Instructor's Solution Manual for An Introduction to Stochastic Modeling (Modelling) Fourth Edition M

An Introduction to Stochastic Modeling Fourth Edition Instructor Solutions Manual

Mark A. Pinsky Department of Mathematics Northwestern University Evanston, Illinois

Samuel Karlin Department of Mathematics Stanford University Stanford, California

AMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Academic Press is an imprint of Elsevier

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Contents

Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

24

Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41

Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

52

Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61

Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65

Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

68

Chapter 1

1 1 2.1 E [1 {A1 }] = Pr {A1 } = 13 . Similarly, E [1 {A1 }] = Pr {Ak } = 13 for k = 1, . . . , 13. Then, because the expected value of a sum 1 1 + · · · + 13 = 1. is always the sum of the expected values, E[N] = E [1 {A1 }] + · · · + E [1 {A13 }] = 13

2.2 Let X be the first number observed and let Y be the second. We use the identity (6xi )2 = 6xi2 + 6i6=j xi xj several times.

1X xi ; N P P X 2 (N − 1) xi2 − i6=j xi xj 1X 2 1 Var[X] = Var[Y] = xi − xi = ; N N N2 P i6=j xi xj E[XY] = ; N(N − 1) P P 2 xi i6=j xi xj − (N − 1) Cov[X, Y] = E[XY] − E[X]E[Y] = N 2 (N − 1) Cov[X, Y] 1 ñX,Y = =− . σX σY N −1 E[X] = E[Y] =

2.3 Write Sr = ξ1 + · · · + ξr where ξk is the number of additional samples needed to observe k distinct elements, assuming that k − 1 district elements have already been observed. Then, defining pk = Pr[ξk = 1} = 1 − k−1 N we have Pr [ξk = n] = pk (1 − pk )n−1 for n = 1, 2, . . . and E [ξk ] = p1k . Finally, E [Sr ] = E [ξ1 ] + · · · + E [ξr ] = p11 + · · · + p1r will verify the given formula. n−1

n−1

z }| { z }| { 2.4 Using an obvious notation, the event {N = n} is equivalent to either HTH . . . HTT or THT . . . THH so Pr {N = n} = n−1 n−1 n−1 n−1 P P 31 2 × 21 × 12 = 12 for n = 2, 3, . . . ; Pr {N is even} = n=2,4,... 12 = 23 and Pr {N ≤ 6} = 6n=2 12 = 32 . n−1 P 21 = 32 .

Pr {N is even and N ≤ 6} 5 m=2,4,6 12   In losses  z  }| { 2.5 Using an obvious notation, the probability that A wins on the 2n + 1 trial is Pr Ac Bc . . . Ac Bc A = [(1 − p)(1 − q)]n p,     P∞ p n = 0, 1, . . . Pr{A wins} = n = 0 [(1 − p)(1 − q)]n p = 1−(1−p)(1−q) . Pr{A wins on 2n + 1 play|A wins} = (1 − π)π n where

P 1+ (1−p)(1−q) 2π 2 n π = (1 − p)(1 − q). E # trials|A wins = ∞ n=0 (2n + 1)(1 − π)π = 1 + 1−π = 1− (1−p)(1−q) = 1− (1−p)(1−q) − 1. n−1

5 2.6 Let N be the number of losses and let S be the sum. Then Pr{N = n, S = k} = 16 6 Pk where p3 = p11 = p4 = p10 = P∞ 1 2 3 n=1 Pr{N = n, S = k} = pk . (It is not a correct argument to 15 ; p5 = p9 = p6 = p8 = 15 and p7 = 15 . Finally Pr{S = k} = simply say Pr{S = k} = Pr{Sum of 2 dice = k|Dice differ}. Compare with Exercise II, 2.1.)

2.7 We are given that (*) Pr{U > u, W > w} = [1 − Fu (u)] [1 − Fw (w)] for all u, w. According to the definition for independence we wish to show that Pr{U ≤ u, W ≤ w} = Fu (u)Fw (w) for all u, w. Taking complements and using the addition law Pr{U ≤ u, W ≤ w} = 1 − Pr{U > u or W > w} = 1 − [Pr{U > u} + Pr{W > w} − Pr{U > u, W > w}] = 1 − [(1 − FU (u)) + (1 − FW (w)) − (1 − Fu (u)) (1 − Fw (w))] = FU (u)FW (w) after simplification.

An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

2

Instructor Solutions Manual

R R R 2.8 (a) E[Y] = E[a + bX] = (a + bx)dFX (x) = a dFX (x) + b xdFX (x) = a + bE[X] = a + bµ. In words, (a) implies that the expected value of a constant times

a random variable is the constant times the expected value of the random variable. So E b2 (X − µ)2 = b2 E (X − µ)2 .

(b) Var[Y] = E (Y − E{Y})2 = E (a + bX − a − bµ)2 = E b2 (X − µ)2 = b2 E (X − µ)2 = b2 σ 2 2.9 Use the usual sums of numbers formula (See I, 6 if necessary) to establish n X k=1 n X

1 k(n − k) = n(n + 1)(n − 1); and 6

k2 (n − k) = n

X

k2 −

X

k=1

k3 =

1 2 n (n + 1)(n − 1), so 12

X 2 1 k(n − k) = (n + 1) n(n − 1) 3 h i X 1 3 k2 (n − k) = n(n + 1), and E X2 = n(n − 1) 6 h i 1 Var[X] = E X 2 − (E[X])2 = (n + 1)(n − 2).

18

2.10 Observe, for example, Pr{Z = 4} = Pr{X = 3, Y = 1} = 12 61 , using independence. Continuing in this manner, E[X] =

z

1

2

3

4

5

6

Pr{Z = z}

1 12

1 6

1 4

1 12

1 6

1 4

2.11 Observe, for example, Pr{W = z} = Pr{U = 0, V = 2} + Pr{U = 1, V = 1} = 61 + 16 + 13 . Continuing in this manner, arrive at w

1

2

3

4

Pr{W = w}

1 6

1 3

1 3

1 6

2.12 Changing any of the random variables by adding or subtracting a constant will not affect the covariance. Therefore, by replacing U with U − E[U], if necessary, etc, we may assume, without loss of generality that all of the means are zero.

Because the means are zero,

Cov[X, Y] = E[XY] − E[X]E[Y] = E[XY] = E UV − UW + VW − W 2 = −E W 2 = −σ 2 . (E[UV] = E[U]E[V] = 0, etc.)

2.13 Pr{v < V, U ≤ u} = Pr{v < X ≤ u, v < Y ≤ u} = Pr{v < X ≤ u} Pr{v < Y ≤ u} (by independence) = (u − v)2 ZZ =

fu,v u0 , v0 du0 dv0

(u0 ,v0 )v<v0 ≤u0 ≤u

=

 Zu Zu v



v0

  fu,v u0 , v0 du0 dv0 . 

The integrals are removed from the last expression by successive differentiation, first w.r.t. v (changing sign because v is a lower limit) than w.r.t. u. This tells us fu,v (u, v) = −

d d (u − v)2 = 2 for 0 < v ≤ u ≤ 1. du dv

Instructor Solutions Manual

3

3.1 Z has a discrete uniform distribution on 0, 1, . . . , 9.

3.2 In maximizing a continuous function, we often set the derivative equal to zero. In maximizing a function of a discrete variable, we equate the ratio of successive terms to one. More precisely, k∗ is the smallest k for which p(k+1) p(k) < 1, or, the smallest k

p n−k for which k+1 1−p < 1. Equivently, (b) k∗ = [(n + 1)p] where [x] = greatest integer ≤ x. for (a) let n → ∞, p → 0, λ = np.

Then k∗ = [λ].

3 2 3 3 5 2 3.3 Recall that eλ = 1 + λ + λ2! + λ3! + · · · and e−λ = 1 − λ + λ2! − λ3! + · · · so that sinh λ ≡ 12 eλ − e−λ = λ + λ3! + λ5! + · · ·

P λk e−λ −λ sinh(λ) = 1 1 − e−2λ .

Then Pr{X is odd} = k! = e 2 k=1,3,5,···

3.4 E[V] =

∞ X k=0

= 3.5

∞

1 λk e−λ e−λ X λk+1 = k + 1 k! λ (k + 1)! k=0

1

1 −λ λ e e − 1 = 1 − e−λ . λ λ

E[XY] = E[X(N − X)] = NE[X] − E X 2

= N 2 p − Np(1 − p) + N 2 p2 = N 2 p(1 − p) − Np(1 − p) Cov[X, Y] = E[XY] − E[X]E[Y] = −Np(1 − p).

3.6 Your intuition should suggest the correct answers: (a) X1 is binomially distributed with parameters M and π1 ; (b) N is binomial with parameters M and π1 + π2 ; and (c) X1 , given N = n, is conditionally binominal with parameters n and p = π1 /(π1 + π2 ).

To derive these correct answers formally, begin with Pr {X1 = i, X2 = j, X3 = k} =

M! i j k π π π ; i + j + k = M. i! j! k! 1 2 3

Since k = M − (i + j) Pr {X1 = i, X2 = j} = (a) Pr {X1 = i} =

X

M! j M−i−j πiπ π ; 0 ≤ i + j ≤ M. i! j! (M − i − j)! 1 2 3

Pr {X1 = i, X2 = j}

j M−i

=

X (M − i)!

M! j M−i−j π1i π π i! (M − i)! j! (M − i − j)! 2 3 j=0

=

M π1i (π2 + π3 )M−i , i = 0, 1, . . . , M. i

(b) Observe that N = n if and only if X3 = M – n. Apply the results of (a) to X3 : Pr{N = n} = Pr {X3 = M − n} =

(c) Pr { X1 = k| N = n} =

M! (π1 + π2 )n π3M−n n! (M − n)!

Pr {X1 = k, X2 = n − k} Pr{N = n}

M! k n−k M−n k!(M−n)!(n−k)! π1 π2 π3 = M! n M−n n!(M−n)! (π1 + π2 ) π3

=

k n−k n! π1 π2 , k = 0, 1, . . . , n. k! (n − k)! π1 + π2 π1 + π2

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Instructor Solutions Manual

3.7 Pr{Z = n} =

= =

n X

Pr{X = k}Pr{Y = n − k}

k=0 n X

µk e−µ υ (n−k) e−υ 1X n! = e−(µ+υ) µk υ n−k k! (n − k)! n! k! (n − k)! n

k=0 e−(µ+υ) (µ + υ)n

n!

k=0

(Using binomial formula.)

Z is Poisson distributed, parameter µ + υ.

3.8 (a) X is the sum of N independent Bernoulli random variables, each with parameter p, and Y is the sum of M independent Bernoulli random variables each with the same parameter p. Z is the sum of M + N independent Bernoulli random variables, each with parameter p. (b) By considering the ways in which a committee n people may be formed from a group comprised of M men and N

of n P M+N N M women, establish the identity = . n k n −k k=0 Then Pr{Z = n} =

n X

Pr{X = k}Pr{Y = n − K}

k=0 n X

M N k N−k pn−k (1 − p)M−n+k = p (1 − p) n−k k k=0

M+N n p (1 − p)M+N−n for n = 0, 1, . . . , M + N. = n Note:

N = 0 for k > N. k 3.9 Pr{X + Y = n} =

n X

n X Pr{X = k, Y = n − k} = (1 − π)π k (1 − π)π n−k

k=0

= (1 − π) π

2 n

n X

k=0

1 = (n + 1)(1 − π)2 π n for n = 0.

k=0

3.10

3.11

k

Binomial n = 10 p = .1

Binomial n = 100 p = .01

Poisson λ=1

0 1 2

.349 .387 .194

.366 .370 .185

.368 .368 .184

Pr{U = u, W = 0} = Pr{X = u, Y = u} = (1 − π)2 π 2u , u ≥ 0.

Pr{U = u, W = w > 0} = Pr{X = u, Y = u + w} + Pr{Y = u, X = u + w} = 2(1 − π)2 π 2u+w ∞

X Pr{U = u} = Pr{U = u, W = w} = π 2u 1 − π 2 .

Pr{W = 0} =

w=0 ∞ X

Pr{U = u, W = 0} = (1 − π)2

.

1 − π2 .

w=0

h .

i Pr{W = w > 0} = 2 (1 − π)2 (1 − π)2 1 − π 2 π w , and Pr{U = u, W = w} = Pr{U = u}Pr{W = w} for all u, w.

Instructor Solutions Manual

5

3.12 Let X = number of calls to switch board in a minute. Pr{X ≥ 7} = 1 −

6 k −4 P 4 e k=0

k!

.111.

3.13 Assume that inspected items are independently defective or good. Let X = # of defects in sample.

Pr{X = 0} = (.95)10 = .599 Pr{X = 1} = 10(.95)9 (.05) = .315 Pr{X = 2} = 1 − (.599 + .315) = .086.

1−p 3.14 (a) E[Z] = 1−p p = 9, Var[Z] = p2 = 90

(b) Pr{Z > 10} = (.9)10 = .349.

2 3.15 Pr{X ≤ 2} = 1 + 2 + 22 e−2 = 5e−2 = .677.

3.16 (a) p0 = 1 − b

∞ P k=1

.

(1 − p)k = 1 − b 1−p p

(b) When b = p, then pk is given by (3.4). p When b = 1−p , then pk is given by (3.5). (c) Pr{N = n > 0} = Pr{X = 0, Z = n} + Pr{X = 1, Z = n − 1} = (1 − α)p(1 − p)n + αp(1 − p)n−1

= (1 − α)p + αp (1 − p) (1 − p)n

So b = (1 − α)p + αp (1 − p).   Z+∞  1 Z+∞ 1  λZ 1 2 1 2 1 2 1 2 e 2 z +λz dz = e 2 λ √ e 2 (z−λ) dz = e 2 λ .

4.1 E e =√   2π 2π −∞

−∞

4.2 (a) PrW > θ1 = e−θ/θ = e−1 = .368 . . . . (b) Mode = 0. h i 4.3 X − θ and Y − θ are both uniform over − 21 , 12 , independent of θ, and W = X − Y = (X − θ) − (Y − θ). Therefore the distribution of W is independent of θ and we may determine it assuming θ = 0. Also, the density of W is symmetric since that of both X and Y are.

