ B ⊇ A, we have P(A ∪ B) ≥ P(A) = 43 . Also, A ∩ B ⊆ B implies P(A ∩ B) ≤ P(B) = 38 . Finally, P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 3 3 9 9 1 4 + 8 − P(A ∪ B) = 8 − P(A ∪ B) ≥ 8 − 1 = 8 .

2.3 Since A ⊆ A∪ B, it follows that 1 = P(A) ≤ P(A∪ B), so that P(A∪ B) = 1.

Likewise, P(A ∪ C) = P(B ∪ C) = 1, and from A ∪ B ⊆ A ∪ B ∪ C, we get 1 = P(A ∪ B) ≤ P(A ∪ B ∪ C), so that P(A ∪ B ∪ C) = 1. Then, from P(A ∪ B) = P(A) + P(B) − P(A ∩ B), we get 1 = 1 + 1 − P(A ∩ B), so that P(A ∩ B) = 1, and likewise, P(A ∩ C) = P(B ∩ C) = 1. Finally, from P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C), we get: 1 = 1 + 1 + 1 − 1 − 1 − 1 + P(A ∩ B ∩ C), so that P(A ∩ B ∩ C) = 1.

2.4 We have: Ac ∩ B = B ∩ Ac = B − A and A ⊂ B. Therefore P(Ac ∩ 5 − 41 = 16 0.167. Likewise, B) = P(B − A) = P(B) − P(A) = 12 c c A ∩ C = C − A with A ⊂ C, so that P(A ∩ C) = P(C − A) = P(C) − 7 − 14 = 13 0.333, B c ∩ C = C − B with B ⊂ C, so that P(A) = 12 7 5 c P(B ∩ C) = P(C − B) = P(C) − P(B) = 12 − 12 = 16 0.167. Next, c c c c c A ∩ B ∩ C = A ∩ (B ∩ C ) = A ∩ (B ∪ C) = A ∩ C c = A − C = ∅, so that P(A ∩ B c ∩ C c ) = 0, and Ac ∩ B c ∩ C c = (A ∪ B ∪ C)c = C c , so that 7 5 = 12 0.417.

P(Ac ∩ B c ∩ C c ) = P(C c ) = 1 − P(C) = 1 − 12 2.5 Clearly, A = “More T ’s than H ’s” = “at least 3 T ’s” ={TTTH, TTHT, THTT, 5 = 0.3125. Also, B = “There are both HTTT, TTTT}, so that P(A) = 16 H’s and T’s and any T precedes every H” = {TTTH, TTHH, THHH}, and 3 = 0.1875.

P(B) = 16

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2.6 The event A is defined as follows: A = “x = 7n, n = 1, . . . , 28,” so that A 28 7 consists of 28 points, and P(A) = 200 = 50 = 0.14. Likewise, B = “x = 63 = 3n +10, n = 1, . . . , 63,” so that B consists of 63 points, and so P(B) = 200 √ 2 2 0.315, and C = “x + 1 ≤ 375” = “x ≤ 374” = “x ≤ 374” = “x ≤ 19”, 19 = 0.095. so that C consists of 19 points, and then P(C) = 200 2.7 The set of all possible outcomes consists of the 36 pairs: S = {(1, 1), . . . , (1, 6), . . . , (6, 1), . . . , (6, 6)}. Then: 4 (i) “Equal to 5” = {(1, 4), (2, 3), (3, 2), (4, 1)}, so that P(“Equal to 5”) = 36 = 1 0.111.

9 (ii) “Divisible by 3” = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}, 1 0.333. so that P(“Divisible by 3”) = 12 36 = 3 2.8 Denote by A, B, and C the events that a student reads news magazines A, B, and C, respectively. Then the required probability is P(Ac ∩ B c ∩C c ). However,

P(Ac ∩ B c ∩ C c ) = P((A ∪ B ∪ C)c ) = 1 − P(A ∪ B ∪ C) = 1 − [P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) −P(B ∩ C) + P(A ∩ B ∩ C)] = 1 − (0.20 + 0.15 + 0.10 − 0.05 − 0.04 − 0.03 + 0.02) = 1 − 0.35 = 0.65.

2.9 (i) Denoting by A and B, respectively, that the student is admitted at college A, and is rejected by college B, the required probability is: P(A ∪ B c ) = P(A) + P(B c ) − P(A ∩ B c ) = p1 + 1 − p2 − P(A ∩ B c ).

However, P(A ∩ B c ) = P((Ac ∪ B)c ) = 1 − P(Ac ∪ B) = 1 − p3 , so that P(A ∪ B c ) = p1 + 1 − p2 − 1 + p3 = p1 − p2 + p3 . (ii) P(A ∪ B c ) = 0.6 − 0.2 + 0.3 = 0.7.

2.10 From the definition of A, B, and C, we have: A = {(0, 4), (0, 6), (1, 3), (1, 5), (1, 9), (2, 2), (2, 4), (2, 8), (3, 1), (3, 3), (3, 7), (4, 0), (4, 2), (4, 6), (5, 1), (5, 5), (6, 0), (6, 4)}, B = {(0, 0), (1, 2), (2, 4), (3, 6), (4, 8)}, C = {(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 0), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 0), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 0), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 0), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (5, 8), (5, 9), (6, 0), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 7), (6, 8), (6, 9)},

Revised Answers Manual to Introduction to Probability

or C c = {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.

Therefore, since the number of points in S is 7 × 10 = 70, we have: 9 5 1 18 = 0.257, P(B) = = 0.071, 70 35 70 14 63 9 63 7 P(C) = = = 0.9, or P(C) = 1 − P(C c ) = 1 − = = 0.9.

70 10 70 70 P(A) =

2.11 In the first place, there are 11 blocks altogether and the number of ways of lining them up would be 11!, if they were distinct. In order to find the number of distinct permutations, we have to divide 11! by the product 2! × 4! × 5! to 11! . Arguing in a similar fashion, we get: obtain 2!×4!×5!

• • • • • • • • • • • R R 9!

9!

11!

The number of distinct arrangements is: 4!×5! , so that 4!×5! / 2!×4!×5! = 2!×9!

1 = 0.018.

11!

55 • • • • • • • • • • • (ii) Y Y Y Y Y 6!

For each position of the yellow blocks, there are 2!×5! distinct arrangements of the remaining blocks. Since the yellow blocks can be placed in 7 ways, so 7×6!

11! / 2!×4!×5! = 7×5!×6! = that to be adjacent, the required probability is: 2!×4!

11!

1 0.015.

66 • • • • • • • • • • • (iii) B B Place 2 blue blocks (any two, since they are indistinguishable) at the two 9! distinct arrangements. ends. Then the remaining 9 blocks produce 2!×2!×5!

Thus, the required probability is: 9!

11!

4!×9!

6 0.109.

2!×2!×5! / 2!×4!×5! = 2!×11! = 55 (i)

6!

6!

2.12 Lining up the 6 chips as they appear, there will be 1!1!2!1!1! = 2! = 360 distinct arrangements. Since only one of these arrangements produces the word 1 0.003.

“OFFICE,” the required probability is: 360 2.13 The required probability, clearly, is:

10 2

30 5 × 15 2,480,625 3 × 4 × 1 = 60 66,661,386

0.037.

10

2.14 The m draftees can be chosen in mn ways. Next, leave out the specified man

not to be drafted, and select the m draftees from the remaining n − 1 in n−1 m n (n−1)!/m!(n−m−1)! = ways. Then the required probability is: n−1 m / m = n!/m!(n−m)! (n−1)!(n−m)! (n−1)!(n−m−1)!(n−m) m = n−m n!(n−m−1)! = (n−1)!n(n−m−1)! n =1− n.

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2.15 In the first place, the 3 numbers can be chosen in 16 3 3! ways. If their product is to be a negative number, there must be either 2 positive numbers and 1 negative number, or all 3 negative numbers. This happens in:

10 6 10 6 2 1 + 0 3 3! = [(45 × 6) + 20]3! = 290(3!) ways. Then the required

10 6 10 6 16 290 29

probability is: 290(3!) = = 0.518, or just 16 2 1 + 0 3 / 3 = 560 56 290 29 560 = 56

3

(3!)

0.518.

ways, and x defective can be cho2.16 The 500 bulbs can be chosen in 2,000 500 1,800 200 sen in x ways, whereas the 500 − x good bulbs can be chosen in 500−x ways. Since the probability having 1,800 of 2,000 exactly x defective bulbs among the / 500 chosen is: 200 500 , the required probability is given by: 200 x1,800500−x

1 25 2,000 .

x=0 x 500−x 500

2.17 Clearly, the required probabilities are given by: 26

4 48 13 13 13 3 1 2 1 P(A1 ) = 52 , P(A2 ) = 52 , P(A3 ) = 1 52 1 , 3 3 3 26 26 26 26 + P(A4 ) = 2 1 52 3 0 .

3 After some calculations, we find: 2 1,128 0.118, P(A2 ) = 0.204, 17 5,525 169 0.099, P(A4 ) = 0.5.

P(A3 ) = 1,700

P(A1 ) =

2.18 Class Freshmen Sophomore Junior Senior Totals

GENDER Male 40 30 24 12 106

Female 60 50 46 28 184

Totals 100 80 70 40 290

The number of ways of selecting 12 students is: 290 12 . Next, the number of ways of selecting the committee members under each one of the requirements in parts (i)–(iv), and the respective probabilities are: 184 106 (i)

7

290 5 12

0.218.

Revised Answers Manual to Introduction to Probability

100 80 70 40 (ii)

3

3290 3

3

0.012.

12

60 40 50 30 46 24 28 12 (iii)

2

1

2

1290 2 12

1

2

1

250

12 (iv) 1 − P(no seniors) = 1 − 290

2.19 (i)

0.0005.

1 − 0.162 = 0.838.

12

50

50! = 5!45! = 46×47×48×49×50 = 2 × 10 × 46 × 47 × 48 = 2,118,760.

2×3×4×5 30 20 30!

20! × 2!18! = 28×29×30 × 19×20 = (5×28×29)×(10×19) = (ii) 3 2 = 3!27!

2×3 2 771,400.

(iii)

5

771,400 19,285 2,118,760 = 52,969

0.364.

2.20 We have: (i)

P(S ∩ L) = 1 − P[(S ∩ L)c ] = 1 − P(S c ∪ L c ) = 1 − [P(S c ) + P(L c ) − P(S c ∩ L c )] = 1 − [1 − P(S) + 1 − P(L) − P(S c ∩ L c )] = P(S) + P(L) + P(S c ∩ L c ) − 1 = 0.25 + 0.35 + 0.45 − 1 = 0.05.

(ii)

P(S ∪ L) = P(S) + P(L) − P(S ∩ L) = 0.25 + 0.35 − 0.05 = 0.55.

2.21 Let N be the number of points in the event described. Then, under: (i) N = 1 × 1 × 26 × 26 × 1 × 1 × 10 = 10 × 262 , and the required probability 1 1 10 × 262 0.00004. = = is 10 × 263 × 103 26 × 103 26,000 (ii) N = 1 × 1 × 25 × 24 × 1 × 1 × 10 = 10 × 24 × 25, and the required 10 × 24 × 25 1 0.00004. probability is = 10 × 26 × 25 × 24 × 103 26,000 2 (iii) N = 1 × 1 × 26 × 26 × 1 × 1 × 7 = 7 × 26 , and the required probability 1 1 7 × 262 = = 0.00005. is 3 10 × 26 × 9 × 8 × 7 8 × 9 × 10 × 26 18,720 (iv) N = 1×1×25×24×1×1×7 = 7×24×25, and the required probability 1 7 × 24 × 25 = 0.00005. is 10 × 26 × 25 × 24 × 9 × 8 × 7 18,720

Section 3.3 3.1 (i) The r.v. X takes on the values: 0, 1, 2, 3.

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(ii) Its p.d.f. is: f (0) = f (3) = 18 = 0.125, f (1) = f (2) = 38 = 0.375. (iii) P(X ≥ 2) = f (2) + f (3) = 48 = 21 = 0.5, P(X ≤ 2) = 1 − P(X > 2) = 1 − f (3) = 1 − 18 = 78 = 0.875.

3.2 (i) For 0 < x ≤ 2, f (x) = d (2c(x 2 − 1 x 3 )) = 2c(2x − x 2 ). Thus, f (x) = dx 3 2c(2x − x 2 ), 0 < x ≤ 2 (and 0 elsewhere).

2 (ii) From 0 2c(2x − x 2 ) d x = 1, we get 8c 3 = 1, so that c = 3/8.

3.3 (i) For 0 < x < 1, f (x) = d (x 3 − x 2 + x) = 3x 2 − 2x + 1, so that dx f (x) = 3x 2 − 2x + 1, 0 < x < 1 (and 0 otherwise).

1 (ii) P(X > 21 ) = 1/2 (3x 2 − 2x + 1)d x = x 3 |11/2 − x 2 |11/2 + x|11/2 = 5 1 1 1 8 = 0.625. Alternatively, P(X > 2 ) = P( 2 < X ≤ 1) = F(1) − F( 2 ) = 5 8 = 0.625.

3.4 (i) F(x) 1 0.8 0.6 0.4 0.2 0

4

5

6

7

8

9

x

(ii) P(X ≤ 6.5) = 0.7, P(X > 8.1) = 1 − P(X ≤ 8.1) = 1 − 0.9 = 0.1, P(5 < X < 8) = P(X < 8) − P(X ≤ 5) = 0.7 − 0.4 = 0.3.

3.5 (i) (ii)

∞ −(c+1) d x = − ∞ d x −c = − 1 ∞ = 1 for any c > 0.

1 cx 1 xc 1 x x For x ≥ 1, F(x) = 1 ct −(c+1) dt = − t1c 1 = 1 − x1c , so that

F(x) =

0, x <1 1 1 − x c , x ≥ 1.

3.6 (i) We need two relations which are provided by: 1 (cx + d) d x = 1 and 0 1 1/2 (cx + d) d x = 1/3, or: c + 2d = 2 and 9c + 12d = 8, and hence c = − 43 , d = 53 . (ii) For 0 ≤ x ≤ 1, F(x) =

x 4 5 2x 2 5x 0 (− 3 t + 3 ) dt = − 3 + 3 . Thus,

F(x) =

⎧ ⎨

0, x <0 2 , 0 ≤ x ≤1 − 2x3 + 5x 3 ⎩ 1, x > 1.

Revised Answers Manual to Introduction to Probability

1/2 3.7 Indeed, ∞ ( 1 )x = 1−1/2 = 1/2 = 1. ∞ x 1/2 ∞ x=1 x2 1 = 1, we get c = 1 − α.

3.8 From x=0 cα = c x=0 α = c × 1−α ∞ x c 3.9 From x=1 c = 1−c = 1, we get c = 21 . ∞ 1 x 1 1 2 c 3c 2 3.10 (i) x=0 c( 3 ) = c[1 + 3 + ( 3 ) + · · · ] = 1− 1 = 2 = 1 and c = 3 .

3

1/33 1 x 2 1 (ii) P(X ≥ 3) = 23 ∞ 0.037. x=3 ( 3 ) = 3 × 2/3 = 27 3.11 (i) From 1 c(1− x 2 ) d x = c(x|1 − x 3 1 ) = c(2− 2 ) = 1, we get c = 3/4. −1 −1 3 −1 3 3

0.9

x 3×1.314 = 3.942 = (ii) P(−0.9 < X < 0.9) = 34 (x|0.9 −0.9 − 3 −0.9 ) = 4 4 0.9855.

3.12 (i)

∞ −cx d x 0 ce

∞

= − 0 de−cx = −e−cx |∞ 0 = −(0 − 1) = 1 for every

c > 0. ∞ −10c ) = e−10c . (ii) P(X ≥ 10) = 10 ce−cx d x = −e−cx |∞ 10 = −(0 − e 1 −10c (iii) P(X ≥ 10) = 0.5 implies e = 2 , so that −10c = −log 2 and c = 1 0.693 0.069.

10 log 2 10 3.13 (i)

∞ 1+α ∞ −2−α ∞ 1 ∞ d x = − 1 d x −1−α = − x 1+α = 1, 1 x 2+α d x = (1 + α) 1 x 1

so that f is, indeed, a p.d.f.

1 ∞ 1 = c1+α . (ii) For c > 1, P(X > c) = − x 1+α c ∞ c ∞ 1 3.14 (i) From j=0 j = c j=0 j = c × 1 1 = 3c = 1, we get c = 2 .

2 3 3 3 1− 3 ∞ 1 1/33 1 2 1 (ii) P(X ≥ 3) = c j≥3 3 j = c × 1 = c × 2×32 = 3 × 2×32 = 313 = 1− 3

1 27

0.037.

1 1 1 1 (iii) P(X = 2k + 1, k = 0, 1, . . .) = c ∞ k=0 32k+1 = c( 3 + 33 + 35 + · · · ) = c × 1/31 = c × 38 = 23 × 38 = 0.25.

1− 9

1 1 1 1 (iv) P(X = 3k + 1, k = 0, 1, . . .) = c ∞ k=0 33k+1 = c( 3 + 34 + 37 + · · · ) = 9 9 3 = 23 × 26 = 13 c × 1/31 = c × 26 1− 27

0.231.

3.15 P(X ≤ 3) is the area defined by the points 1, A, and 3 which is: 21 × 2 × 21 = 0.5; P(1 ≤ X ≤ 2) is the area defined by the points 1, B, and 2 which is: 1 1 2 × 1 × 4 = 0.125; P(X > 2) = 1 − P(X ≤ 2) = 1 − P(1 ≤ X ≤ 2) = 1 − 0.125 = 0.875; P(X > 5) = 0.

3.16 (i) P(no items are sold) = f (0) = 1 = 0.5.

2 1 x+1 1 (ii) P(more than 3 items are sold) = ∞ = ( 21 )5 × 1 1 = 16 = x=4 ( 2 ) 1− 2

0.0625. (iii) P(an odd number of items are sold) = ( 21 )2 + ( 21 )4 + ( 21 )6 + · · · = ( 21 )2 × 1 1 0.333.

1 = 3 1− 4

Hint: Refer to #4 in Table 6 in the Appendix.

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3.17 The decision would be on the probabilities P(X ≥ c) and P(X < c). We have: ∞

−λc , and hence P(X < c) = (i) P(X ≥ c) = c λe−λx d x = −e−λx |∞ c = e −λc −λc −λc 1 − e . Next, e ≥ 1−e if and only if c ≤ logλ 2 , in which case one would bet in favor of x ≥ c. If c > logλ 2 , one would bet in favor of x < c. (ii) For λ = 4 log 2, logλ 2 = 0.25, so that one would bet in favor of x ≥ c if c ≤ 0.25, and in favor of x < c if c > 0.25. −cx |∞ = 1 for all c > 0, the 3.18 (i) Since 0∞ c2 xe−cx d x = −cxe−cx |∞ 0 −e 0 given function is a p.d.f. for all c > 0. (ii) From part (i), −cx ∞ |t = cte−ct + e−ct = (ct + 1)e−ct .

P(X ≥ t) = −c x e−cx |∞ t −e

(iii) Here ct + 1 = 0.2 × 10 + 1 = 3, so that (ct + 1)e−ct = 3e−2 0.405.

5 3.19 (i) x=1 cx = c(1 + 2 + 3 + 4 + 5) = 15c = 1, so that c = 1/15.

1 6 (ii) P(X ≤ 3) = 15 (1 + 2 + 3) = 15 = 25 = 0.4, P(2 ≤ X ≤ 4) = 1 9 3 15 (2 + 3 + 4) = 15 = 5 = 0.6.

3.20 We have: 1 1 P(X > x0 ) = x0 n(1 − x)n−1 d x = − x0 d(1 − x)n = −(1 − x)n |1x0 = (1 − x0 )n , and it is given that this probability is 1/102n . Thus, 1 and x0 = 0.99.

(1 − x0 )n = 1012n , or 1 − x0 = 100 3.21 If A1 ∩ A2 = ∅ there would exist s ∈ A1 ∩ A2 , or s ∈ A1 and s ∈ A2 , so that X (s) ∈ B1 and X (s) ∈ B2 . Accordingly, B1 ∩ B2 = ∅, a contradiction.

3.22 If s ∈ ∪i (X ∈ Bi ), then s ∈ (X ∈ Bi ) for at least one i, so that X (s) ∈ Bi . Then X (s) ∈ (∪i Bi ) and s ∈ (X ∈ (∪i Bi )). Thus, the left-hand side is contained in the right-hand side. Next, let s ∈ (X ∈ (∪i Bi )). Then X (s) ∈ ∪i Bi , so that X (s) ∈ Bi for at least one i. Hence s ∈ (X ∈ Bi ), and then s ∈ ∪i (X ∈ Bi ). Thus, the right-hand side is contained in the left-hand side. The proof is completed.

3.23 For B ⊆ , PX (B) = P(X ∈ B) ≥ 0; for Bi ⊆ , Bi ∩ B j = ∅, i = j, first (X ∈ Bi )∩(X ∈ B j ) = ∅ (by Exercise 3.21), and second PX (∪i Bi ) = P(X ∈ (∪i Bi )) = P(∪i (X ∈ Bi )) (by Exercise 3.22), and this is = ∪i P(X ∈ Bi ) = ∪i PX (Bi ). Finally, PX ( ) = P(X ∈ ) = P(S) = 1, and this completes the proof.

3.24 For x ∈ , FX (x) = PX ((−∞, x)) and hence 0 ≤ FX (x) ≤ 1; for x1 < x2 , we have (−∞, x1 ) ⊂ (−∞, x2 ), so that PX ((−∞, x1 )) ≤ PX ((−∞, x2 )), or FX (x1 ) ≤ FX (x2 ); as n → ∞, xn ↓ x implies (−∞, xn ) ↓ (−∞, x), so that PX ((−∞, xn )) ↓ PX ((−∞, x)) (by Theorem 2), or FX (xn ) ↓ FX (x), which shows that FX is right-continuous; next, as n → ∞ and xn ↑ ∞, it follows that (−∞, xn ) ↑ , so that PX ((−∞, xn )) ↑ PX ( ) = 1 (by Theorem 2, again), or FX (xn ) ↑ 1, or FX (∞) = 1; finally, as xn ↓ −∞, we have (−∞, xn ) ↓ ∅, hence PX ((−∞, xn )) ↓ PX (∅) = 0, or FX (xn ) ↓ 0, or FX (−∞) = 0.

Revised Answers Manual to Introduction to Probability

3.25 Parts (i)–(iii) follow immediately from relation (3.2). As for part (iv), observe that {xi+1 } = {. . . , xi , xi+1 } − {. . . , xi }, so that P(X {xi+1 }) = P(X {. . . , xi , xi+1 }) − P(X {. . . , xi }) = P(X ≤ xi+1 ) − P(X ≤ xi ) = FX (xi+1 ) − FX (xi ), i ≥ 1. Finally, f X (x1 ) = FX (x1 ), sinceP(X < x1 ) = 0.

3.26 There is no contradiction between f X (x) = P(X = x) = 0 for all x ∈ , and f X (x)d x = 1, as = ∪x∈ {x} and this union consists of uncountably many terms. Recall that property (P3) of the Axiomatic Definition of Probability stipulates additivity of probabilities for countably many terms only.

3.27 (i) Clearly, pk = cpk−1 = c2 pk−2 = · · · = ck p0 , and since ∞ k=0 pk = 1, ∞ k p0 = 1, so that p0 = 1−c and pk = (1−c)ck , k = we get p0 k=0 c = 1−c 0, 1, . . .

cn+1 k n+1 , (ii) P(X > n) = P(X ≥ n + 1) = ∞ k=n+1 (1 − c)c = (1 − c) 1−c = c 6 and P(X > 5) = (0.8) = 0.262144 0.262.

Chapter 4 Section 4.1 1.1 (i) From A = (A ∩ B) ∪ (A ∩ B c ), we get: P(A) = P(A ∩ B) + P(A ∩ B c ), or 0.5 = P(A ∩ B) + 0.4, and hence P(A ∩ B) = 0.1.

0.1 1 0.167. (ii) P(A | B) = P(A∩B) P(B) = 0.6 = 6 c c (iii) P(A ∪ B ) = P(A) + P(B ) − P(A ∩ B c ) = 0.5 + (1 − 0.6) − 0.4 = 0.5. c )] P(A∩B) 0.1 1 = P(A∪B (iv) P(B|A ∪ B c ) = P[B∩(A∪B c ) = 0.5 = 5 = 0.2.

P(A∪B c ) P(A) P(A) 1.2 We have: P(A|A ∪ B) = P[A∩(A∪B)] = P(A∪B) = P(A)+P(B) (since A ∩ B = P(A∪B) P(B) ∅), and likewise, P(B|A ∪ B) = P[B∩(A∪B)] = P(A)+P(B) .

P(A∪B) 20 2 1.3 Here: (i) P(blonde) = 50 = 5 = 0.4, and and blue eyes) 15 5 = 15/50 (ii) P(blonde|blue eyes) = P(blonde P(blue eyes) 21/50 = 21 = 7 0.714.

1.4 In obvious notation, we have: 0.30 15 1 ∩b2 ) (i) P(b2 |b1 ) = P(b P(b1 ) = 0.52 = 26

0.26 13 1 ∩g2 ) P(g2 |g1 ) = P(g P(g1 ) = 0.48 = 24

0.577.

(ii) 0.542. (iii) P(b2 ) = P(b1 ∩ b2 ) + P(g1 ∩ b2 ) = 0.30 + P(b2 |g1 )P(g1 ) = 0.30 + 13 (1 − 24 ) × 0.48 = 0.52. (iv) P(b1 ∩ g2 ) = P(g2 |b1 )P(b1 ) = (1 − 15 26 ) × 0.52 = 0.22.

1.5 With obvious notation, we have: 7 (i) P(G 3 |G 1 ∩ G 2 ) = 14 18 = 9 3 1 (ii) P(D3 |G 1 ∩ D2 ) = 18 = 6

0.778.

0.167.

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3 (iii) P(D3 |D1 ∩ G 2 ) = 18 = 16 0.167. (iv) P(D3 |G 1 ∩ D2 or D1 ∩ G 2 ) = P(D3 |(G 1 ∩ D2 ) ∪ (D1 ∩ G 2 )) = P((G 1 ∩D2 ∩D3 )∪(D1 ∩G 2 ∩D3 )) P((G 1 ∩D2 )∪(D1 ∩G 2 )) ∩D2 ∩D3 )+P(D1 ∩G 2 ∩D3 ) = P(G 1P(G , and 1 ∩D2 )+P(D1 ∩G 2 )

P(G 1 ∩ D2 ∩ D3 ) = P(D3 |G 1 ∩ D2 )P(D2 |G 1 )P(G 1 ) 1 4 16 8 = × × = , 6 19 20 15 × 19 P(D1 ∩ G 2 ∩ D3 ) = P(D3 |D1 ∩ G 2 )P(G 2 |D1 )P(D1 ) 1 16 4 8 = × × = , 6 19 20 15 × 19 16 16 4 × = , P(G 1 ∩ D2 ) = P(D2 |G 1 )P(G 1 ) = 19 20 5 × 19 4 16 16 P(D1 ∩ G 2 ) = P(G 2 |D1 )P(D1 ) = × = , 19 20 5 × 19 so that the required probability is:

16/15×19 1 32/5×19 = 6

0.167.

1.6 Parts (i) and (ii) follow without any calculations by using the fact that P(·|B) and P(·|C) are probability functions, or directly as follows: ∩B) (i) P(Ac |B) = P(A = P(B−A∩B) = P(B)−P(A∩B) = 1 − P(A∩B) P(B) P(B) P(B) P(B) = 1 − P(A|B). = P((A∩C)∪(B∩C)) (ii) P(A ∪ B|C) = P((A∪B)∩C) P(C) P(C) c

P(B∩C) P((A∩B)∩C) = P(A∩C)+P(B∩C)−P(A∩B∩C) = P(A∩C) = P(C) P(C) + P(C) − P(C) P(A|C) + P(B|C) − P(A ∩ B|C). (iii) In the sample space S = {H H H , H H T , H T H , T H H , H T T , T H T , T T H , T T T } with all outcomes being equally likely, define the events:

A = “the # of H ’s is ≤ 2” = {T T T , T T H , T H T , H T T , T H H , H T H , H H T }, B = “the # of H ’s is >1” = {H H T , H T H , T H H , H H H }.

Then B c = {H T T , T H T , T T H , T T T }, A ∩ B c = B c , A ∩ B = {H H T , H T H , T H H }, so that: c) P(B c ) P(A∩B) P(A|B c ) = P(A∩B P(B c ) = P(B c ) = 1 and 1 − P(A|B) = 1 − P(B) =

3/8 1 − 4/8 = 1 − 43 = 41 . Thus, P(A|B c ) = 1 − P(A|B). (iv) In the sample space S = {1, 2, 3, 4, 5} with all outcomes being equally likely, consider the events A = {1, 2}, B = {3, 4}, and C = {2, 3}, so that

Revised Answers Manual to Introduction to Probability

A ∩ B = ∅ and A ∪ B = {1, 2, 3, 4}, Then: 2/5 2 1 P(C ∩ (A ∪ B)) = = = , whereas P(A ∪ B) 4/5 4 2 P(A ∩ C) 1/5 1 P(B ∩ C) 1/5 1 P(C|A) = = = , P(C|B) = = = , P(A) 2/5 2 P(B) 2/5 2 so that P(C|A ∪ B) =

P(C|A ∪ B) = P(C|A) + P(C|B).

1.7 Clearly, the four events involved are pairwise disjoint and A ∪ (Ac ∩ B) ∪ (Ac ∩ B c ∩ C) ∪ (A ∪ B ∪ C)c = (A ∪ (Ac ∩ B) ∪ (Ac ∩ B c ∩ C)) ∪ (A ∪ B ∪ C)c = (A∪ B ∪C)∪(A∪ B ∪C)c = S, since A∪(Ac ∩ B)∪(Ac ∩ B c ∩C) = A∪ B ∪C as is easily seen.

1 ∩A2 ) yields P(A1 ∩ 1.8 For n = 2, the theorem is true since P(A2 |A1 ) = P(A P(A1 ) A2 ) = P(A2 |A1 )P(A1 ). Next, assume P(A1 ∩ · · · ∩ Ak ) = P(Ak |A1 ∩ · · · ∩ Ak−1 ) · · · P(A2 |A1 )P(A1 ) and show that P(A1 ∩· · ·∩ Ak+1 ) = P(Ak+1 |A1 ∩ · · · ∩ Ak )P(Ak |A1 ∩ · · · ∩ Ak−1 ) · · · P(A2 |A1 )P(A1 ). Indeed, P(A1 ∩ · · · ∩ Ak+1 ) = P((A1 ∩ · · · ∩ Ak ) ∩ Ak+1 ) = P(Ak+1 |A1 ∩ · · · ∩ Ak )P(A1 ∩ · · · ∩ Ak ) (by applying the theorem for two events A1 ∩ · · · ∩ Ak and Ak+1 ) = P(Ak+1 |A1 ∩ · · · ∩ Ak )P(Ak |A1 ∩ · · · ∩ Ak−1 ) · · · P(A2 |A1 )P(A1 ) (by the induction hypothesis).

P(A∩A ) P(A|A j )P(A j ) 1.9 We have: P(A j |A) = P(A) j = , where P(A|A j )P(A j ) = 5 P(A) 5 j(5− j) 5− j j j(5− j) j=1 P(A|A j )P(A j ) = j=1 225 = 15 × 15 = 225 and P(A) = 4/225 1 20 4 225 (1 × 4 + 2 × 3 + 3 × 2 + 4 × 1) = 225 , so that: P(A1 |A) = 20/225 = 20 = 6/225 1 6 3 5 = 0.2, P(A2 |A) = 20/225 = 20 = 10 = 0.3, P(A3 |A) = P(A2 |A) = 0.3, P(A4 |A) = P(A1 |A) = 0.2, P(A5 |A) = 0.

1.10 With obvious notation, we have: P(1st red and 4th red) = P(R1 ∩ R2 ∩ R3 ∩ R4 ) + P(R1 ∩ R2 ∩ B3 ∩ R4 ) + P(R1 ∩ B2 ∩ R3 ∩ R4 ) + P(R1 ∩ B1 ∩ B2 ∩ R4 ) = P(R4 |R1 ∩ R2 ∩ R3 )P(R3 |R1 ∩ R2 )P(R2 |R1 )P(R1 ) + P(R4 |R1 ∩ R2 ∩ B3 )P(B3 |R1 ∩ R2 )P(R2 |R1 )P(R1 ) + P(R4 |R1 ∩ B2 ∩ R3 )P(R3 |R1 ∩ B2 )P(B2 |R1 )P(R1 )+ P(R4 |R1 ∩ B1 ∩ B2 )P(B2 |R1 ∩ B1 )P(B1 |R1 )P(R1 ) = 7 8 9 10 8 5 9 10 8 9 5 10 9 4 5 10 12 × 13 × 14 × 15 + 12 × 13 × 14 × 15 + 12 × 13 × 14 × 15 + 12 × 13 × 14 × 15 = 1 12×13×14×15 (7 × 8 × 9 × 10 + 5 × 8 × 9 × 10 × 2 + 4 × 5 × 9 × 10) = 9×10×156 3 0.429.

12×13×14×15 = 7 1.11 With obvious notation, we have: m (i) P(R1 ) = m+n . (ii) P(R2 ) = P(R1 ∩ R2 )+ P(B1 ∩ R2 ) = P(R2 |R1 )P(R1 )+ P(R2 |B1 )P(B1 ) m(m+r )+mn m(m+n+r ) m+r m m n × m+n + m+n+r × m+n = (m+n)(m+n+r = m+n+r ) = (m+n)(m+n+r ) = m m m+n ; i.e., P(R2 ) = m+n . (iii) We observe that P(R1 ) = P(R2 ). That P(R2 ) should be equal to P(R1 ) is not unreasonable because no information is given what happened in the first drawing.

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) |B1 )P(B1 ) n (iv) P(B1 |R2 ) = P(R2P(R = mn/(m+n)(m+n+r = m+n+r . m/(m+n) 2)

9 6 3 = 35 = 0.6, P(B1 |R2 ) = 20 = 10 = 0.3. (v) Here P(R1 ) = P(R2 ) = 15

1.12 In obvious notation, we have: (i) P(+) = P(+|D)P(D) + P(+|D c )P(D c ) = 0.95 × 0.002 + 0.01 × 0.998 = 0.0019 + 0.00998 = 0.01188.

0.0019 = 0.01188 0.16. (ii) P(D|+) = P(+|D)P(D) P(+) 1.13 With obvious notation, we have: P(+) = P(+|D)P(D) + P(+|D c )P(D c ) = 0.95 × 0.0001 + 0.05 ×0.9999 = 0.000095 + 0.049995 = 0.05009.

(i)

= 0.000095 (ii) P(D|+) = P(+|D)P(D) P(+) 0.05009

0.002.

1.14 Denote by I = “switch I is closed,” II = “switch II is closed,” S = “signal goes through.” Then: (i) P(S) = P(I ∩ I I ) = P(I I |I )P(I ) = P(I )P(I I ) = 0.8 × 0.6 = 0.48. c |I c )P(I c ) 5 = 1×0.2 0.385. (ii) P(I c |S c ) = P(S P(S c) 0.52 = 13

I )P(I I ) 10 = 1×0.4 (iii) P(I I c |S c ) = P(S |IP(S c) 0.52 = 13 c

c

c

0.769.

1.15 Denoting by W , M, and S that a student is a woman, or a man, or a smoker, respectively, we have: P(S) = P(S ∩ W ) + P(S ∩ M) = P(S|W )P(W ) + P(S|M)P(M) = 0.20 × 0.55 + 0.25 × 0.45 = 0.11 + 0.1125 = 0.2225.

(i)

= 0.1125 (ii) P(M|S) = P(S|M)P(M) P(S) 0.2225

0.506.

1.16 Denoting by F, M, and C, respectively, the events that an individual is female, male, and color-blind, we have: P(F) = 0.52,

P(M) = 0.48,

P(C|F) = 0.25,

P(C|M) = 0.05.

Then the required probability is: P(M|C) =

P(C|M)P(M) , where P(C) = P(C|M)P(M) + P(C|F)P(F).

P(C)

Here: P(C) = 0.05 × 0.48 + 0.25 × 0.52 = 0.024 + 0.13 = 0.154, and 0.024 12 P(M|C) = = 0.156.

0.154 77

Revised Answers Manual to Introduction to Probability

1.17 With obvious notation, we have: 1 2 (i) P(B) = P(B|I )P(I ) + P(B|I I )P(I I ) = b1b+r × 21 + b2b+r × 21 = 1 2

b1 b2 1 2 ( b1 +r1 + b2 +r2 ).

b1 1 b1 +r1 × 2 b1 b2 2 ( b1 +r1 + b2 +r2 )

)P(I ) (ii) P(I |B) = P(B|I = 1 P(B)

60 1 3 5 41 (iii) P(B) = 21 ( 36 48 + 84 ) = 2 ( 4 + 7 ) = 56 3/8 P(I |B) = 41/56 = 21 41

b1 (b2 +r2 ) = b1 (b2 +r .

2 )+b2 (b1 +r1 )

0.732,

0.512.

1.18 With obvious notation, we have: (i) P(D) = P(D|I )P(I ) + P(D|I I )P(I I ) + P(D|I I I )P(I I I ) = 0.04 × 0.30 + 0.03 × 0.30 + 0.02 × 0.40 = 0.029. )P(I ) 12 (ii) P(I |D) = P(D|I = 0.04×0.30 = 0.012 P(D) 0.029 0.029 = 29

(iii) P(I I |D) = 0.03×0.30 = 0.009 0.029 0.029 P(I I I |D) = 0.02×0.40 = 0.008 0.029 0.029

0.414.

0.310, 0.276.

1.19 (i) Let pn = P(A|B) when the number of answers is n. Then pn = P(A∩B) P(B) = P(B|A)P(A) , where P(B|A)P(A) = 1× p = p, and P(B) = P(B|A)P(A) P(B) + P(B|Ac )P(Ac ) = p + n1 (1 − p), so that pn = p1− p . p+ n

(ii) When the number of answers is n + 1, the respective probability pn+1 is p p given by: pn+1 = 1− p . Then pn+1 > pn is equivalent to 1− p > p+ n+1 p+ n+1 p 1 1 or n > n+1 (since p < 1), which is true. So, pn is, indeed, increasp+ 1−n p

ing in n. (iii) The larger n is, the more extensive the list of answers the student will have to choose from. Thus, if the student answers the question correctly, this would imply that it is more likely that he/she, actually, did the homework.

In other words, pn should increase with n. (ii)

∞ −λx d x t λe

∞

−λt . = − t de−λx = −e−λx |∞ t =e P(X > s + t) P(X > s + t, X > s) = P(X > s + t|X > s) = P(X > s) P(X > s) −λ(s+t) e (by part (i)) = e−st = e−λt .

1.20 (i) P(X > t) =

(iii) The conditional probability that X is greater than t units beyond s, given that it has been greater than s, does not depend on s and is the same as the (unconditional) probability that X is greater than t. That is, this distribution has some sort of “memoryless” property.

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1.21 Let Ai be the event that the exam is passed the ith time, i = 1, 2, 3. Then: (i) P(A2 ) = P(Ac1 ∩ A2 ) = P(A2 |Ac1 )P(Ac1 ) = 0.8 × (1 − 0.7) = 0.24. (ii) P(A3 ) = P(Ac1 ∩ Ac2 ∩ A3 ) = P(A3 |Ac1 ∩ Ac2 )P(Ac2 |Ac1 )P(Ac1 ) = 0.9 × (1 − 0.8) × (1 − 0.7) = 0.9 × 0.2 × 0.3 = 0.054.

P(passing the exam) = P[A1 ∪ (Ac1 ∩ A2 ) ∪ (Ac1 ∩ Ac2 ∩ A3 )] (iii) = P(A1 ) + P(Ac1 ∩ A2 ) + P(Ac1 ∩ Ac2 ∩ A3 ) = P(A1 ) + P(A2 ) + P(A3 ) = 0.7 + 0.24 + 0.054 = 0.994.

1.22 For i = 1, 2, 3, let Si = “the card drawn the ith time is a spade.” Then the required probability is: P(S1 |S2 ∩ S3 ) = P(S1 ∩ S2 ∩ S3 )/P(S2 ∩ S3 ).

But P(S1 ∩ S2 ∩ S3 ) = P(S3 |S1 ∩ S2 )P(S2 |S1 )P(S1 ) 11 × 12 × 13 11 12 13 × × = , = 50 51 52 50 × 51 × 52 P(S2 ∩ S3 ) = P(S1 ∩ S2 ∩ S3 ) + P(S1c ∩ S2 ∩ S3 ), and P(S1c ∩ S2 ∩ S3 ) = P(S3 |S1c ∩ S2 )P(S2 |S1c )P(S1c ) 12 × 13 × 39 12 13 39 × × = , = 50 51 52 50 × 51 × 52 so that 11×12×13 50×51×52 P(S1 |S2 ∩ S3 ) = (11×12×13)+(12×13×39) = 50×51×52

1.23 (i)

11 11 × 12 × 13 = = 0.22.

12 × 13(11 + 39) 50

P(at least one match occurs) = P(A1 ∪ A2 ∪ A3 ∪ A4 ) = [P(A1 ) + P(A2 ) + P(A3 ) + P(A4 )] − [P(A1 ∩ A2 ) + P(A1 ∩ A3 ) +P(A1 ∩ A4 ) + P(A2 ∩ A3 ) + P(A2 ∩ A4 ) + P(A3 ∩ A4 )] + [P(A1 ∩ A2 ∩ A3 ) + P(A1 ∩ A2 ∩ A4 ) + P(A1 ∩ A3 ∩ A4 ) + P(A2 ∩ A3 ∩ A4 )] − P(A1 ∩ A2 ∩ A3 ∩ A4 )

4 2!

4 1!

4 0!

4 3! × + × − × = × − 4!

2 4!

3 4!

4 4!

1 4!

3!

4!

2!

4!

1!

0! = × − × + × −1× 1!3! 4! 2!2! 4! 3!1! 4!

4!

1 1 1 = 1 − + − , as was to be seen.

2! 3! 4!

Revised Answers Manual to Introduction to Probability

(ii) The rule for n balls/urns seems to be: P(at least one match occurs) = 1 − 1 1 n+1 1 .

2! + 3! − · · · + (−1) n!

x xk (iii) From #6 in Table 6 in the Appendix, we have: e x = ∞ k=0 k! = 1 + 1! + x2 x3 2! + 3! + · · · , which for x = −1 becomes:

e−1 = 1 −

1 1 1 1 1 + − + · · · , or 1 − + − · · · = 1 − e−1 .

1! 2! 3!

2! 3!

1 Thus, for large n, 1 − 2!1 + 3!1 − · · · + (−1)n+1 n!

1 − e−1 = 0.630.

1.24 It is given that: P(D) = 0.01, P(+|D) = 0.95, P(+|D c ) = 0.005. (i) Then P(+) = P(+∩ D)+ P(+∩ D c ) = P(+|D)P(D)+ P(+|D c )P(D c ) = 0.95 × 0.01 + 0.005 × 0.99 = 0.0095 + 0.00495 = 0.01445.

0.0095 950 = 0.01445 = 1445 = 190 0.057. (ii) Here P(D|+) = P(+|D)P(D) P(+) 289 1.25 (i)

P(A) = P(A ∩ A0 ) + P(A ∩ A1 ) + P(A ∩ A2 ) + P(A ∩ A3 ) = P(A|A0 )P(A0 ) + P(A|A1 )P(A1 ) + P(A|A2 )P(A2 ) + P(A|A3 )P(A3 ) = 0.06 × 0.40 + 0.05 × 0.25 + 0.04 × 0.20 + 0.03 × 0.15 = 0.24 + 0.0125 + 0.008 + 0.0045 = 0.049.

(ii)

0.024 24 P(A|A0 )P(A0 ) = = P(A) 0.049 49 12.5 0.0125 = 0.2551, P(A1 |A) = 0.049 49 8 0.008 = 0.1633, P(A2 |A) = 0.049 49 4.5 0.0045 = 0.0918.