1 Pr{W > w} = Pr{X > Y + w} = (1 − w)2 , 2 So fw (w) = 1 − w

w>0

0 ≤ w ≤ 1 and fw (w) = 1 − |w| for − 1 ≤ w ≤ +1 n o −.010 4.4 µc = .010; σc2 = (.005)2 , Pr{C < 0} = Pr C−.010 < = Pr{Z < −2} = .0228. .005 .005 R ∞ R ∞ −3y 4.5 Pr{Z < Y} = 0 dy 2e−2x dx = 52 . x 3e for

5.1 Pr{N > k} = Pr {X1 ≤ ξ, . . . , Xk ≤ ξ } = [F(ξ )]k , k = 0, 1, . . .

Pr{N = k} = Pr{N > k − 1} − Pr{N > k} = [1 − F(ξ )]F(ξ )k−1 , k = 1, 2, . . .

5.2 Pr{Z > z} = Pr {X1 > z, . . . , Xn > z} = Pr {X1 > z} · · · · · Pr {Xn > z} = e−λz · · · · · e−λz = e−nλz , z > 0.

Z is exponentially distributed, parameter nλ.

5.3 Pr{X > k} =

∞ P

p(1 − p)l = p(1 − p)k+1 , k = 0, 1, . . .

I=k+1

E[X] =

∞ P k=0

Pr{X > k} = 1−p p .

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Instructor Solutions Manual

+ − 5.4 Write V = V + − V − when V + = max{V, 0} and V − = max{−V, 0}. Then Pr V > v = 1 − F (v) and Pr V > v = v + − + − Fv (−v) for v > 0. Use (5.3) on V and V together with E[V] = E V − E V . Mean does not exit if E V + = E V − = ∞.

R∞ R∞ R∞ √ √ 5.5 E W 2 = 0 P W 2 > t dt = 0 1 − Fw t dt = 0 2y [1 − Fw (y)] dy by letting y = t.

R∞ R∞ R∞ 5.6 Pr{V > t} = t λe−λv dv = e−λt ; E[V] = 0 Pr{V > t}dt = λ1 0 λe−λi dt = λ1 .

5.7 Pr{V > v} = Pr {X1 > v, . . . , Xn > v} = Pr {X1 > v} · · · · · Pr {Xn > v} = e−λ1 v · · · · · e−λn v = e−(λ1 +···+λn)v , v > 0.

P V is exponentially distributed with parameter λi .

5.8 Spares

3

2

1

0

1 2λ

1 2λ

1 2λ

1 2λ

A B Mean

1 1 1 1 Expected flash light operating duration = 2λ + 2λ + 2λ + 2λ = λ2 = 2 Expected battery operating durations!

Chapter 2

N P

N! m! pm (1 − p)N−m π k (1 − π)m−k k! (m − k)! m=k m! (N − m)!

N = (πp)k (1 − πp)N−k for k = 0, 1, . . . , N. k

1.1 (a) Pr{X = k} =

(b) E[XY] = E[MX] − E[X]2 ; E[M] = Np;

E M 2 = N 2 p2 + Np(1 − p); E[X|M] = Mπ

E[X] = Nπ p; E X 2 |M = M 2 π 2 + Mπ(1 − π); 2 E X = (Nπp)2 + Nπ 2 p(1 − p) + Nπ p(1 − π),

E[MX] = E{ME[X|M]} = π N 2 p2 + Np(1 − p) Cov[X, Y] = E[X(M − X)] − E[X]E[M − X] = −Np2 π(1 − π).

1.2 Pr{X = x|Y = y} =

1/x [1/y + · · · + 1/N] 2 2v − 1 2 = 2v − 1

1.3 (a) Pr{U = u|V = v} =

for 1 ≤ y ≤ x ≤ N.

for 1 ≤ u < v ≤ 6 for 1 ≤ u = v ≤ 6.

Pr{S = s, T = t}

(b) t/s 0 1 2 3

2 1 36

3

4 1 36

2 36

5

6 1 36

2 36

7

2 36 2 36

2 36 2 36

4 5 1 1 5 N ; E[X] = 20 × × = .

2 4 2 2 20 3 1.5 Pr{X = 0} = = .00317.

4 1.4 E[X|N] =

An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

8 1 36

9

2 36 2 36

2 36 2 36

11

2 36 2 36

2 36 2 36

2 36

10 1 36

12 1 36

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Instructor Solutions Manual

1.6 Pr{N = n} =

n 1 2

n = 1, 2, . . .

n 1 n k 2

∞ P 1 n n Pr{X = k} = 4 n=1 k

∞ P 1 n 1 = Pr{X = 0} = 3 n=1 4

∞ P 1 n 4 Pr{X = 1} = n = .

4 9 n=1

Pr{X = k|N = n} =

E[X|N] =

0≤k≤n

N 1 ; E[X] = E[N] = 1.

2 2

1.7 Pr{True S.F.|Diag. S.F.} =

.30(.85) = .51 .30(.85) + .70(.35)

1 2 2 3 Pr{Both Red} 2 = .

1.8 Pr{First Red|Second Red} = = 1 2 Pr{Second Red} 3 +1 1 2

1.9 Pr{X = k} =

∞ P

3

2

3

Pr{X = k|N = n}Pr{N = n}

n=k−1

∞ P 1 e−1 e−1 1 1 −1 = = = e − for k = 0, 1 . . . (n + 1)! (n + 2)! k! n=k−1 n + 2 n! n=k−1 ∞ P

(Poisson, mean = 1).

1.10 A “typical” distribution of 8 families would break down as follows: Family Children

#1 {G}

#2 {G}

#3 {G}

#4 {G}

#5 {B, G}

#6 {B, G}

#7 {B, B, G}

#8 {B, B, B}

(a) Let N = the number of children in a family 1 1 Pr{N = 1} = , Pr{N = 2} = Pr{N = 3} = .

2 4 (b) Let X = the number of girl children in a family.

1 7 Pr{X = 0} = , Pr{X = 1} = 8 8 (c) Family #8 is three times as likely, and family #7 is twice as likely to be chosen as families #5 and #6.

3 Pr{No sisters} = Pr{#8} = .

7 4 Pr{One sister} = .

7 3 Pr{Two brothers} = Pr{#8} = 7 2 Pr{One brother} = Pr{#7} = .

7 2 Pr[No brothers] = .

7

Instructor Solutions Manual

9

2.1 (a) The bid X1 = x is accepted if x > A; Otherwise, if 0 ≤ x ≤ A, one finds where one started due to the independent and identically distributed nature of the bids.

E[XN |X1 = x] =

x M

if x > A; if 0 ≤ x ≤ A

(b) The given equation results from the law of total probability: Z∞ E[XN ] = E [XN |X1 = x] dF(x).

0

(c) When X is exponentially distributed the conditional distribution of X, given that X > A, is the same as the distribution of A + X. (d) When dF(x) = λe−λx dx, then α=e

−λA

Z∞ Z∞ 1 1 −λx and xλe dx = Aα + α yλe−λy dy = α A + and M = A + . λ λ A

0

2.2 The distribution for the sum is p(2) = p(12) =

1 36

p(5) = p(9) =

4 36

p(3) = p(11) =

2 36

p(6) = p(8) =

5 − (.02)2 36

p(4) = p(10) =

3 36

p(7) =

6 + 2(.02)2 36

Pr{Win} = .4924065.

3.1 (a) v = τ 2 = λ, µ = p σ 2 = p(1 − p) E[Z] = λp; Var[Z] = λp(1 − p) + λp2 = λp. (b) Z has the Poisson distribution.

M P

n!

M! pk (1 − p)n−k qn (1 − q)M−n n! (M − n)! n=k k! (n − k)!

M = (qp)k (1 − qp)M−k for k = 0, 1, . . . , M. k

3.2 Pr{Z = k} =

1−α 2 1−α 1−α ,τ = ; E[Z] = 0 Var[Z] = .

2 α α α

(b) E z3 = 0 E z4 = E N 2 + 2N(N − 1)

(1 − α)2 1 − α 1−α (1 − α)(6 − 5α) + =3 − 2 = 2 2 α α α α2

3.3 (a) µ = 0, σ 2 = 1, v =

3.4 (a) E[N] = Var[N] = λ;

E [SN ] = λµ Var [SN ] = λ µ2 + σ 2 .

λ = λ(1 + λ) p

E [SN ] = λµ Var [SN ] = λ µ2 + σ 2 + λ2 µ2

(b) E[N] = λ,

Var[N] =

(c) For large λ, Var [SN ] ∝ λ in (a), while Var [SN ] ∝ λ2 in (b).

10

Instructor Solutions Manual

3.5 E[Z] = (1 + v)µ;

Var[Z] = vσ 2 + µ2 τ 2 .

R1 1 4.1 E [X1 ] = E [X2 ] = pdp = 2 0 R1 1 E [X1 X2 ] = p2 dp = 3 0 Cov [X1 , X2 ] = E [X1 X2 ] − E [X1 ] E [X2 ] =

1 .

12

4.2 Pr{X = i, Y = j} = Pr{X = i, N = i + j} = Pr{X = i|N = i + j}Pr{N = i + j}

i+j i = p (1 − p) j λi+j e−λ /(i + j)! i )( ( ) λ(1 − p) j e−λ(1−p) (λp)i e−λp , i, j = 0. = i! j!

Pr{X = i} =

∞ P

Pr {X = i, Y = j} =

j=0

(λp)i e−λp , i!

[λ(1 − p)] j e−λ(1−p) j! and Pr{X = i, Y = j} = Pr{X = i}Pr{Y = j} for all i, j.

Pr{Y = j} =

4.3 (a) Pr{X = i} =

R∞ λi e−λ i!

0

(b) f (λ|X = k) =

θe−θ λ dλ =

θ R∞ i −(1+θ )λ λe dλ = i! 0

(1 + θ)k+1 λk e−(1+θ )λ k!

θ 1+θ

1 1+θ

i

, i = 0, 1, . . . (Geometric distribution).

(Gamma).

θ R∞ i+j −(z+θ )λ λ e dλ i! j! 0

i+j

θ 1 i+j , i, j ≥ 0. = i 2+θ 2+θ

n

1 θ n 0 ≤ i ≤ n.

Pr{X = i, N = n} = i 2+θ 2+θ

n n

n P n θ θ 1 2 Pr{N = n} = = , n ≥ 0.

2+θ 2 + θ i=0 i 2+θ 2+θ n 1 n Pr{X = i|N = n} = , 0 ≤ i ≤ n. i 2

4.4 Pr{X = i, Y = j} =

4.5 Pr{X = 0, Y = 1} = 0 6= Pr{X = 0}Pr{Y = 1} so X and Y are NOT independent.

E[X] = E[Y] = 0,

Cov[X, Y] = E[XY] =

4.6 Pr{N > n} = Pr {X0 is the largest among {X0 , . . . , Xn }} = largest. Or, one can integrate

1 1 2 + − = 0.

9 9 9

1 since each random variable has the same chance to be the n+1

Z∞ Z∞ Z1 n n Pr{N > n} = [1 − F(x)] f (x)dx = − [1 − F(x)] d[1 − F(x)] = yn dy = 0

0

0

1 1 1 Pr{N = n} = Pr{N > n − 1} − Pr{N > n} = − = . n n + 1 n(n + 1) ∞ ∞ X X 1 Pr{N > n} = E[N] = = ∞. n+1 n=0

n=0

1 n+1

Instructor Solutions Manual

11

4.7 fX,Z (x, z) = α 2 e−αz fX|Z (x|z) =

for 0 ≤ x ≤ z.

fx,z (x, z) α 2 e−αz 1 = 2 −αz = , fz (z) z α ze

0 < x < z.

X is, conditional on Z = z, uniformly distributed over the interval [0, z].

4.8 The key algebraic step in simplifying the joint divided by the marginal is y − µY Q(x, y) − σY

2

1 = 1 − ñ2

(

x − µX σx

2

x − µX − 2ñ σX

y − µY σY

+ ñ

2

y − µY σY

2 y − µY x − µX − ñ σX σY

2 σX 1

− µ = x − µ − ñ (y ) X Y σY 1 − ñ2 σX2 =

1 1 − ñ2

5.1 Following the suggestion E [Xn+2 |X0 , . . . , Xn ] = E [E {Xn+2 |X0 , . . . , Xn+1 } |X0 , . . . , Xn ] = E [Xn+1 |X0 , . . . , Xn ] = Xn 5.2 E [Xn+1 |X0 , . . . , Xn ] = E [2Un+1 Xn |Xn ] = Xn E [2Un+1 ] = Xn .

5.3 E [Xn+1 |X0 , . . . , Xn ] = E 2e−εn +1 Xn |Xn

= Xn E 2e−εn +1 = Xn .

ξn+1 Xn 5.4 E [Xn+1 |X0 , . . . , Xn ] = E Xn p

ξn + 1 = Xn E = Xn lim n→∞ ρ 5.5 (b) Pr {Xn ≥ N for some n ≥ 0|X0 = i} ≤

Xn = 0.

E (X0 ) i = . (In fact, equality holds. See III, 5.3) N N

2 )

Chapter 3 !

k 1

1.1 P55 = 1, PK,K+1 = α

5−k 1 ! , 5 2 !

k = 1, 2, 3, 4

Pij = 0, otherwise.

1.2 (a) Pr {X0 = 0, X1 = 0, X2 = 0} = P0 P200 = 1(1 − α)2 = (1 − α)2 . (b) Pr {X0 = 0, X2 = 0} = P00 P00 + P01 P10 = (1 − α)2 + α 2 .