P(A3 |A) = 0.049 49

P(A0 |A) =

0.4898,

Section 4.2 2.1 From P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C), we get: (i) P(A ∪ B ∪ C) = 0.4 + 0.2 + 0.3 = 0.9. (ii) P(A ∪ B ∪ C) = 0.9 − 0.4 × 0.2 − 0.4 × 0.3 − 0.2 × 0.3 + 0.4 × 0.2 × 0.3 = 0.664.

2.2 Here P(A) = P(A ∩ A) = P(A)P(A) = [P(A)]2 , and this happens if P(A) = 0, whereas, if P(A) = 0, it happens only if P(A) = 1.

2.3 (i) From P(A ∪ B) = P(A) + P(B) − P(A ∩ B), we get P(A ∩ B) = P(A) + P(B) − P(A ∪ B) ≥ P(A) + P(B) − 1. (ii) Independence requires that P(A ∩ B) = P(A)P(B). By disjointness, P(A ∩ B) = P(∅) = 0, so that P(A)P(B) = 0. Thus, the required condition is that at least one of P(A) and P(B) be 0.

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(iii) As in part (ii), 0 = P(∅) = P(A ∩ B) = P(A)P(B), 0 = P(∅) = P(A ∩ C) = P(A)P(C), 0 = P(∅) = P(B ∩ C) = P(B)P(C), from which it follows that at least two of P(A), P(B), and P(C) are 0. From this, it follows, of course, that 0 = P(A ∩ B ∩ C) = P(A)P(B)P(C).

2.4 Since P(A1 ∩ A2 ) = P(A1 )P(A2 ), we have to show that: P(A1 ∩ (B1 ∪ B2 )) = P(A1 )P(B1 ∪ B2 ), P(A2 ∩ (B1 ∪ B2 )) = P(A2 )P(B1 ∪ B2 ), P(A1 ∩ A2 ∩ (B1 ∪ B2 )) = P(A1 )P(A2 )P(B1 ∪ B2 ).

Indeed, P(A1 ∩ (B1 ∪ B2 )) = P((A1 ∩ B1 ) ∪ (A1 ∩ B2 )) = P(A1 ∩ B1 ) + P(A1 ∩ B2 ) = P(A1 )P(B1 ) + P(A1 )P(B2 ) = P(A1 )P(B1 ∪ B2 ), and similarly for P(A2 ∩ (B1 ∪ B2 )).

Finally, P(A1 ∩ A2 ∩ (B1 ∪ B2 )) = P((A1 ∩ A2 ∩ B1 ) ∪ (A1 ∩ A2 ∩ B2 )) = P(A1 ∩ A2 ∩ B1 ) + P(A1 ∩ A2 ∩ B2 ) = P(A1 )P(A2 )P(B1 ) + P(A1 )P(A2 )P(B2 ) = P(A1 )P(A2 )P(B1 ∪ B2 ).

2.5 (i) It is true that P(A∩ B) = P(A)P(B)(= 41 ), P(A∩C) = P(A)P(C)(= 41 ), P(B ∩ C) = P(B)P(C)(= 41 ), but P(A ∩ B ∩ C) = 16 = 21 × 21 × 1 2 = P(A)P(B)P(C), so that A, B, and C are not independent; they are dependent. (ii) P(Ac ) = 1 − P(A) = 0.5, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 1 1 1 3 c c c 2 + 2 − 4 = 4 = 0.75, P(A ∩ B ) = P((A ∪ B) ) = 1 − P(A ∪ B) = 1−0.75 = 0.25, P(A∪B∪C) = P(A)+ P(B)+ P(C)− P(A∩B)− P(A∩ 11 0.917, P(Ac ∩ B c ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C) = 23 − 43 + 16 = 12 11 1 c c 0.083.

C ) = P((A ∪ B ∪ C) ) = 1 − P(A ∪ B ∪ C) = 1 − 12 = 12 c ∩ C) ∪ (A ∩ B ∩ C c ) ∪ (A ∩ B c ∩ C c ), (i) Clearly, A = (A ∩ B ∩ C) ∪ (A ∩ B 2.6 so that: P(A) = P(A ∩ B ∩ C) + P(A ∩ B c ∩ C) + P(A ∩ B ∩ C c ) 1 5 3 2 + 16 + 16 + 16 = 11 + P(A ∩ B c ∩ C c ) = 16 16 = 0.6875. Likewise, c c B = (A ∩ B ∩ C) ∪ (A ∩ B ∩ C ) ∪ (A ∩ B ∩ C) ∪ (Ac ∩ B ∩ C c ), 1 3 2 1 7 + 16 + 16 + 16 = 16 = 0.4375, and C = (A ∩ so that P(B) = 16 c c B ∩ C) + (A ∩ B ∩ C) ∪ (A ∩ B ∩ C) ∪ (Ac ∩ B c ∩ C), so that 1 5 2 1 9 + 16 + 16 + 16 = 16 = 0.5625.

P(C) = 16 1 11 7 9 (ii) P(A ∩ B ∩ C) = 16 = 16 × 16 × 16 = P(A)P(B)P(C), so that the events A, B, and C are not independent. (iii) From A ∩ B = (A ∩ B ∩ C) ∪ (A ∩ B ∩ C c ), we get P(A ∩ B) = 1 3 4 P(A ∩ B ∩ C) + P(A ∩ B ∩ C c ) = 16 + 16 = 16 , so that P(A|B) = 4/16 P(A∩B) 4 0.5714.

P(B) = 7/16 = 7

Revised Answers Manual to Introduction to Probability

4 7 (iv) Since P(A ∩ B) = 16 = 11 16 × 16 = P(A)P(B), the events A and B are not independent.

2.7 We have ⎛ P⎝

n

⎞

⎛⎛

A j ⎠ = 1 − P ⎝⎝

j=1

n

⎞c ⎞

⎛

A j ⎠ ⎠ = 1 − P⎝

j=1

n

⎞ Acj ⎠ = 1 −

j=1

n

P Acj ,

j=1

since independence of A1 , . . . , An implies independence of Ac1 , . . . , Acn (by Theorem 4(ii)).

2.8 (i) The relevant sample space here is: S = {H H H , H H T , H T H , T H H , H T T , T H T , T T H , T T T }, and the winning event is: A = {H H H , T T T } whose probability is P(A) = p 3 + q 3 (q = 1 − p). (ii) For p = 0.5, P(A) = 2 × (0.5)3 = 0.25, and for p = 0.4, P(A) = (0.4)3 + (0.6)3 = 0.28.

2.9 From the table, we have: 6 4 , P(W ∩ F) = , 16 + x 16 + x 6 x P(M ∩ S) = , P(W ∩ S) = , 16 + x 16 + x 6+x 10 6+x 10 , P(W ) = , P(F) = , P(S) = .

P(M) = 16 + x 16 + x 16 + x 16 + x

P(M ∩ F) =

4 10 10 Then the first relation expressing independence becomes: 16+x = 16+x × 16+x or x = 9. Observe that for this value of x, the remaining three relations also hold.

2.10 (i) (ii)

5 10 0 cx d x + 5 c(10 − x)d x = 25c = 1, so that c = 1/25.

f (x ) 0.2

0

5

10

x

10 5 (10 − x)d x = 0.50, 1 7.5 18.75 P(B) = P(5 < X < 7.5) = 25 5 (10 − x)d x = 50 = 0.375.

1 (iii) P(A) = P(X > 5) = 25

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(iv)

P(A ∩ B) P(X > 5, 5 < X < 7.5) = P(A) P(A) P(B) 0.375 P(5 < X < 7.5) = = = 0.75. = P(A) P(A) 0.50 P(B|A) =

(v) Since P(B|A) = 0.75 = 0.375 = P(B), the events A and B are not independent.

2.11 The student will pass the test if he/she answers correctly at least 5 of the last 10 questions. The probability that the student answers correctly exactly x specified of questions is: (0.2)x (0.8)10−x .

Since the can be

10 x questions 10these x (0.8)10−x = (0.2) chosen in x ways, the required probability is: 10 x=5 x

x 10−x = 1 − 0.96721 = 0.03279.

1 − 4x=0 10 x (0.2) (0.8) 2.12 (a) When the drawing is done with replacement, the required probabilities are given below, where n R + n B + n W = n. (i) ( nnR )3 .

W 3 ) . (ii) 1 − ( n B +n n nW nR nB (iii) n × n × n × 6 = 6n R nn3B n W . (b) When the drawing is done without replacement, we have: n R n R (n R − 1)(n R − 2) . (i) n3 = n(n − 1)(n − 2) 3

n B +n W (n B + n W )(n B + n W − 1)(n B + n W − 2) . (ii) 1 − n3 = 1 − n(n − 1)(n − 2) 3 n R n B

nW 6n R n B n W n R n B nW 1 1 1 ×6= . (iii) = n n(n − 1)(n − 2) n(n − 1)(n − 2) 3 2.13 Let Am and Bm be the events that the two people toss m heads each, where m . . . , n. Then the events Am and Bm are independent and P(Am ) = n = m0, 1, n−m = P(B ) (q = 1 − p). Therefore the required probability is: p q m m n n n m n−m 2

n p q . For m=0 P(Am ∩ Bm ) = m=0 P(Am )P(Bm ) =

2 m=0 m 2n n n 2

2 p = 21 , nm=0 mn p m q n−m = 212n nm=0 mn = (0.5) m=0 m .

Finally, for n = 5, we have:

5 5 5 5 5 5 = 1, = 5, = 10, = 10, = 5, = 1, 0 1 2 3 4 5 so that the probability is: (0.5)10 (1 + 52 + 102 + 102 + 52 + 1) = 252 × (0.5)10 252 × 0.00097656

0.246.

2.14 With obvious notation, we have: (i)

P(B) = P((B1 ∩ B2 ∩ B) ∪ (B1 ∩ W2 ∩ B) ∪ (W1 ∩ B2 ∩ B) ∪ (W1 ∩ W2 ∩ B)) = P(B1 ∩ B2 ∩ B) + P(B1 ∩ W2 ∩ B) + P(W1 ∩ B2 ∩ B) (since W1 ∩ W2 ∩ B = ∅)

Revised Answers Manual to Introduction to Probability

= P(B|B1 ∩ B2 )P(B1 ∩ B2 ) + P(B|B1 ∩ W2 )P(B1 ∩ W2 ) + P(B|W1 ∩ B2 )P(W1 ∩ B2 ) = P(B|B1 ∩ B2 )P(B1 )P(B2 ) + P(B|B1 ∩ W2 )P(B1 )P(W2 ) + P(B|W1 ∩ B2 )P(W1 )P(B2 ) (by independence) n1 m1 n1 n2 1 m2 1 = 1× × + × × + × m 1 + n1 m 2 + n2 2 m 1 + n1 m 2 + n2 2 m 1 + n1 n2 × m 2 + b2 m 1 n 2 + n 1 m 2 + 2n 1 n 2 . = 2(m 1 + n 1 )(m 2 + n 2 ) 1 n 2 +n 1 m 2 +2m 1 m 2 (ii) By symmetry, P(W ) = m2(m .

1 +n 1 )(m 2 +n 2 ) (iii) For the given values of m 1 , m 2 , n 2 and n 2 , we have:

25 × 61 10 × 25 + 15 × 35 + 2 × 15 × 25 = 2 × 25 × 60 2 × 25 × 60 61 0.508, = 120 10 × 25 + 15 × 35 + 2 × 10 × 35 25 × 59 P(W ) = = 2 × 25 × 60 2 × 25 × 60 59 0.492. = 120 P(B) =

2.15 Let I and H be the events of intercept and hit, respectively. (i) Then the probability of a single missile hitting the target is: P(H ) = P(H |I )P(I ) + P(H |I c )P(I c ) = 0 ×

1 3 2 + × = 0.50.

3 4 3

Therefore: (ii) P(4 hits) = [P(H )]4 = (0.50)4 = 0.0625. (iii) P(of at least 1 hit) = 1 − P(no hits) = 1 − (0.50)4 = 1 − 0.0625 = 0.9375. (iv) If n missiles are fired, then P(at least 1 is not intercepted) = 1 − P(all n are intercepted) = 1 − ( 13 )n . Therefore n is defined as the smallest n for which 1 − ( 13 )n ≥ 0.95, or ( 13 )n ≤ 0.05, or n ≥ − log(0.05)/ log 3 2.726. Thus, n = 3. (v) If n missiles are fired, then P(at least 1 hits the target) = 1 − P(none hits the target) = 1 − (0.50)n . Then n is defined as the smallest n for which log(0.01) 1 − (0.50)n ≥ 0.99 or (0.50)n ≤ 0.01, or n ≥ log(0.50) 6.645, so that n = 7.

2.16 Denote by Ci the event that circuit i is closed. Then (by Theorem 4(ii)): P(no circuit is closed) = P(C1c ∩ · · · ∩ Cnc ) = P(C1c ) · · · P(Cnc ) (i) = (1 − p1 ) · · · (1 − pn ).

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(ii)

P(at least 1 circuit is closed) = 1 − P(no circuit is closed) = 1 − (1 − p1 ) · · · (1 − pn ).

(iii)

P(exactly 1 circuit is closed) = P((C1 ∩ C2c ∩ · · · ∩ Cnc ) ∪ (C1c ∩ C2 ∩ c C3c ∩ · · · ∩ Cnc ) ∪ · · · ∪ (C1c ∩ · · · ∩ Cn−1 ∩ Cn )) = P(C1 ∩ C2c ∩ · · · ∩ Cnc ) + P(C1c ∩ C2 ∩ C3c ∩ · · · ∩ Cnc ) + · · · c + P(C1c ∩ · · · ∩ Cn−1 ∩ Cn ) = P(C1 )P(C2c ) · · · P(Cnc ) + P(C1c )P(C2 )P(C3c ) · · · P(Cnc ) + · · · c + P(C1c ) · · · P(Cn−1 )P(Cn ) = p1 (1 − p2 ) · · · (1 − pn ) + (1 − p1 ) p2 (1 − p3 ) · · · (1 − pn ) + · · · + (1 − p1 ) · · · (1 − pn−1 ) pn .

(iv) For p1 = · · · = pn = p, we have: The probability in (i) is: (1 − p)n .

The probability in (ii) is: 1 − (1 − p)n .

The probability in (iii) is: np(1 − p)n−1 . (v) The respective probabilities are here: (0.4)5 = 0.01024, 1 − 0.01024 = 0.98976, 5 × 0.6 × (0.4)4 = 3 × 0.0256 = 0.0768.

2.17 (i) Setting pW = P(Wi ) = 0.9 and p R = P(R j ) = 0.6, we have that the required probability is: c P((W1 ∩ R1c ∩ · · · ∩ Rn−2 ∩ Rn−1 ) c ∪(W1c ∩ W2 ∩ R1c ∩ · · · ∩ Rn−3 ∩ Rn−2 ) c c ∪ · · · ∪ (W1 ∩ · · · ∩ Wn−2 ∩ Wn−1 ∩ Rn )) c = P(W1 ∩ R1c ∩ · · · ∩ Rn−2 ∩ Rn−1 ) c c c + P(W1 ∩ W2 ∩ R1 ∩ · · · ∩ Rn−3 ∩ Rn−2 ) c c + · · · + P(W1 ∩ · · · ∩ Wn−2 ∩ Wn−1 ∩ Rn ) c = P(W1 )P(R1c ) . . . P(Rn−2 )P(Rn−1 ) + P(W1c )P(W2 )P(R1c ) . . . c c P(Rn−3 )P(Rn−2 ) + · · · + P(W1c ) . . . P(Wn−2 )P(Wn−1 )P(Rn )

(by independence of all events involved) = pW (1 − p R )n−2 p R + (1 − pW ) pW (1 − p R )n−3 p R + · · · + (1 − pW )n−2 pW p R = pW p R [(1 − p R )n−2 + (1 − pW )(1 − p R )n−3 + · · · + (1 − pW )n−2 ] = pW p R

n−1

(1 − pW )i−1 (1 − p R )(n−i)−1

i=1

= (0.9 × 0.6)

n−1 n−1

(0.1)i−1 (0.4)n−i−1 = 0.54 (0.1)i−1 (0.4)n−i−1 . i=1

i=1

Revised Answers Manual to Introduction to Probability

(ii) For n = 5, this probability becomes: (0.54)[(0.4)3 + (0.1)(0.4)2 + (0.1)2 (0.4) + (0.1)3 ] = (0.54)(0.064 + 0.016 + 0.004 + 0.001) = (0.54)(0.085) = 0.0459.

2.18 For i ≥ 1, we have P(A1i ) = 28 = 14 and P(Bi ) = 28 = 41 . Then: P(A1 ) =

1 + 4

2 3 1 1 1/4 1 + + ··· = = 1 4 4 3 1− 4

0.333.

2.19 For i = 1, 2, 3, set Ai = “ith switch turns on,” and E =“current flows from A to B.” Then: (i) E = A1 ∪ A2 ∪ A3 and P(E) = [P(A1 ) + P(A2 ) + P(A3 )] − [P(A1 ∩ A2 ) + P(A1 ∩ A3 ) + P(A2 ∩ A3 )] + P(A1 ∩ A2 ∩ A3 ) = ( p1 + p2 + p3 ) − ( p1 p2 + p1 p3 + p2 p3 ) + p1 p2 p3 (by independence). (ii) From part (i), P(E) = 3 p − 3 p 2 + p 3 . (iii) P(E) = (0.90 + 0.95 + 0.99) − (0.90 × 0.95 + 0.90 × 0.99 + 0.95 × 0.99) + 0.90 × 0.95 × 0.99 = 2.84 − 2.6865 + 0.84645 = 0.99995; P(E) = 3 × 0.96 − 3 × (0.96)2 + (0.96)3 = 2.88 − 2.7648 + 0.884736 = 0.999936.

2.20 (i)

P(A ∩ (B ∪ C)) = P[(A ∩ B) ∪ (A ∩ C)] = P(A ∩ B) + P(A ∩ C) − P(A ∩ B ∩ C) = P(A)P(B) + P(A)P(C) − P(A)P(B ∩ C) (by independence) = P(A)[P(B) + P(C) − P(B ∩ C)] = P(A)P(B ∪ C), so that A and B ∪ C are independent.

(ii) Independence of A, B, C, implies (by Theorem 4) independence of Ac , B, C c . So, P(Ac ∩(B∩C c )) = P(Ac ∩B∩C c ) = P(Ac )P(B)P(C c ) = P(Ac )P(B∩ C c ) (by independence of B and C c ), so that Ac and B ∩C c are independent. (iii) By (i), A and B ∪ C, or B ∪ C and A are independent. Hence: P(B ∪C|A) = P(B ∪C) = P(B)+ P(C)− P(B ∩C) = P(B)+ P(C)− P(B)P(C) (by independence of B and C).

2.21 Denote by the same letters A, B, and C the events that the so labeled stocks rise. Then: (i) P(all three stocks rise) = P(A ∩ B ∩ C) = P(A)P(B)P(C) (by independence) = 0.4 × 0.6 × 0.7 = 0.168. (ii) P(none of the stocks rise) = P(Ac ∩ B c ∩ C c ) = P(Ac )P(B c )P(C c ) (by independence, see Theorem 4) = (1 − 0.4) × (1 − 0.6) × (1 − 0.7) = 0.6 × 0.4 × 0.3 = 0.072.

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(iii) P(at least one stock rises) = 1 − P(none of the stocks rise) = 1 − 0.072 = 0.928 (by part (ii)). (iv) P(exactly one stock rises) = P[(A ∩ B c ∩ C c ) ∪ (Ac ∩ B ∩ C c ) ∪ (Ac ∩ B c ∩ C)] = P(A ∩ B c ∩ C c ) + P(Ac ∩ B ∩ C c ) + P(Ac ∩ B c ∩ C) = P(A)P(B c )P(C c ) + P(Ac )P(B)P(C c ) + P(Ac )P(B c )P(C)(by independence, see Theorem 4) = 0.4 × (1 − 0.6) × (1 − 0.7) + (1 − 0.4) × 0.6 × (1 − 0.7) + (1 − 0.4) × (1 − 0.6) × 0.7 = 0.4 × 0.4 × 0.3 + 0.6 × 0.6 × 0.3 + 0.6 × 0.4 × 0.7 = 0.048 + 0.108 + 0.168 = 0.324.

2.22 (i) To show that An ⊃ An+1 , n ≥ 1, where An = {H . . . H whatever}, An+1 =

n

{H . . H H whatever}. Let s ∈ An+1 . Then s = H . . . H H whatever. But . n

n

then s ∈ An , since H . . . H whatever is one of the H . . . H H whatever n

n

or H . . . H T whatever. Thus, s ∈ An+1 implies s ∈ An , so that An ⊃ n

An+1 , n ≥ 1. (ii) The ∞ n=1 An must be A = {H H . . .}, because otherwise there would be a T in some place, for example, the (k + 1)-th place; i.e., there would exist s ∈ A with s = H . . . H T H . . . . But then s ∈ Ak+1 and therefore

k

s∈ ∞ n=1 An , a contradiction. (iii) By independence, P(An ) = p n , and by Theorem 2 in Chapter 3, P(A) = P(limn→∞ An ) = limn→∞ p n = 0 (since 0 < p < 1).

Chapter 5 Section 5.1 1.1 In Exercise 3.1 in Chapter 3: x

0

1

2

3

f (x )

1 8

3 8

3 8

1 8

3 1 12 3 3 +2× +3× = = = 1.5, 8 8 8 8 2 3 1 24 3 2 EX = 1 × + 4 × + 9 × = = 3, so that 8 8 8 8 Var(X ) = 3 − 1.52 = 0.75 and s.d. of X 0.866.

EX = 1 ×

1.2 We have: (i) E X = 0, E X 2 = c2 , and Var(X ) = c2 .

Revised Answers Manual to Introduction to Probability

(ii) P(|X − E X | ≤ c) = P(|X | ≤ c) = P(−c ≤ X ≤ c) = P(X = −c or ) X = c) = 1 = cc2 = Var(X . c2 2

1.3 (i) Here E X = 1 × 0.2 + 2 × 0.4 + 3 × 0.3 + 4 × 0.1 = 2.3, E X 2 = 1 × 0.2 + 4 × 0.4 + 9 × 0.3 + 16 × 0.1 = 6.1, so that Var(X ) = 0.81. (ii) Let Y be the r.v. denoting the quantity described. Then Y = 100 − 5X and therefore EY = 100 − 5 × 2.3 = 88.5, Var(Y ) = Var(5X ) = 25 × 0.81 = 20.25, and s.d. of Y = 4.5.

1.4 Let Y be the r.v. denoting the net loss to the company. Then EY = (0 × 0.8) + (500 × 0.1) + (4,500 × 0.08) + (9,500 × 0.02) = 600. If P is the premium to be charged, the expected profit is: P − EY = P − 600. So, if P − 600 = 100, then P = $700.

1.5 The values of X are: 1, 2, 3, 4, 5, 6, each taken with probability 1/6. Thus: (i) M X (t) =

6

x=1

et x ×

1 1 = (et + e2t + e3t + e4t + e5t + e6t ), t ∈ .

6 6

(ii) d 1 M X (t) = (et + 2e2t + 3e3t + 4e4t + 5e5t + 6e6t ) dt 6 t=0 t=0 21 1 = 3.5; = (1 + 2 + 3 + 4 + 5 + 6) = 6 6 d2 1 E X 2 = 2 M X (t) = (et + 4e2t + 9e3t + 16e4t + 25e5t + 36e6t ) dt 6 t=0 t=0 91 1 , so that = (1 + 4 + 9 + 16 + 25 + 36) = 6 6 √ 2 91 21 105 105 Var(X ) = − , and s.d. of X = 1.708. = 6 6 36 6 EX =

1.6 We have: (i) Var(X ) = E(X − E X )2 = E[X 2 + (E X )2 − 2(E X )X ] = E X 2 + (E X )2 − 2(E X )2 = E X 2 − (E X )2 .

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Also, E[X (X − 1)] = E(X 2 − X ) = E X 2 − E X = Var(X ) + (E X )2 − E X (by part (i)),

(ii)

and hence, solving for Var(X ), Var(X ) = E[X (X − 1)] + E X − (E X )2 .

1.7 We have: (i) E[X (X − 1)] = E(X 2 − X ) = E X 2 − E X , so that: E X 2 = E[X (X − 1)] + E X = 27.5 + 5 = 32.5. (ii) Var(X ) = E X 2 − (E X )2 = 32.5 − 25 = 7.5, and s.d. of X = 2.739.

√ 7.5

1.8 (i) Apply #5 in Table 6 in the Appendix with r = 1/2 to obtain: x x ∞ ∞

1 1 1/2 x = x = = 2, 2 2 (1 − 1/2)2 x=0 x=1 x ∞

1 2(1/2)2 E[X (X − 1)] = x(x − 1) = = 4, 2 (1 − 1/2)3 EX =

x=2

(ii) Var(X ) = E[X (X − 1)] + E X − (E X )2 = 4 + 2 − 4 = 2.

1.9 We have: (i) By #5 in Table 6 in the Appendix, applied with r = 1/3 we have: EX =

∞

x=1

2 x× 3

x x ∞ 1 1/3 2 2 1 1 = x = × = .

2 3 3 3 3 (1 − 1/3) 2 x=1

(ii) By #4 in Table 6 in the Appendix, applied with r = et /3, (t < log 3), we have:

∞ ∞

1 2 1 x 2 et x 2 2 et x × = = × , M X (t) = t = e 3 3 3 3 3 1− 3 − et x=0

t

x=0

provided e3 < 1 or t < log 3.

2 d 2et 1 (iii) E X = dt 3−et t=0 = (3−et )2 t=0 = 2 .

3

Revised Answers Manual to Introduction to Probability

1.10 We have: EX =

2

x × 0.5x d x =

0

E X2 =

2

0.5 3 2 4 x = 3 3 0

x 2 × 0.5x d x =

0

1.333,

0.5 4 2 x = 2, so that 4 0

2 16 = , and Var(X ) = 2 − 9 9 √ 2 s.d. of X = 0.471.

3 2

1

7 , E X2 = 1.11 Here E X = 0 x(3x 2 − 2x + 1)d x = 43 x 4 |10 − 23 x 3 |10 + x2 |10 = 12 1 2 3 5 1 1 4 1 1 3 1 13 2 0 x (3x − 2x + 1)d x = 5 x |0 − 2 x |0 + 3 x |0 = 30 , so that Var(X ) = 13 49 67 0.093.

30 − 144 = 720 0 1 0 1 1.12 From the relation: −2 c1 x d x + 0 c2 x d x = 1 and −2 x × c1 x d x + 0 x × 1 c2 x d x = 13 , we obtain: −4c1 + c2 = 2 and 8c1 + c2 = 1, so that c1 = − 12 5 and c2 = 3 .

1.13 We have:

EX =

∞ 0

= 2λ

x × λ xe 2

∞

−λx

d x = −λx 2 e−λx |∞ 0 + 2λ

xe−λx d x = 2

0

∞

∞

x × λe−λx d x =

0

xe−λx d x

0

2 . λ

1.14 Indeed, (i)

E(X − c)2 = E[(X − μ) + (μ − c)]2 = E[(X − μ)2 + (μ − c)2 + 2(X − μ)(μ − c)] = E(X − μ)2 + (μ − c)2 + 2(μ − c)E(X − μ) = E(X − μ)2 + (μ − c)2 = Var(X ) + (μ − c)2 .

(ii) Clearly, the right-hand side above is minimized for c = μ.

1.15 (i) This is a p.d.f. for any c > 0, since c

|x| dx = 2 −c c

0

|x| dx + 2 −c c

c 0

|x| 1 dx = − 2 c2 c

0 −c c2

xd x +

1 c2

c xd x 0

1 1 x2 0 1 x2 c 1 c2 + 2× = 1. =− 2 × + 2 = 2× c 2 −c c 2 0 c 2 c 2

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(ii) Next, E Xn =

1 c2

c

x n |x|d x =

−c

1 =− 2 c

0

0

1 c2

−c

1 x n+1 d x + 2 c −c

1 c2

x n |x|d x +

c

c

x n |x|d x

0

x n+1 d x

0

1 x n+2 0 1 x n+2 c =− 2 × + 2× c n + 2 −c c n+2 0 =

1 1 1 (−c)n+2 cn+2 + = × × [cn+2 + (−c)n+2 ].

2 2 c n+2 c n+2 (n + 2)c2

Therefore:

EX = n

1 2cn × 2cn+2 = n+2 , (n+2)c2

0, 2

if n is even if n is odd.

2

2

In particular, E X = 0 and E X 2 = 2c4 = c2 , so that Var(X ) = c2 .

1.16 (i) Setting x = tan u = sin u , − π < u < π , and observing that cos u 2 2 d d tan u = du du

sin u cos u

=

cos2 u + cos2 u 1 = , cos2 u cos2 u

we get: ∞

dx = 2 −∞ 1 + x

Thus, Next, (ii)

1 π

π/2

1/ cos2 u 2 du =

−π/2 1 + sin u cos u

∞ 1 −∞ f (x)d x = π

∞ dx −∞ x 1+x 2

∞).

=

1 2π

1.17 Indeed, M X (t) = Ee

−π/2

du =

π! π − − = π.

2 2

∞ dx −∞ 1+x 2 = 1.

∞ d(1+x 2 ) −∞ 1+x 2

tX

π/2

=E

∞ n=0

=

1 2 ∞ 2π log(1 + x ) −∞

(t X )n n!

=

∞

n=0

E

=

1 2π (∞ −

(t X )n n!

(assuming that the interchange of the expectation and the summation sign is allowed) ∞

tn (E X n ) . = n! n=0

Revised Answers Manual to Introduction to Probability

1.18 For the discrete case, X ≥ c means xi ≥ c for all values xi of X . f (x ) ≥ c f (x ) and hence x f (x ) ≥ Then x i X i X i i X i x xi c f X (x i ). But i

x f (x ) = E X and c f (x ) = c f (x ) = c.

Thus, E X ≥ c. i X i X i X i xi xi xi The particular case follows, of course, by taking c = 0. In the continuous case, summation signs are replaced by integrals.

1.19 (i) This part has been, actually, established in Exercise 1.6(i). (ii) Var(cX ) = E[cX − E(cX )]2 = E[cX − cE X ]2 = E[c2 (X − E X )2 ] = c2 E(X − E X )2 = c2 Var(X ).

Also, Var(cX + d) = E((cX + d) − E(cX + d))2 = E(cX + d − cE X − d)2 = E(cX − cE X )2 = c2 E(X − E X )2 = c2 Var(X ). (iii) By setting Y = g(X ), we have, by part (i), Var[g(X )] = Var(Y ) = EY 2 − (EY )2 = Eg 2 (X ) − [Eg(X )]2 .

1.20 (i) P(40 ≤ X ≤ 60) = 0.32 (ii) μ = E X = 10 × 0.13 + 20 × 0.12 + 30 × 0.10 + 40 × 0.13 + 50 × 0.10 + 60 × 0.09 + 70 × 0.11 + 80 × 0.08 + 90 × 0.08 + 100 × 0.06 = 49.6, E X 2 = 3,232, so that Var(X ) = 3,232 − (49.6)2 = 3,232 − 2,460.16 = 771.84 and σ 27.782. (iii) [μ − σ, μ + σ ] = [49.6 − 27.782, 49.6 + 27.782] = [21.818, 77.382], [μ−1.5σ, μ+1.5σ ] = [49.6−41.673, 49.6+41.673] = [7.927, 91.273], [μ − 2σ, μ + 2σ ] = [49.6 − 55.564, 49.6 + 55.564] = [−5.964, 105.164].

It follows that the respective proportions of claims are: 53%, 94%, and 100%.

1.21 (i) From ∞ c d x 4 = c ∞ (1 + x)−4 d(1 + x) = − c x(1 + x)−3 |∞ = c = 1, 0 0 (1+x) 0 3 3 we get c = 3.

4 dx 1 1 1 117 4 (ii) P(1 ≤ X ≤ 4) = 3 1 (1+x) 4 = − (1+x)3 |1 = 8 − 125 = 1,000 = 0.117. ∞ ∞ ∞ (iii) 3xd x EX = =3 (1 + x)−4 xd x = − xd(1 + x)−3 (1 + x)4 0 0 0 ∞ ∞ dx x ∞ =− | + = (1 + x)−3 d x (1 + x)3 0 (1 + x)3 0 0

1 1 1 1 ∞ d(1 + x)−2 = − |∞ = . =− 2 0 2 (1 + x)2 0 2

Section 5.2 2.1

1 1 (i) E X = −1 × 18 + 0 × 89 + 1 × 18 = 0, E X 2 = 19 , so that Var(X ) = 19 and 1 s.d. of X = 3 , So, μ = 0, σ = 1/3.

2 (ii) P(|X − μ| ≥ kσ ) = P(|X | ≥ k3 ) = 18 = 19 0.111, both for k = 2 and k = 3.

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(iii) P(|X − μ| ≥ kσ ) ≤ k σ2 σ 2 = k12 . For k = 2, k12 = 0.25 (more than twice as large the exact probability), and for k = 3, k13 = 19 (exactly the same as the exact probability). √ 2.2 (i) From P(|X − μ| < c) ≥ 1 − σ22 ≥ α, we get c = σ/ 1 − α. c (ii) Here c = √ 1 4.464.

2

0.05

2.3 From Exercise 3.11(i) in Chapter 3, c = 3/4. Next,

2 1 1 (i) x x4 1 EX = c = 0, x(1 − x 2 )d x = c − 2 −1 4 −1 −1

3 1 1 4c 4 3 x x5 1 E X2 = c = = × = x 2 (1 − x 2 )d x = c − 3 −1 5 −1 15 15 4 −1 1 = 0.2, so that Var(X ) = 0.2.

5 (ii) P(−0.9 < X < 0.9) = P(|X | < 0.9) = P(|X − E X | < 0.9) ≥ 0.2 1 − (0.9) 1 − 0.247 = 0.753.

2 In Exercise 3.11(ii) of Chapter 3, it was seen that the exact probability P(−0.9 < X < 0.9) = 0.9855. Thus, the approximate probability 0.753 is about 76.4% of the exact probability.

2.4

) (i) By the Tchebichev inequality, P(|X − μ| ≥ c) ≤ Var(X = 0 for all c > 0. c2 (ii) Consider a sequence 0 < cn ↓ 0 as n → ∞. Then P(|X − μ| ≥ cn ) = 0 for all n, or equivalently, P(|X − μ| < cn ) = 1 for all n, whereas, clearly, {(|X − μ| < cn )} is a nonincreasing sequence of events and its limits are ∩∞ n=1 (|X − μ| < cn ). Then, by Theorem 2 in Chapter 3, 1 = limn→∞ P(|X − μ| < cn ) = P(∩∞ n=1 (|X − μ| < cn )). However, it is (|X − μ| < c ) = (|X − μ| ≤ 0) = (X = μ). Thus, clear that ∩∞ n n=1 P(X = μ) = 1, as was to be seen.

Section 5.3 3.1 (i) E X = 1 x × 3x 2 d x = 3 = 0.75, m = x0.5 3x 2 d x = x 3 = 0.5, so 0.5 0 0 4 that x0.5 = (0.5)1/3 0.794, and mean < median. x 3 = 0.125 = 18 = 213 , so that x0.125 = 21 = 0.5. (ii) 0 0.125 3x 2 d x = x0.125 x

x

x n+1

p 1 3.2 (i) From P(X ≤ x p ) = 0 p x n d x = n+1 x n+1 0 p = n+1 = p, we get 1/(n+1) x p = [(n + 1) p] . (ii) For p = 0.5 and n = 3, we have x0.5 = 21/4 1.189.

3.3 (i) From P(X ≤ x p ) = 0x p λe−λx d x = −e−λx |0x p = 1 − e−λx p = p, we get p) . e−λx p = 1 − p, or x p = − log(1− λ (ii) For λ = 1/10 and p = 0.25, we have: x0.25 = −10 log(0.75) 2.877.

Revised Answers Manual to Introduction to Probability

0 1 0 2 2 −1 c1 x d x + 0 c2 (1 − x )d x = 1 and −1 x × 1 2 0 x × c2 (1 − x )d x = 0, we obtain: c1 + 2c2 = 3 and c1 = c2 .

3.4 (i) From the two relations:

c1 x 2 d x + Thus, c1 = c2 = 1.

3 0 0 (ii) Observe that −1 x 2 d x = x3 −1 = 13 , so that x1/3 = 0.

3.5 (i) By differentiation, we have: f (x) = 43 (2x − x 2 ), 0 < x ≤ 2.

Then, for 0 < x < 2: (ii) f (x) = 0 yields x = 1 and f (1) = − 23 < 0, so that m = 1 is the (unique) mode of f . c 5 , we obtain, (iii) For simplicity, set c = x5/32 . Then, from 0 34 (2x −x 2 )d x = 32 3 2 after some simplifications, 8c − 24c + 5 = 0. Observe that c = 21 is a root of this equation. Thus, x5/32 = 21 . (The other roots are 2.927 and −0.427, outside (0, 2]).

3.6 Here (i) x 2 3 4 5 6 7 8 9 10 11 12 f (x ) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 1 (ii) E X = 36 (2 × 1 + 3 × 2 + 4 × 3 + 5 × 4 + 6 × 5 + 7 × 6 + 8 × 5 + 9 × 4 + 10 × 3 + 11 × 2 + 12 × 1) = 252 36 = 7. (iii) The median and the mode are the same here and equal to 7.

3.7 (i) Mode = 1 and the maximum is 1/2. (ii) Mode = 1 and the maximum is 1 − α; α = 21 . (iii) Mode = 0 and the maximum is 2/3. c −∞ f (x)d x, and set x = c − y, so that d x = c 0 −dy and y ranges from ∞ to 0. Then: −∞ f (x)d x = − ∞ f (c − y)dy = ∞ ∞ 0 f (c − y)dy. Again as suggested, P(X ≥ c) = c f (x)d x, and set ∞ x = c + y, so that d x = dy and y ranges from 0 to ∞. Then: c f (x)d x = ∞ ∞ 0 f (c + y)dy. However, f (c − y) = f (c + y) for all y, so that: 0 f (c − ∞ y)dy = 0 f (c + y)dy, or P(X ≤ c) = P(X ≥ c); hence c is the median of

3.8 As suggested, P(X ≤ c) =

X. 3.9 (i) Given as a Hint. c (ii) Immediate, since, for m < c : c − x ≥ 0 and hence m (c − x) f (x)d x ≥ 0, c m m whereas for m ≥ c: m (c− x) f (x)d x = − c (c− x) f (x)d x = c (x −c) f (x)d x ≥ 0, because x − c ≥ 0.

3.10 We have: (i) p = P(Y ≤ y p ) = P[g(X ) ≤ y p ] = P[X ≤ g −1 (y p )], so that: g −1 (y p ) = x p , or y p = g(x p ).

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x

x

(ii) Here p = 0 p e−x d x = −e−x |0 p = 1 − e−x p , so that x p = − log(1 − p). (iii) y p = g(− log(1 − p)).

Chapter 6 Section 6.1

n n−x 1.1 (i) X ∼ B(n, p) implies that P(X = x) = nx p x q n−x = n−x q × n n n−(n−x) p , since x = n−x . However, this last expression is equal to: P(Y = n − x), where Y ∼ B(n, q). (ii) For n = 20, p = 0.625, and x = 8, we have:

20 20 8 12 (0.375)12 (0.625)8 (0.625) (0.375) = 12 8 = 0.9884 − 0.9657 = 0.0227 (from the binomial tables).

" # n + 1 x (n+1)−x n n 1.2 (i) B(n + 1, p; x) = p q = + p x q n+1−x x x x −1 (by Exercise 4.9 in Chapter 2)

n n x n+1−x x n+1−x = p q + p q x −1 x

n n x n−x x−1 n−(x−1) = p× p q +q × p q x −1 x = p B(n, p; x − 1) + qB(n, p; x). (ii)

B(26, 0.25; 10) = (0.25)B(25, 0.25; 9) + (0.75)B(25, 0.25; 10) = 0.25 × 0.0781 + 0.75 × 0.0416 = 0.019525 + 0.0312 = 0.050725.

1.3 If X is the r.v. denoting the number of tickets to win a prize, then X ∼ B(50, 0.01). Therefore:

49 (i) P(X = 1) = 50 0.306.

1 (0.01)(0.99)

0 50 1 − 0.605 = (ii) P(X ≥ 1) = 1 − P(X = 0) = 1 − 50 0 (0.01) (0.99) 0.395.

1.4 If X is the r.v. denoting the number of those favoring the proposal, then X ∼ B(15, 0.4375) = B(15, 7/16). Thus: 4

15 (0.4375)x (0.5625)15−x (i) P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − x x=0 = 1 − 0.141 = 0.859 (from the binomial tables).

Revised Answers Manual to Introduction to Probability

(ii) P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.6894 = 0.3106.

7 = 105 (iii) E X = 15 × 16 16 = 6.5625, √

7 9 Var(X ) = 15 × 16 × 16 = 9×105 , so that σ (X ) = 3 16105 162

1.92.

1.5 If X is the r.v. denoting the number of successes in 18 tossings, then X ∼ B(18, 1/6). Since the number of failures is 18 − X , we have: (i) P(X > 18 − X ) = P(2X > 18) = P(X > 9) = 1 − P(X ≤ 9) 9 x 18−x

18 1 5 0.000196. = 1− x 6 6 x=0

(ii)

P(X = 2(18 − X )) = P(X = 36 − 2X ) = P(3X = 36) 12 6 5 18 1 = P(X = 12) = 6 6 12 0.000002856.

(iii) P(18 − X = 3X ) = P(4X = 18) = P(X = 4.5) = 0.

1.6 If X is the r.v. denoting the number of times the bull’s eye is hit, then X ∼ B(100, p). Therefore: 100 x 100−x

(i) P(X ≥ 40) = 100 (q = 1 − p). x=40 x p q 100 100 x (ii) P(X ≥ 40) = x=40 x (0.25) (0.75)100−x . (iii) E X = np = 100 p and Var(X ) = npq = 100 pq, and for p = 0.25, we √ have: E X = 25, Var(X ) = 18.75, and s.d. of X = 18.75 4.33.

3 1.7 Here E X = np = 100 4 = 25 and Var(X ) = npq = 25 × 4 = 18.75, so that:

P(|X − 25| < 10) = P(|X − E X | < 10) ≥ 1 − = 1 − 0.1875 = 0.8125.

18.75 σ 2 (X ) =1− 102 100

npq pq X X 1.8 Here E( Xn ) = np n = p and Var( n ) = n 2 = n . Thus, P(| n − p| < √ pq/n 400 400 0.05 pq) ≥ 1 − (0.05√ pq)2 = 1 − n , and suffices 1 − n ≥ 0.95, or n ≥ 8,000, so that n = 8,000.

1.9 Indeed, for x = 0, 1, . . . , n − 1,

n p n! f (x + 1) = p x+1 q n−x−1 = × × p x q n−x x +1 (x + 1)!(n − x − 1)! q n!(n − x) p × p x q n−x = × q x!(x + 1)(n − x)! n! p n−x × × p x q n−x = × q x +1 x!(n − x)!

p n − x n x n−x p n−x p q = × = × f (x). q x +1 x q x +1

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1.10 (i)

EX = =

n

x=0 n

n n x n−x n! p q = x p x q n−x x!(n − x)! x

x×

x=1

x

x=1

(n − 1)!n p p x−1 q n−x (x − 1)!x(n − x)!

n

(n − 1)! p x−1 q (n−1)−(x−1) (x − 1)![(n − 1) − (x − 1)]! x=1

n

n − 1 x−1 (n−1)−(x−1) p q = np x −1 = np

= np

x=1 n−1

y=0

n − 1 y (n−1)−y p q (by setting x − 1 = y) y

= np × 1 = np; i.e., E X = np. Next, as above,

E[X (X − 1)] =

n

x(x − 1) ×

x=2

(n − 2)!(n − 1)n (x − 2)!(x − 1)x(n − x)!

p 2 p x−2 q n−x n

(n − 2)! = n(n − 1) p 2 (x − 2)![(n − 2) − (x − 2)]! p

x=2 x−2 (n−2)−(x−2)

q

= n(n − 1) p

2

n

n−2 x=2

= n(n − 1) p 2

x −2

n−2

n−2 y=0

y

p x−2 q (n−2)−(x−2) p y q (n−2)−y (by setting x − 2 = y)

= n(n − 1) p . Therefore, by Exercise 1.6(ii) in Chapter 5 : 2

(ii)

Var(X ) = E[X (X − 1)] + E X − (E X )2 = n(n − 1) p 2 + np − (np)2 = n 2p 2 − np 2 + np − n 2p 2 = np(1 − p) = npq; i.e., Var(X ) = npq.

1.11 We have: (i)

M X (t) =

n

x=0

n n x n−x n p q ( pet )x q n−x e = x x tx

= ( pet + q)n , t ∈ .

x=0

Revised Answers Manual to Introduction to Probability

(ii)

EX =

d d = ( pet + q)n M X (t) dt d t=0 t=0

= n( pet + q)n−1 × pet |t=0 = np, d2 d E X 2 = 2 M X (t) = [n( pet + q)n−1 pet ] dt dt t=0 t=0 d t n−1 t = np [( pe + q) e ] dt t=0 = np[(n − 1)( pet + q)n−2 × pe2t + ( pet + q)n−1 et ]|t=0 = np[(n − 1)( p + 1)], so that Var(X ) = E X 2 − (E X )2 = np[(n − 1) p + 1] − n 2 p 2 = n 2 p 2 − np 2 + np − n 2 p 2 = np(1 − p) = npq.

1.12 Here (n + 1) p = 101 4 = 25.25 is not an integer. Then there is a unique mode obtained for x = 25. In other words, f (25) is the unique maximum probability among the probabilities f (x) = P(X = x), x = 0, 1, . . . , 100, and one would bet on the value X = 25. Also, f (25) 0.092.

1.13 (i) P(X ≤ 10) = 1 − P(X ≥ 11) = 1 − p ∞ q x−1 = 1 − pq 10 × 1 = x=11

1 − q 10 . (ii) 1 − (0.8)10

1−q

1 − 0.107 = 0.893.

1.14 Here X has the geometric distribution with p = 0.01, so that f (x) = (0.01) (0.99)x−1 , x = 1, 2, . . . then: P(X ≤ 10) = 1 − P(X ≥ 11) = 1 − (0.01)(0.99)10 [1 + 0.99 + (0.99)2 + · · · ] 1 = 1 − (0.99)10 0.096. = 1 − (0.01)(0.99)10 × 1 − 0.99 1.15 If X is the r.v. denoting the number of tosses to the first success, then X has the geometric distribution with parameter p = 16 , so that P(X = x) = 1 5 x−1 , x = 1, 2, . . . then: 6(6)

1 5 2 25 0.116. (i) P(X = 3) = = 6 6 216

4 ∞ ∞ (ii)

5 1 5 x−1 1 5 x−1 1 1 = = × × P(X ≥ 5) = 6 6 6 6 6 6 1 − 56 x=5 x=5 4 5 = 0.482.

6 1.16 If X is the r.v. denoting the smallest number of tosses until the first head appears, then X has the geometric distribution with parameter p. Then:

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(i) The probability that the first head appears by the nth time is nx=1 pq x−1 and we require the determination of the smallest value of n for which: n

pq x−1 ≥ α, or 1 −

x=1

∞

pq x−1 ≥ α, or 1 − q n ≥ α, or

x=n+1

n ≥ log(1 − α)/ log q.

So, n is the smallest integer ≥ log(1 − α)/ log q. (ii) For α = 0.95, and p = 0.25 (q = 0.75) and p = 0.50(=q), we obtain: n ≥ 10.42 and n ≥ 4.33, respectively, so that: n = 11, and n = 5, respectively.

1.17 Here f (x) = pq x−1 , x = 1, 2, . . . , so that: (i)

EX =

∞

x × pq x−1 =

x=1

∞

∞

x=1

x=1

d p x qx xq = p q dq

q p 1 = × 2 = , q p p by #5 in Table 6 in the Appendix, applied with r = q, and E[X (X − 1)] = p

∞

∞

x(x − 1) × q x−1 =

x=2

=

p x(x − 1)q x q x=2

2q 2

p 2q × 3 = 2, q p p

by #5 in Table 6 in the Appendix, applied with r = q. Therefore, by Exercise 1.6(ii) in Chapter 5, 1 1 q 2q (ii) Var(X ) = 2 + − 2 = 2 . p p p p (iii) The unique mode is 1, and the corresponding probability is p.

1.18 (i) Here, with q = 1 − p, M X (t) =

∞

x=1

∞

et x × pq x−1 =

qet p t x p (qe ) = × , q q 1 − qet x=1

provided qet < 1 (by #6 in Table 6 in the Appendix). So, M X (t) = t < − log q.

Then: 0.01et (ii) M X (t) = 1−0.99e t < − log(0.99)( t,

0.01).

pet , 1 − qet

Revised Answers Manual to Introduction to Probability

(iii)

pet et d et d = p = p EX = dt 1 − qet t=0 dt 1 − qet t=0 (1 − qet )2 t=0 p 1 = 2 = ; p p

2 pet et d d et (1 + qet ) 2 EX = 2 = p = p dt 1 − qet t=0 dt (1 − qet )2 t=0 (1 − qet )3 t=0 1+q q = , so that Var(X ) = 2 . p2 p For p = 0.01, we have E X = 100 and Var(X ) = 9,900, so that σ (X ) 99.5.

1.19 f (2) = 2 f (0) means e−λ λ2! = 2e−λ λ0! , or λ2 = 4, hence λ = 2. x 1.20 Since P(X = x) = e−λ λx! , x = 0, 1, . . . , we have: P(X = 0) = e−λ = 0.1, so that −λ = log(0.1) −2.3 and λ 2.3. Then: 2

0

P(X = 5) = e−2.3

(2.3)5 5!

0.054.

1.21 Indeed, f (x + 1) = e−λ

λ λ λx+1 λ × λx λx = e−λ = × e−λ = f (x). (x + 1)! x!(x + 1) x +1 x! x +1

1.22 By the Hint, we have: (i)

EX =

∞

x × e−λ

x=0

∞

λx λ × λx−1 = x e−λ x! (x − 1)!x x=1

∞ ∞

λx−1 λy = λe−λ = λe−λ (x − 1)! y! x=1

−λ

y=0

λ

= λe × e = λe = λ; i.e., E X = λ. Likewise, ∞

λ2 × λx−2 x(x − 1) × e−λ E[X (X − 1)] = (x − 2)!(x − 1)x 0

x=2

= λ2 e−λ

∞

λy y=0

y!

= λ2 e−λ eλ = λ2 ,

so that: (ii) Var(X ) = E[X (X − 1)] + E X − (E X )2 = λ2 + λ − λ2 = λ; i.e., Var(X ) = λ.

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1.23 We have: ∞ (λet )x

tx −λ λx = e−λ −λ λet = eλ(et −1) , t ∈ (i) M X (t) = ∞ x=0 e ×e x=0 x! = e e x! (by #6 in Table 6 in the Appendix). t d λ(et −1) [e ]|t=0 = eλ(e −1) × λet |t=0 = λ, (ii) E X = dt E X2 =

d 2 λ(et −1) d t [e ]|t=0 = [eλ(e −1) λet ]|t=0 dt 2 dt d t = λ [eλ(e −1) et ]|t=0 dt t t = λ[eλ(e −1) λe2t + eλ(e −1) et ]|t=0 = λ(λ + 1), so that

Var(X ) = λ(λ + 1) − λ2 = λ2 + λ − λ2 = λ.

1.24 (i) An appropriate probability model here is the Poisson distribution with parameter 3, P(3). (ii) What we are asked to find here is the mode (or modes) of the distribution.

Since λ = 3, Theorem 2 provides for two modes 3 and 2. So, the number of items which arrive within 1 second with the maximum probability is either 2 or 3. The respective probability is 0.2241.

1.25 Here and in the sequel, as n → ∞, npn = λn → λ, pn = λnn → 0, and qn = 1 − λnn → 1. Then:

n x n−x n(n − 1) · · · (n − x + 1) x n−x pn q n = pn q n x x!

λn n−x n(n − 1) · · · (n − x + 1) λn x 1− = x! n n

x λn n n(n − 1) · · · [n − (x − 1)] λn 1 1− = × × !x nx x! x 1 − λxn

x −1 λnx λn n 1 1 ··· 1 − × 1− = 1× 1− × !x n n x! n 1 − λnn →1×

λx λx −λ e × 1 = e−λ , x! x!

because, e xn −→ e x if xn → x, by #6 in Table 6 in the Appendix. n→∞

1.26 We have: 70 10

70

78,591 = 805 = 156,104 0.503.

5

10 70 10 70 10 23,735,264 = 24,040,016 (ii) 801 70 3 2 + 4 1 + 5 0

(i)

5 0 80 5 5

0.987.

Revised Answers Manual to Introduction to Probability

1.27 (i) The distribution of X is hypergeometric with m = 15 and n = 10, so that: 15 10 x

f (x) = (ii) P(X ≥ 10) =

15 x=10

15 10 x

15−x

25

, x = 5, . . . , 15.

15

15−x 25 . x

(iii) This probability is 0, since there are only 10 specimens from the rock R2 .

1.28 We have:

m

f (x + 1) = 1 = m+n × r

1

= m+n × r

n r −x−1 m+n r

x+1

m! n! × (x + 1)!(m − x − 1)! (r − x − 1)!(n − r + x + 1)! n!(r − x) m!(m − x) × x!(x + 1)!(m − x)! (r − x)!(n − r + x)!(n − r + x + 1)

m! r−x n! m−x × × × x +1 x!(m − x)! n − r + x + 1 (r − x)!(n − r + x)! r m n (m − x)(r − x) (m − x)(r − x) r −x × x m+n f (x). = = (m − r + x + 1)(x + 1) (m − r + x + 1)(x + 1) r 1

= m+n ×

1.29 We have m n x r −x m+n r

m!n!(m + n − r )! (m − x)![n − (r − x)]!(m + n)!

r (m − x + 1) · · · m(n − r + x + 1) · · · n = (m + n − r + 1) · · · (m + n) x

r m(m − 1) · · · [m − (x − 1)] · n(n − 1) · · · [n − (r − x − 1)] . = (m + n) · · · [(m + n) − (r − 1)] x

=

Both the numerator and denominator have r factors. Dividing through by (m + n), we get

m n m m r m 1 x −1 x r −x

··· = − − m+n m+n m+n m+n m+n m+n x r

n 1 n r −x −1 n − ··· − × m+n m+n m+n m+n m+n

#−1 "

r −1 1 ··· 1 − × 1× 1− m+n m+n

r x r −x p q −→ , m,n→∞ x m n since m+n → p and hence m+n → 1 − p = q.

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1.30 We have P(X = x) = nx (0.15)x (0.85)n−x , x = 0, 1, . . . , n, so that P(X = 0) = (0.85)n , P(X = 1) = n(0.15)(0.85)n−1 . Therefore, n(0.15)(0.85)n−1 > 85 5.667, so that n = 6.

(0.85)n , or 0.15n > 0.85, or n > 0.85 0.15 = 15 1.31 Let X 1 and X 2 be the r.v.’s denoting the number of voters favoring propositions #1 and #2, respectively. Then X 1 ∼ B(2n, p1 ) and X 2 ∼ B(2n, p2 ), so that: (i) (ii) (iii) (iv)

E X 1 = (2n) p1 = 2np1 , E X 2 = (2n) p2 = 2np2 . x

2n−x .

P(X 1 ≤ n) = nx=0 2n x p1 (1 − p1 ) 2n x 2n 2n

2n−x .

P(X 2 ≥ n) = x=n x p2x (1− p2 )2n−x = 1− n−1 x=0 x p2 (1− p2 ) E X 1 = 24 × 0.3125 = 7.5, E X 2 = 24 × 0.4375 = 10.5, P(X 1 ≤ 12) = 0.9835, P(X 2 ≥ 12) = 1 − 0.6618 = 0.3382 (from the binomial tables).

1.32 Let X be the r.v. denoting the number of defective items among the 100 sampled.

Then the r.v. X has the Hypergeometric distribution with: m = 3, n = 997, r = 100. Therefore: (i) The probability of accepting the entire shipment is: 3 997 P(X ≤ 1) = P(X = 0) + P(X = 1) = 997

100 01,000

100

3 × 997 538,501 100 = 1,000 + 1,00099 = 553,890 100

100

3×997×100(1,000−100) (1,000)2 (1,000−1)

=

3 997 99 + 11,000

100

0.97.

(ii) E X = 3×100 1,000 = 0.3. (iii) Var(X ) = $ 2,991 111 /10

2,991 100×111

0.269, and σ (X ) =

0.519.

m 3 1.33 Since m+n = 1,000 = 0.003, we have that the required approximate value

100 0 1 99 = (0.997)100 + 100 × is: 0 (0.003) (0.997)100 + 100 1 (0.003) (0.997) 99 99 0.003 × (0.997) = (0.997) (0.997 + 0.3) 0.742 × 1.297 0.9624, which is about 99.2% of the probability 0.97. m m = 0.003, so that r × m+n = 0.3. Therefore the probability 1.34 Here r = 100, m+n of acceptance of the entire shipment is 0.963064 (from the Poisson tables), which is about 99.3% of the probability 0.97.

1.35 Let X be the r.v. denoting the number of homeowners who have earthquake insurance. Then X has the hypergeometric distribution with m = 125, n = 875, r = 25. Hence: mnr (m+n−r ) mr 125×25 = 3.125, σ 2 (X ) = (m+n) 2 (m+n−1) m+n = 1,000 170,625 125×875×25×975 = 2.669, and σ (X ) 1.634.

63,936 1,0002 ×999

(i) E X

=

=

(ii) P(3.125 − 1.634 ≤ X ≤ 3.125 + 1.634) = P(1.491 ≤ X ≤ 4.759) = 125 875 1 4 P(2 ≤ X ≤ 4) = 1,000 x=2 x 25−x , denote it by α.

25

Revised Answers Manual to Introduction to Probability

m 125 (iii) Since m+n = 1,000 = 0.125 = 2/16, the binomial approximation (see 4 25 2 x 14 25−x Exercise 1.29 in this chapter) gives α = x=2 x ( 16 ) ( 16 ) 0.8047 − 0.1623 = 0.6424. Next, 25 × 0.125 = 3.125 and the Poisson tables give: α 0.8153 − 0.1991 = 0.6162 for λ = 3.00, and α 0.7254 − 0.1359 = 0.5895 for λ = 3.50. By linear interpolation, we get α 0.6095. (i) Since E X = Var(X ) = λ, it suffices to determine the value of λ. The 1.36

relation P(X = 2) = 3P(X = 4) becomes: e−λ λ2! = 3e−λ λ4! , or λ = 2.

Hence E X = Var(X ) = 2.

Here P(2 ≤ X ≤ 4) = P(X ≤ 4) − P(X ≤ 1) = 0.9473 − 0.4060 = 0.5413, and P(X ≥ 5) = 1− P(X < 5) = 1− P(X ≤ 4) = 1−0.9473 = 0.0527.

Among the x outcomes, we have exactly r successes if r −1 of the successes occur among the first x − 1 outcomes (so that there are (x − 1) − (r − 1) = x − r failures among the first x − 1 outcomes) and the last success occurs on trial. By independence, the probability of this happening is rlast

x−1the −1 × q x−r × p = x−1 q x−r pr . The second expression follows by p r −1 r −1

x−1 x−1 the fact that rx−1 = −1 (x−1)−(r −1) = x−r .

∞ ∞

x − 1 x−r r pr x − 1 (qet )x et x × q p = r M X (t) = x − r q x − r x=r x=r "

# r −1 pr r r +1 (qet )r + = r (qet )r +1 + (qet )r +2 + · · · 0 q 1 2 "

# r pr (qet )r r + 1 t t 2 1 + (qe = ) + (qe ) + · · · 1 qr 2 pr (qet )r 1 = × , provided qet < 1 or t < − log q, qr (1 − qet )r r

pet . = 1 − qet r −1

d pet pet (1 − qet ) − pet (−qet ) M X (t) = r × dt 1 − qet (1 − qet )2 r −1

t t pe pe ( pet )r =r × = r , and 1 − qet (1 − qet )2 (1 − qet )r +1

d2 r r ( pet )r −1 pet (1 − qet )r +1 M X (t) = 2 t 2r +2 dt (1 − qe )

− ( pet )r (r + 1)(1 − qet )r (−qet )

r = r ( pet )r (1 − qet )r +1 (1 − qet )2r +2

+ (r + 1)qet ( pet )r (1 − qet )r 2

(ii)

1.37 (i)

(ii)

(iii)

4

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r ( pet )r (1 − qet )r [r (1 − qet ) + (r + 1)qet ] (1 − qet )2r +2 r ( pet )r (r + qet ) . = (1 − qet )r +2

=

d r pr r M X (t) = r +1 = = E X , t=0 dt p p 2 r d r p (r + q) r (r + q) M X (t) = = = E X 2, 2 r +2 t=0 dt p p2

(iv)

− rp2 = rq . so that σ 2 (X ) = E X 2 − (E X )2 = r (rp+q) 2 p2 2

, so that: (v) E X = 5p , σ 2 (X ) = 5q p2

5 3.75 p = 0.25: E X = 0.25 = 20, σ 2 (X ) = 5×0.75 = 0.0625 = 60, and 0.252 σ (X ) 7.746.

5 2.5 = 10, σ 2 (X ) = 5×0.5 = 0.25 = 10, and σ (X ) p = 0.5: E X = 0.5 0.52 3.162.

5 1.25 = 20 6.667, σ 2 (X ) = 5×0.25 = 0.5625 = 20 p = 0.75: E X = 0.75 3 9 0.752 2.222, and σ (X ) 1.491.

Section 6.2 2.1 Indeed, with α > 1, ∞

∞ y α−1 e−y dy = − y α−1 de−y = −y α−1 e−y |∞ 0 0 0 ∞ +(α − 1) y α−2 e−y dy 0 ∞ y (α−1)−1 e−y dy = (α − 1) (α − 1). = 0 + (α − 1)

(α) =

0 1 α−1 e 2.2 Since f (x) = (α)β αx

(i)

− βx

, x > 0, we have:

∞ ∞ 1 1 α−1 −x/β x × x e d x = x α e−x/β d x (α)β α 0 (α)β α 0 ∞ 1 x = β α y α e−y βdy (by setting = y) α (α)β 0 β ∞ β β = (α + 1) (by (12)) y (α+1)−1 e−y dy = (α) 0 (α) β = × α (α) (α) = αβ; i.e., E X = αβ. Likewise,

EX =

Revised Answers Manual to Introduction to Probability

∞ ∞ 1 β2 α+1 α+1 −y β y e β dy = y (α+2)−1 e−y dy (α)β α 0 (α) 0 β2 β2 = (α + 2) = × (α + 1)α (α) = α(α + 1)β 2 , (α) (α)

E X2 =

so that Var(X ) = α(α + 1)β 2 − α 2 β 2 = α 2 β 2 + αβ 2 − α 2 β 2 = αβ 2 ; i.e., Var(X ) = αβ 2 . (ii) Here α = 1, β = λ1 , so that E X = λ1 , Var(X ) = λ12 . (iii) Here α = r2 , β = 2, so that E X = r , Var(X ) = 2r .

2.3 We have: (i)

∞ ∞ 1 1 t x− x tx α−1 −x/β e × x e d x = x α−1 e β d x α α (α)β 0 (α)β 0 ∞ (1−βt)x 1 − β = x α−1 e dx (α)β α 0 ∞ x 1 1 − = × x α−1 e β/(1−βt) d x α α (1 − βt) (α)[β/(1 − βt)] 0 1 1 1 = ×1= , provided 1 − βt > 0, or t < (1 − βt)α (1 − βt)α β

M X (t) =

(ii)

d [(1 − βt)−α ] = αβ(1 − βt)−α−1 |t=0 = αβ, dt t=0 d E X 2 = [αβ(1 − βt)−α−1 ] dt t=0 EX =

= −αβ(α + 1)(1 − βt)−α−2 × (−β)|t=0 = α(α + 1)β 2 (1 − βt)−α−2 |t=0 = α(α + 1)β 2 , so that Var(X ) = α(α + 1)β 2 − α 2 β 2 = αβ 2 .

2.4 We have: (i) (a) g(X ) = cX , so that Eg(X ) = cE X = c/λ. (b) g(X ) = c(1 − 0.5e−α X ), so that: ∞

c(1 − 0.5e−αx ) × λe−λx d x = c 0 ∞ e−(α+λ)x d x − 0.5cλ 0

0.5cλ ∞ (α + λ)e−(α+λ)x d x =c− α+λ 0 0.5cλ (α + 0.5λ)c =c− = . α+λ α+λ

Eg(X ) =

∞ 0

λe−λx d x

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(ii) (a) 10; (b) 1.5.

2.5 Here f (x) = λe−λx , x > 0, so that: (i) F(x) = 0 λe−λt dt = −e−λt |0x = 1 − e−λx , and 1 − F(x) = e−λx .

Therefore x

r (x) = (ii)

λe−λx f (x) = −λx = λ, constant for all x > 0.

1 − F(x) e P(X > s + t) P(X > s + t, X > s) = P(X > s) P(X > s) −λ(s+t) e 1 − F(s + t) = = 1 − F(s) e−λs −λt =e independent of s > 0.

P(X > s + t|X > s) =

The independence from s says that the underlying distribution is “memoryless.” This is not a particularly desirable attribute if this distribution is to serve as a lifetime distribution.

2.6 Let T1 be the (waiting) time from time 0 to the occurrence of the first event.

Then, for t > 0, P(T1 > t) = P(number of events occurred in (0, t] = 0) (λt)0 = e−λt = e−λt , 0! so that P(T1 ≤ t) = 1 − e−λt and f T (t) = λe−λt , as was to be seen.

For k ≥ 2 and with tk−1 denoting the time of occurrence of the (k − 1)-st event, we have P(Tk > t) = P(number of events occurred in (tk−1 , tk−1 + t) = 0) = e−λt , since (tk−1 + t) − tk−1 = t, and f Tk (t) = λe−λt , as above.

2.7 By Exercise 2.6, the distribution of the waiting time is a r.v. Y which is distributed as the negative exponential with parameter λt. Here λ = 3 and t = 1 second. Then: (i) P(1st particle will arrive within 1 second) = P(Y ≤ 1) = 0 3e−3x d x = −e−3x |10 = 1 − e−3 0.95. (ii) P(waiting for at least another second, given that we have already waited 1

−6

≥2,T >1) P(T ≥2) e one second) = P(T ≥ 2|T > 1) = P(TP(T = P(T >1) >1) = e−3 = 1 0.05. e3

Revised Answers Manual to Introduction to Probability

2.8 (i) Observe that: ∞ ∞ ∞ β β−1 −αx β −αx β β−1 αβx e dx = e (αβx )d x = − de−αx 0

0

= −e

−αx β

|∞ 0 =−

∞

1 eαx

β

0

= −0 + 1 = 1.

0

(ii) For β = 1 and any α > 0. (iii) For n = 1, 2, . . . , ∞ ∞ β n n β−1 −αx β E X = αβ x ×x e d x = αβ x n+β−1 e−αx d x.

0

0 1/β

Set αx β = t, so that x = αt 1/β , d x = ∞

n+β−1 β

1

1 −1 tβ βα 1/β

d x, and 0 < t < ∞. Then:

−1

tβ E X = αβ × e−t dt n+β−1 1/β βα 0 β α

βn + 1 ∞ βn + 1 1 ( βn +1)−1 −t

t = e dt = α n/β α n/β βn + 1 0 n

t

( n +1)−1 −t

e (t > 0), is the p.d.f. of the gamma distribution

because ( n1+1) t β β

with parameters βn + 1 and 1. Thus, E X n = ( βn + 1)/α n/β . For n = 1

and n = 2, we get E X = ( β1 + 1)/α 1/β and E X 2 = ( β2 + 1)/α 2/β , respectively, so that " 1

2

β2 + 1 β + 1 #2 β2 + 1 − β1 + 1 Var(X ) = − = . α 2/β α 1/β α 2/β

2.9 (i) From Exercise 2.8(i), x β β β αβt β−1 e−αt dt = −e−αt |0x = 1 − e−αx , F(x) = 0

so that, for x > 0, β

αβx β−1 e−αx f (x) = = αβx β−1 . β 1 − F(x) e−αx P(X > s + t, X > s) P(X > s + t) P(X > s + t|X > s) = = P(X > s) P(X > s) β

1 − F(x) = e−αx and r (x) = (ii)

β

=

e−α(s+t) 1 − F(s + t) β β = = e−α[(s+t) −s ] . β 1 − F(s) e−αs

(iii) Here, both the failure rate and the conditional survival probability do depend on the variables involved. This is a desirable characteristic for lifetime distributions.

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x

2.10 (i) From 0 p e−x d x = p, we obtain 1 − e−x p = p and hence x p = − log(1 − p). (ii) Since the function y = e x (x > 0) is strictly increasing, it follows (by Exercise 3.10 in Chapter 5) that the pth quantile y p of the r.v. Y = e x is given by y p = e x p = 1/(1 − p). (iii) For p = 0.5, we get x0.5 = − log(1 − 0.5) = − log 0.5 = − log( 21 ) = 1 log 2 0.693, and y0.5 = ( 0.5 ) = 2.

2.11 In all cases: P(a < Z < b) = P(Z ≤ b) − P(Z ≤ a) = by symmetry:

(b) −

(a). Then,

(i) Is left intact. (ii) P(a < Z < b) = P(Z ≤ b) − P(Z ≤ a) = (b) − (a) = (b) − [1 − (−a)] = (−a) + (b) − 1. (iii) P(a < Z < b) = P(−b < Z < −a) = P(Z ≤ −a) − P(Z ≤ −b) = (−a) − (−b). (iv) P(−c < Z < c) = P(Z ≤ c) − P(Z ≤ −c) = (c) − (−c) = (c) − [1 − (c)] = 2 (c) − 1.

2.12 By part (iv) of Exercise 2.11: (i) P(−1 < Z < 1) = 2 (1) − 1 = 2 × 0.841345 − 1 = 0.68269. (ii) P(−2 < Z < 2) = 2 (2) − 1 = 2 × 0.977250 − 1 = 0.9545. (iii) P(−3 < Z < 3) = 2 (3) − 1 = 2 × 0.998650 − 1 = 0.9973.

2.13 Clearly, (i)

P(X < c) = P(X ≤ c) = 2 − 9[1 − P(X ≤ c)] = −7 + 9P(X ≤ c), so that

X −μ c−μ 7 7 ≤ = = 0.875, P(X ≤ c) = or P 8 σ σ 8

and hence (from the normal tables) c−μ σ = 1.15 or c = μ + 1.15σ . (ii) Here c = 5 + 1.15 × 2 = 7.30.

2.14 We have: P(|X − μ| < kσ ) ≥ 1 −

σ2 1 = 1− 2. (kσ )2 k

Therefore, for k = 1, 2, 3, the respective bounds are: 0 (meaningless); 34 = 0.75, which is about 78.6% of the exact probability (0.9545) in the normal 0.8889, which is about 89.1% of the exact probability (0.9973) in case; 89 the normal case.

2.15 Since X ∼ N (μ, σ 2 ) (approximately) with μ = 105 and σ = 20, we have: !

50−105 = P(Z ≥ −2.75) = P(Z ≤ 2.75) = (i) P(X ≥ 50) = P X −μ σ ≥ 20 (2.75) = 0.997020. !

80−105 (ii) P(X ≤ 80) = P X −μ = P(Z ≤ −1.25) = (−1.25) = σ ≤ 20 1 − (1.25) = 1 − 0.894350 = 0.10565.

Revised Answers Manual to Introduction to Probability

!

X −μ 125−105 = P(−0.5 ≤ Z ≤ (iii) P(95 ≤ X ≤ 125) = P 95−105 ≤ ≤ 20 σ 20 1) = (1) − (−0.5) = (1) − [1 − (0.5)] = (0.5) + (1) − 1 = 0.691462 + 0.841345 − 1 = 0.532807.

Remark: In parts (i) and (ii), without changing the outcome, but in order to take into account the nature of the r.v. X , we have: P(X ≥ 50) = P(0 < X ≤ 50) = 1 − (5.25) + (2.75) = (2.75); P(X ≤ 80) = P(0 < X ≤ 80) = (5.25) − (1.25) = 1 − (1.25).

2.16 Let us term as a success the fact that a tested light bulb turns out to be defective.

Assuming that the testings are independent, we have that the number of defective light bulbs is a r.v. Y ∼ B(25, p), where p = P(X < 1,800). We have

1,800 − 2,000 X −μ ≤ = P(Z ≤ −1) = 1 − P(Z ≤ 1) p=P σ 200 = 1 − (1) = 1 − 0.841345 = 0.158655, and therefore: P(at most 15 light bulbs are defective) 15

25 (0.158655) y (0.841345)25−y = P(Y ≤ 15) = y y=0

15

25 (0.159) y (0.841)25−y . y y=0

Remark: As in Exercise 2.15, P(X < 1,800) = P(0 < X ≤ 1,800) = (100) − (1) = 1 − (1).

2.17 Let X be the r.v. denoting the diameter of a ball bearing. Then the probability p that a ball bearing is defective is given by: p = P(X > 0.5 + 0.0006 or X < 0.5 − 0.0006) = P(X > 0.5006 or X < 0.4994) = P(X > 0.5006) + P(X < 0.4994) = 1 − P(X ≤ 0.5006) + P(X < 0.4994)

0.5006 − 0.5007 X −μ 0.4994 − 0.5007 X −μ ≤ +P < = 1− P σ 0.0005 σ 0.0005 1 − P(Z ≤ −0.2) + P(Z ≤ −2.6) = 1 − (−0.2) + (−2.6) = 1 − [1 − (0.2)] + 1 − = (0.2) − (2.6) + 1 = 0.579260 − 0.995339 + 1 = 0.583921.

(2.6)

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Thus, if Y is the r.v. denoting the number of defective ball bearings from among those in a day’s production, then Y ∼ B(n, p), where n is the number of ball bearings in a day’s production. Then the expected number of defective ball bearings is EY = np and the expected proportion of defective ball bearings is np EY 0.583921, or about 58%. n = n = p Remark: As in Exercise 2.15, P(X > 0.5006) + P(X < 0.4994) = 1 − P(0 < X ≤ 0.5006) + P(0 < X < 0.4994) = 1 − (−0.2) + (1001.4) + (−2.6) − (−1001.4) = 1 + (0.2) − (2.6).

2 , x ∈ . Then: 2.18 Here f (x) = √ 1 exp − (x−μ) 2σ 2 2π σ

2

2

(i) f (μ + y) = √ 1 exp(− y2 ) and f (μ − y) = √ 1 exp(− y2 ), so that 2π σ 2π σ f (y − μ) = f (y + μ), y ∈ .

2 f (x) = 0 if x = μ, and ddx 2 f (x)|x=μ = − σ12 f (x) + (ii) ddx f (x) = − x−μ σ2 ( x−μ )2 f (x)|x=μ = − √ 1 3 < 0, so that x = μ maximizes f (x). σ2 2π σ

xp − μ X −μ (i) 2.19 p = P(X ≤ x p ) = P ≤ σ σ

xp − μ xp − μ )= , = P(Z ≤ σ σ x −μ

so that pσ = −1 ( p) and x p = μ + σ −1 ( p). (ii) Since −1 (0.25) = − −1 (0.75) = −0.675, and −1 (0.5) = 0, we have: x0.25 = μ − 0.675σ, x0.50 = μ, and x0.75 = μ + 0.675σ.

2.20 By differentiating twice the p.d.f. f (x), we find that: % &

2 1 x −μ 2 − (x−μ)

2 e 2σ f (x) = − √ 1− . σ 2π σ 3 2

= (1 − 1)2 = 0, it follows that Since for x = μ ± σ, 1 − x−μ σ

∞ ∞ f (μ ± σ ) = 0. y2 x2 1 1 2.21 (i) Indeed, e− 2 d x √ e− 2 dy I2 = √ 2π −∞ 2π −∞ ∞ ∞ x 2 +y 2 1 = e− 2 d x d y.

2π −∞ −∞ From x = r cos θ, y = r sin θ , we obtain: x 2 + y 2 = r 2 , ∂∂rx = cos θ, ∂∂θx = −r sin θ ; ∂∂ry = sin θ, ∂∂θy = r cos θ , so that the determinant of the partial derivatives is: cos θ ×r cos θ −(−r sin θ )×sin θ = r cos2 θ +r sin2 θ = r .

Therefore: 2π ∞

2π ∞ 2 2 1 1 2 − r2 − r2 I = e × r dr dθ = dθ re dr 2π 0 0 2π 0 0

r2 ∞ 1 × 2π − e− 2 0 = 1 × (0 + 1) = 1. = 2π

Revised Answers Manual to Introduction to Probability

(ii) Here ∞

1

√ 2π σ

−∞

=√

1

e

− (x−μ) 2 2σ

∞

2

dx y2

e− 2 σ dy (by setting

x −μ = y) σ

2πσ −∞ ∞ y2 1 =√ e− 2 dy = 1, by part (i).

2π −∞

2.22 Indeed,

∞ ∞ 1 1 1 =

y 2 −1 e−y dy = y − 2 e−y dy.

2 0 0 2

Set y 1/2 = √t , so that y = t2 , dy = t dt, and the range of t is from 0 to ∞.

2 Then: √

∞ √ ∞ 1 −t 2 /2 1 2 2

= 2 e × t dt = e−t /2 dt 2 t 2 0 −∞ (by symmetry of e−t /2 with respect to 0) ∞ √ √ √ 1 2 = π×√ e−t /2 dt = π × 1 = π .

2π −∞ 2

2.23 (i) We have, for t ∈ ,

∞ x2 x 2 −2t x 1 et x × e− 2 d x = √ e− 2 d x 2π −∞ −∞ ∞ (x−t)2 1 2 2 2 = et /2 × √ e− 2 d x = et /2 × 1 = et /2 , 2π −∞

1 M X (t) = √ 2π

since √1 e− 2π

(x−t)2 2

∞

is the p.d.f. of N (t, 1). This establishes the first part.

(ii) Since X ∼ N (μ, σ 2 ) implies Z = X −μ σ ∼ N (0, 1), we have: et /2 = M Z (t) = M X −μ (t) = M 1 X + −μ (t) σ σ σ t − μt = e σ MX , so that σ

μt t 2 t t2 t = e σ e 2 = eμ σ + 2 .

MX σ 2

Setting σt = r , so that t = σ r , we get: σ 2r 2

M X (r ) = eμr + 2 , or by reverting to t, σ 2t2

M X (t) = eμt+ 2 .

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(iii) Here σ 2t2 d μt+ σ 2 t 2 2 e = eμt+ 2 × (μ + σ 2 t)|t=0 dt t=0 = μ; i.e., E X = μ, and

d μt+ σ 2 t 2 2 (μ + σ 2 t) e E X2 = t=0 dt

EX =

σ 2t2

σ 2t2

= eμt+ 2 (μ + σ 2 t)2 + eμt+ 2 × σ 2 |t=0 = μ2 + σ 2 , so that Var(X ) = μ2 + σ 2 − μ2 = σ 2 ; i.e., Var(X ) = σ 2 . σ 2t2

2.24 Observe that M X (t) = eμt+ 2 , with μ = α and σ 2 = 2β, is the m.g.f. of the N (α, 2β) distribution. Thus, X ∼ N (α, 2β). x2 ∞ 2.25 (i) First, E X 2n+1 = √1 x 2n+1 × e− 2 d x = 0, because the integrand is 2π

−∞

an odd function (or by splitting the integral from −∞ to 0 and 0 to ∞ and changing the variable in the integral over (−∞, 0)).

Next, set m 2n = E X 2n and observe that: ∞ ∞ x2 x2 1 1 x 2n e− 2 d x = − √ x 2n−1 de− 2 m 2n = √ 2π −∞ 2π −∞

∞ 2 x 1 2n − 1 ∞ 2n−2 − x 2 = − √ x 2n−1 e− 2 + √ x e 2 dx 2π 2π −∞ −∞ = 0 + (2n − 1)m 2n−2 . So, m 2n = (2n − 1)m 2n−2 , and likewise, m 2n−2 = (2n − 3)m 2n−4 .. . m2 = 1 × m0 m 0 = 1 (since m 0 = E X 0 = 1).

Multiplying out, we get after the cancellations m 2n = (2n − 1)(2n − 3) · · · 1 (2n − 1)(2n − 3) · · · 1 × (2n)(2n − 2) · · · 2 = (2n)(2n − 2) · · · 2 (2n)!

1 × 2 · · · × (2n − 1)(2n) = n = (2 × 1) · · · [2 × (n − 1)] × (2 × n) 2 [1 · · · (n − 1)n] (2n)! = n .

2 (n!) (ii) Here E X = 0 and E X 2 = 212! = 1, so that Var(X ) = 1.

(1!)

Revised Answers Manual to Introduction to Probability

(iii) We know that, if X ∼ N (μ, σ 2 ), then Z = X −μ σ ∼ N (0, 1). Then:

0 = EZ = E

X −μ σ

=

1 (E X − μ), σ

1 so that E X = μ. Also, 1 = Var(Z ) = Var( X −μ σ ) = σ 2 Var(X ), so that 2 Var(X ) = σ .

2.26 (i) By Exercise 1.17 in Chapter 5, and with t ∈ , ∞ ∞ ∞ n 2k

t 2k+1 n t 2k t M X (t) = + (E X ) = (E X ) (E X 2k+1 ) n!

(2k)!

(2k + 1)! n=0

k=0

k=0

∞ ∞

t 2k (2k)! t 2k = × (E X 2k ) = k (2k)!

2 × k! (2k)!

=

k=0 ∞

k=0

(t 2 /2)k

k=0

k!

= et /2 which is the m.g.f. of N (0, 1).

2

(ii) Thus, X ∼ N (0, 1).

1 2.27 Since f (x) = 2α , for −α < x < α, we have: 3 = 0.75 and α = 2. (i) P(−1 < X < 2) = 2α 2 (ii) P(|X | < 1) = P(−1 < X < 1) = 2α = α1 , P(|X | > 2) = 1 − P(|X | ≤ 4 2) = 1 − P(−2 ≤ X ≤ 2) = 1 − 2α = 1 − α2 , so that α1 = 1 − α2 and α = 3.

2.28 We have: β

1 x2 β 1 β 2 − α2 dx = × = β −α β −α 2 α 2(β − α) α (β − α)(β + α) α+β = = , 2(β − α) 2 β x3 β 1 1 × E X2 = x2 × dx = β −α α β −α 3 α EX =

x×

=

(β − α)(β 2 + αβ + α 2 ) β 3 − α3 = 3(β − α) 3(β − α)

=

β 2 + αβ + α 2 β 2 + αβ + α 2 , so that Var(X ) = − 3 3

=

α 2 + 2αβ + β 2 β 2 + αβ + α 2 − 3 4

α+β 2

2

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1 (4β 2 + 4αβ + 4α 2 − 3α 2 − 6αβ − 3β 2 ) 12 (α − β)2 β 2 + α 2 − 2αβ = ; = 12 12 α+β (α − β)2 i.e., E X = , Var (X ) = .

2 12 =

2.29 In the first place, E X = 1 3 , so that:

1 1 2 x2 1 1 x3 1 2 0 x d x = 2 0 = 2 and E X = 0 x d x = 3 0 =

(i) E(3X 2 − 7X + 2) = 3E X 2 − 7E X + 2 = 3 × 13 − 7 × 21 + 2 = −0.5.

1 (ii) E(2e X ) = 2Ee X = 2 0 e x d x = 2e x |10 = 2(e − 1) 3.44.

2.30 (i) Since X ∼ P(0.2), we have: P(X ≤ 1) = 0.982477 from the Poisson tables. ∞ (ii) Here Y ∼ 0.04e−0.04y , (y > 0), so that: P(Y > 5) = 5 0.04e−0.04y dy = −0.2 0.818. Alternatively, P(Y > 5) = P(0 events −e−0.04y |∞ 5 = e occurred in the interval (0, 5)) = e−0.2 (0.2) 0! above.

0

= e−0.2

0.818 as

2.31 (i) Since Y = (X − d)I(d,α) (X ), we have: α 1 (x − d)I(d,α) (x) d x EY = α 0

1 d 1 α = (x − d)I(d,α) (x)d x + (x − d)I(d,α) (x)d x α 0 α d α α 1 1 = (x − d)d x = (x − d)d(x − d) α d α d (α − d)2 (α − d)2 1 (x − d)2 |αd = ; i.e., EY = . = 2α 2α 2α 2 √ α 2 2 (ii) From (α−d) 2α√ = c × 2 , we get cα = (α − d) , or α c = α − d, or d = (1 − c)α. (iii) For c = 41 , we have d = α2 (and, of course, EY = 41 E X ).

2.32 (i) Let Y be defined by: ⎧ ⎨ C, 0 < X ≤ μ/2 Y = C/2, μ/2 < X ≤ μ ⎩ 0, X > μ.

Then the required expected amount is EY . Since the p.d.f. of X is f (x) =

1 −x/μ e , x > 0, μ

Revised Answers Manual to Introduction to Probability

we have: P(0 < X ≤ μ/2) =

μ/2 0

1 −x/μ μ/2 e d x = −e−x/μ 0 = 1 − e−1/2 , μ μ/2

P(μ/2 < X ≤ μ) = −e−x/μ μ

= e−1/2 − e−1 ,

P(X > μ) = e−1 .

Therefore, EY = C(1 − e−1/2 ) + C2 (e−1/2 − e−1 ) = C2 (2 − e−1/2 − e−1 ). (ii) Observe that the distribution of the r.v. Y is independent of μ. Then for the given value of C, we have: EY = 1,200(2 − e−1/2 − e−1 ) 1,200(2 − 0.607 − 0.368) = 1,230.

1 2.33 (i) From ∞ cx −3 d x = − c × 12 |∞ = 2c × c12 = 2c = 1, we have c = 21 , so c 2 x c 1 1 that f (x) = 2x 3 , x > 2 . m (ii) Here, 21 1 x −3 d x = − 41 × x12 |m1 = − 41 ( m12 − 4) = 1 − 4m1 2 = 0.5, and 2

√

2

hence m = 22 0.707. (iii) Observe that f Y (0) = P(Y = 0) = P(X ≤ c or X ≥ m) = P(X ≥ m) = 1/2, and that c < X < m, or −m < −X < −c, or 0 < m − X < m − c, or 0 < Y < m − c. Then, for 0 < y < m − c, FY (y) = P(Y ≤ y) = P(m − X ≤ y) = P(X ≥ m − y) = 1 − P(X ≤ m − y) = 1 − FX (m − y), so that c . f Y (y) = − f (m − y)(−1) = (m − y)3 Thus, f Y (y) = √

(iv)

c = (m − y)3

m − c = 22 − 21 = m−c

1

4 for 0 < y < !3 = √ 3 ( 2 − 2y) 2 −y

2 2 √ 2−1 1 2 , and f Y (0) = 2 . m−c c

f Y (y)dy =

0

√

0

dy = c (m − y)3

m−c

(m − y)−3 dy

0

c m−c c d(m − y)−2 = |m−c 2 0 2(m − y)2 0

1 1 1 1 1 = ×2= . = − 2 2 4 c m 4 2 =

(v)

m−c

m−c 1 c + EY = 0 × y dy = c y(m − y)−3 dy 2 (m − y)3 0 0 " # m−c

y c c m−c m−c −2 yd(m − y)−2 = | − (m − y) dy = 2 0 2 (m − y)2 0 0

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#

"

m−c c m−c c m−c 1 m−c −1 = | − d(m − y) − 2 c2 2 c2 m−y 0 0 √

3 2−4 c m−c 1 1 = − + = 0.061.

2 c2 c m 4

=

2.34 (i) The median m is determined through the relation m λe−λx d x = 0.5, 0 −λm = 0.5 from where m or −e−λx |m 0.693/λ. For 0 = 0.5, or 1 − e λ = 0.005, we get m 138.6. (ii) Since Y = M if and only if X ≥ M, we have f Y (M) = P(Y = M) = ∞ P(X ≥ M) = M λe−λx d x = e−λM . Also, since Y < M if and only if X < M, we have, for (0 <)y < M : FY (y) = P(Y ≤ y) = P(X ≤ y) = y −λx d x = 1 − e−λy , so that f (y) = d (1 − e−λy ) = λe−λy . Thus, Y 0 λe dy f Y (y) =

λe−λy , if 0 < y < M e−λM , if y = M.