1.3 Pr {X2 = G, X3 = G, X4 = G, X5 = D | X1 = G} = P3GG PGD = α 3 (1 − α).

1.4

0 0 1 P= 2 3

1

2

3

.1 .3 .2 .4 0 .4 .2 .4 0 0 .6 .4 0 0 0 1 (0)

2.1 Observe that the columns of P sum to one. Then, b pi = 1/4 for i = 0, 1, 2, 3 and P 1 1P (n+1) (n) Pki = . by induction pi = 3k=0 pk Pki = 4 4

1 (5) 2.2 P00 = (1 − α)5 + 10(1 − α)3 α 2 + 5(1 − α)α 4 = 1 + (1 − 2α)5 2 (3)

2.3 P11 = .684772 2.4 (0, 0) (0, 1) (1, 0) (1, 1)

(0, 0)

(0, 1)

(1, 0)

(1, 1)

α 0 α 0

1−α 0 1−α 0

0 1−β 0 1−β

0 β 0 β (3)

2.5 Pr {X3 = 0 | X0 = 0, T > 3} = P00 3.1

P=

0

1

2

3

0

0

0

0

1

1

1 15

14 15

0

0

2

0

8 15

7 15

0

3

0

0

3 5

2 5

.

(3) (3) P00 + P01 = .6652.

An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

Instructor Solutions Manual

3.2

P=

13

0

1

2

3

0

0

0

0

1

1

0

0

1 2

1 2

2

0

1 4

1 2

1 4

3

1 8

3 8

3 8

1 8

0

1

2

3.3 (a) 0 P=1 2

.1 .4 .5 .5 .5

0

.1 .4 .5

(b) Long run lost sales per period = (.1) π1 = .0444. . . .

3.4 Pr{Customer completes service | In service} = Pr{Z = k | Z ≥ k} = α.

P01 = β, P00 = 1 − β, Pi,i+1 = β(1 − α) Pi,i−1 = (1 − β) α, Pii = (1 − β) (1 − α) + αβ, i = 1.

3.5

0 1 2

H 1 2

H

1 2

HH HHT

0

P=

HH

HHT

0

0

0

1 2

0

0

0

1 2

1 2

0

0

0

1

3.6 P(a,b),(a,b+1) = 1 − p

a ≤ 3, b ≤ 2

P(a,b),(a+1,b) = p

a ≤ 2, b ≤ 3

P(3,b),A wins = p

b≤3

P(a,3),B wins = 1 − p,

a≤3

PA wins, A wins = PB wins, B wins = 1.

3.7 P0,k = αk for k = 1, 2, . . .

P0,0 = 0 Pk,k−1 = 1

h≥1

3.8 Pk,k−1 = q Nk Pk, k+1 = p N−k N

.

Pk,k = p Nk + q N−k N

14

Instructor Solutions Manual

k N −k 3.9 Pk,k−1 = Pk,k+1 = N N 2 2 N −k k + Pk,k = N N 3.10 (a) ξ

=2 3 4 0 2 1 2 2 n =1 2 3 4 5 6 7 8 Xn = 2 −1 0 4 2 1 2 0

(b) P41 = Pr {ξ = 3} = .2 P04 = Pr {ξ = 0} = .1 4.1 0 = 0

1=H

2 = HH

0 1 2

1 1 2

2

3

0

0

1

1 2

0

1 2

0

2

0

0

1 2

1 2

3

0

0

0

1

0

P=

3 = HHT

 1 1  v0 = 1 + v0 + v1   v0 = 8 2 2      1 1 v1 = 1 + v0 + v2 v1 = 6 2 2       1  v2 = 2 v2 = 1 + v2 2 0=0

1=H

2 = HT

0 1 2

1 1 2

2

3

0

0

1

0

1 2

1 2

0

2

1 2

0

0

1 2

3

0

0

0

1

0

3 = HTH

 1 1  v0 = 1 + v0 + v1   v0 = 10 2 2       1 1 v1 = 1 + v1 + v2 v1 = 8 2 2        1  v2 = 6 v2 = 1 + v0 2

Instructor Solutions Manual

15

4.2 v1 = 1

v1 = 1 1 2

1 1 3 1 1 v3 = 1 + + = 1+ + 3 3 2 2 3

1 v2 = 1 + v1 2 1 1 v3 = 1 + v1 + v2 3 3 .. . vm = 1 +

v2 = 1 +

1 1 1 v1 + v2 + · · · + vm−1 m m m

1 1 1 + + · · · + ∼ Log m 2 3 m 1 for i > j.

Note: wjj = 1 and wij = j+1

Solve to get vm = 1 +

P 1 1 2j − 3 solves vm = 1 + m−1 4.3 We will verify that vm = 2 1 + + ··· + j=1 m2 vj . Change variables to Vk = kvk . To 2 m

1 2 1 − 3k solves Vm = m + (V1 + · · · + Vm−1 ). Using the given Vk , use sums of numshow: Vk = 2 (k + 1) 1 + + · · · + 2 k m bers and interchange order as follows:

m−1 X k=1

Vk =

=

=

m−1 k XX k=1 l=1 m−1 X l=1 m−1 X l=1

m+1 m

m−1

X 1 2(k + 1) − 3 k l k=1

m−1

3m(m − 1) 2X (k + 1) − l 2 k=l

3m(m − 1) 2 m(m + 1) l(l + 1) − − . l 2 2 2

Then, m+

m−1

m−1

k=1

l=1

X1 2X 2(m + 1) Vk = 2(m + 1) − 3m + = Vm m l m

as was to be shown.

As in Problem 4.2, vm ∼ log m.

4.4 Change the matrix to 0

1

2

3

0 1

0

0

0

1 .1

.2

.5

.2

2 0

0

1

0

3 .2

.2

.3

.3

and calculate the probability that absorption takes place in State O: u10 = .1 + .2u10 + .2u30

)

u10 = .2115

u30 = .2 + .2u10 + .3u30 u30 = .3462

16

Instructor Solutions Manual

4.5

1

2 1 2

3

4 1 2

5

6

7

0

0

0

1

0

2

1 3

0

1 3

0

1 3

0

0

3

0

0

1

0

0

0

0

4

1 3

0

0

0

1 3

0

1 3

5

0

1 3

0

1 3

0

1 3

0

6

0

0

1 2

0

1 2

0

0

7

0

0

0

0

0

0

1

0

7 12 3 u23 = 4

2 3 5 u63 = 6

u13 =

u53 =

4.6 v0 = 1 + qv0 + pv1 v1 = 1 + qv0 + pv2 v2 = 1 + qv0 + pv3 v3 = 1 + qv0

u43 =

5 12

q = 1−p

gives v0 = (1 + qv0 )(1 + p + p2 + p3 ) = (1 + p + p2 + p3 ) + v0 (1 − p4 ) So v0 =

1 + p + p2 + p3 1 − p4 = 4 . p (1 − p) 1 − 1 − p4

4.7 The stationary Markov transitions imply that E Now use the law of total probability.

4.8

0

1

2

3

4

5

0

1

0

0

0

0

0

1

1 4

3 4

0

0

0

0

2

0

2 5

3 5

0

0

0

3

0

0

1 2

1 2

0

0

4

0

0

0

4 7

3 7

0

5

0

0

0

0

5 8

3 8

Xn = # of red balls in use at nth draw. v5 = 4 +

7 8 5 + 2 + + = 11.85.

2 4 5

P∞

n n=1 β c (Xn ) |X0 = i, X1 = j

= βhj while E β 0 c (X0 ) |X0 = i = c (i) .

Instructor Solutions Manual

17

0

1

2

3

4

5

0

1

0

0

0

0

0

1

1 8

7 8

0

0

0

0

2

0

2 8

6 8

0

0

0

3

0

0

3 8

5 8

0

0

4

0

0

0

4 8

4 8

0

5

0

0

0

0

5 8

3 8

4.9

Xn = # of red balls in use v5 =

8 8 8 8 8 4 + + + + = 18 = 18.266. . . .

1 2 3 4 5 15

4.10 Let Xn be the number of coins that fall tails in the nth toss, X0 = 5. Make “1” an absorbing state.

0

1

2

3

4

5

0

1

0

0

0

0

0

1

0 1 4 1 8 1 16 1 32

1 1 2 3 8 4 16 5 32

0 1 4 3 8 6 16 10 32

0

0

0

0

0

0

0

0

2 P=

3 4 5

1 8 4 16 10 32

1 16 5 32

0 1 32

U51 = .7235023041 . . .

4.11 Label the states (x, y) where x = # of red balls and y = # of green balls, “win” = (1, 0), “lose” = {(0, 2), (2, 0)} .

(2, 2)

(1, 2) 1 2

(2, 1) 1 2

(1, 1) win

(2, 2)

0

(1, 2)

0

0

0

(2, 1)

0

0

0

(1, 1)

0

0

0

0

win

0

0

0

lose

0

0

0

1 U(2, 2), win = .

3

0 2 3 2 3

0

lose 0

0

1 2 1

1 3 1 3 1 2 0

0

0

1

0 0

18

Instructor Solutions Manual

4.12

0 0 1 win lose

1

win

lose

.3 .2 .5 .1 0 0 0 0

0 .4 1 0

.5 8 0 z0 = U0 ,win = 53 0 1

4.13 Use the matrix 0

1

2

0

1

2

0

0

0

0

.3 .2

.5

1

0

0

0

.5 .1

.4

2

0

0

1

0

0

0

0

.3

.2 .5

0

0

0

1

.5

.1 .4

0

0

0

2

0

0

0

0

1

0

U0,2 = .67669. . .

4.14 The long but sure way to solve the problem is to set up the 36 × 36 transition matrix for the Markov chain Zn = (Xn−1 , Xn ), making states (x, y) where x + y = 5 or x + y = 7 absorbing. However, all that is important prior to ending the game is the last roll, and by aggregating states where possible, we get the Markov transition matrix.

1 or 2 1 3

3 or 4 1 6

5 or 6 1 6

5 1 6

7 1 6

3 or 4

1 6

1 6

1 3

1 6

1 6

5 or 6

1 6

1 3

1 3

0

1 6

5

0

0

0

1

0

7

0

0

0

1

1

U5,5 =

15 51

1 or 2

U1,5 =

22 51

U3 5 =

Pr {5 before 7} = 4.15

1

21 51

1 59 U1,5 + U3 5 + U5,5 = .

3 153

2

3

4

5

1

.96 .04

0

0

0

2

0

.94 .06

0

0

3

0

0

.94 .06

0

4

0

0

0

.96 .04

5

0

0

0

0

v1 = 133

1 3

1

Instructor Solutions Manual

4.16

19

0

1

2

3

4

5

0

1

0

0

0

0

0

1

.2

0

.8

0

0

0

2

0

.4

0

.6

0

0

3

0

0

.6

0

.4

0

4

0

0

0

.8

0

.2

5

0

0

0

0

0

1

U35 =

17 .

32

4.17 Let ϕi (s) = E sT | X0 = i for i = 0, 1 Then ϕ0 (s) = s [.7ϕ0 (s) + .3ϕ1 (s)] ϕ1 (s) = s [.6ϕ1 (s) + .4] which solves to give .4s 1 − .6s

.4s .3s ϕ0 (s) = .

1 − .6s 1 − .7s ϕ1 (s) =

h w (and h → h − 1) or Xn+1 = w − 1 w · pr · w+h w+h (and h → h + 1) where h = 2N − n − 2w is the number of half cigars in the box.

The recursion to verify is

4.18 The transition probabilities are: if Xn = w, then Xn+1 = w

vn (w) =

w · pr ·

w h vn+1 (w) + vn+1 (w − 1) w+h w+h

This is straight forward, and easiest if one first does the algebra to verify 1 1+w n + 2w − 2N vn+1 (w − 1) = vn (w) + w(1 + w)

vn+1 (w) = vn (w) +

n N−1 ∞ NX 1 n 1 4.19 PN = .

1− 2 2 2 n=0

As a function of N, this does NOT converge but oscillates very slowly (cycles ∝ log N) and very slightly about the (Cesaro 1 limit) 2 log 2.

5.1 v0 = 1 + a0 v0 + a1 v1 + a2 v2 v1 = 1 + (a0 + a1 )v1 + a2 v2

v2 = 1 + (a0 + a1 + a2 )v2

1 v2 = 1−(a0 +a = a13 = v0 = v1 .

1 +a2 )

20

Instructor Solutions Manual

5.2 (a) P0 = a1 , q0 = 1 − a1 = (a2 + a3 + · · · ).

Pk = ak+1 /(ak+1 + ak+2 + · · ·), for k ≥ 1. (b) P0 = a1 , q0 = 1 − a1 Pk = ak+1 /(ak+1 + ak+2 + · · ·) , qk = 1 − pk PN−1 = 1

for 1 ≤ k < N − 1.

The above assumes instantaneous replacement.

5.3 0 P=1 2

0

1

2

α 0 1−α

1−α α 0

0 1−α α

5.4 After reviewing (5.8), set up the matrix 0

1 1 6

2 1 6

3 1 6

4 1 6

5 1 6

6 1 6

7

8

9

10

13 ∼ 13

0

0

0

0

0

0

0

0

1

0

0

1 6

1 6

1 6

1 6

1 6

1 6

0

0

0

0

0

2

0

0

0

1 6

1 6

1 6

1 6

1 6

1 6

0

0

0

0

3

0

0

0

0

1 6

1 6

1 6

1 6

1 6

1 6

0

0

0

4

0

0

0

0

0

1 6

1 6

1 6

1 6

1 6

1 6

0

0

5

0

0

0

0

0

0

1 6

1 6

1 6

1 6

1 6

0

1 6

6

0

0

0

0

0

0

0

1 6

1 6

1 6

1 6

0

2 6

7

0

0

0

0

0

0

0

0

1 6

1 6

1 6

1 6

2 6

8

0

0

0

0

0

0

0

0

0

1 6

1 6

1 6

3 6

9

0

0

0

0

0

0

0

0

0

0

1 6

1 6

4 6

10

0

0

0

0

0

0

0

0

0

0

0

1 6

5 6

13

0

0

0

0

0

0

0

0

0

0

0

1

0

∼ 13

0

0

0

0

0

0

0

0

0

0

0

0

1

U0,13 = .1819

1 5.5 If Xn = 0, then Xn+1 = 0 and E[Xn+1 | Xn = 0] = 0 = Xn . If Xn = i > 0 then Xn+1 = i ± 1, each w. pr. so 2 E [X0 ] i E[Xn+1 | Xn = i] = i = Xn . So {Xn } is a martingale. If X0 = i, then Pr {max Xn ≥ N} ≤ = by the maximal inequality.