(iii) Since λM = 0.005 × 200 = 1, we have: 0.005e−0.005y , if 0 < y < 200 if y = 200. e−1 ,

f Y (y) = 2.35 (i) EY = M 0

M −λy dy + Me−λM , and by integration, 0 yλe

yλe−λy dy = −

M

yde−λy = −ye−λy

0

M 0

+

M

e−λy dy

0

M 1 M −λy 1 −λM = −Me − de = −Me−λM − e−λy 0 λ 0 λ !

1 −λM e = −Me−λM − −1 , λ

so that EY = λ1 (1 − e−λM ). (ii) EY = 200(1 − e−1 ).

Chapter 7 Section 7.1 1.1 We have: (i) P(X = 4, Y = 0) = f X ,Y (4, 0) = 0.10. (ii) P(X = 5) = f X ,Y (5, 0) + f X ,Y (5, 1) + f X ,Y (5, 2) = 0.05 + 0.03 + 0.02 = 0.10.

Revised Answers Manual to Introduction to Probability

(iii)

P(Y = 1) = f X ,Y (0, 1) + f X ,Y (1, 1) + f X ,Y (2, 1) + f X ,Y (3, 1) + f X ,Y (4, 1) + f X ,Y (5, 1) = 0.015 + 0.030 + 0.075 + 0.090 + 0.060 + 0.030 = 0.30.

(iv)

P(X ≤ 3, Y ≥ 1) = f X ,Y (0, 1) + f X ,Y (0, 2) + f X ,Y (1, 1) + f X ,Y (1, 2) + f X ,Y (2, 1) + f X ,Y (2, 2) + f X ,Y (3, 1) + f X ,Y (3, 2) = 0.015 + 0.010 + 0.030 + 0.020 + 0.075 + 0.050 + 0.090 + 0.060 = 0.35.

1.2 We have: y /x 0 1 2

0 0.3 0 0

1 0.2 0.2 0

2 0.075 0.15 0.075

P(X = x, Y = y) = P(Y = y|X = x)P(X = x), and observe that: P(X = 0, Y = 1) = P(X = 0, Y = 2) = P(X = 1, Y = 2) = 0. For the remaining values of x and y, we obtain: P(X = 0, Y = 0) = P(Y = 0|X = 0)P(X = 0) = 1 × 0.3 = 0.3, P(X = 1, Y = 0) = P(Y = 0|X = 1)P(X = 1) = 0.5 × 0.4 = 0.2, P(X = 1, Y = 1) = P(Y = 1|X = 1)P(X = 1) = 0.5 × 0.4 = 0.2, P(X = 2, Y = 0) = P(Y = 0|X = 2)P(X = 2) = (0.5)2 × 0.3 = 0.075, P(X = 2, Y = 1) = P(Y = 1|X = 2)P(X = 2) = 0.5 × 0.3 = 0.15, P(X = 2, Y = 2) = P(Y = 2|X = 2)P(X = 2) = (0.5)2 × 0.3 = 0.075.

1.3 Since P(X < Y ) + P(Y < X ) = 1, and the p.d.f. is symmetric with respect to x and y, one would expect that P(X < Y ) = 0.5. Indeed,

P(X < Y ) = =

f X ,Y (x, y)d x d y =

2 1 2 y x 0

1.4

2

(i)

2 0

y

+y×x 0

1 y 0

3 dy = 2

(x + y)d x d y

0

1 0

y 2 dy =

y3 1 3 1 × = = 0.5.

2 3 0 2

1 xy 7 x2 + dy = 2x 2 + x, and (2x 2 + x)d x = , so that 2 6 0

0 1 2 6 7 6 2 xy x + d y d x = × = 1, and f X ,Y is, indeed, a p.d.f.

2 7 6 0 0 7

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Revised Answers Manual to Introduction to Probability

(ii)

1.5 (i)

6 6 2 xy xy 2 x + x + dx dy = dy dx P(X > Y ) = 2 7 (y<x) 2 (x>y) 7

1 x

6 1 x 2 xy 6 1 1 x = x + dy dx = x 2d y d x + x y dy d x 7 0 0 2 7 0 0 2 0 0

6 6 1 1 5 15 = × = + = 0.268.

7 4 16 7 16 56 c y

−x−y e dx dy = e−x−y d x d y P(X ≤ Y ≤ c) = (x≤y≤c) 0 0 y

c c c −y −x −y −x y = e e d x dy = e (−e |0 )dy = e−y (1 − e−y )dy 0 0 0 0 c c c 1 1 1 1 e−y dy − e−2y dy = −e−y |c0 + e−2y = − c + 2c . = 2 2 e 2e 0 0 0

(ii) For c = log 2, we have: P(X ≤ Y ≤ log 2) = 18 = 0.125. x ∞ ∞ x 1.6 (i) We have: P(X ≤ x) = e−t−y dt dy = e−t−y dy dt 0 0 0 0 ∞

x x −t −y e e dy dt = e−t (−e−y |∞ = 0 )dt 0 0 0 x = e−t dt = −e−t |0x = 1 − e−x , x > 0.

0

(ii) P(Y ≤ y) = 1 − e−y , y > 0, either by symmetry or by integrating as in part (i). (iii) Since P(X < Y ) + P(Y < X ) = 1, then symmetry would dictate that P(X < Y ) = 0.5. Indeed, y

∞ ∞ y e−x−y d x d y = e−y e−x d x dy P(X < Y ) = 0 0 0 0 ∞ ∞ ∞ e−y (1 − e−y )dy = e−y dy − e−2y dy = 0

0

0

∞ 1 1 1 = 1 − e−2y = 1 − = = 0.5.

2 2 2 0

(iv) Taking into consideration the graph depicted here, we have: 3−y

3 3 3−y −x−y −y −x e dx dy = e e d x dy P(X + Y < 3) = 0

=

3 0

=

3 0

0

0

3−y e−y (−e−x |0 )dy =

e−y dy − e−3

3 0

3

0

e−y (1 − e y−3 )dy

0

dy = −e−y |30 − 3e−3

= 1 − e−3 − 3e−3 = 1 − 4e−3

0.801.

Revised Answers Manual to Introduction to Probability

y 3 3−x x +y =3

y

x

x

0

3

1.7 From c y 0

0

2 2 dx dy = 2 2 c c

c y 0

0

2 dx dy = 2 c

c

y dy =

0

2 y2 c × c2 2 0

2 c2 = 1, = 2× c 2 we have that c can be any positive constant.

1.8 We have:

∞

xy y 2 ∞ −yx ∞ = −2c × e− 2 x y cye− 2 d x = cy − de 2 y y y ∞ ∞ ∞ 2 xy y2 − y2 = 2ce , so that cye− 2 d x d y = 2c e− 2 dy 0

y

0

∞ √ √ y2 2 = c 2π × √ e− 2 dy = c 2π × 1 = c 2π = 1, 2π 0 1 and c = √ .

2π

c2 c1 c2 c1

c 2 c2 2 c2 c3 x y2 d x d y = y2 x d x dy = 1 y dy = 1 2 = 1, 1.9 2 0 6 0 0 0 0 so that c12 c23 = 6.

1.10 We have:

√

2−x

2−x

For 0 < x ≤ 1, 1−x cxdy = cx, and for 1 < x ≤ 2, cx dy = cx(2−x).

0 2 2c 1 Since 0 cx d x = 2c and cx(2 − x) d x = , we have 2c + 2c 3 = 1 and 3 1 6 c = 7.

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1.11 We have: ∞ y c 0

−y

(y 2 − x 2 )e−y d x d y = c

∞

e−y

0

=c

∞ e 0

−y

" y −y

# (y 2 − x 2 )d x dy

4c 4y 3 dy = × 3 3

∞

y 3 e−y dy.

0

However, ∞ ∞ ∞ ∞ −y −y −y ∞ −y e dy = − de = −e |0 = 1; ye dy = − yde−y 0 0 0 0 ∞ −y ∞ = −ye |0 + e−y dy = 0 + 1 = 1; 0 ∞ ∞ y 2 e−y dy = − y 2 de−y 0 0 ∞ ∞ ∞ 2 −y ∞ = −y e |0 + 2 ye−y dy = 2; y 3 e−y dy = − y 3 de−y 0 0 0 ∞ = −y 3 e−y |∞ + 3 y 2 e−y dy = 3 × 2 = 6. Therefore: 0 0 ∞ y 4c (y 2 − x 2 )e−y d x d y = c × 6 = 8c = 1, so that 3 0 −y c = 1/8.

1.12 The joint p.d.f. of X and Y , f , is given by: f (x, y) = C22 for (x, y) in the triangle O AC; for 0 ≤ x ≤ y ≤ C. Therefore: C y (i) 2 EU = (x 2 + y 2 ) × 2 d x d y C 0 0 C y

C y 2 = 2 x2 dx dy + y2 d x d y C 0 0 0 0

C 2 C4 2C 2 2 1 C 3 y dy + y 3 dy = 2 × = . = 2 C 3 0 C 3 3 0 (ii) EU = 23 .

1.13 Consider the points (x, y) and (xn , yn ) with xn ↑ ∞, yn ↑ ∞, as n → ∞, and let Tn be the rectangle determined by the points (x, y), (xn , y), (xn , yn ), and (x, yn ). Then it is known (see Proposition 1(iv) in Chapter 7) that P[(X , Y ) ∈ Tn ] = P(X ≤ xn , Y ≤ yn ) + P(X ≤ x, Y ≤ y) −P(X ≤ x, Y ≤ yn ) − P(X ≤ xn , Y ≤ y).

Revised Answers Manual to Introduction to Probability

Taking the limits of both sides, as n → ∞, we have P(X > x, Y > y) = 1+ P(X ≤ x, Y ≤ y)− P(X ≤ x)− P(Y ≤ y). Implementing the assumption made, this relationship becomes P(X > x)P(Y > y) = 1 + P(X ≤ x, Y ≤ y) − P(X ≤ x) − P(Y ≤ y), whereas P(X > x)P(Y > y) = [1 − P(X ≤ x)][1 − P(Y ≤ y)] = 1 − P(X ≤ x) − P(Y ≤ y) + P(X ≤ x)P(Y ≤ y).

Thus, 1 − P(X ≤ x) − P(Y ≤ y) + P(X ≤ x)P(Y ≤ y) = 1 + P(X ≤ x, Y ≤ y) − P(X ≤ x) − P(Y ≤ y), or P(X ≤ x, Y ≤ y) = P(X ≤ x)P(Y ≤ y), as was to be seen.

Section 7.2 2.1 We have: f X (0) = 0.05, f X (1) = 0.10, f X (2) = 0.25, f X (3) = 0.30, f X (4) = 0.20, f X (5) = 0.10; f Y (0) = 0.50, f Y (1) = 0.30, f Y (2) = 0.20.

2.2 We have: f X (0) = 0.3, f X (1) = 0.4, f X (2) = 0.3; f Y (0) = 0.575, f Y (1) = 0.35, f Y (2) = 0.075.

2.3 We have: f X (−4) = f X (−2) = f X (2) = f X (4) = 0.25; and f Y (−2) = f Y (−1) = f Y (1) = f Y (2) = 0.25.

2.4 We have: 7 17 12 , f X (2) = , f X (3) = ; 36 36 36 7 14 15 f Y (1) = , f Y (2) = , f Y (3) = .

36 36 36 2 2/36 2 3 (ii) f X |Y (1|1) = = , f X |Y (2|1) = , f X |Y (3|1) = , 7/36 7 7 7 1 10 3 , f X |Y (2|2) = , f X |Y (3|2) = , f X |Y (1|2) = 14 14 14 (i) f X (1) =

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Revised Answers Manual to Introduction to Probability

4 5 6 , f X |Y (2|3) = , f X |Y (3|3) = ; 15 15 15 2 2 3 f Y |X (1|1) = , f Y |X (2|1) = , f Y |X (3|1) = , 7 7 7 2 10 5 f Y |X (1|2) = , f Y |X (2|2) = , f Y |X (3|2) = , 17 17 17 3 3 6 , f Y |X (2|3) = , f Y |X (3|3) = . f Y |X (1|3) = 12 12 12 2.5 We have: f X |Y (1|3) =

3 2 5 6 , f X (1) = , f X (2) = , f X (3) = ; 16 16 16 16 8 f Y (1) = f Y (2) = ; 16 4 2 6 4 f X |Y (0|1) = , f X |Y (1|1) = , f X |Y (2|1) = , f X |Y (3|1) = ; 16 16 16 16 2 2 4 8 f X |Y (0|2) = , f X |Y (1|2) = , f X |Y (2|2) = , f X |Y (3|2) = .

16 16 16 16 1 30 EX = (1 × 2 + 2 × 5 + 3 × 6) = , 16 16 8 24 EY = (1 + 2) = ; 16 16 26 1 (1 × 2 + 2 × 6 + 3 × 4) = , E(X |1) = 16 16 1 34 E(X |2) = (1 × 2 + 2 × 4 + 3 × 8) = ; 16 16 26 8 34 8 30 E[E(X |Y )] = × + × = .

16 16 16 16 16 E X = E[E(X |Y )](= 30 16 ), as it should be.

76 1 2 (1 × 2 + 4 × 5 + 9 × 6) = , EX = 16 16 40 1 EY 2 = (1 × 8 + 4 × 8) = , so that 16 16 2 2 76 30 40 24 79 1 Var(X ) = − , Var(Y ) = − = = .

16 16 64 16 16 4 x

2x 2 f X (x) = = , x = 1, . . . , n; n(n + 1) n(n + 1)

(i) f X (0) =

(ii)

(iii) (iv)

2.6 (i)

f Y (y) = (ii)

f X |Y (x|y) =

y=1 n

2(n − y + 1) 2 = , n(n + 1) n(n + 1) x=y

y = 1, . . . , n.

1 2/n(n + 1) = , 2(n − y + 1)/n(n + 1) n−y+1

1 2/n(n + 1) f Y |X (y|x) = = , 2x/n(n + 1) x

y = 1, . . . , x, x = 1, . . . , n;

y = 1, . . . , x; x = 1, . . . , n.

Revised Answers Manual to Introduction to Probability

(iii)

⎞ ⎛ y−1 n

1 1 ⎝ = E(X |Y = y) = x× x− x⎠ n−y+1 n−y+1 n=y x=1 x=1 " # n(n + 1) (y − 1)y 1 = − n−y+1 2 2 n(n + 1) − (y − 1)y = y = 1, . . . , n; 2(n − y + 1) x x

1 x +1 1 x(x + 1) = , y× = y= E(Y |X = x) = x x 2x 2 n

y=1

y=1

x = 1, . . . , n.

2.7 We have: 1

1 (x + y)dy = x + , 0 < x < 1, and 2 0 1 1 (x + y)d x = y + , 0 < y < 1. f Y (y) = 2 0

f X (x) =

2.8 We have:

6 1 6 2 2 (x + y 2 )dy = x + = (3x + 1), 0 ≤ x ≤ 1; 5 0 5 5 5 1 3 6 6 3 (x + y 2 )d x = y 2 + = (2y 2 + 1), 0 ≤ y ≤ 1. f Y (y) = 5 0 5 5 5

f X (x) =

2.9 We have: (i)

f X (x) = f Y (y) =

x 0∞

ye−x dy =

x 2 e−x , x > 0; 2

ye−x d x = ye−y ,

y > 0.

y

(ii)

(iii)

ye−x = e y × e−x , 0 < y ≤ x < ∞; ye−y 2ye−x 2y f Y |X (y|x) = 2 −x = 2 , 0 < y ≤ x < ∞. x e x ∞ elog 2 × e−x d x = −2e−x |∞ P(X > 2 log 2|Y = log2) = log 4 f X |Y (x|y) =

2 log 2

2 = 2e− log 4 = = 0.5.

4

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2.10 We have: (i)

∞

∞ xe−(x+y) dy = xe−x e−y dy = xe−x , x > 0; 0 0 ∞ ∞ xe−(x+y) d x = e−y xe−x d x = e−y , y > 0. f Y (y) =

f X (x) =

0

0

xe−x × e−y

= e−y , x > 0, y > 0. xe−x ∞ −x ∞ P(X > log 4) = xe−x d x = −xe−x |∞ log 4 − e |log 4

(ii) f Y |X (y|x) = (iii)

log 4

1 + log 4 log 4 1 + = = 4 4 4

0.597.

2.11 We have: ∞

∞

(i) f Y (y) = 0 21 ye−x y d x = 21 0 ye−yx d x = 21 , 0 < y < 2, either by integration or by observing that ye−yx , x > 0 (0 < y < 2) is the p.d.f. of the negative exponential p.d.f. with parameter y. (ii) f X |Y (x|y) = 21 ye−x y / 21 = ye−yx , x > 0, 0 < y < 2, which is the negative exponential p.d.f. with parameter y. Also, f X |Y (x|y = 21 ) = 1 −x/2 , x > 0.

2e ∞ (iii) E(X |Y = y) = 0 x × ye−yx d x = 1y and E(X |y = 21 ) = 2.

2.12 We have: (i)

2

6x 2 2 6 2 xy 3x 2 x + dy = f X (x) = dy + y dy 2 7 0 7 0 0 7 2

3x y2 6x 6x 12x 2 12x 2 + × + = (2x + 1), 0 < x ≤ 1; = 7 7 2 0 7 7 7

1 1 1 6 x3 6 2 xy 3y x 2 f Y (y) = x + dx = × × + 2 7 3 0 7 2 0 0 7 2 3y + , 0 < y ≤ 2; = 14 7

xy 6 2 2x + y 7 x + 2 f Y |X (y|x) = 6x , 0 < x < 1, 0 < y ≤ 2. = 4x + 2 (2x + 1) 7 =

(ii)

EY =

2

y×

0

2

0 2 2 3 y3 y2 3y 2 8 + dy = × + × = ; 14 7 14 3 2 7 2 0 7 2

2

2x y2 y3 2x + y 1 dy = × × + 4x + 2 4x + 2 2 0 4x + 2 3 0 0 4x 8 2 3x + 2 = + = × , 0 < x ≤ 1.

4x + 2 3(4x + 2) 3 2x + 1

E(Y |X = x) =

y×

Revised Answers Manual to Introduction to Probability

(iv)

1

2 3x + 2 6x × × (2x + 1)d x 2x + 1 7 0 3

8 x3 1 4 x2 1 = = EY .

3× = +2× 7 3 0 2 0 7

(iii)

E[E(Y |X )] =

P X < 21 , Y > 21 1 1 , and P Y > X< =

2 2 P X < 21

1 6 1/2 5 P X< = , x(2x + 1)d x = 2 7 0 28

1 1 6 1/2 2 69 xy 2 P X < ,Y > x + = dy dx = , 2 2 7 0 2 448 1/2

1 1 483 X< so that P Y > = = 0.8625.

2 2 560

2.13 We have: c

2 2 2(c − x) dy = 2 × (c − x) = , 0 < x < c; 2 c c c2 x y 2 2 2y d x = 2 × y = 2 , 0 < y < c. f Y (y) = 2 c c c 0

(i)

f X (x) =

(ii)

2/c2 1 = , 0 < x ≤ y < c; 2 2y/c y 2/c2 1 , 0 < x ≤ y < c. f Y |X (y|x) = = 2 2(c − x)/c c−x ' ( 1 1 2(c − x) 2 x2 2c − 1 1 (iii) P(X ≤ 1) = d x = 2 c × x|0 − . = 2 c c 2 0 c2 0 f X |Y (x|y) =

xy

2.14 From Exercise 1.8, f X ,Y (x, y) = √1 ye− 2 , 0 < y < x, and it has been found 2π

in the discussion of Exercise 1.8 that f Y (y) = 2ce−y /2 = √2 e−y /2 , y > 0.

2π Hence 1 y2 y f X |Y (x|y) = ye 2 e− 2 x , 0 < y < x.

2 2

2

2.15 We have: (i) f X (x) = c12 y 2

c2 2 0 x y dy

=

c23 x 3 ,

0 < x < c1 , f Y (y) =

, 0 < y < c2 . , 0 < x < c1 , 0 < y < c2 . (ii) f X |Y (x|y) = 2x c2 2

1

c1 2 0 xy dx

=

e67

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Revised Answers Manual to Introduction to Probability

c23 x c22 c13 c23 c1 x 3 c1 0 x × 3 d x = 3 × 3 0 = 9 ; E(X |Y = y) = 2c13 c1 2c1 2x 2 x 3 c1 0 x × c2 d x = c2 × 3 0 = 3c2 = 3 , 0 < y < c2 .

1 1 1 3 c c2 y 2 c3 c3 c3 c E[E(X |Y )] = 0 2 2c31 × 12 dy = 31 × y3 02 = 19 2 = E X .

(iii) We have: E X =

(iv)

2.16 From Exercise 1.10, we have c = 6/7 and: 6x 6x(2 − x) ; for 1 < x ≤ 2, f X (x) = ; i.e., 7 7 ⎧ 0<x ≤1 ⎨ 6x/7, f X (x) = 6x(2 − x)/7, 1 < x < 2 ⎩ 0, elsewhere.

(i) For 0 < x ≤ 1, f X (x) =

(ii) From part (i), we have: For 0 < x ≤ 1, 6

f Y |X (y|x) = 76 f Y |X (y|x) =

x

= 1, and for 1 < x < 2,

7x 6 1 7x . That is, = 6x(2−x) 2−x 7

⎧ 0<x ≤1 ⎨ 1, f Y |X (y|x) = 1/(2 − x), 1 < x ≤ 2 ⎩ 0, otherwise and always 1 ≤ x + y < 2.

2.17 From Exercise 1.11, we have c = 1/8 and:

y 1 y 2 y 2 e−y e−y 3 y (i) f Y (y) = ×x x (y − x 2 )e−y d x = − 8 −y 8 24 −y −y 1 3 −y = y e , y > 0.

6 (ii) f X |Y (x|y) = 43 × y y−x 3 , −y ≤ x ≤ y, y > 0.

2

2

y

y x −x −y

x

0

x

y

x

Revised Answers Manual to Introduction to Probability

(iii) From the figure, we see that the ranges of x and y are as follows: x < 0, −x ≤ y; x ≥ 0, x ≤ y; and, of course, y > 0, −y ≤ x ≤ y. Then:

1 ∞ 2 (y − x 2 )e−y dy x < 0, f X (x) = 8 −x

1 ∞ 2 −y x 2 ∞ −y = y e dy − e dy, 8 −x 8 −x ∞

∞

∞

and, by part (i): −x e−y dy = e x , −x ye−y dy = −xe x + e x , −x y 2 e−y dy = x 2 e x − 2xe x + 2e x , so that f X (x) =

1 2 x 1 (x e − 2xe x + 2e x − x 2 e x ) = (1 − x)e x .

8 4 ∞

2

∞

Similarly, for x ≥ 0, f X (x) = 18 x y 2 e−y dy − x8 x e−y dy, and, by ∞ ∞ ∞ part (i) again, x e−y dy = e−x , x ye−y dy = xe−x + e−x , x y 2 e−y 1 2 −x −x −x −x dy = x e + 2xe + 2e , so that f X (x) = 4 (1 + x)e . Thus, f X (x) =

1 x 4 (1 − x)e , 1 x 4 (1 + x)e ,

x <0 x ≥ 0.

2.18 (i) Without any calculations, we recognize that f X |Y (·|y) is the Poisson p.d.f. with parameter y. (ii) Now f Y (y) = e−y , y > 0, and hence x x f X ,Y (x, y) = f X |Y (x|y) f Y (y) = e−y yx! × e−y = e−2y yx! , x = 0, 1, . . . (iii) Finally, ∞

1 ∞ −2y x yx e−2y dy = e y dy f X (x) = x! x! 0 0 ∞ 1 (x + 1) y (x+1)−1 e−y/1/2 dy = x+1 2 × x! 0 (x + 1)(1/2)x+1 (x + 1) × 1 (since the integrand is the gamma p.d.f. with = x+1 2 x! parameters x + 1 and 1/2) x! (since (x + 1) = x!) = x+1 2 x!

1 1 = x+1 ; i.e., f X (x) = x+1 , x = 0, 1, . . .

2 2 2.19 (i) First, f X ,Y (x, y) = f X |Y (x|y) f Y (y) = ( xy ) p x q y−x × e−λ λy! (q = 1 − p), so that: y

f X (x) =

∞

y ! x y−x −λ λ y p q e y! x y=x

(the summation over y starting from x, because ( xy ) = 0 for x > y)

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x ∞ p y! λy qy =e q x!(y − x)! y! y=x x ∞ p 1 (λq) y = e−λ q x! y=x (y − x)! x ∞ p 1 (λq)x+t (by setting y − x = t) = e−λ q x! t! t=0 x ∞ (λq)x (λq)t −λ p =e q x! t! t=0 x x p (λq) λq (λ p)x q x λ(1− p) = e−λ e = e−λ e q x! q x x! (by #6 in Table 6 in the Appendix) (λ p)x λ −λ p (λ p)x e e , x = 0, 1, . . . , = e−λ p = e−λ x! x! −λ

as was to be seen. (ii)

f Y |X (y|x) =

( xy ) p x q y−x e−λ λ y /y! e−λ p (λx!p)

x

y! p x q y−x λ y x! x!(y − x)!λx p x y! (λq) y−x ; i.e., = e−λq (y − x)! (λq) y−x f Y |X (y|x) = e−λq , y = x, x + 1, . . . , (y − x)! = e−λ+λ p

so that Y is, conditionally upon X = x, distributed as Poisson with parameter λq over the set: {x, x + 1, . . .}.

2.20 (i) By definition, Var(X |Y = y j ) = E{[X − E(X |Y = y j )]2 |Y = y j }.

But [X − E(X |Y = yi )]2 = X 2 + [E(X |Y = y j )]2 − 2X E(X |Y = yi ), so that Var(X |Y = y j ) = E{X 2 + [E(X |Y = y j )]2 − 2X E(X |Y = y j )|Y = y j } ) * xi2 + [E(X |Y = y j )]2 − 2xi E(X |Y = y j ) f X |Y (xi |y j ) = i

=

i

xi2 f X |Y (xi |y j ) + [E(X |Y = y j )]2

i

f X |Y (xi |y j )

Revised Answers Manual to Introduction to Probability

− 2E(X |Y = y j )

xi f X |Y (xi |y j )

i

= E(X 2 |Y = y j ) + [E(X |Y = y j )]2 × 1 − 2E(X |Y = y j ) × E(X |Y = yi ) = E(X 2 |Y = y j ) − [E(X |Y = y j )]2 . (ii) Repeat the same arguments by replacing the conditioning Y = y j with Y = y and by integrating over x.

2.21

1

y

1

y

1

(i) Indeed, 0 0 8x y d x dy = 8 0 (y 0 x d x)dy = 4 0 y 3 dy = 1.

1 1 (ii) f X (x) = x 8x y dy = 8x x y dy = 4x(1 − x 2 ), 0 < x < 1; y y f Y (y) = 0 8x y d x = 8y 0 x d x = 4y 3 , 0 < y < 1.

1 1 1 (iii) E X = 0 x × 4x(1 − x 2 )d x = 4 0 x 2 (1 − x 2 )d x = 4( 0 x 2 d x − 1 4 1 1 1 5 8 2 2 2 3 0 x d x) = 15 ; E X = 0 x × 4x(1−x )d x = 4( 0 x d x− 0 x d x) = 1 1 64 11 2 2 3 , so that Var(X ) = E X − (E X ) = 3 − 225 = 225 .

1 1 (iv) EY = 0 y × 4y 3 dy = 4 0 y 4 dy = 45 ; 1 1 EY 2 = 0 y 2 × 4y 3 dy = 4 0 y 5 dy = 23 , so that 2 6 Var(Y ) = EY 2 − (EY )2 = 23 − 16 25 = 75 = 225 .

2.22

y (i) f X |Y (x|y) = 8x = 2x , 0 < x ≤ y < 1; 4y 3 y2

8x y 2y f Y |X (y|x) = 4x(1−x 2 ) = 1−x 2 , 0 < x ≤ y < 1.

y y 2 2y 2x 2 0 x × y 2 d x = y 2 0 x d x = 3 , 0 < y < 1; 1 1 2y 2(1−x 3 ) 2 2 E(Y |X = x) = x y × 1−x 2 dy = 1−x 2 x y dy = 3(1−x 2 ) , 0 < x <

(ii) E(X |Y = y) = 1.

2.23

1

8 3 (i) E[E(X |Y )] = 0 2y 3 × 4y dy = 3 (ii) Immediate.

2.24

2.25

1 4 8 0 y dy = 15 = E X .

Var(X |Y = y) = E(X 2 |Y = y) − [E(X |Y = y)]2 2 y 2x 2y = x 2 × 2 dy − (by Exercise 2.22(ii)) y 3 0 y y2 4y 2 y2 2 4y 2 = − = , 0 < y < 1. = 2 x3 dx − y 0 9 2 9 18 1

1

1

2

(i) 0 y 2 d x d y = 2 0 (1 − y)dy = 2(1 − y2 |10 ) = 2(1 − 21 ) = 1. (ii) The shaded area is the region represented by 0 < y < x < 1 and y > x 2 .

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y 1.0 y > x2

y = x2

0.5

0

(iii)

0

0.5

1.0

x

P(Y > X 2 ) = 1 − P(Y ≤ X 2 ) and P(Y ≤ X 2 ) = =2

1 1 0

√

dx dy = 2

y

1 0

(1 −

√

y)dy = 2(1 −

2 3

1

(y≤x 2 )

2 dx dy

dy 3/2 )

0

2 2 1 2 = 2(1 − × y 3/2 |10 ) = 2(1 − ) = 2 × = , so that 3 3 3 3 1 P(Y > X 2 ) = .

3 x

1

f X (x) = 2 0 dy = 2x, 0 < x < 1; f Y (y) = 2 y d x = 2(1 − y), 0 < y < 1; 2 3 1 EY = 0 y × 2(1 − y)dy = 2 × y2 |10 −2 × y3 |10 = 13 . x 2 = x1 , 0 < y < x < 1; E(Y |X = x) = 0 y × x1 dy = (v) f Y |X (y|x) = 2x x 2 , 0 < y < x < 1.

(iv)

(vi) E[E(Y |X )] = E( X2 ) = 21

1 x3 1 1 0 x × 2xd x = 3 |0 = 3 = EY .

2.26 For all three parts, we need the marginal p.d.f.’s f X and f Y . We have: (i) f X (0) = 0.14, f X (1) = 0.36, f X (2) = 0.30, f X (3) = 0.20; f Y (0) = 0.40, f Y (1) = 0.35, f Y (2) = 0.21, f Y (3) = 0.04.

Then (ii) P(X ≥ 2) = f X (2) + f X (3) = 0.50; P(Y ≤ 2) = 1 − P(Y ≥ 3) = 1 − f Y (3) = 1 − 0.04 = 0.96; P(X ≥ 2, Y ≤ 2) = P(X = 2, Y = 0) + P(X = 2, Y = 1) + P(X = 2, Y = 2) + P(X = 3, Y = 0) + P(X = 3, Y = 1) + P(X = 3, Y = 2) = 0.12 + 0.10 + 0.07 + 0.08 + 0.07 + 0.05 = 0.49.

Revised Answers Manual to Introduction to Probability

E X = 1 × 0.36 + 2 × 0.30 + 3 × 0.20 = 1.56,

(iii)

EY = 1 × 0.35 + 2 × 0.21 + 3 × 0.04 = 0.89; E X 2 = 1 × 0.36 + 4 × 0.30 + 9 × 0.20 = 3.36, EY 2 = 1 × 0.35 + 4 × 0.21 + 9 × 0.04 = 1.55, so that: (iv) Var(X ) = 3.36 − (1.56)2 = 0.9264, Var(Y ) = 1.55 − (0.89)2 = 0.7579, and therefore σ (X ) σ (Y ) 0.871.

2.27 (i) We have f (x) = X x3

(ii) (iii) (iv)

0.962,

x 1 3cx 2 3cx 2 0 c(x + y)dy = 2 , 0 < x < 1, and 0 2 d x =

3c c 1 2 × 3 |0 = 2 = 1, so that c = 2.

From part (i), it follows that f X (x) = 3x 2 , 0 < x < 1, and f Y |X (y|x) = 2 (x + y), 0 < y < x < 1.

3x 2 y +y0 )y0 .

Here P(Y ≤ y0 |X = x0 ) = 3x2 2 0 0 (x0 + y)dy = (2x03x 2 0 0 (2×0.15+0.1)×0.1 From part (iii), P(Y ≤ 0.1|X = 0.15) = = 16 27 3×(0.15)2

0.593.

2.28

√ 1/ c

2

(i) From 0 cx 2 d x = 1, we find c = 1/9, so that f X (x) = x9 , 0 < x < 3. x , 0 < x < 3, 0.5x < t < (ii) Clearly, f X ,T (x, t) = f T |X (t|x) f X (x) = 18 2.5x. (iii) On the basis of the Hint made, we have: 2t x 2 d x = 3.84 For 0 < t < 1.5 : f T (t) = 0.4t 18 36 t ; and for 1.5 < t < 7.5 : 3 0.16 2 x f T (t) = 0.4t 18 d x = 0.25 − 36 t . Thus, f T (t) = (iv) By part (iii), P(2 < T < 4) =

3.84 2 36 t ,

0 < t < 1.5 2 , 1.5 ≤ t < 7.5.

0.25 − 0.16 t 36

2 0.16 2 11.26 4 (0.25 − 36 t )dt = 27

0.417.

2.29 (i) f X (0) = 0.30, f X (1) = 0.38, f X (2) = 0.32, f Y (0) = 0.32, f Y (1) = 0.30, f Y (2) = 0.25, f Y (3) = 0.13; 0.15 15 0.07 7 = 30 , f Y |X (1|0) = 0.30 = 30 , f Y |X (0|0) = 0.30 0.06 6 0.02 2 f Y |X (2|0) = 0.30 = 30 , f Y |X (3|0) = 0.30 = 30 ; 13 0.11 11 f Y |X (0|1) = 0.13 0.38 = 38 , f Y |X (1|1) = 0.38 = 38 , 0.09 9 0.05 5 ; f Y |X (2|1) = 0.38 = 38 , f Y |X (3|1) = 0.38 = 38 0.04 4 0.12 12 f Y |X (0|2) = 0.32 = 32 , f Y |X (1|2) = 0.32 = 32 , 10 0.06 6 f Y |X (2|2) = 0.10 0.32 = 32 , f Y |X (3|2) = 0.32 = 32 . (ii) E X = 1.02, EY = 1.19.

25 0.833, E(Y |1) = 44 1.158, E(Y |2) = 50 E(Y |0) = 30 38 32

1.563.

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(iii) E X 2 = 1.6, so that Var(X ) = 1.6 − (1.02)2 = 0.5596, EY 2 = 2.47, so that Var(Y ) = 2.47 − (1.19)2 = 1.0539; 49 46 106 53 2 E(Y 2 |0) = 30 , E(Y 2 |1) = 92 38 = 19 , E(Y |2) = 32 = 16 , so that: 49 25 2 845 169 Var(Y |X = 0) = 30 − ( 30 ) = 900 = 180 0.939, 1,560 92 44 2 390 Var(Y |X = 1) = 38 − ( 38 ) = 1,444 = 361 1.080, 50 2 892 223 Var(Y |X = 2) = 106 0.871.

32 − ( 32 ) = 1,024 = 256

Chapter 8 Section 8.1 1.1 Each of these r.v.’s is distributed as B(15, 1/6), so that E X = EY = 15 × 16 = 2.5, and therefore E(X + Y ) = E X + EY = 5.

1.2 (i) Consider the continuous case, as in the discrete case integrals are simply replaced by summation signs. So, for relation (8.3), we have: ∞ ∞ E[cg(X , Y )] = −∞ −∞ cg(x, y) f X ,Y (x, y)d x d y ∞ ∞ = c −∞ −∞ g(x, y) f X ,Y (x, y)d x d y = cEg(X , Y ); ∞ ∞ E[cg(X , Y ) + d] = −∞ −∞ [cg(x, y) + d] f X ,Y (x, y)d x d y ∞ ∞ ∞ ∞ = c −∞ −∞ g(x, y) f X ,Y (x, y)d x d y + d −∞ −∞ f X ,Y (x, y)d x d y = cEg(X , Y ) + d.

For relation (8.4), we have: ∞ ∞ E[g1 (X , Y ) + g2 (X , Y )] = −∞ −∞ [g1 (x, y) + g2 (x, y)] f X ,Y (x, y) ∞ ∞ ∞ ∞ d x d y = −∞ −∞ g1 (x, y) f X ,Y (x, y)d x d y + −∞ −∞ g2 (x, y) f X ,Y (x, y)d x d y = Eg1 (X , Y ) + Eg2 (X , Y ).

As for relation (8.6), we have: For the discrete case, let xi and y j be the values of the r.v.’s X and Y , respectively, and let f X ,Y be their joint p.d.f. Then, for all xi and y j , we have g(xi , y j ) ≤ h(xi , y j ), hence g(xi , y j ) f X ,Y (xi , y j ) ≤ h(xi , y j ) f X ,Y (xi , y j ), and by summing over xi and y j , we get: Eg(X , Y ) ≤ Eh(X , Y ). In particular, g(X ) ≤ h(Y ) means that, if X (s) = xi and Y (s) = y j , then g(xi ) ≤ h(y j ), hence g(xi ) f X ,Y (xi , y j ) ≤ h(y j ) f X ,Y (xi , y j ), and by summing over xi and y j , we get:

g(xi ) f X ,Y (xi , y j ) = xi g(xi ) y j f X ,Y (xi y j ) = x

i yj ) = E X , xi y j h(y j ) f X ,Y (xi , y j ) = y j xi h(y j ) × xi g(x i ) f X (x i

f X ,Y (xi , yi ) = y j h(y j ) xi f X ,Y (xi , y j ) = y jh(y j ) f Y (y j ) = EY .

The result follows. For the continuous case summations are replaced by integrations. (ii) For relation (8.8), it suffices to justify the second part only. Indeed, Var[cg(X , Y ) + d] = E {[cg(X , Y ) + d] − E[cg(X , Y ) + d]}2

Revised Answers Manual to Introduction to Probability

= E[cg(X , Y ) + d − cEg(X , Y ) − d]2 , + = E c2 [g(X , Y ) − Eg(X , Y )]2 = c2 E[g(X , Y ) − Eg(X , Y )]2 = c2 Var[g(X , Y )].

1.3 Indeed, Mc1 X +d1 ,c2 Y +d2 (t1 , t2 ) = Eet1 (c1 X +d1 )+t2 (c2 Y +d2 ) = E[ed1 t1 +d2 t2 × e(c1 t1 )X +(c2 t2 )Y ] = ed1 t1 +d2 t2 Ee(c1 t1 )X +(c2 t2 )Y = ed1 t1 +d2 t2 M X ,Y (c1 t1 , c2 t2 ), t1 , t2 ∈ .

1.4 For relation (8.16), we have: ∂ ∂ M X ,Y (t1 , t2 )|t1 =t2 =0 = Eet1 X +t2 Y |t1 =t2 =0 ∂t1 ∂t1

∂ t1 X +t2 Y |t1 =t2 =0 =E e ∂t1 = E(X et1 X +t2 Y )|t1 =t2 =0 = E X , and similarly for the second relation.

As for relation (8.17), we have: ∂2 ∂2 M X ,Y (t1 , t2 )|t1 =t2 =0 = Eet1 X +t2 Y |t1 =t2 =0 ∂t1 ∂t2 ∂t1 ∂t2

2 ∂ t1 X +t2 Y |t1 =t2 =0 =E e ∂t1 ∂t2 = E(X Y et1 X +t2 Y )|t1 =t2 =0 = E(X Y ).

Section 8.2 2.1 Indeed, (i)

P(Ac ∩ B c ) − P(Ac )P(B c ) = P(A ∪ B)c ) − P(Ac )P(B c ) = 1 − P(A ∪ B) − [1 − P(A)][1 − P(B)] = 1 − P(A) − P(B) + P(A ∩ B) − 1 + P(B) + P(A) − P(A)P(B) = P(A ∩ B) − P(A)P(B).

(ii)

P(Ac ∩ B) − P(Ac )P(B) = P(B − A ∩ B) − [1 − P(A)]P(B) = P(B) − P(A ∩ B) − P(B) + P(A)P(B) = −[P(A ∩ B) − P(A)P(B)].

(iii)

P(A ∩ B c ) − P(A)P(B c ) = P(A − A ∩ B) − P(A)[1 − P(B)] = P(A) − P(A ∩ B) − P(A) + P(A)P(B) = −[P(A ∩ B) − P(A)P(B)].

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2.2 Apply the exercise cited in the Hint for Z = X −Y with E Z = 0 and Var(Z ) = 0, in order to get that P(Z = 0) = 1 or P(X = Y ) = 1. For the second part, take Z = X + Y , with E Z = 0 and Var(Z ) = 0 in order to obtain that P(Z = 0) = 1 or P(X = −Y ) = 1.

2.3 Utilizing the results in Exercise 2.1 in Chapter 7, we have: (i)

E X = 1 × 0.10 + 2 × 0.25 + 3 × 0.30 + 4 × 0.20 + 5 × 0.10 = 2.8, EY = 1 × 0.30 + 2 × 0.20 = 0.7, E X 2 = 1 × 0.10 + 4 × 0.25 + 9 × 0.30 + 16 × 0.20 + 25 × 0.10 = 9.5, EY 2 = 1 × 0.30 + 4 × 0.20 = 1.1, so that Var(X ) = 9.5 − (2.8)2 = 9.5 − 7.84 = 1.66, Var(Y ) = 1.1 − (0.7)2 = 1.1 − 0.49 = 0.61, E(X Y ) = 1×1× 0.030+1× 2× 0.020+ 2×1× 0.075+ 2×2

(ii)

× 0.050 + 3 × 1 × 0.090 + 3 × 2 × 0.060 + 4 × 1 × 0.060 + 4 × 2 × 0.040 + 5 × 1 × 0.030 + 5 × 2 × 0.020 = 1.96, so that Cov(X , Y ) = 1.96 − 2.8 × 0.7 = 1.96 − 1.96 = 0 and therefore ρ(X , Y ) = 0. (iii) The r.v.’s X and Y are uncorrelated.

2.4 From Exercise 2.2 of Chapter 7, we have: f X (0) = 0.3, f X (1) = 0.4, f X (2) = 0.3; f Y (0) = 0.575, f Y (1) = 0.35, f Y (2) = 0.075.

Therefore: (i) E X = 1 × 0.4 + 2 × 0.3 = 1, EY = 1 × 0.35 + 2 × 0.075 = 0.5, E X 2 = 1 × 0.4 + 4 × 0.3 = 1.6, EY 2 = 1 × 0.35 + 4 × 0.075 = 0.65, so that Var(X ) = 1.6 − 1 = 0.6, Var(Y ) = 0.65 − (0.5)2 = 0.4. (ii) E(X Y ) = 1 × 1 × 0.2 + 2 × 1 × 0.15 + 2 × 2 × 0.075 = 0.8, so that Cov(X , Y ) = 0.8 − 1 × 0.5 = 0.3, and ρ(X , Y ) = √

√ 0.3 = 1.25 0.24 √ 0.6 × 0.4

0.613.

(iii) The r.v.’s X and Y are positively correlated. Thus, the larger the number of children (in the families under consideration) the larger the number of boys, and vice versa.

Revised Answers Manual to Introduction to Probability

2.5 From Exercise 2.3 in Chapter 7, we have: f X (−4) = f X (−2) = f X (2) = f X (4) = 0.25; f Y (−2) = f Y (−1) = f Y (1) = f Y (2) = 0.25.

Y 2 1

* −4

−3

−2

* 0

−1

1

2

−1 *

3

4

X

*

−2

Therefore: E X = (−4 − 2 + 2 + 4) × 0.25 = 0,

(i)

EY = (−2 − 1 + 1 + 2) × 0.25 = 0, Var(X ) = E X 2 = (16 + 4 + 4 + 16) × 0.25 = 10, Var(Y ) = EY 2 = (4 + 1 + 1 + 4) × 0.25 = 2.5.

E(X Y ) = [(−4) × 1 + (−2) × (−2) + 2 × 2 + 4 × (−1)] × 0.25 = 0 × 0.25 = 0, so that

(ii)

Cov(X , Y ) = 0 and hence ρ(X , Y ) = 0; the r.v.’s X and Y are uncorrelated. (iii) ρ(X , Y ) = 0 implies that the r.v.’s X and Y are uncorrelated. This is borne out by the graph, which shows that the four points are irregularly spread in the plane and have no resemblance to any degree of linear arrangement.

2.6 We have: (i)

77 1 (1 × 7 + 2 × 17 + 3 × 12) = , 36 36 1 80 20 EY = (1 × 7 + 2 × 14 + 3 × 15) = = , 36 36 9 1 183 E X2 = (1 × 7 + 4 × 17 + 9 × 12) = , 36 36 198 99 1 (1 × 7 + 4 × 14 + 9 × 15) = = , EY 2 = 36 36 18 EX =

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so that 183 − Var(X ) = 36 (ii)

77 36

2

659 198 − = 2 , Var(Y ) = 36 36

80 36

2 =

728 .

362

1 (1 × 1 × 2 + 1 × 2 × 1 + 1 × 3 × 4 + 2 × 1 × 2 + 2 × 2 36 × 10 + 2 × 3 × 5 + 3 × 1 × 3 + 3 × 2 × 3 + 3 × 3 × 6) 171 = , so that 36 171 77 × 80 4 Cov(X , Y ) = − = − 2 , and 36 362 36 4/362 2 ρ(X , Y ) = − √ = −√ −0.006. √ 2 182 × 659 659 × 728/36 E(X Y ) =

2.7 In Exercise 2.5 of Chapter 7, we have already found that: EX =

24 316 64 30 , EY = , Var(X ) = 2 , and Var(Y ) = 2 .

16 16 16 16

Next, 1 (1 × 1 × 1 + 1 × 2 × 1 + 2 × 1 × 3 + 2 × 2 × 2 16 + 3 × 1 × 2 + 3 × 2 × 4) 47 = , so that 16

E(X Y ) =

Cov(X , Y ) = and

32 47 30 24 − × = 2, 16 16 16 16

√ 32/162 32 2 79 ρ(X , Y ) = √ = √ = √ 79 316 × 64/162 16 79

0.225.

2.8 We have: 1 (−2 − 1 + 1 + 2) = 0, 4 10 1 Var(X ) = E X 2 = (4 + 1 + 1 + 4) = , 4 4 1 10 5 34 1 EY = (4 + 1 + 1 + 4) = = , EY 2 = (16 + 1 + 1 + 16) = , 4 4 2 4 4 34 25 9 1 Var(Y ) = − = , E(X Y ) = (−2 × 4 − 1 × 1 + 1 × 1 + 2 × 4) = 0, 4 4 4 4 EX =

Revised Answers Manual to Introduction to Probability

so that Cov(X , Y ) = 0 and ρ(X , Y ) = 0.

The fact that Y is completely determined by X through the relation Y = X 2 and yet ρ(X , Y ) = 0 seems peculiar from first viewpoint. However, it is entirely correct and appropriate, since ρ(X , Y ) = 0 states that there is no linear relationship between X and Y of whatever degree. Indeed, their relationship is quadratic.

2.9 By means of the marginal p.d.f.’s found in Exercise 2.7 of Chapter 7, we have:

1 1 x3 7 1 1 1 1 1 (i) dx = , x x+ + x2 = + = EX = 2 3 0 4 0 3 4 12 0 7 and by symmetry, EY = 12 .

4

1

1

5 5 (ii) E X 2 = x4 0 + 16 x 3 0 = 41 + 16 = 12 , and EY 2 = 12 , by symmetry.

2 2 Since E X = EY and E X = EY , it follows that Var(X ) = Var(Y ) = 5 7 2 11 12 − ( 12 ) = 144 . (iii) By Exercise 1.3 in Chapter 7,

E(X Y ) =

1 1 0

=

1 1 0

x y(x + y)d x d y

0

x 2 y d x dy +

0

1 1 0

x y2 d x d y =

0

1 2 = .

6 3

1/144 7 2 1 1 Therefore Cov(X , Y ) = 13 − ( 12 ) = − 144 , ρ(X , Y ) = − 11/144 = − 11 . (iv) The r.v.’s X and Y are negatively correlated.

2.10 By employing the marginal p.d.f.’s found in Exercise 2.8 of Chapter 7, we obtain:

3 2 1 x(3x + 1)d x = , EX = 5 0 5

3 1 3 EY = y(2y 2 + 1)dy = .

5 0 5

(i)

2 1 2 13 EX = , x (3x + 1)d x = 5 0 30 1 3 11 EY 2 = , y 2 (2y 2 + 1)dy = 5 0 25

(ii)

2

so that Var(X ) =

13 9 11 11 9 2 − = , Var(Y ) = − = .

30 25 150 25 25 25

Next, by means of the joint p.d.f., we get:

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7 6 1 1 x y(x + y 2 )d x d y = E(X Y ) = , so that 5 0 0 20 9 1 1/100 7 − =− , ρ(X , Y ) = − √ Cov(X , Y ) = √ 20 25 100 22/5 150 √ 3,300 −0.131. =− 440

(iii)

(iv) The r.v.’s X and Y are negatively correlated.

2.11 We have: E(X Y ) =

1 y

x y · 8x y d x dy y

y2 x 2 d x dy

0

=8

0 1 0

1

0

8 4 y 5 dy = , 3 0 9 Cov(X , Y ) = E(X Y ) − (E X )(EY ) 4 8 4 4 = − × = 9 15 15 225 =

0.018.

∞ 2.12 (i) For only fixed x, we have E(Y 2 |X = x) = −∞ y 2 f Y |X (y|x)dy, so that ∞ 2 E(Y |X ) = y 2 f Y |X (y|X )dy, −∞

and

∞ ∞ [ y 2 f Y |X (y|x)dy] f X (x)d x −∞ −∞ ∞ ∞ = y 2 f Y |X (y|x) f X (x)d y d x −∞ −∞ ∞ ∞ = y 2 f X ,Y (x, y)d y d x −∞ −∞ ∞ ∞ = y 2 f X ,Y (x, y)d x d y −∞ −∞ ∞ ∞ = y2[ f X ,Y (x, y)d x]dy −∞ −∞ ∞ = y 2 f Y (y)dy

E[E(Y 2 |X )] =

−∞ 2

= EY .

Revised Answers Manual to Introduction to Probability

(ii) Likewise, ∞ y f Y |X (y|X )dy] E[X E(Y |X )] = E[X −∞ ∞ ∞ = [x y f Y |X (y|x)dy] f X (x)d x −∞ −∞ ∞ ∞ = x y f Y |X (y|x) f X (x)d y d x −∞ −∞ ∞ ∞ = x y f X ,Y (x, y)d x d y = E(X Y ), −∞ −∞

∞ E[E(X Y |X )] = E[ X y f Y |X (y|X )dy] −∞ ∞ ∞ = [ x y f Y |X (y|x)dy] f X (x)d x −∞ −∞ ∞ ∞ = x y f Y |X (y|x) f X (x)d y d x −∞ −∞ ∞ ∞ = x y f X ,Y (x, y)d x d y = E(X Y ). −∞ −∞

So, E(X Y ) = E[E(X Y |X )] = E[X E(Y |X )]. x 2.13 (i) Since Y |X = x ∼ U (0, x), it follows that E(Y |X = x) = 2 , and therefore α 1 α x × dx = .

EY = E[E(Y |X )] = α 4 0 2

α x α 1 1 1 1 (ii) EY 2 = E[E(Y 2 |X )] = dx = × x 3d x y 2 × dy x α 3α x 0 0 0 α 3 2 2 2 α α α 7α 2 1 α = , so that Var(Y ) = − = = x 2d x = 3α 0 9α 9 9 16 9 × 16 2 7α . = 144 α 1 x (iii) E(X Y ) = E[E(X Y |X )] = E[X E(Y |X )] = x × × dx 2 α 0 2 2 1 α α α x = × f racx 3 3|α0 = , so that Cov(X , Y ) = − × 2α 6 6 2 4 α2 . = 24 ,Y ) (iv) ρ(X , Y ) = Cov(X σ (X )σ (Y ) =

√ √ 2 √ α /24√ = 12 √12 = 1484 (α/ 12)×(α 7/12) 24 7

0.655.

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Section 8.3 8 3.1 In Exercise 2.21 in Chapter 7, we have already found that: E X = 15 , EY = 45 , 11 6 4 Var(X ) = 15 2 , Var(Y ) = 152 and E(X Y ) = 9 . Therefore: √

2

8 √ = √4 (i) Cov(X , Y ) = 49 − 15 × 45 = 1542 and ρ(X , Y ) = √4/15 = 2 3366 11 6 15 × 15

66

0.492. The r.v.’s X and Y are positively correlated. (ii) Var(X + Y ) = Var(X ) + Var(Y ) + 2Cov(X , Y ) 6 4 11 + +2× = 225 225 225 1 25 = . = 225 9 3.2 For convenience, set E X = μ and Var(X ) = σ 2 . Then EY = aμ + b, Var(Y ) = a 2 σ 2 , and E(X Y ) = E(a X 2 + bX ) = a E X 2 + bμ = a(σ 2 + μ2 ) + bμ = aσ 2 + aμ2 + bμ.

Thus, Cov(X , Y ) = aσ 2 + aμ2 + bμ − μ(aμ + b) = aσ 2 + aμ2 + bμ − aμ2 − bμ = aσ 2 , and a aσ 2 = . So, |ρ(X , Y )| = 1, ρ(X , Y ) = σ |a|σ |a| and ρ(X , Y ) = 1 if and only if a > 0 (so a = |a|), and ρ(X , Y ) = −1 if and only if a < 0 (so that a = −|a|).

3.3 Indeed, (i)

P(U V < 0) = P[(X + Y )(X − Y ) < 0] = P(X 2 − Y 2 < 0) = P(X 2 < Y 2 ) = P(|X | < |Y |).

(ii) E(U V ) = E(X 2 − Y 2 ) = E X 2 − EY 2 = 0. (iii) Cov(U , V ) = E(U V ) − (EU )(E V ) = E(X 2 − Y 2 ) − (E X + EY )(E X − EY ) = E X 2 − EY 2 − [(E X )2 − (EY )2 ] = [E X 2 − (E X )2 ] − [EY 2 − (EY )2 ] = Var(X ) − Var(Y ) = 0, so that U and V are uncorrelated.

Revised Answers Manual to Introduction to Probability

3.4 We have: E[Y − (α X + β)]2 = E[Y 2 + (α X + β)2 − 2Y (α X + β)] = E(Y 2 + α 2 X 2 + β 2 + 2αβ X − 2α X Y − 2βY ) = EY 2 + α 2 E X 2 + β 2 + 2αβ E X − 2α E(X Y ) − 2β EY = g(α, β) (a function of α and β to be minimized with respect to α and β).

Then:

∂ g(α, β) = (2E X 2 )α + (2E X )β − 2E(X Y ) = 0 ∂α ∂ g(α, β) = 2β + (2E X )α − 2EY = 0, or: ∂β (E X 2 )α + (E X )β = E(X Y ) (E X )α + β = EY .

Solving for α and β, we get: α̂ = =

E(X Y ) E X EY 1 E X2 E X EX 1

=

E(X Y ) − (E X )(EY ) Cov(X , Y ) = 2 2 E X − (E X ) σ X2

ρ(X , Y )σ X σY σY = ρ(X , Y ); 2 σ σX X

β̂ = EY − α̂ E X . ∂ ∂ ∂ 2 Next, ∂α 2 g(α, β) = 2E X , ∂α ∂β g(α, β) = 2E X , ∂β ∂α g(α, β) = 2E X , 2

2

2

∂2 g(α, β) = 2, and the 2 × 2 matrix M of the partial derivatives is: ∂β 2

M=

2E X 2 2E X 2E X 2

=4

E X2 E X EX 1

.

Since the 1 × 1 and 2 × 2 determinants E X 2 and E X 2 − (E X )2 = Var(X ) are > 0, it follows that the values α̂ and β̂ do, indeed, minimize the function g(α, β). Thus, Ŷ = α̂ X + β̂ is the required best linear predictor of Y .

Alternatively, with λ1 , λ2 in with λ21 + λ22 = 0,

λ1 E X2 E X = λ21 E X 2 + 2λ1 λ2 E X + λ22 (λ1 λ2 ) EX 1 λ2 = [λ21 E X 2 − λ21 (E X )2 ] + [λ21 (E X )2 + 2λ1 λ2 E X + λ22 ] = λ21 [E X 2 − (E X )2 ] + [λ1 (E X ) + λ2 ]2 = λ21 Var(X ) + [λ1 (E X ) + λ2 ]2 > 0, so that the required matrix is positive definite.

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Chapter 9 Section 9.1 1.1 (i) From 0∞ 0∞ 0∞ c3 e−c(x1 +x2 +x3 ) d x1 d x2 d x3 = ( 0∞ ce−cx1 d x1 ) × ∞ ∞ ( 0 ce−cx2 d x2 ) × ( 0 ce−cx3 d x3 ) = 1 × 1 × 1 = 1, we get that c can be any positive constant. (ii) From part (i), f X 1 (x1 ) = ce−cx1 , for x1 > 0, and by symmetry, f X 2 (x2 ) = ce−cx2 , x2 > 0, f X 3 (x3 ) = ce−cx3 , x3 > 0. f

(x1 ,x2 ,x3 )

3 −cx1 −cx2 −cx3

(iii) f X 1 ,X 2 |X 3 (x1 , x2 |x3 ) = X 1 ,X 2f,XX 3 (x3 ) = c e ce−cx3 = c2 e−c(x1 +x2 ) , 3 for x1 > 0, x2 > 0. (iv) In the first place, from part (i), f X 2 ,X 3 (x2 , x3 ) = c2 e−c(x2 +x3 ) , for x2 > 0, x3 > 0. Then f X 1 |X 2 ,X 3 (x1 |x2 , x3 ) = =

f X 1 ,X 2 ,X 3 (x1 , x2 , x3 ) f X 2 ,X 3 (x2 , x3 ) c3 e−cx1 −cx2 −cx3 = ce−cx1 , for x1 > 0. c2 e−cx2 −cx3

1.2 We have: M X 1 ,X 2 ,X 3 (t1 , t2 , t3 ) = Eet1 X 1 +t2 X 2 +t3 X 3 ∞ ∞ ∞ = et1 x1 +t2 x2 +t3 x3 × c3 e−c(x1 +x2 +x3 ) d x1 d x2 d x3 0 0 0 " ∞ #" ∞ #" ∞ # 3 −(c−t1 )x1 −(c−t2 )x2 −(c−t3 ) =c e d x1 e d x2 e d x3 0 0 0 " #" # ∞ ∞ 1 1 = c3 − − e−(c−t1 )x1 e−(c−t2 )x2 c − t1 c − t2 0 0 " ∞# 1 e−(c−t2 )x3 × − c − t3 0 =

c3 , provided all t1 , t2 , and t3 are < c. (c − t1 )(c − t2 )(c − t3 )

1.3 It should be stated at the outset that all m.g.f.’s used here are assumed to be finite for all of their arguments. Then, we proceed as follows. For the direct part, it suffices to determine the m.g.f. of c1 X 1 + · · · + cn X n . Indeed, for t ∈ , Mc1 X 1 +···+cn X n (t) = Eet(c1 X 1 +···+cn X n ) = Ee(c1 t)X 1 +···+(cn t)X n = M X 1 ,...,X n (c1 t, . . . , cn t), which is known.

For the converse part, we have to determine the joint M X 1 , . . . , X n . To this end, let t1 , . . . , tn in . Then M X 1 ,...,X n (t1 , . . . , tn ) = Eet1 X 1 +···+tn X n . However,

Revised Answers Manual to Introduction to Probability

t1 X 1 + · · · + tn X n = ( tt1 )t X 1 + · · · + ( ttn )t X n (for any t = 0) = t(c1 X 1 ) + · · · + t(cn X n ), where c1 = tt1 , . . . , cn = ttn . Therefore: M X 1 ,...,X n (t1 , . . . , tn ) = Eet(c1 X 1 )+···+t(cn X n ) = Mc1 X 1 ,...,cn X n (t), which is known.

1.4 Indeed, ⎛ ⎞⎤ ' ⎞ ⎡' ⎞ (⎛ n (⎛ n m n m m

Cov ⎝ Xi , Yj⎠= E ⎣ Xi ⎝ Y j ⎠⎦ − E X i ⎝E Yj⎠ i=1

j=1

i=1

j=1

i=1

j=1

⎛ ⎞ ' ( n

m m n

⎝ ⎠ =E Xi Y j − E Xi EY j i=1 j=1

= = =

m n

i=1 j=1 m n

E(X i Y j ) −

i=1 m n

j=1

(E X i )(EY j )

i=1 j=1

[E(X i Y j ) − (E X i )(EY j )]

i=1 j=1 m n

Cov(X i , Y j ).

i=1 j=1

1.5 (i) As in Exercise 2.32 in Chapter 6, let X i be the lifetime of the ith equipment, and define Yi by: ⎧ 0 < X i ≤ μ/2 ⎨ C, Yi = C/2, μ/2 < X i ≤ μ ⎩ 0, X i > μ, i = 1, . . . , n. Then to be refunded is: n the expected amount n −1/2 − e−1 ).

Yi ) = i=1 EYi = n EY1 = nC E( i=1 2 (2 − e 100 (ii) Here E( i=1 Yi ) = 120,000(2 − e−1/2 − e−1 ) 120,000(2 − 0.607 − 0.368) = 123,000.

Section 9.2 2.1 We have:

n − n1 n − n 1 − · · · − n k−1 n ··· n2 nk n1 (n − n 1 )! (n − n 1 − · · · − n k−1 )! n! × × = n 1 !(n − n 1 )! n 2 !(n − n 1 − n 2 )! n k !(n − n 1 − · · · − n k−1 − n k )! n! = n 1 !n 2 ! · · · n k !(n − n 1 − · · · − n k−1 − n k )!

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=

n! , since (n − n 1 − · · · − n k−1 − n k )! = 0! = 1. n 1 !n 2 ! · · · n k !

2.2 Let X 1 , X 2 , and X 3 be the r.v.’s denoting the number of customers buying brand A, brand B, and just browsing, respectively. Then these r.v.’s have jointly the multinomial distribution with parameters n = 10, p1 = 0.25, p2 = 0.40, and p3 = 0.35. Therefore: 10!

(0.25)2 × (0.40)3 × (0.35)5 (i) P(X 1 = 2, X 2 = 3, X 3 = 5) = 2!3!5!

2,520 × 0.000021 = 0.05294 0.053. (ii) The conditional joint distribution of X 1 and X 2 , given X 3 = 6, is multino5 0.40 8 mial with parameters 10 − 6 = 4, and 0.25 0.65 = 13 , 0.65 = 13 . Therefore: 1 3 8 5 4!

P(X 1 = 1, X 2 = 3|X 3 = 6) = 4 × 0.0896 1!3! 13 13 = 0.3584 0.358.

2.3 This is an application of Theorems 2 and 2 with k = 4, and s and t suitably chosen.

P(X A = x A , X B = x B , X AB = x AB , X O = x O ) (i) n! p x A p x B p x AB p x O = x A !x B !x AB !x O ! A B AB O n!

(0.40)x A(0.10)x B(0.05)x AB (0.45)x O . = x A !x B !x AB !x O ! (ii) Take s = 3, Y = n − (X A + X B + X AB ), r = x A + x B + x AB , and q = 1 − ( p A + p B + p AB ). Then: P(X A = x A , X B = x B , X AB = x AB ) n! p x A p x B p x AB q n−r = x A !x B !x AB !(n − r )! A B AB n!

(0.40)x A (0.10)x B (0.05)x AB (0.45)n−r , = x A !x B !x AB !(n − r )! same as in part (i). (iii) Take s = 2, Y = n − (X A + X B ), r = x A + x B , and q = 1 − ( p A + p B ).

Then: n! p x A p x B q n−r P(X A = x A , X B = x B ) = x A !x B !(n − r )! A B n!

(0.40)x A (0.10)x B (0.50)n−r . = x A !x B !(n − r )! (iv) Take s = 1, Y = n − X A , r = x A , and q = 1 − p A . Then: P(X A = x A ) =

n! n! p x A q n−r = (0.40)x A (0.60)n−r . x A !(n − r )! A x A !(n − r )!

Revised Answers Manual to Introduction to Probability

(v) Take s = 1, t = 3, r = x O , and q = 1 − p O . Then: P(X A = x A , X B = x B , X AB = x AB |X O = x O ) x A x B

pB p AB x AB pA (n − r )! = x A !x B !x AB ! q q q

x A x B

0.10 0.05 x AB 0.40 (n − r )! = x A !x B !x AB ! 0.55 0.55 0.55 x A x B x AB 2 1 8 (n − r )! = . x A !x B !x AB ! 11 11 11 (vi) Take s = 2, t = 2, r = x AB + x O , and q = 1 − ( p AB + p O ). Then: P(X A = x A , X B = x B |X AB = x AB , X O = x O )

(n − r )! p A x A p B x B = x A !x B ! q q

x A

0.10 x B (n − r )! 0.40 (n − r )!

(0.8)x A (0.2)x B . = = x A !x B ! 0.50 0.50 x A !x B ! (vii) Take s = 3, t = 1, r = x B + x AB + x O , and q = 1 − ( p B + p AB + p O ).

Then: P(X A = x A |X B = x B , X AB = x AB , X O = x O )

(n − r )! p A x A x A! pA xA = = = 1, since x A! q x A! pA n − r = n − (x B + X AB + x O ) = x A and q = p A . (viii) For the given values of n and x A , x B , x AB , and x O , we have: For (i): P(X A = 8, X B = 2, X AB = 1, X O = 9) 20! = (0.40)8 (0.10)2 (0.05)1 (0.45)9 8!2!1!9!

0.014.

For (ii): P(X A = 8, X B = 2, X AB = 1) is the same as in part (i).

For (iii): P(X A = 8, X B = 2) =

20!

(0.40)8 (0.10)2 (0.50)10 8!2!10!

0.053.

20!

For (iv): P(X A = 8) = 8!12!

(0.40)8 (0.60)12 For (v):

P(X A = 8, X B = X AB = 1|X O = 9) =

0.180.

8 2 1 2 1 8 11!

8!2!1! 11 11 11 0.116.

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For (vi): P(X A = 8, X B = 2|X AB = 1, X O = 9) =

10!

(0.8)8 (0.2)2 8!2!

0.302.

For (vii): P(X A = 8|X B = 2, X AB = 1, X O = 9) = 1.

2.4 Take the partial derivative of the m.g.f. in (9.12) with respect to ti to get n( p1 et1 + · · · + pk etk )n−1 × p × eti , and evaluate it at t1 = · · · = tk = 0 in order to obtain E X i = npi . Likewise, E X j = np j . Next, take the partial derivatives of the same m.g.f. with respect to ti and t j in order to get: n( p1 et1 + · · · + pk etk )n−1 × pi eti × (n − 1)( p1 et1 + · · · + pk etk )n−2 × p j et j , and evaluate it at t1 = · · · = tk = 0 in order to obtain E(X i X j ) = n(n−1) pi p j .

Combining these quantities, we have: Cov(X i , X j ) = n(n − 1) pi p j − npi × np j = n 2 pi p j − npi p j − n 2 pi p j = −npi p j .

Next, take the second order partial derivative of the m.g.f. in (9.12) with respect to ti to obtain: n(n − 1)( p1 et1 + · · · + pk etk )n−2 pi2 e2ti + n( p1 et1 + · · · + pk etk )n−1 × pi eti , and evaluate it at t1 = · · · = tk = 0 to get: E X i2 = n(n − 1) pi2 + npi . Then : Var(X i ) = n(n − 1) pi2 + npi − n 2 pi2 = n 2 pi2 − npi2 + npi − n 2 pi2 = npi (1 − pi ), and likewise, Var(X j ) = np j (1 − p j ). Then the expression for the correlation coefficient ρ(X i , X j ) in (9.12 ) follows.

2.5 We have: E X A = np A = 20 × 0.40 = 8, E X B = np B = 20 × 0.10 = 2, E X AB = np AB = 20 × 0.05 = 1, E X O = np O = 20 × 0.45 = 9; Var(X A ) = np A (1 − p A ) = 20 × 0.40 × 0.60 = 4.8, Var(X B ) = np B (1 − p B ) = 20 × 0.10 × 0.90 = 1.8, Var(X AB ) = np AB (1 − p AB ) = 20 × 0.05 × 0.95 = 0.95, Var(X O ) = np O (1 − p O ) = 20 × 0.45 × 0.55 = 4.95;

Revised Answers Manual to Introduction to Probability

Cov(X A , X B ) = −np A p B = −20 × 0.40 × 0.10 = −0.8, Cov(X A , X AB ) = −np A p AB = −20 × 0.40 × 0.05 = −0.4, Cov(X A , X O ) = −np A p O = −20 × 0.40 × 0.45 = −3.6; 0.8 −0.272, √ 4.8 × 1.8 0.4 −0.187, ρ(X A , X AB ) = − √ √ 4.8 × 0.95 3.6 ρ(X A , X O ) = − √ −0.739. √ 4.8 × 4.95

ρ(X A , X B ) = − √

20 10 2 4 1 1 32 6

8,075 = 75,516 0.10693. !

6!

10 4 5 1 0.08941. (ii) P(X 1 = 4, X 2 = 1, X 3 = 1) = 4!1!1!

16 16 16

2.6 (i) P(X 1 = 4, X 2 = 1, X 3 = 1) =

(iii) The probability 0.08941 is about 83.6% of the probability 0.10693.

2.7 Let X i , i = 1, 2, 3, 4 be the number of students among the 20 who are, respectively, freshmen, sophomore, junior, and senior. Then: 20!

(0.20)4 (0.23755 ) (i) P(X 1 = 4, X 2 = 5, X 3 = 6, X 4 = 5) = 4!5!6!5!

6 5 (0.3125) (0.25) 0.01075. (ii) The (marginal) distribution of X 4 is B(20, 0.25), so that:

20 P(X 4 = 5) = (0.25)5 (0.75)15 5 5 15 12 20 4 = = 0.2024 (from the binomial tables).

16 16 5

(iii) Let X be the r.v. denoting the total number of students among the 20 who are either junior or senior. Then X ∼ B(20, 0.5625) = B(20, 9/16), and the required probability is: 11 9 9 11 7 9 20 7 20 9 = = 0.1759 P(X = 11) = 16 16 16 16 9 11 (iv)

(from the binomial tables).

20

20 9 x 7 20−x P(X ≥ 11) = x 16 16 =

x=11 20

x=11

=

20 20 − x

7 16

20−x

9 16

9

20 7 y 9 20−y y=0

y

16

16

x

(by setting y = 20 − x)

= 0.6350 (from the binomial tables).

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2.8 (i) The r.v.’s X 1 , X 2 , X 3 have the multinomial distribution with parameters n and p1 = p 2 , p2 = 2 p(1 − p), p3 = (1 − p)2 .

15! [(0.75)2 ]8 (2 × 0.75 × 0.25)6 × (ii) P(X 1 = 8, X 2 = 6, X 3 = 1) = 8!6!1! [(0.25)2 ]1 = 45,045 × (0.5625)8 × (0.375)6 × (0.0625) 0.07847. (iii) Since X 2 ∼ B(15, = B(15, 6/16), we have: 0.375) 6 6 10 9 = 0.2025 (from the binomial tables).

P(X 2 = 6) = 15 6 16 16 0.07847 2 =6,X 3 =1) (iv) P(X 1 = 8, X 3 = 1|X 2 = 6) = P(X 1 =8,X P(X 2 =6) 0.2025 0.38751.

2.9 The r.v.’s X 1 , X 2 , X 3 have the multinomial distribution with parameters n = 25, k = 3, and p1 = 0.625 (= 10/16), p2 = 0.3125 (= 5/16), p3 = 0.0625 (= 1/16). Then: 25! (i) P(X 1 = 16, X 2 = 7, X 3 = 2) = 16!7!2!

(0.625)16 (0.3125)7 (0.0625)2 0.04533. (ii) Here P(X 1 = 15, X 2 = 8) = P(X 1 = 15, X 2 = 8, X 3 = 2) = 25!

15 8 2 0.04533.

15!8!2! (0.625) (0.3125) (0.0625) (iii) Since X 3 ∼ B(25, 0.0625), we have: 1 2 15 23 = 0.2836 (from the binomial tables), or P(X 3 = 2) = 25 2 ( 16 ) ( 16 ) 0.26559 by direct calculation. (iv) Since X 1 ∼ B(25, 0.625), X 2 ∼ B(25, 0.3125), X 3 ∼ B(25, 0.0625), we have:

E X 1 = 25 × 0.625 = 15.625, σ 2 (X 1 ) = 25 × 0.625 × 0.375 = 5.859375, and σ (X 1 ) E X 2 = 25 × 0.3125 = 7.8125,

2.421;

σ 2 (X 2 ) = 25 × 0.3125 × 0.6875 = 5.37109375, and σ (X 2 ) E X 3 = 25 × 0.0625 = 1.5625,

2.318;

σ 2 (X 3 ) = 25 × 0.0625 × 0.9375 = 1.46484375, and σ (X 3 ) 1.218.

2.10 Let X 1 , X 2 , and X 3 be the r.v.’s denoting the numbers, among the 15 selected, watching news, a documentary program, and other programs, respectively. Then these r.v.’s have the multinomial distribution with parameters n = 15, k = 3, p1 = 0.3125 (= 5/16), p2 = 0.25 (= 4/16), p3 = 0.4375 (= 7/16).

Therefore: 15! (i) P(X 1 = 5, X 2 = 4, X 3 = 6) = 5!4!6!

(0.3125)5 (0.25)4 (0.4375)6 0.05148. (ii) Here X 3 ∼ B(15, 0.25) = B(15, 4/16), so that: P(X 3 ≥ 3) = 1 − P(X 3 ≤ 2) = 1 − 0.2361 = 0.7639 (from the binomial tables).

Revised Answers Manual to Introduction to Probability

(iii) Let X be the r.v. denoting the number of those among the 15 selected, who do not watch the news. Then X = 15 − X 1 , so that: P(X ≤ 5) = P(15 − X 1 ≤ 5) = P(X 1 ≥ 10) = 1 − P(X 1 ≤ 9) = 1 − 0.9949 = 0.0051 (from the binomial tables), since X 1 ∼ B(15, 0.3125) = B(15, 5/16).

2.11 (i) The joint distribution of the r.v.’s X i , i = 1, . . . , k is multinomial with parameters n and pi , i = 1, . . . , k. (ii) By symmetry, we have: p1 = p8 = P(X > 3) = 1 − P(X ≤ 3) = 1 −

(3)

= 1 − 0.998650 = 0.00135, p2 = p7 = P(2 ≤ X < 3) = (3) − (2) = 0.998650 − 0.977250 = 0.0214, p3 = p6 = P(1 ≤ X < 2) = (2) − (1) = 0.977250 − 0.841345 = 0.135905, p4 = p5 = P(0 ≤ X < 1) = = 0.341345.

(1) − 0.5 = 0.841345 − 0.5

Section 9.3 3.1 The second line in (16) follows from the first line by adding and subtracting the 1 2 quantity ρ 2 ( x1 σ−μ ) . The expression in the following line follows by the fact 1 that the first three terms on the previous line form a perfect square. What follows is obvious and results from a suitable regrouping of the entities involved.

3.2 Indeed, ∞ ∞ E(X Y ) = x y f X ,Y (x, y)d x d y −∞ −∞ ∞ ∞ = x y f Y |X (y|x) f X (x)d x d y −∞ −∞ ∞ ∞ = x y f Y |X (y|x) f X (x)d y d x −∞ −∞ " ∞ # ∞ = x f X (x) y f Y |X (y|x)dy d x −∞ −∞ " # ∞ ρσ2 = x f X (x) μ2 + (x − μ1 ) d x σ1 −∞ 1 (because, by (9.18), Y |X = x ∼ N (bx , (σ2 1 − ρ 2 )2 ), ρσ2 (x − μ1 )) so that E(Y |X = x) = bx = μ2 + σ1

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" ∞ # ∞ ρσ2 2 = μ2 x f X (x)d x + x f X (x)d x − μ1 x f X (x)d x σ1 −∞ −∞ −∞ ! ρσ2 E X 2 − μ21 = μ2 + σ1 ρσ2 = μ2 + × σ12 = μ2 + ρσ1 σ2 . σ1 ∞

3.3 In Exercise 3.2, it was found that E(X Y ) = μ1 μ2 + ρσ1 σ2 , where μ1 = E X , μ2 = EY , σ1 = s.d. of X and σ2 = s.d. of Y . Then ρ(X , Y ) =

Cov(X , Y ) (μ1 μ2 + ρσ1 σ2 ) − μ1 μ2 = = ρ. σ1 σ2 σ1 σ2

Cov(c1 X , c2 Y ) = E{[c1 X − E(c1 X )][c2 Y − E(c2 Y )]}

3.4 (i)

= E[c1 (X − E X ) × c2 (Y − EY )] = c1 c2 E[(X − E X )(Y − EY )] = c1 c2 Cov(X , Y ). (ii)

E(c1 X + c2 Y ) = c1 E X + c2 EY (by (8.3) and (8.4) in Chapter 8); Var(c1 X + c2 Y ) = σ 2 (c1 X ) + σ 2 (c2 Y ) + 2Cov(c1 X , c2 Y ) = c12 σ12 + c22 σ22 + 2c1 c2 Cov(X , Y ) = c12 σ12 + c22 σ22 + 2c1 c2 σ1 σ2 ρ (by Theorem 3 in Chapter 8, and part (i)).

(iii) In (ii) here, the quantities μ1 , μ2 , σ1 , σ2 , and ρ will have to be computed by using the p.d.f. f X ,Y (and its marginals), whereas in Example 4(i), they are read out of the formula for f X ,Y given there.

3.5 Setting g(t1 , t2 ) for the exponent; i.e., g(t1 , t2 ) = μ1 t1 + μ2 t2 +

1 2 2 σ t + 2ρσ1 σ2 t1 t2 + σ22 t2 , 2 1 1

we have: ∂ g(t1 , t2 ) = μ1 + σ12 t1 + ρσ1 σ2 t2 , ∂t1 ∂ g(t1 , t2 ) = μ2 + σ22 t2 + ρσ1 σ2 t1 , ∂t2 ∂2 ∂2 2 g(t , t ) = σ , g(t1 , t2 ) = σ22 , 1 2 1 ∂t12 ∂t22 ∂2 g(t1 , t2 ) = ρσ1 σ2 . ∂t1 ∂t2

Revised Answers Manual to Introduction to Probability

Then:

∂ M X ,Y (t1 , t2 ) = M X ,Y (t1 , t2 ) μ1 + σ12 t1 + ρσ1 σ2 t2 , ∂t1

∂ M X ,Y (t1 , t2 ) = M X ,Y (t1 , t2 ) μ2 + σ22 t1 + ρσ1 σ2 t1 , ∂t2 2

∂2 M X ,Y (t1 , t2 ) = M X ,Y (t1 , t2 ) μ1 + σ12 t1 + ρσ1 σ2 t2 2 ∂t1 +M X ,Y (t1 , t2 ) × σ12 , 2

∂2 M X ,Y (t1 , t2 ) = M X ,Y (t1 , t2 ) μ2 + σ22 t2 + ρσ1 σ2 t1 2 ∂t2 + M X ,Y (t1 , t2 ) × σ22 ,

∂2 M X ,Y (t1 , t2 ) = M X ,Y (t1 , t2 ) μ1 + σ12 t1 + ρσ1 σ2 t2 ∂t1 ∂t2

× μ2 + σ22 t2 + ρσ1 σ2 t1 +M X ,Y (t1 , t2 ) × ρσ1 σ2 .

Evaluating these quantities at t1 = t2 = 0, we get: ∂ M X ,Y (t1 , t2 )|t1 =t2 =0 = E X = μ1 , ∂t1 ∂ M X ,Y (t1 , t2 )|t1 =t2 =0 = EY = μ2 , ∂t2 ∂2 M X ,Y (t1 , t2 )|t1 =t2 =0 = E X 2 = μ21 + σ12 , ∂t12 ∂2 M X ,Y (t1 , t2 )|t1 =t2 =0 = EY 2 = μ22 + σ22 , ∂t22 ∂2 M X ,Y (t1 , t2 )|t1 =t2 =0 = E(X Y ) = μ1 μ2 + ρσ1 σ2 . ∂t1 ∂t2 Therefore, σ 2 (X ) = μ21 + σ12 − μ21 = σ12 , σ 2 (Y ) = μ22 + σ22 − μ22 = σ22 , 1 σ2 Cov(X , Y ) = μ1 μ2 + ρσ1 σ2 − μ1 μ2 = ρσ1 σ2 , and ρ(X , Y ) = ρσ σ1 σ2 = ρ.

3.6 (i) For t ∈ , we have: Mc1 X +c2 Y (t) = Eet(c1 X +c2 Y ) = Ee(c1 t)X +(c2 t)Y = exp μ1 (c1 t) + μ2 (c2 t) 2

1 + σ12 (c1 t)2 + 2ρσ1 σ2 (c1 t)(c2 t) + σ22 (c2 t)2 2 # "

1 = exp (c1 μ1 + c2 μ2 )t + c12 σ12 + 2c1 c2 ρσ1 σ2 + c22 σ22 t 2 , 2

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which is the m.g.f. of a r.v. distributed as normal with parameters c1 μ1 + c2 μ2 and c12 σ12 + 2c1 c2 ρσ1 σ2 + c22 σ22 . Thus, c1 X + c2 Y has this distribution. (ii) For t1 , t2 ∈ , we have: M X ,Y (t1 , t2 ) = Eet1 X +t2 Y = Ee(t1 X +t2 Y )×1 = Mt1 X +t2 Y (1). σ 2t2

On the other hand, if Z ∼ N (μ, σ 2 ), then M Z (t) = eμt+ 2 and M Z (1) = σ2

eμ+ 2 . Since it is assumed that t1 X + t2 Y is normally distributed, we have then: M X ,Y (t1 , t2 ) = Mt1 X +t2 Y (1)

2 [Var(t1 X + t2 Y )]2 = exp E(t1 X + t2 Y ) + .

2

But E(t1 X + t2 Y ) = μ1 t1 + μ2 t2 , Var(t1 X + t2 Y ) = σ12 t12 + σ22 t22 + 2ρσ1 σ2 t1 t2 , where μ1 = E X , μ2 = EY , σ12 = Var(X ), σ22 = Var(Y ), and ρ = ρ(X , Y ). Therefore

σ12 t12 + 2ρσ1 σ2 t1 t2 + σ22 t22 , M X ,Y (t1 , t2 ) = exp μ1 t1 + μ2 t2 + 2 which is the m.g.f. of the bivariate normal distribution with parameters μ1 , μ2 , σ12 , σ22 , and ρ.

Therefore the joint distribution of X and Y is the bivariate normal with parameters just described.

3.7 (i) First, this function cannot be a bivariate normal p.d.f. because, for x and 1 y outside the interval [−1, 1], the function becomes 2π exp (− x +y 2 ), which coincides with the p.d.f. of the bivariate normal distribution with μ1 = μ2 = 0, σ1 = σ2 = 1, and ρ = 0. However, for x = y = −1, e.g., the function becomes f (−1, −1) = π1e which is different from 2π1 e , the above cited p.d.f. evaluated at x = y = −1. Next, the given function is, indeed, a p.d.f. To see this, set I for the rectangle [−1, 1] × [−1, 1], and 2 2 1 1 3 3 exp(− x +y also let g(x, y) = 2π 2 ), h(x, y) = 2π e x y . Then: ∞ ∞

f (x, y)d x d y = g(x, y)d x d y 2

−∞ −∞

Ic

[g(x, y) + h(x, y)]d x d y

+

I

g(x, y)d x d y +

= Ic

g(x, y)d x d y +

I

h(x, y)d x d y I

2

Revised Answers Manual to Introduction to Probability

= =

∞ ∞ −∞ −∞ ∞ ∞ −∞

g(x, y)d x d y +

1 1 h(x, y)d x d y

−1 −1 1 1

1 − x 2 +y 2 2 dx dy + e −∞ 2π

−1

1 3 3 x y dx dy 2π e −1

x4 1 1 y4 1 × =1+ × = 1 + 0 = 1.

2π e 4 −1 4 −1 (ii) As for the marginals, we have: f 1 (x) =

1 2π

−1 −∞

e−

x 2 +y 2 2

dy +

1 2π

1 −1

e−

x 2 +y 2 2

dy +

1 2π e

1 −1

x 3 y 3 dy

∞ ∞ 1 x 2 +y 2 x 2 +y 2 x4 1 1 y3 × + e− 2 dy = e− 2 dy + 2π 1 2π −∞ 2π e 4 −1 ∞ 2 2 2 y y x 1 1 1 = √ e− 2 × √ e− 2 d x + 0 = √ e− 2 × 1 2π 2π −∞ 2π 1 − y2 = √ e 2, 2π so that Y ∼ N (0, 1). Likewise, X ∼ N (0, 1).

3.8 (i) The joint m.g.f. of X and Y is given by: # "

1 2 2 2 2 M X ,Y (t1 , t2 ) = exp μ1 t1 + μ2 t2 + σ1 t1 + 2ρσ1 σ2 t1 t2 + σ2 t2 , 2 t1 , t2 ∈ .

Therefore: MU ,V (t1 , t2 ) = Eet1 U +t2 V = Eet1 (X +Y )+t2 (X −Y ) = Ee(t1 +t2 )X +(t1 −t2 )Y = M X ,Y (t1 + t2 , t1 − t2 ) 1 = exp μ1 (t1 + t2 ) + μ2 (t1 − t2 ) + σ12 (t1 + t2 )2 2 2

+ 2ρσ1 σ2 (t1 + t2 )(t1 − t2 ) + σ22 (t1 − t2 )2 # "

1 = exp (μ1 + μ2 )t1 + (μ1 − μ2 )t2 + τ12 t12 + 2ρ0 τ1 τ2 t1 t2 + τ22 t22 , 2 since σ12 (t1 + t2 )2 + 2ρσ1 σ2 (t1 + t2 )(t1 − t2 ) + σ22 (t1 − t2 )2 = σ12 t12 + 2σ12 t1 t2 + σ12 t22 + 2ρσ1 σ2 t12 − 2ρσ1 σ2 t22 + σ22 t12 − 2σ22 t1 t2 + σ22 t22

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= σ12 + 2ρσ1 σ2 + σ22 t12 + σ12 − 2ρσ1 σ2 + σ22 t22 + 2 σ12 − σ22 t1 t2 σ 2 − σ22 = τ12 t12 + 2 1 τ1 τ2 t1 t2 + τ22 t22 τ1 τ2 = τ12 t12 + 2ρ0 τ1 τ2 t1 t2 + τ22 t22 .

However, the last expression above, namely, " #

1 2 2 2 2 exp (μ1 + μ2 )t1 + (μ1 − μ2 )t2 + τ1 t1 + 2ρ0 τ1 τ2 t1 t2 + τ2 t2 2 is the m.g.f. of the bivariate normal distribution with parameters μ1 + μ2 , μ1 − μ2 , τ12 , τ22 , and ρ0 . Therefore this is the joint distribution of the r.v.’s U and V . (ii) That U ∼ N (μ1 +μ2 , τ12 ) and V ∼ N (μ1 −μ2 , τ22 ) follows by the fact that the distributions of U and V are the marginal distributions of the bivariate normal distribution as given above. (iii) The r.v.’s U and V are uncorrelated if and only if ρ(U , V ) = 0. However, ρ(U , V ) = ρ0 = (σ12 − σ22 )/τ1 τ2 . Therefore ρ(U , V ) = 0 if and only if σ12 = σ22 .

1 2 2 2 3.9 Since Y |X = x ∼ N (μ2 + σ σ1 (x −μ1 ), (σ2 1 − ) ), we have: Y |X = 84 ∼ N (μ, σ 2 ) where μ = 91.35, σ 2 = 35.4375. Therefore, setting U ∼ N (μ, σ 2 ), we have:

92 − 91.35 P(Y > 92|X = 84) = P(U > 92) = 1 − √ 35.4375 1 − (0.11) = 1 − 0.543795 = 0.456205.

P(X > Y ) = P(X − Y > 0) = P(U > 0), where U ∼ N (μ, σ 2 ) with μ = −8, σ 2 = 46. Therefore :

8 = 1 − (1.18) P(X > Y ) = P(U > 0) = 1 − √ 46 = 1 − 0.881 = 0.119.