N N i Of course, (5.13) asserts Pr {max Xn ≥ N} = .

N

Instructor Solutions Manual

6.1 v1 = 1 + .7v2

21

)

v1 = 170 93 = 1.827957 v2 = 1 + .1v1 3 1 10 80 ñ1 = , ñ2 = , 81 = , 82 = 7 21 7 63 81 + 82 v1 = = 1.827957.

1 + ñ1 + ñ2  6.2 v0 = 1 + αv0 + βv2    1 + β + β2 v1 = 1 + αv0 . v0 =  β3   v2 = 1 + αv0 + βv1 3abc . The answer is unknown for four urns. a+b+c 7.1 Observe that each state is visited at most a single time so that Wij = Pr {Ever visit j |X0 = i } . Clearly then Wi,j = 1 and Wi,j = 0 for j > i.

6.3 For three urns, v(a, b, c) = E [T] =

Wi,i−1 = Pi,i−1 =

1 i

Wi,i−2 = Pi,i−2 + Pi,i−1 Wi−1,i−2 =

1 1 + i i

1 i−1

=

Wi,i−3 = Pi,i−3 + Pi,i−2 Wi−2,i−3 + Pi,i−1 Wi−1,i−3

1 1 1 1 = 1+ + = . i i−2 i−2 i−2

1 . i−1

1 Continuing in the manner, we deduce that Wi,j = j+1 for 1 ≤ j < i.

7.2 Observe that each state is visited at most a single time so that Wij = Pr {Ever visit j |X0 = i } . Clearly, then Wi,i = 1 and Wi,j = 0 for j > i. We claim that k+1 Wk,j = 2 k

j for 1 ≤ j < k. ( j + 1) ( j + 2)

satisfies the first step equation Wk,j = Pk,j +

k−1 P

Pk,l Wl,j .

l=j+1

Evaluating the right side with the given Pk,l and Wl,j we get

k−1 X 2j 2l l+1 j R.H.S. = 2 + 2 l k k2 ( j + 1) ( j + 2) l=j+1

=

k−1 X 2j 2j 2 + (l + 1) k2 k2 ( j + 1) ( j + 2) l=j+1

2j 2j 2 k (k + 1) ( j + 1) ( j + 2) + − 2 2 k2 k2 ( j + 1) ( j + 2)

j k+1 =2 = Wk,j as claimed.

2 ( j + 1) ( j + 2) =

7.3 Observe that for j transient and k absorbing, the event {Xn−1 = j, Xn = k} is the same as the event {T = n, XT−1 = j, XT = k}, whence ∞ ∞ X X (n−1) Pr {XT−1 = j, XT = k |X0 = i } = Pr {Xn−1 = j, Xn = k |X0 = i } = Pi,j Pjk = Wi,j Pjk . n=1

n=1

22

Instructor Solutions Manual

1 7.4 (a) Wi,j = j+1 for 1 ≤ j < i. (See the solution to 7.1). Using 7.3, then,

1 (b) Pr {XT−1 = j |X0 = i } = Wi,j Pj,0 = (j+1)j , 1 ≤ j < i, and Pr {XT−1 = i |X0 = i } = Pi,0 = 1i . h i π 7.5 vk = BN,k (N + 1)2 − (N − 2k)2 2

where BN,k = 2−2N

2 (N − k) N −k

2k . k

8.1 No adults survive if a local catastrophe occurs, which has probability 1 − β, and if independently, all N dispersed offspring fail to survive, which has probability (1 − α)N . Another example in which looking at mean values alone is misleading.

8.2 A first step analysis yields E [Z] = 1 + µ E [Z] whence E [Z] = 1/(1 − µ), µ < 1.

8.3 Possible families: GG GB BG BBG BBBG . . .

1 1 1 1 1 ...

Probability: 4 4 4 8 16

Let N = Total children, X = Male children k 1 3 (a) for k = 3.

Pr {N = 2} = , Pr {N = k} = 4 2 1 Pr {X = 1} = , 2

(b)

1 Pr {X = 0} = , 4

k+1 1 Pr {X = k} = , k = 2.

2

1 1 8.4 E [Zn+1 | Z0 , . . . , Zn ] = µn+1 E [ξ1 + +ξx | Xn = µn Zn ] = µn+1 µXn = Zn . Suppose 1 Z0 = X0 = 1. Pr {Xn is ever greater than kµn } ≤ . k 9.1 ξ = # of male children

k= 0

1

2

3

9 3 11 9 32 32 32 32 11 9 9 3 ϕ(s) = + s + s2 + s3 .

32 32 32 32

Pr {ξ = k} =

u = ϕ(u) has smallest solution u∞ = .76887.

9.2 ξ = # of male children k= 0

1

2

3

1 3 3 3 1 + 32 32 4 32 32 1 27 3 2 1 3 ϕ (s) = + s+ s + s .

32 32 32 32 Pr {ξ = k} =

u = ϕ(u) has smallest solution u∞ = .23607.

9.3 Following the hint Pr {X = k} = 9.4 ϕ (1) =

Z

π (k |λ )f

P∞

k=0 pk = 1,

(λ)

α k 0 (k + α) θ 1 dλ = , k! 0 (α) 1 + θ 1+θ

p0 = ϕ (0) = 1 − p.

Refering to Equations I, (6.18) and (6.20). ∞ X β ϕ (s) = 1 − p (−s)k , k k=0

k = 0, 1, . . . .

Instructor Solutions Manual

so

23

β (1 − β) . . . (k − 1 − β) β pk = −p ≥ 0 because 0 < β < 1. (−1)k = p k k! n=1

2 3 4

1 1 1 2 1 1 2 1 4 1 1 2 1 4 1 8 Pr {All red} = 4 4 4 4 4 4 4 4 4 4 2n−1 1 Pr {All red at generations} = .

4 1 2 1 1 (b) Pr {Culture dies out} = Pr {Red cells die out} ϕ (s) = + s + s2 . Smallest solution to u = ϕ(u) is u∞ = .

12 3 4 3

9.5 (a)

9.6 s = ϕ (s) is the quadratic 0 = as2 − (a + c)s + c whose smallest solution is u∞ = ac < 1 if c < a.

9.7 Family

GG GB BG BBG BBBG 1 1 1 1 1 Probability 4 4 4 8 16 Let X = # of male children.

1 1 Pr {X = 1} = 4 2 k+1 1 Pr {X = k} = for k = 2.

2 ∞ k+1 X 1 1 1 s2 1 1 1 sk = + s + . ϕ (s) = + s + 4 2 2 4 2 4 2−s Pr {X = 0} =

k=2

1 1 2s (2 − s) + s2 5 E [X] = ϕ 0 (1) = . + 2 4 (2 − s)2 4 k+1 1 3 Pr {X > k} = Pr {X > 0} = 4 2 ∞ X 5 E [X] = Pr {X > h} = 4 ϕ 0 (s) =

k=0

9.8 ϕ (s) = (1 − c)

∞ P k=0

(cs)k =

1−c .

1 − cs

s = ϕ (s) is a quadratic whose smallest solution is u∞ =

1 1−c provided c > . c 2

9.9 (a) Let X = # of male children.

1 3 1 5 Pr {X = 0} = + = 4 4 2 8

3 1 k+1 Pr {X = k} = for k = 1.

4 2

∞ s k 5 3P 5 3 s (b) ϕ (s) = + = + .

8 8 k=1 2 8 8 2−s u0 = 0 un = ϕ (un−1 ) u5 = .9414 9.10 (a) U∞ is smallest solution to u = f [g(u)].

n/2 (b) f 0 (1)g0 (1) when n = 0, 2, 4, . . . f 0 (1)(n+1)/2 g0 (1)(n−1)/2 when n = 1, 3, 5, . . .. (c) Yes, both change (unless f 0 (1) = g0 (1) in b).

Chapter 4

1.1 Let Xn be the number of balls in Urn A. Then {Xn } is a doubly stochastic Markov chain having 6 states, whence limn→∞ Pr{Xn = 0|X0 = i} = 16 .

1.2 Let Xn be the number of balls in Urn A.

0 1

0

1

2

3

4

5

0 1 5

1

0 4 5

0

0

0

0

0

0

0

2

0

2 5

0

3 5

0

0

3

0

0

3 5

0

2 5

0

4

0

0

0

5

0

0

0

π0 = π5 =

1 ; 32

π1 = π4 =

P=

4 5 0

1 5 0

0 1

π0 =

5 ; 32

π2 = π3 =

10 32

1 32

1.3 The equations for the stationary distribution simplify to: π0 = (α1 + α2 + α3 + α4 + α5 + α6 ) π0 π1 = (α2 + α3 + α4 + α5 + α6 ) π0 π2 = (α3 + α4 + α5 + α6 ) π0 π3 = (α4 + α5 + α6 ) π0 π4 = (α5 + α6 ) π0 π5 = α6 π0 !

6 X 1= kαk π0

X

αi = 1

k=1

π0 = P6

1

k=1 kαk

=

1 Mean of α distribution

1.4 Formally, one can look at the Markov chain Zn = (Xn , Xn+1 ) and ask for limn→∞ Pr{Zn = (k, m)} = lim Pr{Xn = k, Xn+1 = m} = πk Pkm .

An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

Instructor Solutions Manual

A 0

A 1.5 P =

B C D

B 1 2

1 3 0 1 2

2 πA = ; 8

0 1 1 2

25

C

1 3 0

D 1 2 1 3 0

0

0

0

3 πB = ; 8

πC =

1 8

πD =

2 8

Note: πNODE ∝ # arcs at NODE.

1.6 lim Pr Xn+1 = j X0 = i = πj . n→∞

lim Pr Xn = k, Xn+1 = j X0 = i = πk Pkj . n→∞

1.7 π0 =

6 3 6 4 ; π1 = ; π2 = ; π3 = .

19 19 19 19

1.8 P8 has all positive entries. π0 = .2529 π1 = .2299 π2 = .3103 π3 = .1379 π4 = .0690 4 1 π2 = π3 = .

10 10 1.10 The matrix is doubly stochastic, whence 1 πk = for k = 0, 1, . . . , N. N +1 1.9 π0 = π1 =

62 117 = .3087 π3 = = .1636 379 379 143 = .3773. (b) π2 + π3 = 379 (c) π0 (P02 + P03 ) + π1 (P12 + P13 ) = .1559 Q Q Q Q Q Q Q Q Q Q Q n Q 1.12 (a) P = P = 2 = . Then Q2 = (P − )(P − ) = P2 − P − P + 2 = P2 − . Similarly, P = Pn =

Q Q Q Qn+1 Q Pn − = Pn+1 − . = and Qn+1 = P − n n 1 1 0 − 2 2 1 0 0 0 (b) Qn = 2 n n 1 1 − 0 2 2 1.11 (a) π0 =

π2 P21 6 × .2 11 = = .1714 π0 = 1.13 lim Pr Xn−1 = 2 Xn = 1 = n→∞ π1 7 24

1 ξ1 = 3

Period 4 3 q=2 1 0 −1 0

1

2 ξ2 = 4

3 ξ3 = 0

LOST

SALES 2

π1 =

4 ξ4 = 2

3

7 24

π2 =

6 .

24

5

4

5

26

Instructor Solutions Manual

(a) X0 = 4

0 1 (b) P = 2 3 4

X1 = 1 X2 = 0 X3 = 2 0

1

2

3

.6 .3 .1 .3 .1

.3 .3 .2 .3 .2

.1 .3 .3 .3 .3

0 0 .1 0 .3 .1 .1 0 .3 .1

X4 = 2

4

2.1 (c) π0 + π1 + π2 = .3559 + .2746 + .2288 = .8593 2.2 State is (x, y) where x = # machines operating and y = # days repair completed.

(2, 0) (2, 0) (1, 0) (a) P = (1, 1) (0, 0) (0, 1)

(1 − α)2 0 (1 − β) 0 0

(1, 0) 2α(1 − α) 0 β 0 1

(1, 1)

(0, 0)

(0, 1)

0 1−β 0 0 0

α2

0 β 0 1 0

0 0 0 0

(b) π(2,0) + π(1,0) + π(1,1) = .6197 + .1840 + .1472 = .9508.

2.3 (a) π0 = .2549; π1 = .2353; π2 = .3529; (b) π2 + π3 = .5098 (c) π0 (P02 + P03 ) + π1 (P12 + P13 ) = .2235.

0 0 1 2.4 P = 2 3

1

2

π3 = .1569

3

.1 .3 .2 .4 1 0 0 0 0 1 0 0 0 0 1 0

E[ξ ] = 2.9 π0 =

10 1 9 6 4 = ; π1 = ; π2 = ; π3 = .

29 E[ξ ] 29 29 29

(4)

(4)

2.5 (a) P(s,s)(s,s) + P(s,s),(c,s) = .3421 + .1368 = .4789. (b) π(s,s) + π(c,s) = .25 + .15 = .40.

2.6 To establish the Markov property and to get the transition probability matrix, use Pr{N = n|N = n} = β. π1 = 2.7 P00 = p;

β . p+β P0,1 = q = 1 − p;

Pii = αp + βq;

Pi,i+1 = aq

Pi,i−1 = βp for i = 1.

1 2.8 π(1,0) = 1+2p .

p = .01 π(1,0) = .9804

.02 .9615

.05 .9091

.10 .8333

Instructor Solutions Manual (0)

27

(1)

(k)

f00 = ab(1 − b)k−2 ,

3.1 (a) f00 = 0 f0,0 = 1 − a

k = 2.

(b) We are asked do show: (n)

P00 =

n X (k) (n−k) f00 P00

where n = 1 and

k=0 (n)

P00 =

b a + (1 − a − b)n . a+b a+b

Some preliminary calculations: (i)

n

n

k=2

k=2

i b X (k) b X ab h f00 = ab(1 − b)k−2 = 1 − (1 − b)n−1 a+b a+b a+b

n i a X (k) ab h (ii) f00 (1 − a − b)n−k = (1 − b)n−1 − (1 − a − b)n−1 a+b a+b k=2

Then

n X

(k) (n−k)

f00 P00

(1) (n−1)

= f00 P00

k=0

+

n X (k) (n−k) f00 P00 k=2

i ab h a b n−1 + + (1 − a − b) 1 − (1 − a − b)n−1 = (1 − a) a+b a+b a+b

b (1 − a)a ab = + − (1 − a − b)n−1 a+b a+b a+b

=

a b + (1 − a − b)n = Pn00 . a+b a+b

as was to be shown.

3.2 For a finite state a periodic irreducible Markov chain, Pij (n) → πj > 0 as n → ∞ for all i, j. Thus for each i, j, there exists Nij such that Pij (n) > 0 for all n > Ni,j . Because there are only a finite number of states N = maxi,j Ni,j is finite, and for n > N, we have Pi,j > 0 for all i, j.

3.3 We first evaluate n=1 (n)

P00 = 0 (n)

Because P00 = 1, (3.2) may be rewritten (n)

(n)

f00 = P00 −

n−1 X (k) (n−k) f00 P00 . k=1

(1)

(1)

Finally f00 = P00 = 0 (2)

(2)

(1) (1)

f00 = f00 − f00 P00 =

1 4

1 8

3 1 1 5 (4) f00 = − = 8 4 4 16

7 1 1 1 1 5 (5) f00 = − − = 32 8 4 4 8 32 (3)

f00 =

2 1 4

3 1 8

4 3 8

5 7 32

6 17 64

28

Instructor Solutions Manual

4.1 (a) π0 + π1 = 1 and (β, α)P = (β, α). (b) A first return to 0 at time n entails leaving 0 on the first step, staying in 1 for n − 2 transitions, and then returning to 0, (n) whence f00 = αβ(1 − β)n−2 for n = 2. (c) m0 =

∞ X

(n)

nf00 =

n=1

= 1+

∞ X

αβ

k=2

∞ X n ∞ X ∞ X X (n) (n) f00 = f00

n=1 k=1 ∞ X

k=1 n=k

(1 − β)n−2 = 1 + α

∞ X (1 − β)k−2

n=k

k=2

α+β 1 α = . = 1+ = β β π0 4.2 π0 = .1507 π1 = .3493 π2 = .1918 π3 = .3082 4.3 The period of the Markov chain is d = 2. While there is no limiting distribution, there is a stationary distribution. Set p0 = qN = 1. Solve. π0 = π0

π0 = π0

π0 = q1 π1

π1 =

π1 = p0 π0 + q2 π2

1 π2 = (π1 − p0 π0 ) = q2

p0 p1 p2 π3 = π0 q1 q2 q3

π2 = p1 π1 + q3 π3

1 π0 = q1

p0 π0 q1

p0 p1 π0 q1 q2

.. . πk = pk−1 πk−1 + qk+1 πk+1

πk+1 = ρk+1 π0

Where ρk+1 =

p0 p1 · · · · · pk . Upon adding q1 q2 · · · · · qk+1

1 = π0 + · · · + πN = (ρ0 + ρ1 + · · · · · ρN )π0 π0 =

1 1 = P Q k 1 + ρ1 + ρ2 + · · · · · +ρN 1+ N k=1

l=1

pl−1 ql

and πk = ρk π0 . ∞ P

4.4 π0 = α1 π0 + π1 π1 = α2 π0 + π2

π0 = π0 = αk π0 k=1 ∞ P π1 = (1 − α1 ) π0 = αk π0 k=2 ∞ P π2 = (1 − α1 − α2 ) π0 = αk π0 k=3

π2 = α3 π0 + π3 πn =

∞ P

! αk π0

k=n+1 ∞ X n=1

πn = 1

1 implies π0 = P∞ P∞ n=0

k=n+1 αk

P P∞ P∞ P∞ Let ξ be a random variableP with Pr(ξ = k) = αk . Then ∞ n=0 k=n+1 αk = n=0 Pr(ξ > n) = E[ξ ] = k=1 kαk . In order that π0 > 0 we must have ∞ kα = E[ξ ] < ∞.

Note: The Markov chain here models the remaining life in a renewal k k=1

Instructor Solutions Manual

29

process. The result says that, under the conditions α1 > 0, α2 > 0, a renewal occurs during a particular period, in the limit, with probability 1/mean life of a component. See Chapter VII. 4.5 Recall that a return time is always at leastP1. (a) Straight forward (b) To simplify, use πi = 1 and X

πi Pik = πk to get

i

X

πi mij = 1 +

i

and subtract

P

X

πk mkj = 1 +

k6=j

X

πi mij

i6=j

πi mij from both sides to get πi mij = 1.

i6=j

4.6

n o (n) n ≥ 1 : P00 > 0 = {4, 5, 8, 9, 10, 12, 13, 14, . . .} (19)

(20)

d(0) = 1 P−1,−4 = 0,

Pij

> 0 for all i, j.

4.7 Measure time in trips, so there are two trips each day. Let Xn = 1 if car and person are at the same location prior to the nth trip; = 0, if not. The transition probability matrix is

P=

0 1

0

1

0 1−p

1 p

p(1−p) In the long run he is not with car for π0 = 1−p 2−p fraction of trips, and walks in rain π0 p = 2−p fraction of trips. The fraction

of days he/she walks in rain is 2p(1−p) 2−p .

With two case, let Xn be the number of cars at the location of the person.

0 P=1 2

0

1

2

0 0 1−p

0 1−p p

1 p 0

2p(1−p) π0 = 1−p 3−p and fraction of days walk in rain is 2pπ0 = 3−p . Note that the person never gets wet if p = 0 or p = 1.

4.8 The equations are 1 1 1 1 π0 = π0 + π1 + π2 + π3 + · · · 2 3 4 5 1 1 1 1 π1 = π0 + π1 + π2 + π3 + · · · = π0 2 3 4 5 1 1 1 1 π2 = π1 + π2 + π3 + · · · = π0 − π0 3 4 5 2 1 1 1 1 π3 = π2 + π3 + · · · = π0 − π0 − π1 4 5 2 3 1 1 1 1 π4 = π3 + · · · = π0 − π0 − π1 − π2 5 2 3 4 Starting with the equation for π1 and solving recursively we get 1 1 πk = πk−1 − πk−2 = for k = 0. k k!

P e−1 πk = 1 ⇒ π0 = e−1 and πk = ,k = 0 k!

(Poisson, θ = 1).

30

Instructor Solutions Manual

5.1 The stationary distributions for PB and Pc are respectively, (π3 , π4 ) = hitting probabilities from the transient states to the recurrent classes are 3−4

5−7

.4569 .5431 .1638 .8362 .4741 .5259

0 U=1 2 This gives

0

P¥ =

1

0 1 2 3

2

3

O

4 5 6 7

4

5

6

7

.2284 .2284 .2296 .1288 .0819 .0819 .3534 .1983 .2371 .2371 .2223 .1247 1 1 2 2 O 1 1 2 2 .4227 .2371 O .4227 .2371 .4227 .2371

.1848 .2845 .1789

5.2 Aggregating states {3, 4} and {5, 6, 7}, 0

1

2 3−4 5−7

.1 0 0 1 .5 2 3−4 0 5−7 0

.2 .1 0 0 0

.1 .2 0 0 0

0 U=1 2

3−4

5−7

.44 .23 .52

.56 .77 .48

.3π3 + .6π4 = π3 π3 + π4 = 1 (∞)

P04 = U0,3−4 0

P∞ =

.3 .1 .3 1 0

.3 .6 .2 0 1

6 = .46 13 ⇒ 7 π4 = = .54 13 π3 =

π4 = .44 × .54 = .24 3

4

5

6

7

0 1 2

.20 .10 .24

.24 .12 .28

.25 .35 .21

.15 .20 .12

.16 .22 .14

3 4

.46 .46

.54 .54 .45 .45 .45

.26 .26 .26

.29 .29 .29

5 6 7

1

2

.3402 .3402 .3402

1 1 2, 2

and (π5 , π6 , π7 ) = (.4227 .2371 .3402). The

Chapter 5

1.1 Following the hint: Z1 Pr{X = k} =

k k−1 −λ(1−x) λ x

e

(k − 1)!

0

−λx

e

λk e−λ dx = (k − 1)!

Z1

xk−1 dx =

λk e−λ k!

0

1.2 The number of defects in any interval in Poisson distributed, from Theorem 1.1, and the numbers of defects in disjoint intervals are independent random variables because its truth for both major and minor types.

1.3 gX (s) =

∞ X µk e−µ k=0

k!

sk = e−µ

∞ X (µs)k k=0

k!

= e−µ eµs = e−µ(1−s) , |s| < 1.

X+Y

= E sX sY 1.4 gN (s) = E s

= E sX E sY (using II, (1, 10), (1, 12)). = gX (s)gY (s).

In the Poisson case (Problem 1.3) gN (s) = e−α(1−s) e−β(1−s) = e−(α+β)(1−s) .

1.5 (a)

(Poisson, θ = α + β).

1 3 3 λh − 21 λ2 h2 + 3! λ h 1 − p0 (h) 1 − e−λh = = h h h 1 1 = λ − λ2 h + λ2 h2 − · · · → λ as h → 0.

2 3!

(b)

p1 (h) λhe−λh = = λe−λh → λ as h → 0. h h

(c)

1 2 2 −λh λ h e p2 (h) 1 = 2 = λ2 He−λh → 0 as h → 0. h h 2

1.6 Pr{X(t) = k, X(t + s) = n} = Pr{X(t) = k, X(t + s) − X(t) = n − k} =

(λt)k e−λt (λs)n−k e−λs k! (n − k)!

(λt)k e−λt [λs]n−k e−λs k! (n − k)!

Pr{X(t) = k|X(t + s) = n} = [λ(t + s)]n e−λ(t+s) n!

k

t s n−k n = k t+s t+s An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

Binomial, p =

t t+s

32

Instructor Solutions Manual

1.7 Pr{Survive at time t} =

∞ X

Pr{Survive k shocks} Pr{k shocks} =

k=0

∞ X k=0

αk

e−λt (λt)k = e−λt(1−α) . k!

1 1 1 1.8 eλt = 1 + λt + (λt)2 + (λt)3 + (λt)4 + · · · 2 3!

4!

1 1 1 e−λt = 1 − λt + (λt)2 − (λt)3 + (λt)4 − · · · 2 3!

4!

1 1 λt 1 e − e−λt = λt + (λt)3 + (λt)5 + · · · 2 3!

5!

1 1 λt e − e−λt = 1 − e−2λt .

Pr{X(t) is odd} = e−λt 2 2 h i E X(T)2 | T = t = λt + λ2 t2 .

1.9 (a) E[X(T)|T = t] = λt Z1 (b)

E[X(T)] =

1 λtdt = λ = 1 2

when λ = 2.

0

h i Z1 1 1 2 E X(T) = λt + λ2 t2 dt = λ + λ2 .

2 3 0

h i 1 1 1 Var[X(T)] = E X(T)2 − E[X(T)]2 = λ + λ2 − λ2 2 3 4 1 1 4 = λ + λ2 = when λ = 2.

2 12 3 1.10 (a) K  T  Z ZT 1 (b) cE  X(t)dt = c λtdt = λcT 2 .

2 0

(c) A.C. =

0

1 1 k + λcT 2 .

T 2

d (d) A.C. = O ⇒ T ∗ = dT

r

2K . λc

Note: In (a), could argue for Dispatch Cost = K1{X(T) > 0} and E[Dispatch Cost] = K 1 − e−λT 1.11 The gamma density fk (x) =

λk xk−1 e−λx , (k − 1)!

1.12 (a) Pr X 0 (t) = k =

Z∞

x>0

(θt)k e−θ t −θ e dθ k!

0

tk = k!

Z∞ θ k e−(1+t)θ dθ 0

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33

k Z∞ 1 t 1 = xk e−x dx k! 1 + t 1+t 0

=

t 1+t

k

1 1+t

(b) Pr X 0 (t) = j, X 0 (t + s) = j + k =

(See I, (6, 4)).

Z∞

(θt) j e−θt (θs)k e−θs −θ e dθ j! k!

0

=

j+k f

t 1+s+t

j

s 1+s+t

k

1 .

1+s+t

λ n 2.1 Pr{X(n, p) = 0} = (1 − p)n = 1 − → e−λ , n → ∞ n (n − k)p λ Pr{X(n, p) = k + 1} = → , n → ∞, λ = np.

Pr{X(n, p) = k} (k + 1)(1 − p) k+1 √ 2.2 In a small sample (sample size ∞ #tags) there are many potential pairs, and a small probability for each particular pair to be drawn. The probability that pair i is drawn is approximately independent of the probability that pair j is drawn.

N 2n n 2N(2N − 2) · · · · · (2N − 2N + 2) 2.3 Pr{A} = = .

2N(2N − 1) · · · · · (2N − n + 1) 2N n = .7895 n−1 Y 1− i=1

i 2N − i

when n = 10, N = 100. ( n−1 ) n−I X i Y i − 2N−i ∼ ∼ . e = exp − = 2N i=1

i=1

−n(n − 1) = .7985 = exp 4N

when n = 10, N = 100.

2.4 XN , the number of points in [0, 1) when N points are scattered over [0, N) is binomially distributed N, p = 1/N.

k

1 1 N−k N → e−1 k! , k = 0.

Pr {XN = k} = 1− k N N 2.5 Xr , the number of points within radius 1 of origin is binomially distributed with parameters N and p = ππr2 = Nλ . Pr{X, = k} =

N−k k e−λ N λ k 1 − Nλ → λ k! , k = 0, 1, . . . .

N k k 1

N! 1 1 2 N−k−l 1− 2.6 Pr Xi = k, Xj = l = k! l! (N − k − l)! M M M

1 k+1 −2λ (Multinomial) → λ e = k! l!

k −λ l −λ λ e λe . k! l!

Fraction of locations having 2 or more accounts → 1 − e−λ − λe−λ .

2.7 Pr{k bacteria in region of area a} −

N−k k e−c N a k N → ∞ Na 1 − Aa as , → c. → c k!

A k a→o A

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2.8 p1 = .1, p2 = .2, p3 = .3, p4 = .4

P 2 pi = .30

k

Pr {Sn = k}

e−1 k!

Diff .