X + Y = U ∼ N (μ, σ 2 ) with μ = 172, σ 2 = 316. Therefore :

4 1 − (0.45) P(X + Y > 180) = P(U > 180) = 1 − √ 79 = 1 − 0.673645 = 0.326355.

3.10 (i) E X = 3.2, EY = 12, σ 2 (X ) = 1.44, σ 2 (Y ) = 16, ρ(X , Y ) = 0.7, and Cov(X , Y ) = ρσ1 σ2 = 0.7 × 1.2 × 4 = 3.36. (ii) E(X |Y = 10) = μ + ρσ1 (y − μ ) = 3.2 + 0.7 × 1.2 (10 − 12) = 2.78, 1 2 σ2 4 ρσ2 0.7 × 4 (3.8 − 3.2) = 13.4, (x − μ1 ) = 12 + E(Y |X = 3.8) = μ2 + σ1 1.2

Revised Answers Manual to Introduction to Probability

1 σ 2 (X |Y = 10) = (σ1 1 − ρ 2 )2 = σ12 (1 − ρ 2 ) = 1.44 × 0.51 = 0.7344, 1 σ 2 (Y |X = 3.8) = (σ2 1 − ρ 2 )2 = σ22 (1 − ρ 2 ) = 16 × 0.51 = 8.16. (iii) X ∼ N (3.2, 1.44), Y ∼ N (12, 16). (iv) P(0.8 < X < 4.2) = (1) − (−2) = (1) + (2) − 1 = 0.818595, P(Y > 14) = 1 − (0.5) = 0.308538. (v) By part (ii), X |Y = 10 ∼ N (2.78, 0.7344), Y |X = 3.8 ∼ N (13.4, 8.16). (vi) P(X > 3.2|Y = 10) = 1 − (0.49) = 0.312067, P(Y < 12|X = 3.8) = (−0.49) = 1 − (0.49) = 0.312067.

1 2 2 2 3.11 We know that Y |X = x ∼ N (μ2 + σ σ1 (x − μ1 ), (σ2 1 − ) ). Therefore: E(Y |X = 68) = 69 + 0.8×1 2 (69 − 68) = 69.4, the conditional s.d. of Y |X = √ 2 69 is: 1 × 1 − 0.8 = 0.6, P(Y > 70|X = 69) = P(U > 70), where 70−69.4 = 1 − (1) = U ∼ N (69.4, 0.62 ), so that P(U > 70) = 1 − 0.6 1 − 0.841345 = 0.158655. By Exercise 3.6 above, we have that: c1 X + c2 Y ∼ N (c1 μ1 + c2 μ2 , c12 σ 2 + 2c1 c2 ρσ1 σ2 + c22 σ22 ) for any constants c1 and c2 . √ Taking c1 = 1, c2 = −1, we get that X − Y ∼ N (−1, ( 1.8)2 ), so that: P(Y > X ) = P(X − Y < 0) = P(U < 0), where U ∼ N (μ, σ 2 ) with μ = −1 and σ 2 = 1.8. Therefore: P(Y > X ) = P(U < 0) = ( √1 ) 1.8 (0.75) = 0.773373.

3.12 (i) σ 2 (U ) = σ 2 (X + cY ) = σ 2 (X ) + c2 σ 2 (Y ) + 2cCov(X , Y ) = σ12 + c2 σ22 + 2cρσ1 σ2 . (ii) By setting g(c) = σ12 + σ22 c2 + 2ρσ1 σ2 c, we have g (c) = 2σ22 c + 2ρσ1 σ2 = 0, so that c = −ρσ1 /σ2 . Also, g (c) = 2σ22 > 0 for all c, and hence σ 2 (U ) is minimized for c = −ρσ1 /σ2 . (iii) minc σ 2 (U ) = σ12 + (ρσ1 /σ2 )2 σ22 + 2(−ρσ1 /σ2 )ρσ1 σ2 = σ12 + ρ 2 σ12 − 2ρ 2 σ12 = (1 − ρ 2 )σ12 . (iv) X and Y are uncorrelated (independent) if and only if ρ = 0. Then c = 0, so that σ 2 (U ) = σ12 = minc σ 2 (U ).

3.13 Clearly, f (x, y) is maximized if and only if the exponent of e is minimized, and since 1/(1 − ρ 2 ) is a constant, concentrate on the quantity in the brackets. y−μ2 2 1 By setting x−μ σ1 = u, σ2 = v, the quantity in the brackets is g(u, v) = u − ∂ ∂ 2ρuv + v 2 . The equations ∂u g(u, v) = 2u − 2ρv = 0, ∂v g(u, v) = −2ρu + ∂ ∂ ∂ 2v = 0 yield u = v = 0. Also, ∂u 2 g(u, v) = 2 = ∂v 2 g(u, v), ∂u∂v g(u, v) = −2ρ, so that the 2 × 2 matrix is

2 −2ρ 1 −ρ =4 . −2ρ 2 −ρ 1 2

2

2

1 −ρ = 1−ρ 2 are both positive, the values −ρ 1 u = v = 0 minimize g(u, v). Then the values x = μ1 , y = μ2 minimize the exponent, and hence maximize f (x, y).

Since the determinants |1| and

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An alternative way to show minimization of g(u, v) by u = v = 0, is to show 1 −ρ that the matrix is positive definite. Indeed, with λ1 , λ2 not both 0, −ρ 1

1 −ρ

λ1 = λ21 + λ22 − 2ρλ1 λ2 . λ1 λ2 −ρ 1 λ2 Next, −1 < ρ < 1 or −1 < −ρ < 1, and if λ1 λ2 > 0, then −λ1 λ2 < −ρλ1 λ2 , so that λ21 + λ22 − 2ρλ1 λ2 > λ21 + λ22 − 2λ1 λ2 = (λ1 − λ2 )2 > 0.

If λ1 λ2 < 0, then from −ρ < 1, we get −ρλ1 λ2 > λ1 λ2 , so that λ21 + λ22 − 2ρλ1 λ2 > λ21 + λ22 + 2λ1 λ2 = (λ1 + λ2 )2 > 0.

3.14 (Proof by differentiation). (i) We have Ŷ = a + b(X − μ1 ), so that Ŷ − Y = (a − bμ1 ) + (b2 X − Y ), and (Ŷ − Y )2 = (a − bμ1 )2 + 2b(a − bμ1 )X + bX 2 + 2(bμ1 − a)Y + Y 2 − 2bX Y . Since E X = μ1 , E X 2 = σ12 + μ21 , EY = μ2 , EY 2 = σ22 + μ22 , E(X Y ) = Cov(X , Y ) + E(X )E(Y ) = ρσ1 σ2 + μ1 μ2 , we get: def E(Ŷ − Y )2 = a 2 − 2μ2 a + σ12 b2 − 2ρσ1 σ2 b + (μ22 + σ22 ) = h(a, b). ∂h 2 Then: ∂h ∂a = 2a −2μ2 , ∂b = 2σ1 b−2ρσ1 σ2 , and by equating them to 0, we

∂ h ∂ h ∂ h ∂ h 2 2 obtain: a = μ2 , b = ρσ σ1 . Next, ∂a 2 = 2, ∂a∂b = 0, ∂b∂a = 0, ∂b2 = 2σ1 , and

! λ 1 0

λ1 1 2 = λ = λ21 + σ12 λ22 > 0 σ λ λ1 λ2 1 1 2 0 σ12 λ2 λ2 2

2

2

2

2 for λ21 + λ22 = 0. It follows that the values a = μ2 and b = ρσ σ1 minimize

the E(Ŷ − Y )2 .

2 (ii) For the above values of a and b : Ŷ = a + b(X − μ1 ) = μ2 + ρσ σ1 (X − μ1 ), and we recognize it as the E(Y |X = x).

3.15 We get an intersection as described by setting f (x, y) = c, some constant c, or equivalently,

y − μ2 y − μ2 2 x − μ1 x − μ1 2 + − 2ρ = c0 σ1 σ1 σ2 σ2 1 (= −2(1 − ρ 2 ) log(2π σ1 σ2 c 1 − ρ 2 )), or 1 ρ 1 (x − μ1 )2 − 2 (x − μ1 )(y − μ2 ) + 2 (y − μ2 )2 = c0 .

2 σ1 σ2 σ1 σ2 However, it is known from analytic geometry that the above equation represents an ellipse if and only if the discriminant ( σ1ρσ2 )2 − σ11 × σ12 < 0 or ρ 2 − 1 < 0 or ρ 2 < 1 which is, indeed, the case. Furthermore, the equation = 0 or ρ = 0. represents a circle if and only if σ2ρ 1 σ2

Revised Answers Manual to Introduction to Probability

Chapter 10 Section 10.1 3 1.1 For x = 0 and y = 1, we have: f X ,Y (0, 1) = 18 , f X (0) = 16 , and f Y (1) = 21 .

3 Since f X ,Y (0, 1) = 18 = 16 × 21 = f X (0) f Y (1), the r.v.’s X and Y are not independent.

1.2 In Exercise 1.1 of Chapter 7, we have the joint p.d.f. f X ,Y , and in Exercise 2.1 of Chapter 7, we have the marginal p.d.f.’s f X and f Y . For the r.v.’s X and Y to be independent, it suffices to have:

f X ,Y (x, y) = f X (x) f Y (y), x = 0, 1, 2, 3, 4, 5, and y = 0, 1, 2.

Checking these relations, we obtain: (0, 0): 0.025 = 0.05 × 0.50 (0, 1): 0.015 = 0.05 × 0.30 (0, 2): 0.010 = 0.05 × 0.20 (1, 0): 0.050 = 0.10 × 0.50 (1, 1): 0.030 = 0.10 × 0.30 (1, 2): 0.020 = 0.10 × 0.20 (2, 0): 0.125 = 0.25 × 0.50 (2, 1): 0.075 = 0.25 × 0.30 (2, 2): 0.050 = 0.25 × 0.20

(3, 0): 0.150 = 0.30 × 0.50 (3, 1): 0.090 = 0.30 × 0.30 (3, 2): 0.060 = 0.30 × 0.20 (4, 0): 0.100 = 0.20 × 0.50 (4, 1): 0.060 = 0.20 × 0.30 (4, 2): 0.040 = 0.20 × 0.20 (5, 0): 0.050 = 0.10 × 0.50 (5, 1): 0.030 = 0.10 × 0.30 (5, 2): 0.020 = 0.10 × 0.20.

So, X and Y are independent.

1.3 The points at which f X ,Y ,Z is not zero (and, in fact, equals 1/4) are the points: (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1).

Then: (i)

f X ,Y (1, 0) = f X ,Y (0, 1) = f X ,Y (0, 0) = f X ,Y (1, 1) = 1/4; f X ,Z (1, 0) = f X ,Z (0, 1) = f X ,Z (0, 0) = f X ,Y (1, 1) = 1/4; f Y ,Z (1, 0) = f Y ,Z (0, 1) = f Y ,Z (0, 0) = f Y ,Z (1, 1) = 1/4.

(ii) From part (i): f X (1) = f X (0) =

1 ; 2

f Y (1) = f Y (0) =

1 ; 2

f Z (1) = f Z (0) =

1 .

2

(iii) From parts (i) and (ii), we have that, for all x, y, and z, f X (x) f Y (y) = f X ,Y (x, y); f X (x) f Z (z) = f X ,Z (x, z); and f Y (y) f Z (z) = f Y ,Z (y, z), so that the pairs X , Y ; X , Z ; and Y , Z consist of independent r.v.’s. (iv) The r.v.’s X , Y , and Z are not independent, because, for example, f X (1) f Y (1) f Z (1) =

1 1 = = f X ,Y ,Z (1, 1, 1).

8 4

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1.4 In Exercise 2.8 of Chapter 7, we found that: f X (x) =

2 (3x + 1), 0 ≤ x ≤ 1; 5

f Y (y) =

3 (2y 2 + 1), 0 ≤ y ≤ 1.

5

By taking, e.g., x = y = 0.1, we get: f X ,Y (0.1, 0.1) = 0.132,

f X (0.1) = 0.52,

f Y (0.1) = 0.612.

Since 0.132 = 0.52 × 0.612, the r.v.’s X and Y are not independent, they are dependent.

2 2 1.5 By Example 4 in Chapter 7, f X ,Y (x, y) = 21 2 x y, 0 < x < y < 1, and by 21 2 4 Example 7 in the same chapter, f X (x) = 4 x (1 − x ), 0 < x < 1, f Y (y) = 21 2 √ y, 0 < y < 1. The relation f X ,Y (x, y) = f X (x) f Y (y) need not be true 6 y 7 245 for all x and y. For example, for x = 13 and y = 41 , f X ,Y ( 13 , 41 ) = 24 = 3,888 = 1 1 f X ( 3 ) f Y ( 4 ), so that the r.v.’s X and Y are not independent.

1.6 We have: 1

(i) f X (x) = 65 0 (x 2 + y)dy = 65 (x 2 × y|10 + 21 y 2 |10 ) = 65 (x 2 + 21 ), 0 ≤ x ≤ 1, 1 and f Y (y) = 65 0 (x 2 + y)d x = 65 ( 13 x 3 |10 + y × x|10 ) = 65 ( 13 + y) = 6 1 5 (y + 3 ), 0 ≤ y ≤ 1.

(ii) X and Y are not independent, because, e.g., f X ,Y 21 , 41 = 35 , f X 21 = 1 9 7 3 9 7 10 , f Y 4 = 10 , and 5 = 10 × 10 .

1.7 We have: 1

1

c

2

(i) f X (x) = 0 dy = 1, 0 < x < 1; f Y (y) = 0 d x = 1, 0 < y < 1. (ii) Clearly, f X ,Y (x, y) = f X (x) f Y (y) for all x and y, so that X and Y are independent. c c c−y (iii) y2 c dx dy = (c − y)dy = c2 − P(X + Y < c) = 2 0 0 0 0 2 2 c c = , provided 0 < c ≤ 1. = c2 − 2 2

1

A

B

C 0

e100

0

x

1

c

2

Revised Answers Manual to Introduction to Probability

For 1 < c ≤ 2, x + y = c and y = 1, so that x = c − 1. Then the length of AB is 1 − x = 1 − (c − 1) = 2 − c, so that the area of ABC is (2 − c)2 /2.

2 2 Thus, P(X + Y < c) = 1 − (2−c) = 4c−c2 −2 . So, P(X + Y < c) = 2

c2 /2, 0 < c ≤ 1, P(X + Y < c) = 4c−c2 −2 , 1 < c ≤ 2.

1 = 0.03125. (iv) P X + Y < 14 = 32 2

1.8 (i) Here f X (x) = 8x( 1 y dy)( 1 z dz) = 8 × y 2 1 × z 2 1 = 2x, 0 < x < 1; 0 0 2 0 2 0 and likewise, f Y (y) = 2y, 0 < y < 1;

f Z (z) = 2z, 0 < z < 1.

(ii) Clearly, for all x, y, and z : f X ,Y ,Z (x, y, z) = f X (x) f Y (y) f Z (z), so that the r.v.’s X , Y , and Z are independent. (iii) There are 3! = 6 arrangements of X , Y , and Z all of which have the same probability, 16 , by symmetry. Hence the result. By calculations, 1 z y P(X < Y < Z ) = 8 x yz d x dy dz 0 0 0

1 z x2 y y× dy dz =8 z 2 0 0 0

1 y4 z z6 1 1 z× dz = =4 = .

4 6 6 0 0 0 ∞ ∞ 1.9 (i) We have: −∞ −∞ f X ,Y (x, y)d x d y = c (x 2 +y 2 ≤9) d x d y = c × 9π , because the integral is the area of a circle centered at the origin and having radius r = 3,√so that the area is πr 2 = 9π . Hence c = 1/9π . √ 9−x 2 2 9 − x 2 , 0 ≤ x ≤ 3, and likewise f Y (y) = (ii) f X (x) = c −√9−x 2 dy = 9π 1 2 2 9π 9 − y , 0 ≤ y ≤ 3.

1 (iii) For example, for x = y = 3, f X (3) = f Y (3) = 0 and f X ,Y (3, 3) = 9π , so that f X (3) f Y (3) = f X ,Y (3, 3), and the r.v.’s X and Y are dependent.

1.10 (i) Clearly ∞ ∞ ∞

c3 e−c(x+y+z) d x d y dz ∞ ∞

ce−ex d x ce−cy dy ce−cz dz = 1 × 1 × 1 = 1,

0

=

0∞ 0 0

0

0

so that c can be any positive constant. ∞ (ii) Also, f X ,Y (x, y) = c2 e−cx−cy ( 0 ce−cz dz) = c2 e−cx−cy , x > 0, y > 0, and likewise, f X ,Z (x, z) = c2 e−cx−cz , x > 0, z > 0; f Y ,Z (y, z) = c2 e−cy−cz ,

y > 0, z > 0.

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(iii) From part (ii), f X (x) = ce−cx f Y (y) = ce−cy ,

∞ −cy dy = ce−cx , x > 0; and likewise, 0 ce

f Z (z) = ce−cz , z > 0.

y > 0;

(iv) From parts (ii) and (iii), we have that, for any x, y, and z: f X (x) f Y (y) = f X ,Y (x, y) ; f X (x) f Z (z) = f X ,Z (x, z); f Y (y) f Z (z) = f Y ,Z (y, z), so that the pairs X , Y ; X , Z ; and Y , Z consist of independent r.v.’s.

Finally, from part (iii) and for all x, y, and z, we have f X (x) f Y (y) f Z (z) = f X ,Y ,Z (x, y, z), so that the r.v.’s X , Y , and Z are independent.

1.11 (i) For the continuous case: ∞ ∞ ∞ f X ,Y (x, y)dy = g(x)h(y)dy = g(x) h(y)dy, f X (x) = −∞ −∞ −∞ ∞ ∞ f Y (y) = g(x)h(y)d x = h(y) g(x)d x, −∞

and

−∞

∞ ∞

f X ,Y (x, y)d x d y = −∞ −∞ ∞ ∞ = g(x)d x × h(y)dy.

1=

−∞

∞ ∞ −∞ −∞

g(x)h(y)d x d y

−∞

∞

∞

∞

That is, −∞ g(x)d x × −∞ h(y)dy = 1, and so, if we set −∞ h(y)dy = c, ∞ then −∞ g(x)d x = 1/c. Thus, f X (x) = cg(x),

1 f Y (y) = h(y) c

c=

∞ −∞

h(y)dy .

For the discrete case, replace integrals by summation signs. (ii) From the last expression above, it follows that: f X (x) f Y (y) = g(x)h(y) = f X ,Y (x, y), for all x, y, so that the r.v.’s X and Y are independent. (iii) If g(x) = h(x), then, for the continuous case again, 1=

∞ −∞

f X (x)d x = c

∞ −∞

h(x)d x = c × c = c2 ,

so that c = 1, and then f X (x) = g(x), f Y (y) = g(y); i.e., X and Y are identically distributed.

Revised Answers Manual to Introduction to Probability

(iv) By symmetry, P(X > Y ) = P(X < Y ) and since P(X > Y ) + P(X < Y ) = 1, the result follows.

By calculations, one would have:

g(x)g(y)d x d y = g(x)g(y)d y d x P(X > Y ) = (y<x) (y<x) " x # ∞ = g(x) g(y)dy d x −∞ −∞ ∞ x = g(x)G(x)d x, where G(x) = g(y)dy −∞ −∞ ∞ = G(x) dG(x), by assuming that g is continuous −∞

= =

1

y dy, by setting y = G(x), so that 0 < y < 1

0 2 y 1

2 0

= 1/2 = 0.5.

1 1.12 (i) Here E X = λ1 = 0.005 = 200 days. (ii) By independence,

1 1 M X +Y (t) = M X (t)MY (t) = , t < 0.005, = 2 t 2 (1 − 200t) 1− λ which is the m.g.f. of the gamma distribution with α = 2 and β = 200.

Thus, X + Y has this distribution, and therefore its p.d.f. is given by: f X +Y (t) =

1 te−t/200 = (0.005)2 te−0.005t, t > 0.

(2)2002

(iii) From part (ii), and setting λ for 0.005, we have: P(X + Y ≥ 500) ∞ ∞ 2 −λt = λ te dt = −λ td e−λt 500 500 ∞ = −λte−λt |∞ + λ e−λt dt = 500λe−500λ − e−λt |∞ 500 500 500

= 500λe−500λ + e−500λ = 2.5e−2.5 + e−2.5

0.287.

1.13 (i) For y > 0, FY (y) = P(Y ≤ y) = P(− log X ≤ y) = P(log X ≥ −y) = P(X ≥ e−y ) = 1 − P(X < e−y ) = 1 − P(X ≤ e−y ) = 1 − e−y ; i.e., FY (y) = 1 − e−y , for y > 0, and FY (y) = 0, for y ≤ 0.

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d Hence, for y > 0, f Y (y) = dy (1 − e−y ) = e−y , so that Y has the negative exponential distribution with parameter λ = 1. Or, to put it differently, it has the gamma distribution with α = 1 and β = 1/λ. (ii) At this point, recall that, if the r.v. U has the gamma distribution with parameters α and β, then its m.g.f. is:

MU (t) =

1 1 , t< . α (1 − βt) β

In the case of the negative exponential distribution with parameter λ = 1 (i.e., α = 1 and β = λ1 = 1), the m.g.f. becomes: MU (t) =

1 , t < 1.

1−t

Therefore, by independence of the r.v.’s Y1 , . . . , Yn , we have: M Z (t) =

1 , t < 1, (1 − t)n

which is the m.g.f. of gamma with α = n and β = 1.

1.14 We have: σ12 (at)2

(aσ1 )2 t 2

(i) MU (t) = Ma X +b (t) = ebt M X (at) = ebt ×eμ1 (at)+ 2 = e(aμ1 +b)t+ 2 which is the m.g.f. of N (aμ1 +b, (aσ1 )2 ), so that U ∼ N (aμ1 +b, (aσ1 )2 ).

Likewise for V. (ii) For t1 , t2 ∈ and by independence of X and Y, we have: MU ,V (t1 , t2 ) = Eet1 U +t2 V = Ee(at1 )X +(bt1 )+(ct2 )Y +dt2 = E[ebt1 +dt2 × e(at1 )X × e(ct2 )Y ] = ebt1 +dt2 Ee(at1 )X × Ee(ct2 )Y = ebt1 +dt2 M X (at1 )MY (ct2 ) = ebt1 +dt2 × e(aμ1 )t1 + = e(aμ1 +b)t1 +

(aσ1 )2 t12 2

× e(cμ2 )t2 +

(aσ1 )2 t12 (cσ )2 t 2 +(cμ2 +d)t2 + 22 2 2

(cσ2 )2 t22 2

.

(iii) From the last expression on the right-hand side of part (ii), we have: (aσ1 )2 t12

MU ,V (t1 , t2 ) = e(aμ1 +b)t1 + 2 × e(cμ2 +d)t2 + = MU (t1 )MV (t2 ),

(cσ2 )2 t22 2

since MU (t1 ) = Ma X +b (t1 ) = ebt1 M X (at1 ) = ebt1 × e(aμ1 )t1 + = e(aμ1 +b)t1 +

(aσ1 )2 t12 2

,

(aσ1 )2 t12 2

Revised Answers Manual to Introduction to Probability

and likewise for MV . The factorization MU ,V (t1 , t2 ) = MU (t1 )MV (t2 ), t1 , t2 ∈ implies independence of the r.v.’s U and V. 1.15 (i) The joint p.d.f. of X and Y is: f X ,Y (x, y) = λ2 e−λ(x+y) , for x > 0, y > 0.

Therefore, for t > 0:

FT (t) = P(T ≤ t) = P(X + Y ≤ t) = λ2 e−λ(x+y) d x d y (x+y≤t)

t t−x

λ2 e−λx × e−λy d y d x t−x

t −λx −λy = λe λe dy d x 0 0

t t t−x −λx −λy λe λe dy d x = λe−λx (−e−λy |t−x = 0 )d x 0 0 0 t t = λe−λx (1 − e−λt+λx )d x = λe−λx d x − λte−λt =

0

0

0

0

= −e−λx |t0 − λte−λt = 1 − e−λt − λte−λt .

That is, FT (t) = 1 − e−λt − λte−λt (t > 0), so that f T (t) = λe−λt − λe−λt + λ2 te−λt = λ2 te−λt ; i.e., f T (t) = λ2 te−λt , t > 0, which is, actually, gamma distribution with α = 2 and β = 1/λ.

x+y=t

t

y=t −x

0

x

t

t

(ii) The m.g.f. of the r.v.’s X and Y is given by: M X (t) = MY (t) =

1 , t < λ, 1 − λt

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and by independence: MT (t) = M X +Y (t) = Eet(X +Y ) = E(et X × etY ) = (Eet X ) × (EetY ) 1 = M X (t) × MY (t) = 2 , 1 − λt which is the m.g.f. of gamma with α = 2 and β = 1/λ. Hence the distribution of T = X + Y is gamma with these parameters. (iii) The p.d.f. of T for λ = 13 is, by part (i): f T (t) =

1 −t/3 te , t > 0.

9

Then: 6

1 −t/3 1 6 dt = − t de−t/3 P(T ≤ 6) = te 3 0 0 9 6

6 1 1 6 −t/3 = − te−t/3 + e dt = −2e−2 − de−t/3 3 3 0 0 0 = −2e−2 − e−t/3 0 = −2e−2 − (e−2 − 1) = 1 − 3e−2 ; i.e., 6

P(T ≤ 6) = 1 − 3e−2

0.595.

1.16 We have: M X̄ (t) = M 1 n n

i=1

n X i (t) = M i=1 X i

" #n n t t t = = M M Xi . n n n i=1

1.17 (i) Since X and Y are B(15, 1/6), we have: Var(X ) = Var(Y ) = 15× 1 × 5 = 6 6 25 25 12 , so that, by independence: Var(X + Y ) = Var(X ) + Var(Y ) = 2 × 12 = (ii)

√ 5 6 25 , and s.d. of (X + Y ) = 2.041.

6 6 Since E X = EY = 15 × 16 = 2.5 and E(X + Y ) = 2 × 2.5 = 5, we have:

P(X + Y ≤ 10) = P(0 ≤ X + Y ≤ 10) = P[0 − 5 ≤ (X + Y ) − E(X + Y ) ≤ 10 − 5] = P[−5 ≤ (X + Y ) − E(X + Y ) ≤ 5] = P[|(X + Y ) − E(X + Y )| ≤ 5] 1 5 Var(X + Y ) 25/6 =1− = 0.833. ≥1− = 1− 2 5 25 6 6 1.18 Since E X i = p and Var(X i ) = pq, we have: (i) Always E X̄ = n1 × np = p, and by independence Var( X̄ ) = n12 × npq = pq n .

Revised Answers Manual to Introduction to Probability

√ 1 √ (ii) Since P(| X̄ − p| < 0.1 pq) ≥ 1 − (0.1pq/n = 1 − 0.01n = 1 − 100 n , it pq)2 suffices for 1 − 100 n ≥ 0.99, or n to be the smallest integer ≥ 10,000, so that n = 10,000.

1.19 From X i ∼ P(λ), it follows that E X i = Var(X i ) = λ. Then: (i) E X̄ = n1 × nλ = λ, and by independence, Var( X̄ ) = n12 × nλ = λn .

X̄ ) (ii) From P(| X̄ − λ| < c) ≥ 1 − Var( = 1 − λ/n = 1 − ncλ2 , it suffices for c2 c2 λ 1 − ncλ2 ≥ 0.95, or for n to be the smallest integer which is ≥ 0.05c 2 = 2 20λ/c . √ √ (iii) For c = λ, n = 20; for c = 0.1 λ, n = 2,000.

1.20 (i) Clearly, f X (−1) = 2α + β, f X (0) = 2β, f X (1) = 2α + β; f Y (−1) = 2α + β, f Y (0) = 2β, f Y (1) = 2α + β. (ii) By part (i), E X = EY = (−1)(2α +β)+1×(2α +β) = 0; and E(X Y ) = (−1) × (−1) × α + (−1) × 1 × α + 1 × (−1) × α + 1 × 1 × α = 0. (iii) From part (ii), Cov(X , Y ) = E(X Y ) − (E X )(EY ) = 0 − 0 = 0. (iv) The r.v.’s X and Y are not independent, because, for example, f (0, 0) = 0 = (2β) × (2β) = f X (0) f Y (0).

1.21 It was seen in Exercise 1.10 that the r.v.’s X , Y , and Z are independent and they all have the negative exponential distribution with parameter c. Then: E X = EY = E Z = 1/c and Var(X ) = Var(Y ) = Var(Z ) = 1/c2 .

By independence, it follows then that: E(X Y ) = (E X )(EY ) = 1/c2 ; E(X Y Z ) = (E X )(EY )(E Z ) = 1/c3 ; Var(X + Y ) = Var(X ) + Var(Y ) = 2/c2 ; Var(X + Y + Z ) = Var(X ) + Var(Y ) + Var(Z ) = 3/c2 .

1.22 (i) Always, E X̄ = 1 × nμ = μ, and by independence, Var( X̄ ) = 12 × nσ 2 = n n σ 2 /n.

2 X̄ ) (ii) From P(| X̄ − μ| < kσ ) ≥ 1 − Var( = 1 − σk 2 σ/n2 = 1 − nk1 2 , it suffices for (kσ )2

1 100 1 − nk1 2 ≥ 0.99, or for n to be the smallest integer which is ≥ 0.01k 2 = k2 .

For k = 1, n = 100; for k = 2, n = 25; and for k = 3, n = 12.

1.23 Since E X̄ = μ and Var( X̄ ) = σ 2 /n, the Tchebichev inequality gives: P(| X̄ − μ| < 0.5σ ) ≥ 1 −

4 σ 2 /n 1 =1− . =1− (0.5σ )2 0.25n n

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Thus, it suffices 1−

4 ≥ 0.99, or n ≥ 400, so that n = 400. n

1.24 (i) Always, E X̄ = n1 × nμ = μ, and by independence, Var( X̄ ) = 12 × n nσ 2 = σn .

2

X̄ ) (ii) Since P(| X̄ − μ| < cσ ) ≥ 1 − Var( = 1 − σc2 σ/n2 = 1 − nc12 , it suffices (cσ )2 2

1 1 − nc12 ≥ α, or for n to be the smallest integer which is ≥ (1−α)c 2.

1 100 = 1−α . Therefore: (iii) For c = 0.1, we obtain: n ≥ (0.1)21(1−α) = 0.01(1−α)

For α = 0.90, we have n = 1,000; for α = 0.95, we have n = 2,000; and for α = 0.99, we have n = 10,000.

1.25 The r.v.’s U and V have the bivariate normal distribution with correlation coefficient ρ0 = (σ12 − σ22 )/τ1 τ2 . Then, by Corollary 1 to Proposition 1, the r.v.’s U and V are independent if and only if ρ0 = 0, which happens if and only if σ12 = σ22 .

1.26 By Theorem 1(iii), it suffices to show that: MY ,Z (t1 , t2 ) = MY (t1 )M Z (t2 ) for all t1 , t2 belonging in a non-degenerate interval containing 0. Indeed, MY ,Z (t1 , t2 ) = Eet1 Y +t2 Z = Eet1 g(X i1 ,...,X im )+t2 h(X j1 ,...,X jn ) = E[et1 g(X i1 ,...,X im ) × et2 h(X j1 ,...,X jn ) ] ∞ ∞ ··· et1 g(xi1 ,...,xim ) × et2 h(x j1 ,...,x jn ) f X 1 ,...,X k (x1 , . . . , xk )d x1 . . . d xk = −∞

−∞

(for the continuous case) ∞ ∞ ··· et1 g(xi1 ,...,xim ) × et2 h(x j1 ,...,x jn ) f X 1 (x1 ) · · · f X k (xk )d x1 . . . d xk = −∞

−∞

(by independence of the X i ’s) ∞ ∞ = ··· [et1 g(xi1 ,...,xim ) f X i1 (xi1 ) . . . f X im (xim ) × et2 h(x j1 ,...,x jn ) × −∞

−∞

f X j1 (x j1 ) . . . f X jn (x jn )]d xi1 . . . d xim d x j1 . . . d x jn # " ∞ ∞ t1 g(xi 1 ,...,xi m ) = ··· e f X i1 (xi1 ) . . . f X im (xim )d xi1 . . . d xim × −∞ −∞ m integrals

" ∞

∞

··· e −∞ −∞

t2 h(x j1 ,...,x jn )

# f X j1 (x j1 ) . . . f X jn (x jn )d x j1 . . . d x jn

n integrals t1 g(X i 1 ,...,X i m )

= Ee × Eet2 h(X j1 ,...,X jn ) t1 Y t2 Z = Ee × Ee = MY (t1 )M Z (t2 ),

Revised Answers Manual to Introduction to Probability

as was to be seen. The integrals are replaced by summation signs in the discrete case.

1.27 (i) M X i (t) = 1/(1 − λt ), t < λ, i = 1, . . . , n, so that MU (t) =

1 , (1 − λt )n

which is the m.g.f. of gamma with α = n and β = λ1 . Therefore U is distributed as gamma with parameters α = n and β = λ1 . (ii) Since E X i = λ1 and σ 2 (X i ) = λ12 , it follows that EU = n/λ, σ 2 (U ) = n/λ2 . Alternatively, since U is gamma-distributed with parameters α = n and β = λ1 , it follows that EU = αβ = n/λ and σ 2 (U ) = αβ 2 = n/λ2 .

2 . (iii) For λ = 1/2, we have MU (t) = 1/(1 − 2t)2n/2 which is the m.g.f. of χ2n 2 .

Therefore U ∼ χ2n 2 , so that P(U > 20) = (iv) For λ = 1/2 and n = 15, we have that U ∼ χ30 2 0.914 (by χ 2 -tables and linear 1 − P(U ≤ 20) = 1 − P(χ30 ≤ 20) interpolation).

1.28 (i) U ∼ N (0, 2). (ii) Var(U ) = Var(X + Y ) = σ 2 (X ) + σ 2 (Y ) = 2 (by independence); 2 Cov(X , U ) = Cov(X , X + Y ) = √Cov(X , X ) + Cov(X , Y ) = σ (X ) = Cov(X ,Y ) 1 1; ρ(X , Y ) = σ (X )σ (Y ) = √ = 2/2. (iii) M X ,U (t1 , t2 ) Ee(t1 +t2 )X +t2 Y

(t1 +t2 )2 2

t22

t 2 +2t22 +2t1 t2 ), t1 , t2 ∈ .

2 2 t12 +2t22 +2t1 t2 ∂ = (t1 + t2 )exp( ) and ∂t∂1 ∂t2 M X ,U ∂t1 M X ,U (t1 , t2 ) 2 t 2 +2t 2 +2t t t 2 +2t 2 +2t t (t1 , t2 )|t1 =t2 =0 = exp( 1 22 1 2 )+(t1 +t2 )×exp( 1 22 1 2 )|t1 =t2 =0 =

independence) = e

(iv)

2

= Eet1 X +t2 U = Eet1 X +t2 (X +Y ) = (t +t )X t Y = Ee 1 2 × Ee 2 = M X (t1 + t2 )MY (t2 ) (by × e 2 = exp( 1

1. But the left-hand side above equals E(XU ). Thus, Cov(X , U ) = E(XU ) = 1, as it should be.

1.29 (i) By independence of X and Y , f X ,Y (x, y) = λ1 λ2 e−λ1 X −λ2 Y , x > 0, y > 0, so that: ∞ y λ1 λ2 e−λ1 x−λ2 y d x d y P(X < Y ) = 0 0 y ∞ −λ2 y λ2 e ( λ1 e−λ1 x d x)dy = 0 0 ∞ y λ2 e−λ2 y (−e−λ1 x |0 )dy = 0 ∞ λ2 e−λ2 y (1 − e−λ1 y )dy = 0

e109

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∞ λ2 (λ1 + λ2 )e−(λ1 +λ2 )y dy λ1 + λ2 0 λ2 λ1 = 1− ; i.e., P(X < Y ) = . λ1 + λ2 λ1 + λ2

= 1−

λ (ii) Here P(X < Y ) = 4λ = 41 = 0.25.

1 1 1.30 (i) From c −1 d x −1 dy = 1, we get c = 41 and f X ,Y (x, y) = 41 , −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.

1 (ii) f X (x) = 41 −1 dy = 21 , −1 ≤ x ≤ 1, and similarly, f Y (y) = 21 , −1 ≤ y ≤ 1. (iii) Since f X ,Y (x, y) = 41 = 21 × 21 = f X (x) f Y (y) for all x and y as specified above, it follows that X and Y are independent. (iv) The probability P(X 2 + Y 2 ≤ 1) is simply the area of the circle of radius π 2 2 1 centered at 0, divided by 4; i.e., 4 . In more!detail, P(X + Y ≤ 1) = √ 1 1−x 2 1 1 d x d y = 4 −1 −√1−x 2 dy d x = I = P(X 2 + Y 2 ≤ 1) = 4

(x 2 +x 2 ≤1) √ 1 1 2 2 −1 1 − x d x.

Set x = sin θ, − π2 ≤ θ ≤ π2 . Then:

π π π 1 2 1 2 1 2 2 cos θ d sin θ = cos θ dθ ; i.e., I = cos2 θ dθ. But I = 2 − π2 2 − π2 2 − π2 π π π π 2 2 2 2 2 cos θ dθ = cos θ d sin θ = sin θ cos θ |− π + sin2 θ dθ − π2

− π2

π

=

2

− π2 π 2

− π2

2

π

(1 − cos2 θ )dθ = π −

cos2 θ dθ =

2

− π2

− π2

cos2 θ dθ, so that

π π . It follows that I = .

2 4

Section 10.2 2.1 Since the r.v.’s X and Y are independent and distributed as B(15, 1/6), it 30 1 t follows that X + Y ∼ B(30, 1/6) and P(X + Y ≤ 10) = 10 t=0 t 6 × !30−t 5 .

6 2.2 (i) By independence, X + Y ∼ B(m + n, p). (ii) Here X + Y ∼ B (20, 0.25), and hence (from the binomial tables): P(5 ≤ X + Y ≤ 15) =

15

20 (0.25)t (0.75)20−t t t=5

= 1.0000 − 0.4148 = 0.5852.

Revised Answers Manual to Introduction to Probability

2.3 (i) By independence, Sn ∼ B(n, p). (ii) Since X i ∼ B(1, p), it follows that E X i = p, Var(X i ) = pq (q = 1 − p), i = 1, . . . , n. (iii) Always, E Sn = np, and by independence, Var(Sn ) = npq.

2.4 (i) Clearly, p = P(X 1 ∈ I ) = I f (x)d x, for the continuous case, and p = x∈I f (x), for the discrete case. (ii) Define the r.v.’s Y1 , . . . , Yn as follows: Yi = 1 if X i ∈ I , and Yi = 0 otherwise. Then the r.v.’s Y1 , . . . , Yn are also independent and their common distribution is B(1, p). Thus, if Y = Y1 + · · · + Yn , then Y ∼ B(n, p), and the question is rephrased as follows: P(at least k of X 1 , . . . , X k take values n y n−y (q = in I ) = P(Y ≥ k) = 1 − P(Y ≤ k − 1) = 1 − k−1 y=0 y p q 1 − p). ∞ −1 = 1/e, and hence the (iii) Here p = 1/λ λe−λx d x = −e−λx |∞ 1/λ = e n−y k−1 n 1 y 1 − 1e . probability in part (ii) is equal to: 1 − y=0 y e (iv) The required probability is:

1 1 4 1 3 −4 1− 1− 1− e e e

0.469.

2.5 Let X be the r.v. denoting the breakdown voltage of the diode in question. Then X ∼ N (40, 1.5). Therefore: X −μ (i) P(39 < X < 42) = P( 39−40 < 42−40 P(−0.67 < Z < 1.5 < σ 1.5 ) 1.33) = (1.33) − (−0.67) = (1.33) + (0.67) − 1 = 0.908241 + 0.748571 − 1 = 0.656812. (ii) If X i , i = 1, . . . , 5 are independent copies of X , then define the r.v.’s Yi , i = 1, . . . , 5 by:

Yi = 1 if X i > 42 and Yi = 0 if X i ≤ 42.

Then the r.v.’s Y1 , . . . , Y5 are independent and distributed as B(1, p), where p = P(X 1 > 42), and Y = Y1 + · · · + Y5 ∼ B(5, p). However, p = P(Z > 42−40 P(Z > 1.33) = 1 − (1.33) = 1 − P(X 1 > 42) 1.5 ) 0.908241 0.091759 and therefore P (at least one has voltage exceeding 42) = P(Y ≥ 1) = 1 − P(Y = 0) = 1 − 05 (0.091759)0 (0.908241)5 = 1 − (0.908241)5 1 − 0.618 = 0.382.

Remark: To be precise, one should write: p = P(X 1 > 42) = 1 − P(X 1 ≤ 42) = 1 − P(0 ≤ X 1 ≤ 42) = 1 − P(0 < X 1 ≤ 42) = 1 − (1.33) + (−26.67) = 1 − (1.33), which leaves the result unchanged.

2.6 (i) Since X i ∼ B(1, p), i = 1, . . . , n, and these r.v.’s are independent, it follows that X = X 1 + · · · + X n ∼ B(n, p).

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(ii) From f (x) = nx p x q n−x from the Poisson tables: f (0) f (1) f (2) f (3) f (4)

−2 2 , x = 0, 1, . . . , we get e−np (np) x! = e x! x

0.1353 0.4060 − 0.1353 0.6767 − 0.4060 0.8571 − 0.6767 0.9473 − 0.8571

x

= 0.1353 = 0.2707 = 0.2707 = 0.1804 = 0.0902.

2.7 (i) Here X = X 1 + · · · + X n ∼ P(λ1 + · · · + λn ). (ii) Since E X i = Var(X i ) = λi , i = 1, . . . , n, we have: E X̄ =

1 1 (λ1 + · · · + λn ), Var( X̄ ) = 2 (λ1 + · · · + λn ). n n

(iii) For λ1 = · · · = λn = λ, say, E X̄ = λ and Var( X̄ ) = λ/n.

2.8 Let X i be the r.v. denoting the number of no-shows in the ith flight, and assume (reasonably enough) that the r.v.’s X 1 , . . . , X 5 are independent. Then the r.v.

X = X 1 + · · · + X 5 ∼ P(5λ), and since λ = 2, we have that X ∼ P(10).

Therefore, we obtain from the Poisson tables: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)

P(X = 0) = e−10 0.

P(X ≤ 5) = 0.0671.

P(X = 5) = 0.0671 − 0.0293 = 0.0378.

P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.0293 = 0.9707.

P(X ≤ 10) = 0.538.

P(X = 10) = 0.5838 − 0.4579 = 0.1259.

P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − 0.4579 = 0.5421.

P(X ≤ 15) = 0.9513.

P(X = 15) = 0.9513 − 0.9165 = 0.0348.

P(X ≥ 15) = 1 − P(X ≤ 14) = 1 − 0.9165 = 0.0835.

2.9 (i) For x1 = 0, 1, . . . , t, we have:

n X i=t P X 1 = x1 , i=1 P(X 1 = x1 , T = t) = P(X 1 = x1 |T = t) = P(T = t) P(T = t)

n P X 1 = x1 , i=2 X i = t − x1 = P(T = t)

n P(X 1 = x1 )P i=2 X i = t − x 1 . = P(T = t) n However, T ∼ P(λ) and i=2 X i ∼ P(λ − λ1 ). Therefore the last expression above becomes:

−λ x1 e 1 λ1 /x1 ! [e−(λ−λ1 ) (λ − λ1 )t−x1 /(t − x1 )!] = e−λ λt /t!

Revised Answers Manual to Introduction to Probability

x1

λ − λ1 t−x1 λ1 t! × e−λ1 −λ+λ1 +λ x1 !(t − x1 )! λ λ

x1 λ1 λ1 t−x1 t 1− = , as was to be seen. λ λ x1 =

(ii) It follows immediately from the derivations in part (i) and symmetry. (iii) In this case, λ = nc and therefore the conditional distribution is B(t, n1 );

i.e., P(X 1 = x1 |T = t) = xt1 ( n1 )x1 (1 − n1 )t−x1 .

2.10 By Exercise 1.14(i) here, −Y ∼ N (−μ2 , σ22 ). Then: (i) By independence of X and −Y , we have that X −Y ∼ N (μ1 −μ2 , σ12 +σ22 ).