0 1 2 3 4

.3024 .4404 .2144 .0404 .00024

.3679 .3679 .1840 .0613 .0153

−.0655 .0725 .0304 −.0209 −.0151

2.9 p1 = p2 = p3 = .1, p4 = .2

P 2 pi = .07

k

Pr {Sn = k}

k 1 1 2 /k!

2 e

Diff .

0 1 2 3 4

.5832 .3402 .0702 .0062 .0002

.6065 .3033 .0758 .0126 .0016

−.0233 .0369 −.0056 −.0064 −.0014

2.10 One need only to observe that |Pr {Sn ∈ I} − Pr{X(µ) ∈ I}| ≤ Pr {Sn 6= X(µ)} ≤

n X

Pr {ε ( pi ) 6= X (pi )} .

k=1

2.11 From the hint, Pr{X in B} ≤ Pr{Y in B} + Pr{X 6= Y}. Similarly Pr{Y in B} ≤ Pr{X in B} + Pr{X 6= Y}.

Together Pr{X in B} − Pr{Y in B} ≤ Pr{X 6= Y}.

2.12 Most older random number generators fail this test, the pairs (U2N , U2N+1 ) all falling in a region having little or no area.

3.1 To justify the differentiation as the correct means to obtain the density, look at Z∞Z∞

fw1 ,w2 w01 , w02 dw01 dw02 = [1 + λ (w2 − w1 )]e−λw2 . w1 w2

Zw2 3.2 fw2 (w2 ) = fw1 ,w2 (w1 , w2 ) dw1 = λ2 w2 e−λw2 , w2 > 0 0

fw1 /w2 (w1 /w2 ) =

λ2 e−λw2 1 = w2 λ2 w2 e−λw2

for

0 < w1 < w2

(Uniform on (0, w2 ]) Theorem 3.3 conditions on an event of positive probability. Here Pr {W2 = w} = 0 for all w. {W2 = w2 } is NOT the same event as X (w2 ) = 2.

3.3 fs0 ,s1 (s0 , s1 ) = fw1 ,w 2 (s0 , s0 + s1 ) (Jacobean = 1)

= λ2 exp{−λ (s0 + s1 )} = λe−λs0 λe−λs1 .

Compare with Theorem 3.2. for another approach.

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Z∞ 3.4 fw1 (w1 ) = λ2 e−λw2 dw2 = λe−λw1 , w1 > 0. w1

Zw2 fw2 (w2 ) = λ2 e−λw2 dw1 = λ2 w2 e−λw2 , w2 > 0 0

3.5 One can adapt the solution of Exercise 1.5 for a computational approach. For a different approach, Pr{X(T) = 0} = Pr {T < W1 } =

θ (Review I, 5.2). λ+θ

Using the memoryless property and starting afresh at time W1 Pr{X(T) > 1} = Pr {W2 ≤ T} = Pr {W1 ≤ T} Pr {W2 ≤ T |W1 ≤ T } =

1 λ+θ

2

.

Similarly, Pr{X(T) > k} =

λ λ+θ

k+1

and Pr{X(T) = k} =

θ λ+θ

λ λ+θ

k

,

k≥0

RT 3.6 T = WQ , the waiting time for the k th arrival, whence E[T] = Qλ . 0 N(t)dt = Total customer waiting time = Area under N(t) up to T = WQ = 1S1 + 2S2 + · · · + (Q − 1)SQ−1 (Draw picture) so  T  Z 1 Q(Q − 1) E  N(t)dt = [1 + 2 + · · · + Q − 1] = . λ 2λ 0

3.7 The failure rate is λ = 2 per year. Let X = # of failures in a year If Stock(Spares)

Inoperable unit if

Pr{Inoperable}

0

X=1

.8647

1

X=2

.5940

2

X=3

.3233

3

X=4

.1429

4

X=5

.0527

5∗

X=6

.0166

3.8 We use the binomical distribution in Theorem 3.3.

Pr {Wr > s|X(t) = n} = Pr{X(s) < r|X(t) = n} =

r−1 X s n−k n s k 1− k t t k=0

fwr |N(t)=n = (s) = −

r−1 s r−1 1 d X n s k s n−k n! s n−r 1− = 1− . k ds t t (r − 1)! (n − r)! t t t k=0

(1) (2) 3.9 (a) Pr W1 < W1 =

λ1 λ1 + λ2

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Instructor Solutions Manual

n o n o n o (1) (2) (1) (2) (2) (1) (2) (b) Pr W1 < W2 = Pr W2 < W1 + Pr W1 < W2 < W2

2 2

λ2 λ1 λ1 λ2 + + = λ1 + λ2 λ1 + λ2 λ1 + λ2 λ1 + λ2

2

λ1 λ2 = 1+2 λ1 + λ2 λ1 + λ2

λ1 λ1 + λ2

2

3.10 Xn+1 = 2n+1 exp{− (Wn + sn )} = Xn 2e−sn

E [Xn+1 |Xn ] = Xn 2E e−sn = Xn 2

Z∞ e−2x dx = Xn .

0

4.1 fw1 ,...,wk−1 ,wk+1 ,...wn |X(1)=n,wk =w (w1 , . . . , wk−1 , wk+1 , . . . , wn ) k−1

n−k 1 1 n! = (k − 1)! (n − k)! = n! w 1−w wk−1 (1 − w)n−k (k − 1)! (n − k)!

4.2 Appropriately modify the argument leading to (4.7) so as to obtain the joint distribution of X(t) and Y(t). Answer: X(t) and Y(t). are independent Poisson random variables with parameters λ

Zt

[1 − G(v)] dv and λ

Zt

0

0

4.3 The joint distribution of U =

V=

G(v) dv, respectively.

1 − W3 1 − W2

W1 W2

and W = W2 is

fU,V,W (u, v, w) = 6w(1 − w) for

0 ≤ u, v, w ≤ 1

whence fU,V (u, v) =

Z1 6w(1 − w) dw = 1

for

0 ≤ u, v ≤ 1

0

4.4 Let Z(t) = min W1 + Z1 , . . . , Wx(t) + Zx(t)

Pr{Z(t) > z} =

∞ X

Pr{Z(t) > z|X(t) = n}Pr{X(t) = n}

n=0

=

∞ X

Pr{U + ξ > z}n Pr{X(t) = n}

n=0

= e−λt exp λt Pr{U + ξ > z} (U is unif. (0, t])   Zt   1 = exp −λt + λt [1 − F(z − u)] du   t 0

  Zz   = exp −λ F(v) dv   z−t

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Let t → ∞   Zz   F(v) dv Pr{Z > z} = exp −λ   −∞

w n βw n 4.5 Pr {W1 > w|N(t) = n} = Pr {U1 > w, . . . Un > w} = 1 − = 1− t n if n = βt → e−βw if t → ∞, n → ∞, n = βt.

Under the specified conditions, W is exponentially distributed, but not with rate λ.

1 4.6 (a) E [W1 |X(t) = 2 ] = t.

3 3 1 (b) E [W3 |X(t) = 5 ] = t = t.

6 2

(c) fw2 |X(t) =5 (w) = 20w(1 − w)3 t5 for 0 < w < t.   " n # X(t) ∞ X X X   4.7 E f (Wi ) = E f (Wi ) |X(t) = n Pr{X(t) = n} i=1

n=0

i=1

= λt E[ f (U)] U is unif. (0, t] =λ

Zt

f (u) du.

0

4.8 This is a generalization of the shot noise process in Section 4.1.

Zt E[Z(t)] = E[ξ ]λ

e−αu du = E[ξ ]

0

λ 1 − e−αu . α

∞

X 4.9 E W1 W2 · · · WN(t) = E [W1 · · · Wn |N(t) = n] Pr{N(t) = n}

= =

n=0 ∞ X

E [U1 , . . . Un ] Pr{N(t) = n}

n=0 ∞ X n=0

t 2

n

n −λt (λt)

e

n!

1 −λt 1− 2 t

=e

.

4.10 The generalization allows the impulse response function to also depend on a random variable ξ so that, in the notation of Section 4.1 I(t) =

X(t) X

h (ξh , t − Wk ),

k=1

the ξk being independent of the Poisson process 4.11 The limiting density is f (x) = c

= c 1 − loge x 

for 0 < x < 1; for 1 < x < 2; Zx

= c1 − log x + .. .

 log(t − 1)  dt , for 2 < x < 3; t

2

where c = exp{−Euler’s constant} = .561459 . . .

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Instructor Solutions Manual

In principal, one can get the density for any x from the differential equation. In practice? x 2 x 5.1 f (x) = 2 for 0 < x < R because F(x) = Pr{X ≤ x|N = 1} = .

R R2 5.2 It is homogeneous Poisson process whose intensity is proportional to the mean of the Poisson random variable N. Note: If X = R cos 2 and Y = R sin 2, then (X, Y) is uniform on the disk.

5.3 For i = 1, 2, . . . , n2 let ξi = 1 if two or more points are in box i, and ξi = 0 otherwise. Then the number of reactions is ξ1 + · · · + ξn2 and p = Pr{ξ = 1} = 21 λ2 d2 + o d2 .

1 1 n2 p = λ2 (td)2 + · · · → λ2 µ2 .

2 2 The number of reactions is asymptotically Poisson with parameter 21 λ2 µ2 . h i 5.4 The probability that the centre of a sphere is located between r and r + dr units from the origin is 2d 43 π r3 = 4λπr2 dr.

R∞ The probability that such a sphere covers the origin is r f (x) dx. Then mean number of spheres that cover the origin is R R∞ R∞ 3 4 2 ∞ 0 4λπr r f (x) dxdr = 3 λπ 0 r f (r) dr. The distribution is Poisson with this mean.

5.5 FD (x) = Pr{D ≤ x} = 1 − Pr{D > x} = 1 − Pr{No particles in circle of radius x} n o = 1 − exp −vπx2 , x > 0.

5.6 1 − FR (x) = Pr{R > x} = Pr{No stars in sphere, radius x}

4 = exp −λ π x3 , 3 Whence fR (x) =

4 d 3 FR (x) = 4λπx2 e 3 λπx , dx

x>0

5.7 The hint should be sufficient for (a). For (b)  x  Z∞ Z∞ Z∞ Z∞ Z∞ Z  λ(r) dr = 2πλ r f (x) dxdr = 2πλ rdr f (x) dx = πλ x2 f (x) dx.   0

0

r

0

0

0

6.1 Nonhomogeneous Poisson, intensity λ(t) = λG(t). To see this, let N(t) be the number of points ≤ t in the relocated process = # point in .

6.2 The key observation is that, if 2 is uniform or [0, 2π), and Y is independent with an arbitrary distribution, then Y + 2 (mod 2π) is uniform.

6.3 From the shock model of Section 6.1, modified for the discrete nature of the damage process, we have ∞

E[T] =

1 X (n) G (a − 1) λ n=0

#

∞ X a−1 X 1 n+k−1 n k = 1+ p (1 − p) (See I, (3.6)) k λ "

n=1 k=0

(∞ )# a−1 X X n−1+k 1 k n−1 = 1+ p(1 − p) p n−1 λ "

k=0

n=1

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39

# " a−1 X 1 1+ p(1 − p)k (1 = p)−k−1 (See I, (6.21)) = λ k=0

1 ap = 1+ . λ 1−p 6.4 One can use the results of I, Section 5.2 or, since T1 is exponentially distributed, µ Pr {X (T1 ) = k} =

Z∞

(λt)k e−λt −µt µe dt k!

0

=

µ λ+µ

µ λ+µ

k

,

k = 0, 1, . . .

(See I, (6.4)).

Ta has a gamma density and a similar integration applies. See the last example in II, Section 4.

6.5 12 T is exponentially distributed with rate parameter 2µ, whence Z∞

(λt)k e−λt 1 T =k = 2µe−2µt dt Pr X 2 k!

0

=

2µ λ + 2µ

2λ λ + 2µ

k

,

k = 0, 1, . . .

(See I, (6.4))

6.6 Instead of representing Wk + Xk , and Wk + Yk as distinct points on a single axis, write them as a single point (Wk + Xk , Wk + Yk ) in the plane. The result is a nonhomogeneous Poisson point process in the positive quadrant with intensity λ(x, y) = λ

min{x,y} Z

f (x − u) f (y − u) du

0

where f (·) is the common density for X, Y. 6.7 Refer to Exercise 6.3. a

Pr{Z(t) > z|N(t) > 0} =

e−λz t − e−λt ; 1 − e−λt

0 < z < 1.

Let Y(t) = t1/a Z(t). For large t, we have o n y Pr{Y(t) > y|N(t) > 0} = Pr Z(t) > 1/α |N(t) > 0 t α

=

e−λy − e−λt α −−−→ e−λy −λt t→∞ 1−e

(Weibull).

6.8 Following the remarks in Section 1.3 we may write N(t) = M[3(t)] where M(t) is a Poisson process of unit intensity.

Pr{Z(t) > z} = Pr min Y1 , . . . YM[3(t)] > z

= exp{−G(z)3(t)} = exp −zα 3(t) . n o n z o Pr t1/α Z(t) > z = Pr Z(t) > 1/α t

3(t) α = exp −zα −−−→ e−θ z , z > 0 (Weibull) t→∞ t

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Instructor Solutions Manual

6.9 To carry the methods of this section a little further, write N(dt) = N(t + dt) − N(t) so N(dt) = 1 if and only if t < Wk ≤ t + dt for some Wk . Then N(t) X

(Wk ) = 2

Zt

k=1

 E

N(t) X

x2 N(dx) so

0

 t  Z Zt 2 2 (Wk ) = E x N(dx) = x2 E[N(dx)] 

k=1

0

=λ

Zt

0

x2 dx =

λt3 .

3

0

6.10 (a) Pr{Sell asset} = 1 − e−(1−θ ) . (b) Because the offers are uniformly distributed, the expected selling price, given that it is sold, is E[Return] =

i 1+θ h 1 − e−(1−θ ) 2

θ ∗ = .2079,

max E[Return] = .330425 θ

Rt − {1−θ (u)}du

(c) Pr{Sell in(t, t + dt]} = e Z1 E[Return] =

2 9

[1 − θ(t)] dt.

 

 Zt  1 + θ(t) exp − [1 − θ(u)] du [1 − θ(t)] dt   2 0

0

=

0

Z1 0

1 (2 − t) dt = , a miniscule improvement over (b).

3

1+θ 2

.

Chapter 6

Z1 1.1 Pr{X(U) = k} =

−βu k−1

−βu

1−e

e

1−e Z −β

xk−1

du =

0

0

k dx 1 1 − e−β . = β βk

1.2 λk = α + βk λ0 λ1 · · · · · λn−1 = β n

α α α +1 ····· +n−1 β β β

(λ0 − λk ) · · · · · (λk−1 − λk ) = β k (−1)k k! (λn − λk ) · · · · · (λk+1 − λk ) = β n−k (n − k)k !

Pn (t) = λ0 · · · · · λn−1

n X

Bk,n e−λk t

k=0

=

= Note:

α β

α β +1

·····

α β +n−1

n! ! α n + n − 1 −αt β e 1 − e−βt n

−αt

e

n X n k=0

k

−e−βt

k

for n = 0, 1, . . .

P∞

n=0 Pn (t) = 1 using I, (6.18)–(6.20).

1.3 The probabilistic rate of increase in the infected population is jointly proportional to the number who can give the disease, and the number available to catch it. λk = αk(N − k),

k = 0, 1, . . . , N.

1.4 (a) λk = α + θk = 1 + 2k.

1 −t 1 −3t 1 −5t (b) P2 (t) = 2 e − e + e 8 4 8 1 −1 1 −3 1 −5 P2 (1) = e − e + e = .0688.

4 2 4 1.5 The two possibilities X(w2 ) = O or X(w2 ) = 1 give us Pr {W1 > w1 , W2 > w2 } = Pr {X(w1 ) = 0, X(w2 ) = 0} + Pr {X(w1 ) = 0, X(w2 ) = 1} = P0 (w1 ) P0 (w2 − w1 ) + P0 (w1 ) P1 (w2 − w1 )

1 1 = e−λ0 w2 + λ0 e−λ0 w1 e−λ0 (w2 −w1 ) + e−λ1 (w2 −w1 ) λ1 − λ0 λ0 − λ1 λ1 λ0 = e−λ0 w2 + e−λ1 w2 e−(λ0 −λ1 )w1 λ1 − λ0 λ0 − λ1 Z∞ Z∞

= fw1 ,w2 w01 , w02 dw01 , dw02 , w1 w2 An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

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Instructor Solutions Manual

hence fw1 ,w2 (w1 , w2 ) =

∂ ∂ Pr {W1 > w1 , W2 > w2 } = λ0 λ1 e−λ0 w1 e−λ1 (w2 −w1 ) ∂w2 ∂w1

Setting s0 = w1 , s1 = w2 − w1 (Jacobean = 1) fs0 ,s1 (s0 , s1 ) = (λ0 e−λ0 s0 )(λ1 e−λ1 s1 ).

1.6 E [Sk ] = (1 + k)−e ;

E [W∞ ] =

∞ X

(1 + k)−e < ∞ when Q > 1.

k=1 −λ0 t

1.7 (a) Pr {S0 ≤ t} = 1 − e

; Pr {S0 > t} = e−λ0 t .

Zt h i Pr {S0 + S1 ≤ t} = 1 − e−λ1 (t−x) λ0 e−λ0 x dx = 1 − e−λ0 t − 0

h i λ0 t e−λ1 t 1 − e−(λ0 −λ1 ) λ0 − λ1

λ0 λ1 e−λ1 t − e−λ0 t = 1 − Pr {S0 + S1 > t} . λ0 − λ1 λ1 − λ0

Zt λ0 λ0 Pr {S0 + S1 + S2 ≤ t} = 1− e−λ1 (t−x) − e−λ0 (t−x) λ2 e−λ2 x dx λ0 − λ1 λ1 − λ0 = 1−

0

= 1−

(b) P2 (t) =

λ0 λ2 λ1 λ2 λ0 λ1 e−λ1 t − e−λ0 t = e−λ2 t . (λ0 − λ1 ) (λ2 − λ1 ) (λ1 − λ0 ) (λ2 − λ0 ) (λ0 − λ2 ) (λ1 − λ2 )

λ0 λ2 λ1 λ2 λ0 λ1 − e−λ1 t + e−λ0 t + e−λ2 t (λ0 − λ1 ) (λ2 − λ1 ) (λ1 − λ0 ) (λ2 − λ0 ) (λ0 − λ2 ) (λ1 − λ2 )

λ0 λ1 e−λ1 t − e−λ0 t (λ0 − λ1 ) (λ1 − λ0 )

1 1 1 −λ0 t −λ1 t −λ2 t . e + e + e = λ0 (λ1 − λ0 ) (λ2 − λ0 ) (λ0 − λ1 ) (λ2 − λ1 ) (λ0 − λ2 ) (λ1 − λ2 ) −

1.8 Equations (1.2) become P00 (t) = −βP0 (t) P0n (t) = −βPn (t) + αPn−1 (t),

n = 2, 4, 6, . . .

P0n (t) = −αPn (t) + βPn−1 (t),

n = 1, 3, 5, . . .

Seeming over n = 0, 2, 4 . . . and n = 1, 3, 5 . . . gives P0even (t) = αPodd (t) − βPeven (t) = α [1 − Peven (t)] − βPeven (t) P0even (t) = α − (α − β)Peven (t). β α (α+β)t Let Qeven (t) = e(α+β)t Peven (t), then Q0even (t) = α e(α+β)t and since Qeven (0) = 1, the solution is Qeven (t) = α+β + α+β e β α and Peven (t) = e−(α+β)t Qeven (t) = α+β + α+β e−(α+β)t .

1.9 Equations (1.2) become P00 (t) = −βP0 (t) P0n (t) = −βPn (t) + αPn−1 (t),

n = 2, 4, 6, . . .

P0n (t) = −αPn (t) + βPn−1 (t),

n = 1, 3, 5, . . .

Multiply the nth equation by n, sum, collect terms and simplify to get M 0(t) = αPodd (t) + βPeven (t) = α + (βα)Peven (t),

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43

and (See Problem 1.8 above)

h i 2αβ β −α β M(t) = t+ 1 − e−(α+β)t . α+β β +α α+β

n N −(N−n)λt 1.10 It is easily checked through (1.5) that Pn (t) = e 1 − e−λt , by working with Qn (t) = eλn t Pn (t) = n

Rt N (1 − e−λt )n ; Q0 (t) ≡ 1; Qn (t) = λn−1 0 e−λx Qn−1 (x) dx (from (1.5)). n

1.11 P1 (t) = λ0 e

−λ1 t

Zt

eλ1 x e−λ0 x dx =

0

P2 (t) = λ1 e

−λ2 t

Zt 0

= λ0 λ1

λ0 λ0 e−λ1 t + e−λ0 t λ0 − λ1 λ1 − λ0

λ0 λ0 e−λ1 x + e−λ0 x eλ2 x dx λ0 − λ1 λ1 − λ0

1 1 1 1 e−λ1 t + e−λ0 t + + e−λ2 t (λ0 − λ1 ) (λ2 − λ1 ) (λ1 − λ0 ) (λ2 − λ0 ) (λ0 − λ1 ) (λ1 − λ2 ) (λ1 − λ0 ) (λ0 − λ2 )

and 1 1 1 + = (λ0 − λ1 ) (λ1 − λ2 ) (λ1 − λ0 ) (λ0 − λ2 ) (λ0 − λ2 ) (λ1 − λ2 ) P3 (t) = λ0 λ1 λ2 e

−λ2 t

Zt 0

= λ0 λ1 λ2

1 1 e−λ0 t + e−λ1 t (λ3 − λ0 ) (λ2 − λ0 ) (λ1 − λ0 ) (λ3 − λ1 ) (λ2 − λ1 ) (λ0 − λ1 ) +

1.12 P2 (t) = λ1 e

−λ2 t

Zt 0

1 1 1 e−λ1 x + e−λ0 x + e−λ2 x eλ3 x dx (λ0 − λ1 ) (λ2 − λ1 ) (λ1 − λ0 ) (λ2 − λ0 ) (λ0 − λ2 ) (λ1 − λ2 )

1 1 e−λ2 t + e−λ3 t (λ3 − λ2 ) (λ1 − λ2 ) (λ0 − λ2 ) (λ0 − λ3 ) (λ1 − λ3 ) (λ2 − λ3 )

λ0 λ0 e−λ1 x + e−λ0 x eλ2 x dx λ0 − λ1 λ1 − λ0

1 1 1 − e−(λ1 −λ2 )t + 1 − e−(λ0 −λ2 )t (λ0 − λ1 ) (λ1 − λ2 ) (λ1 − λ0 ) (λ0 − λ2 )

e−λ0 t e−λ1 t e−λ2 t = λ0 λ1 + + (λ1 − λ0 ) (λ2 − λ0 ) (λ2 − λ1 ) (λ0 − λ1 ) (λ0 − λ2 ) (λ1 − λ2 ) = λ0 λ1 e−λ2 t

1 1 1 Note: (λ0 −λ1 )(λ + (λ1 −λ0 )(λ = (λ0 −λ2 )(λ .

1 −λ2 ) 0 −λ2 ) 1 −λ2 )

1.13 Let Qn (t) = eλt Pn (t). Then Q0 (t) ≡ 1 and (1.5) becomes Q0n (t) = λQn−1 (t) which solves to give Q1(t) = λt; Q2 (t) = (λt)n (λt)n e−λt 1 2 .

2 (λt) . . . Qn (t) = n! and Pn (t) = n!

2.1 Using the memoryless property (Section I, 5.2) Pr{X(T) = 0} = Pr{T > Wn } =

N Y i=1

Pr T > Wi T > Wi−1

N Y µi = . µi + θ i=1

44

2.2

Instructor Solutions Manual

PN (t) = −e−θ t PN−1(t) = θt e−θ t

Pn (t) = (θt)N−n e−θt (N − n)!

P0 (t) = 1 −

N X

for n = 1, 2, . . . , N

Pn (t).

n=1

2.3 Refer to Figure 2.1 to see that Area = WN + · · · + W1 Hence E[Area] = E [WN ] + · · · + E [W1 ] = NE [SN ] + (N − 1)E [SN−1 ] + · · · + E [S1 ] = 2.4 (a)

µN = NMθ µN−1 = (N − 1)(M − 1)θ µk = k(M − (N − k))θ,

(b) E [WN ] =

N X

E [Sk ] =

k=1

1 = θ

N X k=1

N X k=1

PN

n n=1 µn .

k = 0, 1, . . . , N. 1 k[M − (N − k)]θ

1 1 − (M − N)k (M − N)(M − N + k)

( N ) N X1 X 1 1 − = θ(M − N) k M−N +k k=1

∼

k=1

1 {log N − log M + log(M − N)}. θ(M − N)

2.5 Breakdown rule is “exponential breakdown”.

NL NL = k sinh µk = kK k k

1

N X

Z N X 1 dx 1 1

∼ E [WN ] = NL = (Riemann approx.) = L k L N k sinh x sinh sinh k X k=1 k=1 N

2.6 (a) E[T] = E [WN ] =

N X k=1

R/N

0

N

1X1 .

E [Sk ] = α k k=1

Z∞ Z1 Z∞ Z∞h dy i 1 −αt N (b) E[T] = Pr{T > t} dt = [1 − P0 (t)] dt = 1 − yN 1− 1−e dt = α 1−y 0

0

0

0

Z1 i 1 h 1 1 1 1 2 N−1 = 1 + y + y + ··· + y dy = 1 + + + ··· + . α α 2 3 N 0

3.1 Pr{X(t + h) = 1|X(t) = 0} = λh + o(h) so λ0 = λ.

Pr{X(t + h) = 0|X(t) = 1} = λ(1 − α)h + o(h) so µ1 = λ(1 − α).

The Markov property requires the independent increments of the Poisson process.

3.2 If we assume that the sojourn time on a single plant is exponentially distributed then X(t) is a birth and death process.

Reflecting behavior µ0 = 0, λ0 = m10 , λN = 0, µN = m1N might be assumed. In an actual experiment, and these have been

Instructor Solutions Manual

45

done, one must allow for the escape or other loss of the beetle. We add a state 1 = Escaped (1 is called the “cemetery”) and assume (0 < α < 1) α h + o(h) 2mk α Pr{X(t + h) = k − 1|X(t) = k} = h + o(h) 2mk 1−α h + o(h) Pr{X(t + h) = 1|X(t) = k} = mk Pr{X(t + h) = 1|X(t) = 1} = 1.

Pr{X(t + h) = k + 1|X(t) = k} =

This is a general finite state continuous time Markov process—See Section 6. It is also a birth and death process with “killing”.

3.3

Pr{V(t) = 1} = π for all t (See Exercise 3.3) E[V(s)V(t)] = Pr{V(s) = 1, V(t) = 1} = Pr{V(s) = 1} × Pr{V(t) = 1|V(s) = 1} = πP11 (t − s) = π[1 − P10 (t − s)] Cov[V(s)V(t)] = E[V(s)V(t)] − E[V(s)]E[V(t)] = πP11 (t − s) − π 2 = π(1 − π)e−(α+β)(t−s) .

3.4 Because V(0) = 0, E[V(t)] = P01 (t), and Zt E[S(t)] =

P01 (u) du =

0

Zt

(π − π e−τ u ) du = π t −

0

π (1 − e−τ t ), τ

τ = α + β.

4.1 Single repairman R = 1 λk = 2

for k = 0, 1, 2, 3, 4, µk = k for k = 0, 1, 2, 3, 4, 5.

4 2 4 θ0 = 1, θ1 = 2, θ2 = 2, θ3 = , θ4 = , θ5 = .

3 3 15 5 X 218 109 θk = = 30 15 k=0

Two repairman R = 2 λ0 = λ1 = λ2 = λ3 = 4; λ4 = 2 µk = k θ0 = 1, θ1 = 4, θ2 = 8, θ3 = 5 X k=0

θk =

32 64 32 , θ4 = , θ5 = , 3 3 15

579 193 = .