Therefore: " (ii) (X − Y ) − (μ1 − μ2 ) $ > P(X > Y ) = P(X − Y > 0) = P σ12 + σ22 # μ1 − μ2 −$ σ12 + σ22

μ1 − μ2 μ1 − μ2 =1− −$ . = P Z > −$ σ12 + σ22 σ12 + σ22 (iii) When μ1 = μ2 , then

(− $μ12−μ22 ) = σ1 +σ2

P(X > Y ) = 0.5 (regardless of σ1 and σ2 ).

2.11

σ12

(0) = 0.5, and hence

σ22

(i) Here X̄ ∼ N (μ1 , m ), Ȳ ∼ N (μ2 , n ) and X̄ , Ȳ are independent, so that σ2

σ2

X̄ − Ȳ ∼ N (μ1 − μ2 , m1 + n2 ). Therefore: P( X̄ > Ȳ ) = P( X̄ − Ȳ > 0) ⎤ ⎡ ( X̄ − Ȳ ) − (μ − μ ) − μ μ 1 2 1 2 ⎦ $ 2 > −$ 2 = P⎣ σ1 σ22 σ1 σ22 + n m + n ⎛ ⎞ ⎛m ⎞ μ1 − μ2 ⎠ μ1 − μ2 ⎠ = P⎝Z > − $ 2 = 1 − ⎝− $ 2 . σ1 σ22 σ1 σ22 m + n m + n (ii) For μ1 = μ2 , 3μ12−μ22 = 0, regardless of the values of m, n and σ1 , σ2 , so σ1 σ2 m + n

that: P( X̄ > Ȳ ) = 1 −

(0) = 1 −

1 1 = = 0.5.

2 2

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2.12 (i) Here X i ∼ N (μ, σ 2 ) implies αi X i ∼ N (αi μ, (αi σ )2 ), and the r.v.’s αi X i , i = 1, . . . , n are independent, so that:

n n

2 2 X∼N μ αi , σ αi ; i=1

and likewise,

i=1

n n

2 2 Y ∼N μ βj, σ βj . j=1

j=1

(ii) For t1 , t2 ∈ , M X ,Y (t1 , t2 )

n n

αi X i + t2 βi X i = E exp(t1 X + t2 Y ) = E exp t1 = E exp

" n

#

i=1

(αi t1 + βi t2 )X i = E

i=1

= = =

n

i=1 n

i=1 n

i=1

n

i=1

exp(αi t1 + βi t2 )X i

i=1

E exp(αi t1 + βi t2 )X i (by independence of the X i ’s) M X i (αi t1 + βi t2 ) # " 1 exp μ(αi t1 + βi t2 ) + σ 2 (αi t1 + βi t2 )2 2 σ 2t2

(by the fact that the m.g.f. of N (μ, σ 2 ) is: eμt+ 2 ) % n & n

1 2 2 = exp μ(αi t1 + βi t2 ) + σ (αi t1 + βi t2 ) 2 i=1 i=1

"

n n n 1 2 2 2 σ αi t1 + μ βi t2 + αi t1 = exp μ 2 i=1 i=1 i=1

#2

n n +2 σ2 αi βi t1 t2 + σ 2 βi2 t22 . i=1

i=1

At this point, set: μ

n

i=1

αi = μ1 , μ

n

i=1

βi = μ2 , σ 2

n

i=1

αi2 = σ12 , σ 2

n

i=1

βi2 = σ22 .

Revised Answers Manual to Introduction to Probability

Then: σ2

n

αi βi

i=1

" 4 1/2 1/2 # 1/2 n n n n = αi βi αi2 × βi2 αi2 σ i=1

i=1

i=1

4 n

1/2

n 2 ×σ βi =

" n i=1

αi βi

i=1

αi2

1/2 n

i=1

= ρσ1 σ2 , where ρ =

n i=1

i=1

βi2

1/2 # σ1 σ2

i=1

αi βi

4 n

αi2

i=1

1/2 n

1/2 βi2

.

i=1

Thus,

# "

1 2 2 2 2 M X ,Y (t1 , t2 ) = exp μ1 t1 + μ2 t2 + σ1 t1 + 2ρσ1 σ2 t1 t2 + σ2 t2 .

2

(iii) This is the m.g.f. of the bivariate normal distribution with parameters μ1 , μ2 , σ1 , σ2 , and ρ(X , Y ) = ρ. (See, e.g., relation (11) in Chapter 6 of the book A Course in Mathematical Statistics, 2nd edition (1997), Academic Press, by G. G. Roussas.) n (iv) Then the r.v.’s X and Y are independent if and only if ρ = 0, or i=1 αi βi = 0, from the definition of ρ, as was to be seen.

2

2.13 From X ∼ N (0, σ 2 ), we have that σX ∼ N (0, 1) and σX2 ∼ χ12 , and likewise, Y2 ∼ χ12 . Then, by independence, σ2

X2 + Y 2 ∼ χ22 . Therefore: σ2 (i) For 1

X 2 + Y 2 ≤ r = P(X 2 + Y 2 ≤ r 2 )

2 r2 r2 X + Y2 2 = P χ . ≤ ≤ =P 2 σ2 σ2 σ2

r > 0, P(R ≤ r ) = P

(ii) For σ = 1, P(R ≤ r ) = P(χ22 ≤ r 2 ). Then, for the given values of r , we get: r2 r2

2.772, r 2 4.605, r 2 5.993, r 2 = 7.377, 9.211, and r 2 10.595.

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Then the χ 2 -tables provide the respective probabilities: 0.75, 0.90, 0.95, 0.975, 0.99, and 0.995.

2.14 (i) Since P(| X̄ − μ| < cσ ) ≥ 1 − σ 2 ( X̄ n ) = 1 − 1 , it suffices 1 − 1 ≥ p n (cσ )2 nc2 nc2 1 1 , and n is the smallest integer ≥ c2 (1− . or n ≥ c2 (1− p) p)

(ii) For p = 0.95 and c = 1, we have n = 20, whereas for c = 0.5, we get: n = 80, and for c = 0.25, we have n = 320.

2.15 (i) By independence, M X (t) = 1/(1 − βt)α , α = α1 + α2 + α3 (t < 1/β), and this is the m.g.f. of the gamma distribution with parameters α and β.

It follows that E X = αβ and Var(X ) = αβ 2 . (ii) Here E X = 2(10 + 100 + 1,000) = 2,220, σ 2 (X ) = 4 × 1,110, and √ σ (X ) = 2 1,110 66.634.

2.16 We have X ∼ N (μ, 2σ 2 ), Y ∼ N (μ, σ 2 ) independent, so that X + Y ∼ 3σ N (2μ, 3σ 2 ) and X +Y 2 ∼ N (μ, 4 ). Therefore

X +Y − μ ≤ 1.5σ P 2 X +Y = P(−1.5σ ≤ − μ ≤ 1.5σ ) 2

#

" X +Y 2 × 1.5σ 2 −2 × 1.5σ −μ ≤ ≤ √ =P √ √ 2 σ 3 σ 3 σ 3 √ √ = P(− 3 ≤ Z ≤ 3), where Z ∼ N (0, 1), √ = 2 ( 3) − 1 2 (1.73) − 1 = 2 × 0.958185 − 1 = 0.91637.

2

Section 10.3 3.1 (i) The first equation follows from the definition of S 2 . As for the second, we have: nS 2 =

n

(X i − X̄ )2 =

i=1

=

n

i=1 n

i=1

=

n

i=1

X i2 − 2 X̄ X i + X̄ 2

X i2 − 2 X̄

n

X i + n X̄ 2

i=1

X i2 − 2 X̄ × n X̄ + n X̄ 2 =

n

i=1

X i2 − n X̄ 2 .

Revised Answers Manual to Introduction to Probability

(ii) Indeed n

(X i − μ)2 =

i=1

n

[(X i − X̄ ) + ( X̄ − μ)]2 i=1

n

= [(X i − X̄ )2 + 2( X̄ − μ)(X i − X̄ ) + ( X̄ − μ)2 ] i=1 n

= (X i − X̄ )2 + 0 + n( X̄ − μ)2 i=1

since ×

n

2( X̄ − μ)(X i − X̄ ) = 2( X̄ − μ)

i=1

n

X i − n X̄

= 2( X̄ − μ)(n X̄ − n X̄ ) = 0

i=1

= nS 2 + n( X̄ − μ)2 (by the definition of S 2 ). (iii) From part (ii) and the definition of S 2 , we have: 1 1 (X i − μ)2 = (X i − X̄ )2 + ( X̄ − μ)2 . n n n

n

i=1

i=1

n

2 2 i=1 (X i − μ) ] = σ and E X̄ = μ, so that σ2 2 E( X̄ − μ) = Var( X̄ ) = n . Therefore, taking the expectations in the

By Proposition 4, E[ n1

above identity, n we have: 2 n 2 (X i − X̄ ) ] + σn , or nσ 2 = E[ i=1 (X i − X̄ )2 ] + σ 2 , or σ 2 = E[ n1 i=1 n

n 1 2 E[ i=1 (X i − X̄ )2 ] = (n − 1)σ 2 , and hence E[ n−1 i=1 (X i − X̄ ) ] = 2 σ .

Chapter 11 Section 11.1 1.1 The r.v. Y takes on the values y = x 3 , x = 0, 1, . . . ; i.e., y = 0, 1, 23 , 33 , . . . , 1/3 and f Y (y) = P(Y = y) = P(X = x) = P(X = y 1/3 ) = (1 − α)α y .

1.2 (i) In the first place, by a number of ways, we have that X ∼ N ( μ−160 , 25σ 2 ).

9

81

(ii) 0.95 = P(90 ≤ Y ≤ 95) = P(90 ≤ 95 X + 32 ≤ 95) = P( 290 9 ≤ X ≤ 315 ) and a 32.222, b = 35.

9 (iii) pk = P(μ − kσ ≤ Y ≤ μ + kσ ) = P(μ − kσ ≤ 95 X + 32 ≤ μ + kσ ) = −160 −160 ≤ X ≤ 5μ+5kσ ), and ak = 5μ−160 − k 5σ P( 5μ−5kσ 9 9 9 9 , bk =

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1.3 (i) (ii)

5μ−160 + k 5σ 9 9 . We could arrive at the same result directly from the relation pk = P(E X − kτ ≤ X ≤ E X + kτ ), where τ = s.d. of X and observing 5μ−160 , s.d. of X = 5σ that E X = 59 EY − 160 9 = 9 9 .

1 b Here u = ax + b and hence x = a u − a , so that ddux = a1 . Then fU (u) = 1 u−b a f X ( a ), u ∈ T , the image of S under the transformation u = ax + b.

Now S = (0, ∞) and T = (b, ∞). Then, for u > b, fU (u) = a1 x × ∞ λ −λu b λ λ a du λ exp(−λ u−b = b ae a ) = exp( a ) × a exp(− a u). Since − λa u ∞ −b/a −e = e , it follows that fU is the negative exponential p.d.f., b

over the interval (b, ∞), with parameter λ/a.

1 , for x ∈ S = (c, d), we have: (iii) If X ∼ U (c, d), so that f X (x) = d−c 1 1 fU (u) = a(d−c) = (ad)−(ac) , for ac + b < u < ad + b, or for u ∈ T = (ac + b, ad + b), so that U ∼ U (ac + b, ad + b).

1.4 From y = e x , we have x = log y, ddyx = 1y , y > 1, and hence f Y (y) =

1.5

1 λ λ −λ log y = −(λ+1) , y > 1; from z = log x, we have λ = y×y λ = λy y × λe yelog y z z x = e z , ddzx = e z , and hence f Z (z) = e z × λe−λe = λe z−λe , z ∈ .

1 From y = e x , we get x = log y, ddyx = 1y , and then f Y (y) = (β−α)y , eα < y < ez eβ . From z = log x, we get x = e z , ddzx = e z , and f Z (z) = β−α , log α < z <

log β.

1.6 (i) From y = −2 log x, we have x = e−y/2 , ddyx = − 21 e−y/2 , so that f Y (y) =

1 −y/2 , y > 0, which is the p.d.f. of a χ22 -distributed r.v.

2e (ii) From part (i), Yi ∼ χ22 , i = 1, . . . , n, and they are independent. Hence n 2 i=1 Yi has the χ2n distribution by Theorem 5 in Chapter 10.

1.7 From y = x1 (x = 0), we have x = 1y , ddyx = − y12 , and therefore f Y (y) = 2

y 1 × √1 y 2 e− 2 y2 2π

= √1 e−y /2 , y ∈ , which is the p.d.f. of the N (0, 1) 2

2π

distribution. $ $ √ 1 , so that, 1.8 From y = 21 mx 2 , we have x = m2 y, ddyx = m2 × 2√1 y = √2my 3 for y > 0: 2y −y/m 2 1 2 × e × f Y (y) = √ = √ y 1/2 e−y/m π m m πm 2my 3 1 −y/m = 3 y 2 −1e , since 2 m 3/2

√

1 1 1 3 π = 1+ = =

2 2 2 2 2 (by relations (6.14) and (6.16) in Chapter 6).

Thus, f Y (y) =

3 1 y 2 −1 e−y/m , y > 0, 3 3/2 2 m

which is the gamma distribution with parameters α = 23 and β = m.

Revised Answers Manual to Introduction to Probability

√ √ 1.9 For y > 0, FY (y) = P(Y ≤ y) = P(X 2 ≤ y) = P(− y ≤ X ≤ y) = √ √ √ √ P(X ≤ y) − P(X ≤ − y) = FX ( y) − FX (− y), so that d √ √ d √ √ d FY (y) = f X ( y) ( y) − f X (− y) (− y) dy dy dy 1 −y/2 1 −y/2 1 1 = √ ×√ e + √ ×√ e 2 y 2 y 2π 2π 1 1 = √ e−y/2 = √ 1/2 y 1/2−1 e−y/2 π2 2π y 1 = y 1/2−1 e−y/2 (1/2)21/2

f Y (y) =

√ (since π = (1/2), by relation (6.16) in Chapter 6), and this is the p.d.f. of a χ12 -distributed r.v.

1.10 For y > 0,

1 2 mX ≤ y FY (y) = P(Y ≤ y) = P 2 ' 3 3 ( 2y 2y 2y 2 )= P − ≤X≤ = P(X ≤ m m m ' 3 ( 2y =P X≤ (since X ≥ 0 with probability 1) m '3 ( '3 ( 2y 2y d d FY (y) = FX = FX , so that f Y (y) = m dy dy m '3 ( 3 2y −y/m 2y 2 1 1 × e fX = √ =√ m m 2my 2my π = √ =

3 2 1 √ −y/m ye = 1√ y 2 −1 e−y/m √ 3/2 πm m π m 2

3 1 y 2 −1 e−y/m (3/2)m 3/2

(since by relations (6.14) and (6.16) in Chapter 6, ( 23 ) = ( 23 −1)× ( 23 −1) = 1 1 1√ 3 2 ( 2 ) = 2 π ), and this is the p.d.f. of gamma with α = 2 and β = m.

1 x 1.11 (i) Since f X (x) = t −t , t1 ≤ x ≤ t2 , and FX (x) = t −t , t1 ≤ x ≤ t2 , we 2 1 2 1 have, for y > 0: 1 1 ≤ y) = P(X ≥ ) X y 1 1 = 1 − P(X < ) = 1 − P(X ≤ ) y y

FY (y) = P(Y ≤ y) = P(

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1 1 , and therefore : = 1 − FX ( ) = 1 − y (t2 − t1 )y d 1 f Y (y) = FY (y) = . dy (t2 − t1 )y 2 (ii) Here, FY (y) =

0, y≤0 , 1 , y>0 1 − 10y

f Y (y) =

1 , 10y 2

0,

y>0 . y≤0

Section 11.2 2.1 The sample space here consists of 36 pairs (x, y), x = 1, . . . , 6, y = 1, . . . , 6, each occurring with probability 1/36, and Z takes on the values 2, . . . , 12 as follows: 2: (1, 1) 3: (1, 2), (2, 1) 4: (1, 3), (2, 2), (3, 1) 5: (1, 4), (2, 3), (3, 2), (4, 1) 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) 9: (3, 6), (4, 5), (5, 4), (6, 3) 10: (4, 6), (5, 5), (6, 4) 11: (5, 6), (6, 5) 12: (6, 6).

z−1 13 − z , z = 2, . . . , 7 and f Z (z) = , z = 8, . . . , 12.

36 36 2.2 (i) From the transformations: u = x + y, v = xy , we get x = vy, u = vy + y = u uv uv and then x = 1+v . So, x = 1+v , y = (1 + v)y and hence y = 1+v u 1+v , u > 0, v > 0. Hence f Z (z) =

v ∂x u 1 ∂y u ∂x ∂y = , = = , =− ; , ∂u 1 + v ∂v (1 + v)2 ∂u 1 + v ∂v (1 + v)2 so that J=

v 1+v 1 1+v

u (1+v)2 u − (1+v) 2

=−

u . Also, x + y = u.

(1 + v)2

u Therefore fU ,V (u, v) = e−u , u > 0, v > 0.

2 (1 + v) ∞ 1 1 1 (ii) ue−u du = ×1= , v > 0; f V (v) = 2 2 (1 + v) 0 (1 + v) (1 + v)2 ∞ ∞ dv −u = ue (1 + v)−2 dv fU (u) = ue−u (1 + v)2 0 0 1 ∞ = −ue−u = ue−u ; i.e., fU (u) = ue−u , u > 0.

1+v 0

Revised Answers Manual to Introduction to Probability

(iii) Clearly, for all u and v, fU ,V (u, v) = fU (u) f V (v), so that U and V are independent.

2.3 (i) From u = 21 (x + y), v = 21 (x − y), we get: x + y = 2u, x − y = 2v and hence: x = u + v, y = u − v. Since x > 0 and y > 0, we have v > −2u and v < u, whereas u(= 21 (x + y)) > 0. Thus, the range of u and v is defined by: u > 0, and for each such u, −u < v < u.

Next, J=

1 1 = −2, and therefore : 1 −1

fU ,V (u, v) = 2 × e−(u+v) × e−(u−v) = 2e−2u , −u < v < u, u > 0. (ii) Since for each u > 0, we have −u < v < u, it follows that: u dv = 4e−2u , u > 0. fU (u) = 2e−2u −u

Split the relation −u < v < u into two as follows: 0 < v < u and −u < v < 0 or 0 < −v < u. Then, for v > 0, ∞ −2v f V (v) = 2e−2u du = −e−2u |∞ , and for v < 0, v =e v ∞ 2v 2e−2u du = −e−2u |∞ f V (v) = −v = e , as was to be seen. −v

u b 2.4 (i) Here u = ax + b, v = cy + d, (x, y) ∈ S, so than x = u−b a = a − a, y = v−d v d c = c − c , (u, v) ∈ T , the image of S under these transformations. ∂x 1 ∂x Next, ∂u = a , ∂v = 0, ∂∂uy = 0, ∂∂vy = 1c , so that the Jacobian is 1/ac. It follows that:

u−b v−d 1 f X ,Y , fU ,V (u, v) = |ac| a c

u−b v−d 1 fX fY , (u, v) ∈ T . = |ac| a c

(ii) Here S = T = 2 and 2 4 # " 1 1 u−b fU ,V (u, v) = × − μ1 exp − 2σ12 |ac| 2π σ1 σ2 a " 2 4 # v−d 2 × exp − μ2 2σ2 c

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Revised Answers Manual to Introduction to Probability

= √

1

exp{−[u − (aμ1 + b)]2 /2(aσ1 )2 } 2π |a|σ1 1 ×√ exp{−[v − (cμ2 + d)]2 /2(cσ2 )2 }, 2π |c|σ2

so that U and V are independent and distributed as N (aμ1 + b, (aσ1 )2 ) and N (cμ2 + d, (cσ2 )2 ), respectively.

2.5 (i) From independence of X and Y , we have U ∼ N (0, 2), V ∼ N (0, 2). (ii) Consider the transformations: u = x + y, v = x − y, from which: x=

1 1 (u + v), y = (u − v), and J = 2 2

1 2 1 2

1 2 1 −2

1 = − , so that : 2

2 1 1 1 2 2 fU ,V (u, v) = × exp − [(u + v) + (u − v) ] 2 2π 8 2 2 u v 1 exp − exp − = 2 × 2π 4 4 # # " " 1 1 u2 v2 = √ √ exp − √ × √ √ exp − √ .

2π 2 2( 2)2 2π 2 2( 2)2 Then U and V are also independent. (iii) Therefore P(U < 0, V > 0) = P(U < 0)P(V > 0) = 41 = 0.25. √ √ 2.6 (i) From u = √1 (x + y), v = √1 (x − y), we get: x + y = 2u, x − y = 2v, 2

and hence x =

2 √ 2 (u + v), y = 22 (u − v). Furthermore, 2

√

J =

√

√

√

√

2 2 2 2

2 2

= −1, so that

− 22

" "√ #2 " √ #2 2# 1 2 2 1 fU ,V (u, v) = exp − (u + v) + (u − v) 2π 2 2 2 " # 1 1 exp − (u 2 + v 2 ) = 2π 2 1 2 1 2 1 1 = √ e− 2 u × √ e− 2 v , u, v ∈ .

2π 2π (ii) From the last expression above, it follows that U ∼ N (0, 1) and V ∼ N (0, 1). (iii) From the factorization in the same expression above, it also follows that U and V are independent.

Revised Answers Manual to Introduction to Probability

(iv) By independence of X and Y , we have directly: 1 X + Y ∼ N (0, 2) and hence U = √ (X + Y ) ∼ N (0, 1). Likewise, 2 1 X − Y ∼ N (0, 2) and hence V = √ (X − Y ) ∼ N (0, 1).

2 2

2

2

2.7 X ∼ N (0, σ 2 ) implies σX ∼ N (0, 1) and σX2 ∼ χ12 . Likewise, Yσ 2 ∼ χ12 and σX2 ,

2.8

2 2 Y2 are independent, so that X σ+Y = V ∼ χ22 . Thus, f V (v) = 21 e−v/2 , v > 0.

2 σ2 dv Now, set U = σ 2 V . Then from u = σ 2 v, we have du = σ12 and fU (u) = 2 2 1 × 21 e−u/2σ = 2σ1 2 e−u/2σ , as was to be seen. σ2 √ Consider the r.v. Z = + Z 2 and a r.v. with values in (0, 2π ]. Then: X =

Z cos , Y = Z sin . Look at the transformations: x = z cos θ, y = z sin θ from which we get: ∂∂zx = cos θ, ∂∂θx = −z sin θ ; ∂∂zy = sin θ, ∂∂θy = z cos θ , so cos θ −z sin θ = z, and therefore f Z , (z, θ ) = π1 × z, 0 ≤ z ≤ that J = sin θ z cos θ 2π 1, 0 < θ ≤ 2π . Then: f Z (z) = 0 πz dθ = πz × 2π = 2z, 0 ≤ z ≤ 1. Finally, for 0 < z ≤√1, set U = Z 2 and consider the transformation u = z 2√from which dz = 2√1 u . Therefore fU (u) = 2 u × 2√1 u = we get z = u, 0 < u ≤ 1, and du 1. So, fU (u) = f Z 2 (u) = 1, 0 ≤ u ≤ 1.

2.9 The p.d.f. f X r of X r is given by:

r +1 1 2 f X r (t) = √ (r +1)/2 . r × 2 πr 2 1+ t r

Since

r +1 #

"

t2 2 t 2 r 1/2 t 2 1/2 2 1+ = 1+ × 1+ → et /2 r r r

(as r → ∞), it suffices to consider the first term in the product above. At this point, employ the Sterling formula, which states that: (n) → 1 as n → ∞, √ 2π n (2n−1)/2 e−n r and apply it for n = r +1 2 and n = 2 successively in order to get: r +1 r /2 − r +1

√ r +1 r /2 − r +1

r +1 e 2 e 2 r +1 2 / 2π 2 2 r2 = × . r −1 r −1 √

r 2 r 2 − r2 r2 / 2π r2 2 e− 2 e 2

The first factor on the right-hand side above converges to 1. Thus, it suffices to study the second factor. We have: r +1 r /2 − r − 1 r +1 r /2 − r +1 √ # " e 2 e 2e 2 r 1 r 1/2 2 2 = = 1 + . √ r r /2 r −1/2 r r −1 − r 1/2 −r /2 r 2e 2 e 2 e 2 2 2

e123

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Revised Answers Manual to Introduction to Probability

√ √ The r cancels with the r which appears in f X r (t) and [(1+ r1 )r ]1/2 −→ e1/2 , r →∞

which cancels with the e1/2 in the denominator above. Thus, f 2.10 We have here: f X r (t) =

( r +1 ) √ 2 r πr ( 2 )

×

1 , t ∈ .

2 (1+ tr )(r +1)/2

1 (i) For r = 1, the p.d.f. reduces to f X 1 (t) = π1 × 1+t 2 , t ∈ , because √ 1 ( 2 ) = π, by relation (6.16) in Chapter 6. Then, it is known (see Exercise ∞ dt 1 1.16 in Chapter 5) that −∞ t × π1 × 1+t 2 = 2π (∞ − ∞), so that the E X 1 does not exist. ( r +1 ∞ tdt 2 ) (ii) For r ≥ 2, E X r = √πr ( and r t 2 (r +1)/2 ) −∞ (1+ r )

2

2 d 1 + tr

2 (r +1)/2 −∞ 1 + t r

r +1

t2 − 2 r ∞ t2 1+ = d 1+ 2 −∞ r r − r +1 +1 ∞ 2 2 t r d 1+ = r +1

r 2 − 2 + 1 −∞

∞

t dt r =

2 (r +1)/2 t 2 −∞ 1 + r

∞

∞

=−

1 r × = 0, r − 1 1 + t 2 r −1 2 −∞ r

so that E X r = 0. ∞

r +1

∞

r +1

(iii) Next, look at the integral −∞ t 2 ×(1+ tr )− 2 dt = 2 0 t 2 (1+ tr )− 2 dt, by symmetry with regard to the region of integration and the fact that √ 2 the integrand is an even function. Then set: tr = x, so that t = r x 2

2

√

rdx , and 0 < x < ∞. Therefore the integral equals: (t > 0), dt = 2√ x

∞ 2

r x(1 + x)

0

− r +1 2

√ ∞ √ r +1 r x 1/2 (1 + x)− 2 d x. √ dx = r r 2 x 0

dy 1 At this point, set 1+x = y, so that x = 1−y y , d x = − y 2 , and y ∈ (0, 1).

Also, 1 + x = 1y . Then the integral equals: √ r r

− r +1 1 2 √ r −4 (1 − y)1/2 1 dy × × = r r (1 − y)1/2 y 2 dy 1/2 2 y y y 0 0 1 √ r −2 3 =r r y 2 −1 (1 − y) 2 −1 dy.

1

0

2

1 −t X r (t) −→ √2π e 2 . r →∞

Revised Answers Manual to Introduction to Probability

3 Now, apply the relation given in the Hint with α = r −2 2 and β = 2 to obtain:

3 1 √ √ r −2

r −2 3 −1 −1 2 r r y 2 (1 − y) 2 dy = r r r +1 2 , provided r ≥ 3.

2 0

Then:

3 √ r −2

2

r +1 2 r ×r r πr 2 2 r −2 1 1 × r = √ r 2 2r −22 π 2 −1 2

E X r2 = √

r +1 2

(by the usual recursive relation of the gamma function) √

√ 1 r π = π since = √ 2 π (r − 2) r r . So, E X r2 = Var(X r ) = , r ≥ 3. = r −2 r −2 2.11 Using f for the argument, the p.d.f. of X r1 ,r2 is given by:

r1 r1/2 2 r1 +r f 2 −1 2 (r1 /r2 ) f X r1 ,r2 ( f ) = × (r +r )/2 , f > 0. r21 r22 1 + r1 f 1 2 r2

For k = 1, 2, . . . , we have: r1 +r2 r1 r1

− r1 +r2 2 ∞ r +2k 2

r1 1 2 r2 k −1 2

E X r1 ,r2 = 1+ f f df. r1 r2 r2 2 2 0 Set rr21 f = x, or f = rr21 x, d f = rr21 d x. Hence:

− r1 +r2 ∞ 2 r1 +2k r1 −1 1+ f 2 f df r2 0 ∞ r1 +2k 2 −1 r1 +2k r1 +r2 r2 r2 = x 2 −1 (1 + x)− 2 × d x r r1 1 0 r1 +2k ∞ 2 r1 +2k r1 +r2 r2 = x 2 −1 (1 + x)− 2 d x. r1 0 dy 1 = y, or x = 1−y Set 1+x y , d x = − y 2 , y ∈ (0, 1). Then the last integral

above is:

r1 +2k r1 +2k r1 +r2 1 2 −1 × y − 2 +1 × y 2 × y −2 dy 0 (1 − y)

=

1 r2 −2k −1 2 0 y

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Revised Answers Manual to Introduction to Probability

(1 − y)

r1 +2k 2 −1

dy =

r −2k

( 2 2

r +2k

) ( 1 2

r +r ( 1 2 2 )

Therefore:

)

, provided r2 − 2k > 0, or r2 > 2k.

r1 +r2 r1 r1

r1 +2k r1 +2k 2 2

r2 −2k

r2 2 r2 k 2 r1 r2 × r1 +r2 2 × E X r1 ,r2 = r1 2 2 2 r1 +2k r2 −2k k 2 2 r2 , r2 > 2k. = × r1 r21 r22

For k = 1 (and r2 > 2, or r2 ≥ 3),

r1 r1 r2 r21 + 1 r22 − 1 2 2 −1 r2 r2 2

E X r1 ,r2 = × × = r1 r1 r21 r22 r21 r22 − 1 r22 − 1 r2 = . r2 − 2 For k = 2 (and r2 > 4, or r2 ≥ 5),

2 r1 +2 r −4 2 r12+4 r22−4 r2 r2 2 + 1 22 2 =

× E X r1 ,r2 = r1 r1 r21 r22 r21 r22 2 r1 +2 r1 +2 r2 −4

r2 = 2r1 r2 2 r22−2 r1 2 2 − 1 2 2 2 r1 +2 r1 r1 r2 −4 r2 r1 (r1 + 2) r2 2 × 2 2 2 . = =

r1 r21 r22 − 1 r22−4 r22−4 r1 (r2 − 2)(r2 − 4) 1 (r1 +2) 2 Therefore Var(X r1 ,r2 ) = E X r21 ,r2 −(E X r1 ,r2 )2 = ( rr21 )2 (r2r−2)(r −( r2r−2 )2 = 2 −4)

2r22 (r1 +r2 −2) . r1 (r2 −2)2 (r2 −4)

So, 2r 2 (r1 +r2 −2) 2 E X r1 ,r2 = r2r−2 for r2 ≥ 3, and Var(X r1 ,r2 ) = r (r2 −2) 2 (r −4) for r 2 ≥ 5.

1

2

2

Section 11.3 3.1 We have: k k

i=1

2 c ji y j − μi

j=1

=

k " k

i=1

=

c ji y j

+ μi2 − 2μi

j=1

k " k

i=1

2

j=1

k

# c ji y j

j=1

c ji y j

k l=1

cli yl

+ μi2 − 2μi

k

j=1

# c ji y j

Revised Answers Manual to Introduction to Probability

=

k " k k

i=1

= =

j=1 l=1

k k k

c ji cli y j yl +

k

i=1

i=1

k

i=1

k k

k k

k

c ji cli y j yl + k

c ji cli +

i=1 k

2 =

k k

μ2j − 2

ci j μ j

= =

yi −

i=1

k

c ji μi y j

k k

c ji μi y j

j=1 i=1

c ji μi y j

i=1 c ji cli = 1 if j = l, and

k

k k

yi2 +

i=1

ci j μ j

k k

2 ci j μ j

−2

j=1

k

j=1

μ j μl

k

cil μl

k k

=

k k k

k

μ2j

j=1

i=1 ci j cil = 1 if j = l, and = 0 if j = l). Therefore:

2 ci j μ j

=

j=1

k

yi2 +

i=1 k

j=1

k

μ2j − 2

j=1

y 2j +

k

j=1

k k

j=1

ci j μ j yi

i=1 j=1

μ2j − 2

k k

c ji μi y j ,

j=1 i=1

and this was seen to be equal to the left-hand side of (11.27).

3.2 The row sectors have norm 1, because: c12 j =

ci j cil μ j μl

k

=

k

ci j μ j yi ,

i=1 j=1 l=1

ci j cil =

i=1

= 0 if j = l).

i=1 j=1

l=1

j=1 l=1

(because, by (11.17 ),

μi2 − 2

k

k k

i=1

j=1

j=1 i=1

i=1

j=1

k

c ji y j

i=1 j=1

i=1

j=1

ci j μ j

μi2 − 2

i=1

y j yl

y 2j +

2

k

μi

k

j=1

and k k

k

i=1

(because, by orthogonality, Also, yi −

μi2 − 2

k k k

j=1

k

k

i=1 j=1 l=1

j=1 l=1

=

# c ji y j

j=1

j=1 l=1 i=1

=

k

c ji cli y j yl + μi2 − 2μi

k

1 2 1 = k × = 1, and for i = 2, . . . , k, √ k k j=1

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Revised Answers Manual to Introduction to Probability

k

ci2j =

j=1

i−1

j=1

=

1 √ i(i − 1)

2

# " −(i − 1) 2 + √ i(i − 1)

(i − 1)2 1 i −1 i −1 + = + = 1. i(i − 1) i(i − 1) i i

Also, any two distinct row vectors are orthogonal, because: For i = 2, . . . , k,

k i−1

1 1 1 −(i − 1) +√ ×√ c1 j ci j = √ ×√ i(i − 1) i(i − 1) k k j=1 j=1 i −1 i −1 = √ √ −√ √ = 0, k i(i − 1) k i(i − 1) whereas for l = 2, . . . , k(l = i), k

ci j cl j =

j=1

i−1

1 i −1 1 1 ×√ −√ ×√ , if i < l, √ i(i − 1) l(l − 1) i(i − 1) l(l − 1) j=1

i −1 i −1 = √ −√ = 0, √ √ i(i − 1) l(l − 1) i(i − 1) l(l − 1) and k

j=1

ci j cl j =

l−1

1 1 −(l − 1) 1 ×√ +√ ×√ , if i > l, √ i(i − 1) l(l − 1) i(i − 1) l(l − 1) j=1

l −1 l −1 = √ −√ = 0. √ √ i(i − 1) l(l − 1) i(i − 1) l(l − 1) 3.3 (i) The transformation matrix is: √ √ ⎛ ⎞ The norm of the row vectors −1/√2 1/ √2 0√ is 1, because: C = ⎝ −1/√ 3 −1/√ 3 1/√3 ⎠ 1 2 1 2 −√ + √ = 21 + 21 = 1; 1/ 6 1/ 6 2/ 6 2 2

1 2 1 2 1 2 1 1 1 −√ + −√ + √ = + + = 1; 3 3 3 3 3 3

2 2 2 1 1 2 1 1 4 + √ + √ = + + = 1. √ 6 6 6 6 6 6 Next, any two vectors are orthogonal, because:

1 1 1 1 2 1 1 −√ −√ + √ −√ +0 √ = √ − √ = 0; 2 3 2 3 6 6 6

1 1 1 1 2 1 1 −√ + √ +0 √ = − √ + √ = 0; √ √ 2 6 2 6 6 6 6

Revised Answers Manual to Introduction to Probability

1 1 1 1 2 1 −√ + √ −√ √ √ √ 3 6 3 6 3 6 1 1 2 = − √ − √ + √ = 0.

3 2 3 2 3 2 The independence of Y1 , Y2 , Y3 (as well as the specification of the parameters) follows from Theorem 6. Alternatively, EY1 = − √1 μ1 + √1 μ2 =

(ii)

2 2 √1 (−μ1 + μ2 ); EY2 = − √1 μ1 − √1 μ2 + √1 μ3 = √1 (−μ1 − μ2 + 2 3 3 3 3 μ3 ); EY3 = √1 μ1 + √1 μ2 + √2 μ3 = √1 (μ1 + μ2 + 2μ3 ). Also, 6 6 3 6 Var(Y1 ) = 21 σ 2 + 21 σ 2 = σ 2 ; Var(Y2 ) = 13 σ 2 + 13 σ 2 + 13 σ 2 = σ 2 ; Var(Y3 ) = 16 σ 2 + 16 σ 2 + 46 σ 2 = σ 2 .

If μ1 = μ2 = μ3 = 0, then ( Yσi )2 , i = 1, 2, 3, are independently distributed as χ12 , so that σ12 (Y12 + Y22 + Y32 ) ∼ χ32 .

3.4 The joint p.d.f. of X and Y is given by: f X ,Y (x, y) = where q=

1 1 − ρ2

"

x − μ1 σ1

1 1

2π σ1 σ2 1 − ρ 2

2

− 2ρ

x − μ1 σ1

e−q/2 ,

y − μ2 σ2

+

y − μ2 σ2

2 # .

y−μ2 1 From the transformations: u = x−μ σ1 , v = σ2 , we get:

x = μ1 + σ1 u, y = μ2 + σ2 v, so that: ∂x ∂y ∂y ∂x σ 0 = σ1 σ2 . = σ1 , = 0; = 0, = σ2 ; and J = 1 0 σ2 ∂u ∂v ∂u ∂v Replacing the variables x and y in f X ,Y in terms of u and v, and cancelling out σ1 σ2 , we obtain 1 1 1 e−q/2 , q = (u 2 − 2ρuv + v 2 ), fU ,V (u, v) = 2 1 − ρ2 2π 1 − ρ which is as described. For the converse, start out with fU ,V and replace u and y−μ2 ∂u 1 ∂u ∂v ∂v 1 1 v by x−μ σ1 and σ2 , respectively. Here ∂ x = σ1 , ∂ y = 0; ∂ x = 0, ∂ y = σ2 ,

y−μ2 1 so that J = 1/σ1 σ2 . Then, replacing the u and v in fU ,V by x−μ σ1 and σ2 , respectively, and multiplying by J = 1/σ1 σ2 , we arrive at f X ,Y .

3.5 Set u = cx, v = dy, so that x = 1c u, y = d1 v and ∂∂ux = 1c , ∂∂vx = 0; ∂∂uy = 0, ∂∂vy = d1 ; J = 1/cd. Therefore the joint p.d.f. f X ,Y (x, y) = √1 2 e−q/2 , 2π

1 2 2 where q = 1−ρ 2 (x − 2ρx y + y ), because:

fU ,V (u, v) =

1 1 e−q/2 , where 2π |c||d| 1 − ρ 2

1−ρ

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q=

2

2

u 1 u v v2 v2 u v 1 u = , × + + − 2ρ − 2ρ 0 1 − ρ 2 c2 c d d2 |c| |d| d 2 1 − ρ02 c2

and ρ0 = ρ if cd > 0, and ρ0 = −ρ (to retain the “−” sign in front of the cross product) if cd < 0.

3.6 (i) From the transformations: u = x + y, v = x − y, we get: x = 21 (u +v), y = ∂y 1 ∂x ∂x 1 ∂y 1 2 (u − v), ∂u = ∂v = 2 , ∂u = − ∂v = 2 , so that J = −1/2. Then the joint p.d.f. of X and Y , which is: f X ,Y (x, y) = √1 2 e−q/2 , q = 2π 1−ρ 1 1 2 −2ρx y + y 2 ), becomes: f √1 (x (u, v) = × e−q/2 , where U ,V 2 2 1−ρ 2π 1−ρ 2

q=

1 1 − ρ2

"

u+v 2

2

− 2ρ

u+v 2

u−v 2

+

u−v 2

2 #

1 u 2 (1 − ρ) + v 2 (1 + ρ) u 2 (1 − ρ) + v 2 (1 + ρ) = × 2 1−ρ 2 2(1 − ρ)(1 + ρ) #2 " #2 " u v = √ + √ .

2(1 + ρ) 2(1 − ρ)

=

Therefore: fU ,V (u, v) =

1 √ 2π 2(1 + ρ) 2(1 − ρ) #2 " #2 2# " " 1 u v , × exp − + √ √ 2 2(1 + ρ) 2(1 − ρ) √

which is the bivariate normal with parameters 0, 0, 2(1 + ρ), 2(1 − ρ) and correlation coefficient = 0. (ii) Since the correlation coefficient is 0 (and U , V have jointly the bivariate normal distribution), it follows that U and V are independent. (iii) From the joint p.d.f. fU ,V , we get for the marginals: U ∼ N (0, 2(1 + ρ)), V ∼ N (0, 2(1 − ρ)).

3.7 (i) The r.v.’s U and V have the bivariate normal distribution with parameters 0, 0, l, l, and ρ; this is so by Exercise 3.4. (ii) The joint distribution of U + V and U − V is the bivariate normal with parameters 0, 0, 2(1 + ρ), 2(1 − ρ), and 0; this is so by Exercise 3.6(i). That U + V and U − V are independent and distributed as indicated follows from Exercise 3.6(ii).

Y −μ2 1 (iii) From the transformations: U = X −μ σ1 and V = σ2 , we get, for σ1 = σ2 = σ :X = μ1 + σ U , Y = μ2 + σ2 V , so that: X + Y = (μ1 + μ2 ) + σ (U + V ), X − Y = (μ1 − μ2 ) + σ (U − V ). However, independence of U + V and U − V implies independence of X + Y and X − Y .

3.8 (i) By independence of X̄ and S 2 , P(aμ < X̄ < bμ, 0 < S 2 < cσ 2 ) = P(aμ < X̄ < bμ)P(0 < S 2 < cσ 2 ).

Revised Answers Manual to Introduction to Probability

However, √ √ √ n( X̄ −μ) nμ(b−1) = P[k(a − < < P(aμ < X̄ < bμ) = P nμ(a−1) σ σ σ √ √ √ √ n( X̄ −μ) 1) n < < k(b − 1) n] (since μ = kσ ) = [k(b − 1) n] − σ

2 √ [k(a − 1) n], and P(0 < S 2 < cσ 2 ) = P 0 < (n−1)S < c(n − 1) = σ2

2 P χn−1 < c(n − 1) . Therefore: √ √ P(aμ < X̄ < bμ, 0 < S 2 < cσ 2 ) = { [k(b − 1) n] − [k(a − 1) n]} 2 ×P[χn−1 < c(n − 1)]. (ii) For the given numerical values, we have: √ √ k(b − 1) n = 1.5 × 0.5 × 4 = 3, k(a − 1) n = 1.5 × (−0.5) × 4 = −3, c(n − 1) = 1.487 × 15 = 22.305. Therefore : (3) − (−3) = 2 (3) − 1 = 2 × 0.998650 − 1 = 0.9973, and 2 P(χ15 < 22.305) = 0.90, so that the required probability is: 09973 × 0.90 = 0.89757.

3.9 In matrix notation, relations (1.17) and (1.17 ) are written as follows, respectively: CC = I and C C = I, where it is recalled that “ ” denotes transpose and I is the k × k unit matrix. We wish to show equivalence of (1.17) and (1.17 ); i.e., CC = I implies C C = I and vice versa. Indeed, CC = I implies C−1 (CC ) = C−1 I = C−1 or (C−1 C)C = C−1 or IC = C−1 or C = C−1 .

But then C C = C−1 C = I, as was to be seen. Next, let C C = I. Then (C C)C−1 = IC−1 = C−1 or C (CC−1 ) = C−1 or C I = C−1 or C = C−1 again. Therefore CC = CC−1 = I, as was to be seen.

Section 11.4 4.1 (i) By Theorem 10, Y ∼ U (0, 1). The transformation z = −2 log(1 − y) gives 1 −z/2 y = 1 − e−z/2 , z > 0, so that dy . Then f Z (z) = 1 × 21 e−z/2 = dz = 2 e 1 −z/2 2 , z > 0, which is the p.d.f. of the χ2 .

2e (ii) It follows from part (i), because the r.v.’s Z i , i = 1, . . . , n are independently distributed as χ22 .

Section 11.5 5.1 (i) For x ≥ 1, F(x) =

x −(c+1) dt = − x dt −c = −t −c |x = 1 − 1 . So, 1 1 ct 1 xc

F(x) = 0 for x ≤ 1, and F(x) = 1 −

1 for x > 1. xc

(ii) In formulas (11.30) and (11.31), take a = 1, b = ∞ to obtain. n−1 1 fU (u) = n × × cu −(c+1) = n cu −n c−1 , u > 1, uc

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and

1 f V (v) = n 1− c v

n−1

×cv −(c+1) = n c(v c −1)n−1 v −n c−1 , v > 1.

5.2 Here g1 (y1 ) = n (1 − y1 )n−1 , 0 < y1 < 1; gn (yn ) = nynn−1 , 0 < yn < 1.

Therefore: 1 1 EY1 = y × n(1 − y)n−1 dy = − yd(1 − y)n = −y(1 − y)n |10 0

+

1

0

(1 − y)n dy = −

0

1 1 1 1 y n+1 = ; i.e., EY1 = . n+1 n + 1 n + 1 0

Also, EYn =

1 1 y × ny n−1 dy = n y n dy = 0

n i.e., EYn = . n+1

0

1 n n y n+1 = ; n+1 n+1 0

Then limn→∞ EY1 → 0, and limn→∞ EYn = 1.

5.3 (i) From Example 11, we have, for n = 3 and n = 4, respectively: 3! , α < y1 < y2 < y3 < β, (β − α)3 4! g(y1 , y2 , y3 , y4 ) = , α < y1 < y2 < y3 < y4 < β. (β − α)4 g(y1 , y2 , y3 ) =

Therefore:

y3 3!

3×2 g13 (y1 , y3 ) = dy2 = (y3 − y1 ), 3 (β − α) y1 (β − α)3 y4 y4 y3 4!

4! g14 (y1 , y4 ) = dy dy = × (y3 − y1 )dy3 2 3 (β − α)4 y1 y1 (β − α)4 y1 4! (y3 − y1 )2 y4 4! (y4 − y1 )2 × = × (β − α)4 2 (β − α)4 2 y1 4×3 = (y4 − y1 )2 . (β − α)4

=

(ii) The pattern of integration should now be clear from part (i). Specifically, we have: g1n (y1 , yn ) yn yn−1 y4 y3 n! ··· dy2 dy3 · · · dyn−2 dyn−1 = (β − α)n y1 y1 y1 y1 yn yn−1 y4 n! = ··· (y3 − y1 )dy3 · · · dyn−2 dyn−1 n (β − α) y1 y1 y1

Revised Answers Manual to Introduction to Probability

= =

n!

2 × (β − α)n n!

2 × (β − α)n

yn yn−1 y1

···

y1

y4

(y3 − y1 )2

y1

yn yn−1

y4

dy4 · · · dyn−2 dyn−1

y1

y5

··· (y4 − y1 )2 dy4 · · · dyn−2 dyn−1 y1 y1 yn yn−1 y5 y5 n! = ··· (y4 − y1 )3 dy5 · · · dyn−2 dyn−1 n 2 × 3 × (β − α) y1 y1 y1 y1 yn yn−1 y6 n! = ··· (y5 − y1 )3 dy5 · · · dyn−2 dyn−1 2 × 3 × (β − α)n y1 y1 y1 yn yn−1 y6 y6 n! = ··· (y5 − y1 )4 d y6 . . . d yn−2 d yn−1 n 2 × 3 × 4 × (β − α) y1 y1 y1 y1 yn yn−1 y7 n! = ··· (y6 − y1 )4 d y6 · · · d yn−2 d yn−1 2 × 3 × 4 × (β − α)n y1 y1 y1 y1

= ···························································· yn yn−1 n! = ... (yn−2 − y1 )n−4 d yn−2 d yn−1 n 2 × 3 × 4 × · · · × (n − 4)(β − α) y1 y1 yn yn−1 n! = (yn−2 − y1 )n−3 d yn−1 n 2 × 3 × 4 × · · · × (n − 4)(n − 3)(β − α) y1 y1 yn n! = (yn−1 − y1 )n−3 d yn−1 n 2 × 3 × 4 × · · · × (n − 4)(n − 3)(β − α) y1 =

yn n! n−2 (y − y ) n−1 1 2 × 3 × 4 × · · · × (n − 4)(n − 3)(n − 2)(β − α)n y1

n! (yn − y1 )n−2 (n − 2)!(β − α)n n(n − 1) = (yn − y1 )n−2 , α < y1 < yn < β. (β − α)n =

5.4 For α = 0 and β = 1, we have from Exercise 5.3(ii): g1n (y1 , yn ) = n(n − 1)(yn − y1 )n−2 , 0 < y1 < yn < 1.

Then: E(Y1 Yn ) =

1 yn

y1 yn × n(n − 1)(yn − y1 )n−2 d y1 d yn 0 0 # 1 " yn n−2 = n(n − 1) yn y1 (yn − y1 ) d y1 d yn .

0

However, yn y1 (yn − y1 )n−2 d y1 = −

0

yn 1 y1 d(yn − y1 )n−1 n − 1 0 0 yn yn 1 1 y1 (yn − y1 ) + =− (yn − y1 )n−1 d y1 n+1 n − 1 0 0

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1 = n−1 =−

yn 0

1 n(n − 1)

(yn − y1 )n−1 d y1 yn d(yn − y1 )n = − 0

yn 1 ynn (yn − y1 )n = . n(n − 1) n(n − 1) 0

Then: E(Y1 Yn ) =

1 0

ynn+1 d yn =

1 1 1 ynn+2 = ; i.e., n+2 n+2 0

1 . On the other hand, by Exercise 5.2, n+2 1 n and EYn = , so that: EY1 = n+1 n+1 n 1 1 Cov(Y1 , Yn ) = − . = n + 2 (n + 1)2 (n + 1)2 (n + 2) E(Y1 Yn ) =

5.5 (i) The joint p.d.f. g1n of the r.v.’s Y1 and Yn is, by Exercise 5.3(ii), for α = 0 and β = 1: g1n (y1 , yn ) = n(n − 1)(yn − y1 )n−2 , 0 < y1 < yn < 1.

We have R = Yn − Y1 , set Y = Yn , and consider the transformations: r = yn − y1 , y = yn from which we get: y1 = y − r , yn = y. Then the Jacobian is: −1 1 = −1, and therefore : J= 0 1 f R,Y (r , y) = n(n − 1)r n−2 , 0 < r < y < 1 (as follows from 0 < y1 and y1 < yn , or r < y and r > 0, or 0 < r < y < 1).

Then: 1 n(n − 1)r n−2 dy = n(n − 1)r n−2 (1 − r ), 0 < r < 1. f R (r ) = r

(ii)

Next, E R = n(n − 1)

1

r × r n−2 (1 − r )dr

0

1

= n(n − 1)

r n−1 (1 − r )dr

0

1

= n(n − 1)

r 0

n−1

dr −

1

n

r dr 0

1 1 1 1 = n(n − 1) × rn − × r n+1 n n+1 0 0

n−1 1 1 = = n(n − 1) − . n n+1 n+1

Revised Answers Manual to Introduction to Probability

1 1 5.6 Indeed, from z = ny1 , we have y1 = n1 z and dy dz = n . Therefore, for z > 0, 1 −(nλ)×z/n −λz = λe . f Z (z) = n (nλ)e 5.7 Clearly, P(system works beyond t) = P[min(X , Y ) > t]. So, setting T = min(X , Y ), Example 12 gives f T (t) = (2λ) × e−(2λ)t , t > 0 Hence: ∞

−2λt . (i) P(T > t) = t (2λ)e−(2λ)x d x = −e−(2λ)x |∞ t =e (ii) Since the distribution is the negative exponential with parameter 2λ, we have E T = 1/2λ. (iii) For λ = 13 , P(T > t) = e−2t/3 (and for t = 3, e.g., this probability is e−2 0.135), and E T = 1.5.

5.8 (i) In Example 13, it was seen that F(x) = 1−e−λx , x > 0. Therefore relation (11.31) gives: n−1

(ii)

gn (yn ) = n(1 − e−λyn )n−1 λe−λyn = nλe−λyn (1 − e−λyn ) , yn > 0. ∞ y × 2λe−λy (1 − e−λy ) dy For n = 2, EY2 = 0 ∞ ∞ 1 3 2 −λy = . yλe dy − y × 2λe−2λy dy = − =2 λ 2λ 2λ 0 0 ∞ For n = 3, EY3 = y × 3λe−λy (1 − e−λy )2 dy 0 ∞ y × λe−λy (1 − 2e−λy + e−2λy )dy =3 0 ∞ ∞ ∞ −λy −2λy yλe dy − 3 y × 2λe dy + y × 3λe−3λy dy =3 0

0

0

3 3 1 11 = − + = . λ 2λ 3λ 6λ 5.9 (i) Indeed, (1 − e−λy )n−1 = =

n−1

n−1 (−e−λy )n−1−r r r =0 n−1

(−1)n−r −1

r =0

Then: EYn =

∞ 0

= nλ

n − 1 −λ(n−r −1)y e . r

y × nλe−λy (1 − e−λy )n−1 dy

∞ n−1

n−1 (−1)n−r −1 ye−λy e−λ(n−r −1)y dy r 0 r =0

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= nλ = nλ

∞ n−1

n−1 (−1)n−r −1 ye−λ(n−r )y dy r 0 r =0 n−1

(−1)n−r −1

r =0

× =n =

n λ

∞

0 n−1

n−1 1 r λ(n − r )

y × λ(n − r )e−λ(n−r )y dy

(−1)

r =0 n−1

n−r −1

n−1 n −r n−1

(−1)n−r −1

r =0

×

r

r

(n − r )2

1 λ(n − r ) .

(ii) Applying this formula for n = 2 and n = 3,we get:

1 2 2 3 3 3 1 2 3 11 11 − +1 = × = ; − +1 = × = . λ 4 λ 4 2λ λ 9 4 λ 18 6λ 5.10 As indicated in the Hint, there are n − 2 successive integrations with respect to yn−1 , yn−2 , . . . , y2 over the respective regions (yn−2 , yn ), n−2 1 )] . Multiplying (yn−3 , yn ), . . . , (y1 , yn ), which produce the term: [F(yn )−F(y (n−2)! it by the constant (for the above integration) term: n! f (y1 ) f (yn ), we obtain the n! = n(n − 1); i.e., g1n (y1 , yn ) = n(n − 1)[F(yn ) − desired result, since (n−2)! n−2 F(y1 )] f (y1 ) f (yn ), a < y1 < yn < b.

5.11 Here f (y) = 1, 0 < y < 1, and F(y) = y, 0 < y < 1. Therefore: g1n (y1 , yn ) = n(n − 1)(yn − y1 )n−2 , 0 < y1 < yn < 1.

Chapter 12 Section 12.1 1.1 For any x ∈ , take n 0 > x. Then Fn (x) = 0 for all n ≥ n 0 , and hence, trivially, Fn (x) −→ 0 for all x ∈ . n→∞

P

1.2 Since for every ε > 0, P(|X n | > ε) = P(X n = 1) = pn , then X n −→ 0 if and only if P(X n = 1) −→ 0; i.e., if and only if pn −→ 0. n→∞

n→∞

n→∞ λt

1.3 We know that M X n (t) = ( pn et + qn )n (qn = 1 − pn ) and M X (t) = eλ(e −1) , for all t ∈ . Set npn = λn , so that pn = λnn and qn = 1 − λnn . Then, for t ∈ ,

" #n λn t λn n λn M X n (t) = e +1− = 1 + (et − 1) n n n

Revised Answers Manual to Introduction to Probability

# " λn (et − 1) n t = 1+ −→ eλ(e −1) = M X (t), n→∞ n since λn (et − 1) −→ λ(et − 1) (see #6 in Table 6 in the Appendix). n→∞

1.4 We have seen that the p.d.f.’s of Y1,n and Yn,n are given, respectively, by: g1 (x) = n(1 − x)n−1 , 0 < x < 1;

gn (y) = ny n−1 , 0 < y < 1.

Therefore, for 0 < ε(< 1): (i) P(|Y1,n | > ε) = P(Y1,n > ε) = −(1 − x)n |1ε = (1 − ε)n −→ 0.

1 n−1 d x = − 1 d(1 − x)n = ε n(1 − x) ε

n→∞

(ii) P(|Yn,n − 1| > ε) = 1 − P(|Yn,n − 1| ≤ ε) and P(|Yn,n − 1| ≤ ε) = 1 P(1 − ε ≤ Yn,n ≤ 1 + ε) = P(1 − ε ≤ Yn,n ≤ 1) = 1−ε ny n−1 dy = y n |11−ε = 1 − (1 − ε)n −→ 1, so that P(|Yn,n − 1| > ε) −→ 0. n→∞

1.5 (i)

n→∞

From the transformation u = nx, we get x = n1 u, ddux Therefore:

= n1 , 0 < u < n.

u n−1 1 u n−1 fUn (u) = × n 1 − = 1− , 0 < u < n. n n n dy Also, from the transformation v = n(1 − y), we get y = 1 − nv , dv = 1 − n , 0 < v < n. Therefore:

f Vn (v) =

v n−1 1 v n−1 ×n 1− = 1− , 0 < v < n. n n n

That is, Un and Vn , as defined above, have the same distribution. (ii) The d.f. of Un is, for 0 < u < n: FUn (u) =

u 0

1−

x n

n−1

Thus, FUn (u) =

u x n dx = − d 1− n 0 n u

x u n = − 1− =1− 1− . n n 0

⎧ ⎨

0, n u ≤ 0 1 − 1 − un , 0 < u < n ⎩ 1, u ≥ n;

and likewise for the r.v. Vn . At this point, recall that the d.f. of U is given by: FU (u) = 0, for u ≤ 0, and FU (u) = 1 − e−u , for u > 0. For an arbitrary but fixed u > 0, let n 0 be a positive integer such that u < n 0 . Then it is clear that u < n for all n ≥ n 0 , so that FUn (u) = 1 − (1 − un )n , n ≥ n 0 .

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Since (1 − un )n −→ e−u , we have FUn (u) −→ 1 − e−u = FU (u). Since n→∞

n→∞ d FUn (u) = FU (u) for u ≤ 0, it follows that Un −→ U , as was to be seen. Of

course, the same holds for Vn , since Vn has the same distribution as Un .

1.6 Indeed, since E X̄ n = μ, we have E( X̄ n − μ)2 = Var( X̄ n ) = σn −→ 0.

2

n→∞

1.7 (i) By the Hint, (x + y)2 = x 2 + y 2 + 2x y ≥ 0, or −x y ≤ 21 (x 2 + y 2 ), or − 21 (x 2 + y 2 ) ≤ x y. Also, (x − y)2 = x 2 + y 2 −2x y ≥ 0, or x y ≤ 21 (x 2 + y 2 ). Hence |x y| ≤ 21 (x 2 + y 2 ). Set x = X /E 1/2 |X |2 , y = Y /E 1/2 |Y |2 X Y to get: E 1/2 |X|X|2 YE|1/2 |Y |2 ≤ 21 ( E|X + E|Y ). Take the expectations of both |2 |2 2

2

1 sides to obtain: E 1/2 |X |21E 1/2 |Y |2 E|X Y | ≤ 21 ( E|X E X 2 + E|Y1 |2 EY 2 ) = |2

1 1/2 |X |2 E 1/2 |Y |2 . That |E(X Y )| ≤ E|X Y | is 2 (1 + 1) = 1, or E|X Y | ≤ E

an immediate consequence of properties of sums or integrals, as the case may be. (ii) Take X 1 = X and X 2 = 1 to obtain: |E X | ≤ E|X | ≤ E 1/2 |X |2 .

1.8 Show that E(Yn − X )2 −→ 0. To this end, E(Yn − X )2 = E[(Yn − X n ) + n→∞

(X n − X )]2 = E(Yn − X n )2 + E(X n − X )2 + 2E[(Yn − X n )(X n − X )], and E(Yn − X n )2 −→ 0, E(X n − X )2 −→ 0 by assumption. By the Cauchyn→∞

n→∞

Schwartz inequality |E[(Yn − X n )(X n − X )]| ≤ E 1/2 |X n − Yn |2 × E 1/2 |X n − X |2 −→ 0. The proof is completed. n→∞

Section 12.2 2.1 The assumption that X 1 , . . . , X n are i.i.d. implies that the r.v.’s X 1k , . . . , X nk are also i.i.d. (although their distribution is different from that of the X i ’s). Their n sample mean is n1 i=1 X ik = X̄ n(k) . Then the WLLN applies and gives the result.

∞ 1 1−α tx x t x 2.2 (i) Here M X (t) = ∞ x=0 e × cα = c x=0 (αe ) = c × 1−αet = 1−αet , t provided αe < 1 or t < − log α. t d d 1−α (ii) Now E X = dt M X (t)|t=0 = dt ( 1−αet ) t=0 = α(1−α)e = α(1−α) = (1−αet )2 t=0 (1−α)2 α .

1−α n (iii) Then, with X̄ n = n1 i=1 Xi , " #n n t t 1−α

n = MX M X̄ n (t) = M i=1 = .

Xi n n 1 − αet/n According to the Hint, et/n = 1 + nt + R( nt ), where nt R( nt ) −→ 0 for each fixed t. Then: n→∞

1−α 1−α 1 t = = , αt t/n αt/(1−α)+[α/(1−α)]n R( nt ) 1 − αe 1 − α − n − αR n 1− n

Revised Answers Manual to Introduction to Probability

so that

n 1−α 1 − αet/n

4

2 αt/(1 − α) + [α/(1 − α)]n R nt n 1− n

= 1

α

−→ 1/e−αt/(1−α) = e 1−α t ,

n→∞

and this the m.g.f. of the mean α/(1 − α).

150

x 150−x .

2.3 (i) Here P(X ≤ 80) = 80 x=0 x (0.6) (0.4) 150 (ii) For this part, look at X as the sum i=1 X i of independent r.v.’s X 1 , . . . , X 150 distributed as B(1, 0.6), so that E X = 150 × 0.6 = 90, σ 2 (X ) = 150 × 0.6 × 0.4 = 36, and σ (X ) = 6. Then the CLT applies and gives: P(X ≤ 80) = P(−0.5 < X ≤ 80)

90.5 80 − 90 − − 6 6 = (−1.67) − (−15.8) = (−1.67) = 1 − (1.67) = 1 − 0.952540 = 0.04746.

2.4 Here X ∼ B(1,000, p), and therefore: (i) P(1,000 p − 50 ≤ X ≤ 1,000 p + 50) x 1,000−x 1,000 p+50 p q (q = 1 − p). = x=1,000 p−50 1,000 x For p = 21 , and p = 41 , this formula becomes, respectively: P(450 ≤ X ≤ 550) = =

550

1,000 x=0

x

550

1,000 (0.5)1,000 x

x=450

(0.5)

1,000

449

1,000 (0.5)1,000 , − x x=0

300

1,000 (0.25)x (0.75)1,000−x P(200 ≤ X ≤ 300) = x x=200

300 199

1,000 1,000 x 1,000−x (0.25) (0.75) (0.25)x (0.75)1,000−x . = − x x x=0

x=0

√ (ii) Since E X = 1,000 p, Var(X ) = 1,000 pq, and σ (X ) = 10 10 pq, we have, by using (12.12) and (12.11), P(1,000 p − 50 ≤ X ≤ 1,000 p + 50) = P([1,000 p − 50] < X ≤ 1,000 p + 50)

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if 1,000 p − 50 = integer (where [x] = integer part of x), the probability is

50 [1,000 p − 50] − 1,000 p − √ √ 10 10 pq 10 10 pq

√

[1,000 p − 50] − 1,000 p 10 pq − = , and √ 2 pq 10 10 pq P(1,000 p − 50 ≤ X ≤ 1,000 p + 50) = P(1,000 p − 51 < X ≤ 1,000 p + 50) √

10 pq if 1,000 p − 50 = integer, the probability is 2 pq √ √

√

1.02 10 pq 10 pq 1.02 10 pq = + − 1. − − 2 pq 2 pq 2 pq For p = 21 and p = 41 , the last formula above becomes, respectively: √ √ ( 10) − (1.02 10) − 1 (3.16) + (3.22) − 1 = 0.999211 + 0.999359 − 1 = 0.99857; ' √ ( ' √ ( 2 30 2.04 30 + −1 (3.65) + (3.72) − 1 3 3 = 0.999869 + 0.9999 − 1 = 0.999769.

100 2.5 Setting X = i=1 X i , we have X ∼ B(100, p) with E X = 100 p and Var(X ) = 100 pq (q = 1 − p). Then: 50 50 (i) P(X = 50) = 100 50 p q . (ii) Clearly, P(X = 50) = P(49 < X ≤ 50), and since E X = 100 p, √ σ 2 (X ) = 100 pq, so that σ (X ) = 10 pq, relations (12.12) and (12.11) give:

49 − 100 p 50 − 100 p − P(49 < X ≤ 50) √ √ 10 pq 10 pq without continuity correction. With continuity correction, we have: P(X = 50) = P(49 < X ≤ 50) = P(49.5 < X ≤ 50.5)

49.5 − 100 p 50.5 − 100 p − . √ √ 10 pq 10 pq (iii) For p = 0.5, the probabilities in part (ii) become: (0) − (0.1) −

(−0.2) =

(0.2) − 0.50 = 0.579260 − 0.50 = 0.07926,

(−0.1) = 2 (0.1) − 1 = 2 × 0.539828 − 1 = 0.079656.

Revised Answers Manual to Introduction to Probability

2.6 Consider the r.v.’s X i , i = 1, . . . , 50 given in the Hint. Then: 21 7 1 (1 + 2 + 3 + 4 + 5 + 6) = = = 3.5, 6 6 2 1 91 E X i2 = (1 + 4 + 9 + 16 + 25 + 36) = , 6 6 E Xi =

so that 35 91 49 − = and σ (X i ) = Var(X i ) = 6 4 12

√

105 .

6

Then P(150 ≤ X ≤ 200) = P(149 < X ≤ 200), and by relations (12.12) and (12.11) this is ' ( ' ( 200 − 175 149 − 175 √ √ − √ √ 105 105 6 × 50 6 × 50 '√ ( ' ( √ 210 1.04 210 = − − 7 7 '√ ( ' ( √ 1.04 210 210 = + −1 7 7 (2.07) + = 0.964996.

(2.15) − 1 = 0.980774 + 0.984222 − 1

With continuity correction, we have: P(149 < X ≤ 200) = P(149.5 < X ≤ 200.5) ( ' ( ' √ √ 1.02 210 1.02 210 − − 7 7 ' ( √ 1.02 210 =2 − 1 2 (2.11) − 1 7 = 2 × 0.982571 − 1 = 0.965142.

2.7 As usual, associate the r.v. X i with the ith draw, where X i = 1 when an ace appears and X = 0 otherwise. Then the r.v.’s X 1 , . . . , X 1,000 will be assumed to 4 1 = 13 . be independent and their common distribution is the B(1, p) with p = 52 1,000 1 4 48 12 Furthermore, X = i=1 X i . Since E X i = 13 , Var(X i ) = 52 × 52 = 132 , √

12×1,000 and σ (X ) = 201330 . Therefore, by we have E X = 1,000 13 , Var(X ) = 132 relations (12.12) and (12.11),

P(65 ≤ X ≤ 90) = P(64 < X ≤ 90)

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Revised Answers Manual to Introduction to Probability

' '

= =

90 − 1,000 13 √ 20 30 13

(

√ ( 17 30 − 60 ' √ ( 17 30 + 60

(1.55) + = 0.876421.

' − '

64 − 1,000 13

(

√ 20 30 13

√ ( 16.8 30 − 60 ' √ ( 16.8 30 −1 60

(1.53) − 1 = 0.939429 + 0.936992 − 1

With continuity correction, we have: P(64 < X ≤ 90) = P(64.5 < X ≤ 90.5) ' ' √ ( √ ( 17.65 30 16.15 30 − − 60 60 ' ' √ ( √ ( 17.65 30 16.15 30 = + −1 60 60 (1.61) + (1.47) − 1 = 0.946301 + 0.929219 − 1 = 0.87552.

2.8 According to the Hint, X ∼ B(1,000, 0.03) and 5% of 1,000 = 50. Since E X = 1,000 × 0.03 = 30 and Var(X ) = 30 × 0.97 = 29.1, the required probability is, by relations (12.12) and (12.11): P(X ≤ 50) = P(−0.5 < X ≤ 50)

30.5 50 − 30 − −√ √ 29.1 29.1 ' √ ( ' ( √ 20 29.1 30.5 29.1 = + −1 29.1 29.1 (3.71) + = 0.999896.

(5.65) − 1 =

(3.71)

With continuity correction, we have: P(−0.5 < X ≤ 50) = P(0 < X ≤ 50.5)

30 50.5 − 30 − −√ √ 29.1 29.1 ' ( ' √ ( √ 20.5 29.1 30 29.1 = + −1 29.1 29.1 (3.80) + (5.56) − 1 = = 0.999928.

(3.80)

Revised Answers Manual to Introduction to Probability

n 2.9 (i) Since X = i=1 X i , where the independent r.v.’s X 1 , . . . , X n are distributed as B(1, p), so that E X i = p, Var(X i ) = pq, we have that Xn = X̄ and hence, dispensing with the adjustment to bring it in the form provided by relation (12.12), we have :

X √ √ − p < 0.05 pq = P | X̄ − p| < 0.05 pq P n √ √ = P(−0.05 pq < X̄ − p < 0.05 pq) √ √ √ √ # √ " 0.05 n pq 0.05 n pq n( X̄ − p) =P − < < √ √ √ pq pq pq √ " # √ √ n( X̄ − p) < 0.05 n = P −0.05 n < √ pq √ √ (0.05 n) − (−0.05 n) √ √ = 2 (0.05 n) − 1 = 0.95 or (0.05 n) = 0.975, √ so that 0.05 n = 1.96 and n = 1,536.64, so that n = 1,537. (ii) Since

X √ √ − p < 0.05 pq = P(| X̄ − p| < 0.05 pq) P n Var( X̄ ) 1 pq/n =1− =1− √ (0.05 pq)2 0.0025 pq 0.0025n 400 400 , it suffices for 1 − ≥ 0.95 or n = 8,000. =1− n n

≥1−

(iii) The sample size determined through the CLT is only about 19.2% of that determined by way of the Tchebichev inequality.

2.10 The observed relative frequency of the voters favoring the proposal, among the sampled ones, is Xn , and the required probability should satisfy the condition:

X − 0.53 ≤ 0.02 ≥ 0.99.

P n n Clearly, Xn = n1 i=1 X i = X̄ , E X i = 0.53, Var(X i ) = 0.53 × 0.47 = 0.2491, and therefore, by setting 0.53 = p, 0.47 = q:

X − 0.53 ≤ 0.02 = P(| X̄ − E X̄ | ≤ 0.02) n √ # √ √ " 0.02 n 0.02 n n( X̄ − p) ≤ √ = P(−0.02 ≤ X̄ − p ≤ 0.02) = P − √ ≤ √ pq 0.2491 0.2491 √ √ √

0.02 n 0.02 n 0.02 n =2 − 1 = 0.99, P −√ ≤Z≤ √ √ 0.2491 0.2491 0.2491

P

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or

√

√

0.2491

0.2491

×(0.2491) ( √0.02 n ) = 0.995 and hence √0.02 n = 2.58, or n = (2.58)(0.02) = 2 2

1292 × 0.2491 = 4,145.2731, so that n = 4,146.

2.11 Clearly, E X i = 0, and Var(X i ) = E X i2 = 1. Therefore E X = 0, Var(X ) = √ 1,000 and σ (X ) = 10 10, and by means of (12.12) and (12.11), P(X ≥ 10) = P(10 ≤ X ≤ 1,000) = P(9 < X ≤ 1,000)

9 1,000 − √ √ 10 10 10 10 √ √ = (10 10) − (0.09 10) 1 − (0.28) = 1 − 0.610261 = 0.389739.

2.12 Since E X i = Var(X i ) = λ, setting Sn = (12.12) and (12.11): (i)

n

i=1 X i , we have, by means of

P(Sn ≤ λn) = P(−0.5 < Sn ≤ λn )

Sn − λn λn − λn −0.5 − λn < √ ≤ √ =P √ λn λn λn

−0.5 − λn Sn − λn =P < √ ≤0 √ λn λn

# "

1 0.5 + λn −0.5 − λn = − 1− (0) − √ √ 2 λn λn

0.5 + λn = − 0.5. √ λn

(ii) Let [λn] stand for the integer part of λn. Then, if λn is not an integer: P(Sn ≥ λn) = 1 − P(Sn < λn) = 1 − P(Sn ≤ [λn]) and P(Sn ≤ [λn]) = P(−0.5 < Sn ≤ [λn ])

Sn − λn [λn] − λn −0.5 − λn < √ ≤ √ =P √ λn λn λn

[λn] − λn −0.5 − λn − √ √ λn λn

λn − [λn] −0.5 − λn = − √ − √ λn λn

0.5 + λn λn − [λn] = − , so that √ √ λn λn

P(Sn ≥ λn)

1+

λn − [λn] √ λn

−

0.5 + λn √ λn

.

Revised Answers Manual to Introduction to Probability

If λn is an interger: P(Sn ≥ λn) = 1 − P(Sn < λn) = 1 − P(Sn ≤ λn − 1) and P(Sn ≤ λn − 1) = P(−0.5 < Sn ≤ λn − 1)

Sn − λn −1 −0.5 − λn < √ ≤√ =P √ λn λn λn

−1 −0.5 − λn − √ √ λn λn

0.5 + λn 1 = − , so that √ √ λn λn

0.5 + λn 1 − .

P(Sn ≥ λn) 1 + √ √ λn λn

λn 3λn 3λn λn P ≤ Sn ≤ = P Sn ≤ − P Sn < .

2 4 4 2

(iii)

Now, let λn/2 be non-integer. Then:

" # 3λn λn 3λn λn P Sn ≤ − P Sn < = P Sn ≤ − P Sn ≤ 2 2 4 2 ' ( ' λn √ ( − λn Sn − λn λn Sn − λn =P ≤− ≤ 2√ −P √ √ 4 λn λn λn (as the adjustments, involving −0.5, cancel out) ' ' ( √ ( λn − λn Sn − λn λn Sn − λn =P ≤− ≤− √ 2 −P √ √ 4 λn λn λn ' √ ( ' '√ ( ' λn ( λn ( λn − 2 λn − 2 λn λn − = − − − √ ; √ 4 4 λn λn i.e.,

P

λn 3λn ≤ Sn ≤ 2 4

'

( λn − λn − √ 2 λn

'√

λn 4

( .

Finally, if λn/2 is an integer, then:

3λn λn 3λn λn − P Sn < = P Sn ≤ − P Sn ≤ −1 P Sn ≤ 2 2 4 2 ' √ (

Sn − λn λn 2 + λn Sn − λn ≤− ≤− √ = P −P √ √ 4 λn λn 2 λn ' √ ( '√ (

2 + λn λn λn 2 + λn − = − − − √ √ 4 4 2 λn 2 λn (as the adjustments, involving −0.5, cancel out); i.e.,

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Revised Answers Manual to Introduction to Probability

P

λn 3λn ≤ Sn ≤ 2 4

2 + λn √ 2 λn

'√

λn 4

−

( .

(iv) For (i): P(Sn ≤ 100) (10.05) − 0.5 0.5.

(0.1) = 0.539828.

For (ii): P(Sn ≥ 100) 1 + (0.1) − (10.5) (5.1) − (2.5) 1 − (2.5) = 1 − For (iii): P(50 ≤ Sn ≤ 75) 0.993790 = 0.00621.

−100 100x .

2.13 (i) Here P(X ≤ 116) = 116 x=0 e x! (ii) For n sufficiently large (so that the CLT will apply, and with λ = 100/n), consider the independent r.v.’s X 1 , . . . , X n distributed as P(λ). Then X = n X ∼ P(100) with E X = Var(X ) = 100. Therefore, by (12.12) and i i=1 (12.11): P(X ≤ 116) = P(−0.5 < X ≤ 116) 0.5 + 100 116 − 100 ) − (− ) ( 10 10 = (1.6) − (−10.05) = (1.6) = 0.945201.

50 2.14 The total lifetime of 50 tubes is the r.v. X = i=1 X i , where X i has the negative exponential distribution with parameter λ = 1/1,500. Since E X i = 1 1 2 λ = 1,500, Var(X i ) = λ2 = 1,500 , then E X = 50 × 1,500 and Var(X ) = 2 50 × 1,500 .

Therefore: # " 80,000 − 50 × 1,500 X − EX 1 ≥ P(X ≥ 80,000) = P σ (X ) 50 × 1,5002 P(Z ≥ 0.47) = 1 − (0.47) = 1 − 0.680822 = 0.319178. ∞

−λx d x = −e−λx |∞ = 2.15 (i) The required probability is: P(X ≥ t) = t λe 1 −λt e . (ii) Since E X = 1/λ, Var(X ) = 1/λ2 and therefore EY = n/λ, Var(Y ) = n/λ2 , we have: # " t2 − n/λ t1 − n/λ Y − EY ≤ √ ≤√ P(t1 ≤ Y ≤ t2 ) = P √ n/λ n/λ Var(Y ) " #

λt1 − n λt2 − n λt1 − n λt2 − n Y − EY P − . ≤ √ ≤√ √ √ √ n n n n Var(Y )

(iii) Since −λt = log(0.9)

−0.105, we have:

P[X ≥ − log(0.9)λ] = e−0.105

0.9.

(iv) Since λt√2 −n = 1 and λt√1 −n = −1, we have: n n P(300 ≤ Y ≤ 420)

(1) − (−1) = 2 (1) − 1 = 2 × 0.841345 − 1 = 0.68269.

Revised Answers Manual to Introduction to Probability

2.16 (i) If X ∼ U (0, 1), then E X = 1/2 and Var(X ) = 1/12. Therefore: √ √ # "√ n(a − 1/2) n( X̄ − E X ) n(b − 1/2) ≤ ≤ P(a ≤ X̄ ≤ b) = P √ √ σ (X ) 1/12 1/12 √ " # √ n( X̄ − E X ) √ ≤ 3n(2b − 1) = P 3n(2a − 1) ≤ σ (X ) √ √ [(2b − 1) 3n] − [(2a − 1) 3n]. √ √ (ii) Here (2b − 1) 3n = 0.75 and (2a − 1) 3n = −0.75, so that the above approximate probability becomes: (0.75) −

(−0.75) = 2 (0.75) − 1 = 2 × 0.773373 − 1 = 0.546746.

2.17

2 If X ∼ U (0, θ ), then E X = θ2 and Var(X ) = θ12 .

Therefore: √ 3θ

# √

"√ θ

12 4 − θ2 12 4 − θ2 n( X̄ − E X ) θ 3θ 1 1 =P P < < X̄ < < 4 4 σ (X ) θ 2 /12 θ 2 /12 √ " # n( X̄ − E X ) = P −3< <3 (3) − (−3) σ (X ) = 2 (3) − 1 = 2 × 0.99865 − 1 = 0.9973.

2.18 We have: P(| X̄ − μ| ≤ 0.0001) = P(−0.0001 ≤ X̄ − μ ≤ 0.0001) √ √ √ # " 0.0001 n 0.0001 n n( X̄ − μ) P − ≤ ≤ 0.0005 σ 0.0005 √ √ √ P(−0.2 n ≤ Z ≤ 0.2 n) = 2 (0.2 n) − 1 = 0.99, √ √ or (0.2 n) = 0.995, hence 0.2 n = 2.58 and n = 166.41, so that n = 167.

2 4 1 = 25 = 0.04, we have: 2.19 Since E X̄ = μ and Var( X̄ ) = σn = 100 √ # " 10c n ( X̄ − μ) 10c ≤ ≤ P(| X̄ − μ| ≤ c) = P(−c ≤ X̄ − μ ≤ c) = P − 2 σ 2 P(−5c ≤ Z ≤ 5c) = 2 (5c) − 1 = 0.90, or (5c) = 0.95, hence 5c = 1.645 and c = 0.329.

2.20 (i) It is assumed implicitly that n is large enough, so that the CLT applies.

Then: P(| X̄ n − μ| < kσ ) = P(−kσ < X̄ n − μ < kσ ) √ " # √ √ √ n( X̄ n − μ) <k n = P −k n < (k n) − σ √ √ 1+ p = 2 (k n) − 1 = p or , (k n) = 2

√ (−k n)

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and hence: √ k n=

−1

1+ p 2

" or n =

1 k

−1

1+ p 2

#2 .

Thus, n is the smallest integer which is ≥ [ k1 −1 ( 1+2 p )]2 . (ii) Since E X̄ n = μ and Var( X̄ n ) = σ 2 /n, we have:

σ /n P(| X̄ n − μ| < kσ ) ≥ 1 − (kσ = 1 − nk1 2 . Then it suffices for 1 − nk1 2 ≥ p )2 2

or n ≥ (1−1p)k 2 . Thus, n is the smallest integer which is ≥ 1/(1 − p)k 2 . (iii) For p = 0.90, For p = 0.95, For p = 0.99,

−1 ( 1+ p ) = 2 −1 ( 1+ p ) = 2 −1 ( 1+ p ) = 2

−1 (0.950) = 1.645 (by interpolation).

−1 (0.975) = 1.96. −1 (0.995) = 2.575 (by interpolation).

Then, for the various values of k, the respective values of n are given in the following table for part (i) k\p 0.50 0.25 0.10

0.90 11 44 271

0.95 16 62 385

0.99 27 107 664

For the Tchebichev inequality, the values of n are given by the entries of the table below. k\p 0.50 0.25 0.10

0.90 40 160 1,000

0.95 80 320 2,000

0.99 400 1,600 10,000

2.21 Whether the Yi ’s are approximately normal or not, the CLT applies for Ȳ . Then by the given Hint, we have: 0.95 = P(Ȳ > 2,000) = 1 − P(Ȳ ≤ 2,000), or 0.05 = P(Ȳ ≤ 2,000) = √ √ √ √ n(Ȳ −2,200) n(2,000−2,200) ] P(Z ≤ − n) = (− n) = 1 − P[ √ 200 200 √≤ √ ( n), or ( n) = 0.95, hence n = 1.65, or n = 2.7225, so that n = 3.

So, if in repeated samples of size 3, the averages are > 2,000 about half of the times, the claim is discredited, whereas the claim is supported, if the averages are > 2,000 at about 95% of the times.

2.22 (i) Set Z i = X i − Yi , i = 1, . . . , n. Then these r.v.’s are independent with E Z i = E X i − EYi = μ − μ = 0, Var(Z i ) = Var(X i ) + Var(Yi ) = 2σ 2 and Z̄ = X̄ − Ȳ . Then the CLT applies and gives P(| X̄ − Ȳ √ | ≤ 0.25σ )= √ 0.25σ n n Z̄ P(| Z̄ | ≤ 0.25σ ) = P(−0.25σ ≤ Z̄ ≤ 0.25σ ) = P(− √ ≤ √ ≤ σ 2 √ √ √ √ σ 2 0.25σ n 0.25 n 0.25 n 0.25 n √ √ √ √ ) P(− )≤ Z ≤( )=2 ( ) − 1 = 0.95, or σ 2 √ 2 2 2 √ 0.25 n 0.25 n ( √ ) = 0.975, so that √ = 1.96, n = 122.9312, and n = 123.

2 2

Revised Answers Manual to Introduction to Probability

2

(ii) Since E Z̄ = 0 and Var( Z̄ ) = Var( X̄ ) + Var(Ȳ ) = 2σn , we have: P(| X̄ − Ȳ | ≤ 0.25σ ) = P(| Z̄ | ≤ 0.25σ ) 2 2σ 2 /n =1− . ≥ 1− 2 (0.25σ ) 0.0625n 2 ≥ 0.95, or n = 640.

Thus, it suffices for 1 − 0.0625n Thus, the price we pay for the possibility that the X i ’s and the Yi ’s do not have the same distribution (although they do have the same mean and variance) is 517 additional samplings from each population.

2.23 (i) Independence of the X i ’s and Yi ’s implies independence of X̄ and Ȳ . Since E X̄ = μ1 and E Ȳ = μ2 , we have E( X̄ − Ȳ ) = μ1 − μ2 . Also, by the fact 2 2 that Var( X̄ ) = σn , Var(Ȳ ) = σn , and independence of X̄ and Ȳ , we have 2

Var( X̄ − Ȳ ) = 2σn . (ii) By setting Z i = X i −Yi , the r.v.’s Z 1 , . . . , Z n are i.i.d. with E Z i = μ1 −μ2 √

d n[ Z̄ −(μ1 −μ2 )] √ −→ Z ∼ N (0, 1). σ 2 n→∞ √ d −μ2 )] √ 1 Replacing Z̄ by X̄ − Ȳ , we obtain n[( X̄ −Ȳ )−(μ −→ N (0, 1). σ 2 n→∞ 2

and Var(Z i ) = 2σn . Then, by the CLT,

2.24 We have: EX i√= 1/λ, Var(X i ) = 1/λ2 , so that ESn = n/λ, Var(Sn ) = n/λ2 , and σ (Sn ) = n/λ. Therefore: n/λ n P − n/λ (i) P(Sn ≤ n P) = P(0 ≤ Sn ≤ n P) P(− √ ≤Z≤ √ ) n/λ n/λ √ √ = P[− n ≤ Z ≤ n(λP − 1)], where Z ∼ N (0, 1), √ √ P[Z ≤ n(λP − 1)] = [ n(λP − 1)], since n is expected to be large, √ 1 1 = p, so that n(λP − 1) = −1 ( p) and P = [1 + √ −1 ( p)]. λ n 1 × 2.33) = 1, 023.3. (ii) For the values given, P = 1,000(1 + 100 1 2.25 (i) Here μ = E X 1 = λ1 = 0.2 = 5, 1 1 1 2 2 = 25, so that σ = 5. Therefore: σ = σ (X 1 ) = λ2 = (0.2)2 = 0.04

P(| X̄ n − μ| ≤ c) = P(−c ≤ X̄ n − μ ≤ c) √ √ # " √ n( X̄ n − μ) c n c n =P − ≤ ≤ σ σ σ √ √ √ c n c n c n =2 − 1 = p, or P − ≤Z≤ 5 5 5 √

√ c n 1+ p 5 −1 1 + p = and n = .

5 2 c 2

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Revised Answers Manual to Introduction to Probability

So, n is the smallest integer ≥ [ 5c

−1

(ii) For p = 0.950, 1+2 p = 0.975, so that 1.96)2 = 96.04, so that n = 97.

!

1+ p ].

2 −1 (0.975) = 1.96 and n = (5 ×

1 2.26 We have EX = 0 and Var(X ) = 12 , so that

P(| X̄ 100 | ≤ 0.1) = P(−0.1 ≤ X̄ 100 ≤ 0.1) √ √ P(−0.1 × 10 12 ≤ Z ≤ 0.1 × 10 12) (Z ∼ N (0, 1)) √ √ √ = P( 12 ≤ Z ≤ 12) = 2 ( 12) − 1 = 2 (3.46) − 1 = 2 × 0.99973 − 1 = 0.99946.

Section 12.3 3.1 That the CLT holds means the following: If X 1 , . . . , X n is a random sample of √

size n with mean μ finite and variance σ 2 ∈ (0, ∞), then n( X̄σn −μ) −→ Z ∼ n→∞ n X i . Since √σn −→ 0, part (ii) of Theorem 6 applies N (0, 1), where X̄ n = n1 i=1 n→∞ in √ a trivial manner and gives that: d d n( X̄ n −μ) × √σn = X̄ n − μ −→ Z × 0 = 0, or X̄ n −→ μ, which implies σ n→∞

P

X̄ n −→ μ, by Theorem 1.

d

n→∞

n n 2 2 (X i − X̄ n )2 = 3.2 By Exercise 3.1(i) in Chapter 10, i=1 i=1 X i − n X̄ n ,

n n 1 n 1 n 2 2 = 2 so that n−1 i=1 (X i − X̄ n ) i=1 X i − n−1 X̄ n . However, n−1 × n

P P n X i2 −→ E X 12 = σ 2 + μ2 , X̄ n −→ μ, so that, by by the WLLN, n1 i=1 n→∞

n→∞ n→∞ P n 2 2 Theorem 5(i), X̄ n −→ μ , and by Theorem 6(ii) (and Theorem 1), n−1 × n→∞

P P n 1 n 2 −→ σ 2 , n−1 X̄ n2 −→ μ2 . Finally, by Theorem 5(ii), i=1 X i n→∞ n n→∞

1 2 n n P × Xi − X̄ n2 −→ σ 2 + μ2 − μ2 = σ 2 . n→∞ n−1 n n−1 n

i=1