15 15

R=1 R=2

(a) R=1 R=2

π0

π1

π2

π3

π4

π5

15 109 15 579

30 109 60 579

30 109 120 579

20 109 160 579

10 109 160 579

4 109 64 579

X

(b)

kπk

1.93 3.01

1X kπk N .39 .06

(c) π5 = .0367 2π5 + π4 = .4974

46

Instructor Solutions Manual

4.2 θ0 = 1, θ1 = µλ , θ2 = ∞ X

θk =

k=0

2 λ µ

∞ k X λ k=0

µ

=

, . . . , θk =

λ k k

1

1−λ µ

if λ < µ(= ∞ if λ ≥ µ)

k If λ < µ, then πk = 1 − µλ µλ for k = 0. This is the geometre distribution. The model corresponds to the M/M/1 queue.

If λ ≥ µ, then πk = 0 for all k.

4.3 Repairman Model M = N = 5, R = 1, µ = .2, λ = .5 0

1

2

3

4

5

λk = .5

.5

.5

.5

.5

0

µk =

.2

.4

.6

.8

.10

k =

0

5 25 125 625 628 2 8 48 384 768 768 1920 2400 2000 1250 625 πk = 8963 8963 8963 8963 8963 8963 θk =

1

= .086 .214 .268 .223 .139 .070 Fraction of time repairman is idle = π5 = .07

4 = 6 possible links, (a) λk = α(6 − k) and µk = βk for 0 ≤ k ≤ 6, θ0 = 1, θ1 = 6 βα ; 4.4 There are 2

k 6.5 α 2 α 6 θ2 = , . . . θk = k 1.2 β β 6 X k=0

πk =

α 6 α+β 6 θk = 1 + = β β

6 k 6−k k α β 6 α β 6 = ,0 ≤ k ≤ 6 k β α+β k α+β α+β

Binomial distribution. See derivation of (4.8).

4.5 If X(t) = k, there are N – k unbonded A molecules and an equal number of unbonded B molecules. λk = α(N − k)2 , µk = βk.

4.6 This model can be for mulated as a “Repairman model”. k =0

1

2

3

λk = λ

λ

λ

0

µk = 0

µ

λ µ

2µ 2 1 λ 2 µ

2µ 3 1 λ 4 µ

θk = 1

The computer is fully loaded in states k = 2 and 3, 2 3 1 λ 1 λ + 2 µ 4 µ π2 + π3 = 2 3 . λ 1 λ 1 λ 1+ µ + 2 µ + 4 µ

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47

4.7 The repairman model with N = 3, R = 2 M=2 k λk µk θk πk P

1 = .2, 5 0 1 2 2 4 4 1 0 5 5 1 2 64 160 549 549

µ=

θk = 1 +

1 = .25 4 2 3 1 0 4 2 2 5 5 25 125 8 64 200 125 549 549

λ=

5 25 125 549 + + + .

2 8 64 64

Long run avg. # Machines operating = 0π0 + 1π1 + 2π2 + π3 = 810 549 = 1.48; Avg. Output = 148 items per hour.

4.8 θ0 = 1

1 α θ1 = 2 β

1 α 2 θ2 = 3 β k 1 α θk = , k+1 β

k = 0, 1, . . . h i P 1 k Because log(1 − x) = − x + 21 x2 + 31 x3 + · · · = − ∞ k=1 k x for |x| < 1, we can evaluate ∞ X

θk =

k+1 ∞ α β 1 βX 1 , = log α k+1 β α 1−α β

0 < α < β,

k=0

k=0

and πk = C

1 k+1

k+1 α , β

C=

1

. log 1−1 α β

This is named the “logarithmic distribution”.

5.1 Change K into an absorbing state P by setting λK = µK = 0. Then Pr{Absorption in 0} = Pr{Reach 0 before K}. Because %i = 0 for i = K, (5.7) becomes µm =

K−1 i=m ñi P , as desired.

1+ K−1 i=1 ñi

5.2 %0 = 1, %1 = 4, %2 = 6, %3 = 4, %4 = 1, %5 = 0 u2 =

ñ2 + ñ3 + ñ4 11 = 1 + ñ1 + ñ2 + ñ3 + ñ4 16

(a) Alternatively, solve  4 1 15 + u2  u1 =    5 5 16     3 2  11 u2 = u1 + u3   u2 = 5 5 16 2 3  u = 5 u3 = u2 + u4   3 16 5 5      1 1   u4 = u4 = u3 16 5 u1 =

48

Instructor Solutions Manual

 1 1  + w2    5 5     2  1 3 1  w2 = + w1 + w3   w1 = w4 = 5 5 5 3 3  2 1 2  w3 = + w2 + w4   w2 = w3 =  5 5 5  3     1 1   w4 = + w3 5 5

(b) w1 =

6.1 For i 6= j, Pr{X(t + h) = j|X(t) = i} = λPij h + o(h) for h ≈ 0, The independent increments of the Poisson process are needed to establish the Markov property.

h

i βB βC A 6.2 Fraction of time system operates = αAα+β 1 − . αB +βB αC +βC A 6.3 Pr{Z(t + h) = k + 1|Z(t) = k} = (N − k)λh + o(h), Pr{Z(t + h) = k − 1|Z(t) = k} = kµh + o(h).

6.4 The infinitesimal matrix is

(2, 0) (1, 0) (1, 1) (0, 0) (0, 1) (0, 2)

(2, 0)

(1, 0)

(1, 1)

−2µ α β

2µp −(α + µ)

2µq

(0, 0)

(0, 1)

(0, 2)

µp

µq µp

µq

−(β + µ) α β

−α α β

−(α + β) −β

7.1 (a) (1 − π)P01 (t) + πP11 (t) = (1 − π)π(1 − e−π t ) + π[π + (1 − π)e−π t ] = π − π 2 + π 2 = π . R t (b) Let hR N(dt) =i 1 Rif there is an Reven in (t, t + dt], and zero otherwise. Then N((0, t]) = 0 N(ds) and E[N((0, t])] = t t t E 0 N(ds) = 0 E[N(ds)] = 0 πλds = πλt.

7.2 γ (t) > x if and only if N((t, t + x]) = 0. (Draw picture) 7.3 T > t if and only if N((0, t]) = 0. Pr{T > t} = Pr{N((0, t]) = 0} = f (t, λ) hence φ(t) = − dtd f (t; λ) = c+ µ+ e−µ,t + c− µ− e−µ−t . When α = β1 and λ = 2, then √

√

√ √ √ 1 e− 2t + e 2t 2 ∓ 2 and φ(t) = e−2t = e−2t cosh 2t. µ± = 2 ± 2, c± = 4 2 R∞ R∞ 7.4 E[T] = 0 Pr{T > t}dt = 0 f (t; λ)dt = µc++ + µc−− . when α = β1 and λ = 2, then ( √ √ ) 1 2− 2 2+ 2 12 3 E[T] = = = . √ + √ 4 2+ 2 2− 2 8 2 Since events occurs on average, at a rate of πλ per unit time, the mean duration between events must π1λ (= 1 when α = β = 1 and λ = 2). The discrepancy between the mean time to the first event, and the mean time between events, is another example of length biased sampling.

7.5 Pr {N((t, t + s]) = 0 |N((0, t]) = 0 } =

f (t + s; λ) c+ e−µ+(t+s) + c− e−µ−(t+s) = f (t; λ) c+ e−µ+ t + c− e−µ−t

=

c+ e−(µ+ −µ− )t c− e−µ+ s + e−µ− s → e−µ− s as t → ∞. −(µ −µ −(µ −µ− )t + c )t + − + c+ e + c− c+ e −

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7.6

49

R ∞ −st −µt c ce dt = µ+s hence 0 e c+ c− c+ µ− + c− µ+ + s + = µ+ + s µ− + s (µ+ + s) (µ− + s)

q 1 µ± = (λ + τ ) ± (λ + τ )2 − 4πτ λ 2

ϕ(s; λ) =

c+ µ− + c− µ+ = τ + (1 − π)λ. (µ+ + s) (µ− + s) = s2 + (τ + λ)s + πτ λ.

R∞ 1 = 0 e−st e−π λt dt (a) limτ →∞ φ(s; λ) = π λ+s

)λ 1 (b) limτ →0 φ(s; λ) = s+(1−π = π + (1 − π) 1s .

2 λ+s s +λs 7.7 When α = β = 1 and λ = 2(1 − θ) then µ± = (z − θ) ± R,

c± =

1 1 ∓ 2 2R

and

1 1 1 1 1 −Rt Rt −(2−θ )t g(t; θ) = e cosh Rt + sinh Rt . − e + + e =e 2 2R 2 2R R

t 1 dg 1 dr −(2−θ )t −(2−θ )t t sinh Rt + cosh Rt − 2 sinh Rt =e t cosh Rt + sinh Rt + e dθ R R dθ R √ √ dR 2 1 R θ =0 = 2, . = −√ = dθ θ =0 2 2 −(2−θ )t

√ √ √ √ √ √ dg 2 1√ 1√ 1 −2t −2t = e t cosh 2t + 2 sin 2t − e t sin 2t + 2 cosh 2t − sinh 2t dθ θ =0 2 2 2 2 ) ( √ √ √ 1 −2t 2 = e t cosh 2t + sinh 2t = Pr {N((0, t]) = 1} .

2 2 2

d g Students with symbol manipulating computer skills may test them on Pr{N((0, t]) = 2} = 21 dθ 2 θ =0 .

Pr{λ(t) = 0 and N((0, t]) = 0} f0 (t) a− = → as t → ∞.

7.8 Pr λ(t) = 0 N((0, t]) = 0 = P{N((0, t]) = 0} f (t) c− 7.9 The initial conditions are easy to verify: f0 (0) = a+ + a− = 1 − π, f1 (0) = b+ + b− = π . We wish to check that f 0 0 (t) = −af0 (t) + βf1 (t) f 0 1 (t) = af0 (t) − (β + λ)f1 (t) Now f00 (t) = −a+ µ+ e−µ+t − a− µ− e−µ−t While −αf0 (t) = −aa+ e−µ+ t − aa− e−µ−t βf1 (t) = βb+ e−µ+ t + βb− e−µ−t

50

Instructor Solutions Manual ?

?

Upon equating coefficients, we want to check that −a+ µ+ = −aa+ + βb+ and −a− µ− = −aa− + βb− . Starting with the ?

first, is 0 = a+ (µ+ − α) + βb+

(λ − α + β)(λ + α + β) 1 − (α + β + λ) a+ (µ+ − α) = (1 − π) (λ − α + β) + R − 4 R

1 (λ − α + β)(λ + α + β) = (1 − π) −2α + R − 4 R

λ−α−β α 1 , π= βb+ = πβ 1 + 2 R α+β 2R (α− + β) [a+ (µ+ − α) + βb+ ] = βR2 − β(λ − α + β)(λ + α + β) + 2αβ(λ − α − β) h i = β (α + β + λ)2 − 4aλ − (λ − α + β)(λ + α + β) + 2α(λ − α − β) = β[(α + β + λ){α + β + λ − λ + α − β} − 4αλ + 2α(λ − α − β)] = β[2α(α + β + λ) − 4αλ + 2α(λ − α − β)] = βα[2(α + β) + 2(−α − β)] = 0 ?

To check: 0 = a− (µ− − α) + βb−

α+β +λ 1 [(λ − α + β) − R] a− (µ− − α) = (1 − π) 1 + 4 R

1 β (α + β + λ) (λ − α + β) λ−α+β −R+ −α−β −λ = 4 α+β R

(α + β + λ)(λ − α + β) 1 β −2α − R + = 4 α+β R

1 αβ λ−α−β βb− = 1− 2 α+β R

2α(λ − α − β) 1 β 2α − = 4 α+β R

1 β (α + β + λ)(λ − α + β) 2α(λ − α − β) a− (µ− − α) + βb− = −R + − 4 α+β R R

h i 1 β = −(α + β + λ)2 + 4αλ + (α + β + λ)(λ − α + β) − 2α(λ − α − β) 4R α + β

1 β = [4αλ − 2α(α + β + λ) − 2α(λ − α − β)] = 0 4R α + β ?

?

Verifying the second differential equation reduces to checking if 0 = aa+ + b+ (µ+ − β − λ) and 0 = aa− + b− (µ− − β − λ).

The algebra is similar to that used for the first equation.

7.10 (a) Pr{N(0, h] > 0, N(h, h + t] = 0} = Pr{N(h, h + t] = 0} − Pr{N(0, h] = 0, N(h, h + t] = 0} = f (t; λ) − f (t + h; λ). (b) lim Pr{N(h, h + t] = 0|N(0, h] > 0} h↓0

f (t;λ)−f (t+h;λ) h 1−f (h;λ) h↓0 h

= lim

=

d 0 (t;λ) − dt f (t;λ) = ff 0 (0;λ) . d − dt f (t;λ)|t=0

Instructor Solutions Manual

(c)

51

f 0 (t; λ) −c+ µ+e −µ+t − c− µ− e−µ−1 = = p+ e−µ− t p− e−µ− t . f 0 (0; λ) −c+ µ+ − c− µ− −µ−t

where E[τ | Event at time 0]= µp++ + µp−− . √ √ When α = β = 1 and λ = 2, we have µ± = 2 ± 2, c± = 14 (2 ∓ 2); p± = 12 and E[τ | Event at time 0] = 1. = Long run mean time between events.

(d) Pr{τ > t| Event at time 0} = p+ e−µ+t + p−e

7.11 V(u) plays the role of λ(u), and S(t), that of 3(t).

7.12 This model arose in an attempt to describe a scientist observing butterflies. When the butterfly is at rest, it can be observed perfectly. When it is flying, it may fly out-of-range (the “event”). The observation time is random, and the model may guide us in summarizing the data. The stated properties should be clear from the picture.

Chapter 7

1.1 Pr{N(t − x) = N(t + y)} = =

∞ X k=0 ∞ X

Pr{N(t − x) = k = N(t + y)} Pr {Wk ≤ t − x, Wk+1 > t + y}

k=0

= [1 − F(t + y)] +

∞ X

Pr {Wk ≤ t − x, Wk + Xk+1 > t + y}

k=1 t−x

= [1 − F(t + y)] +

∞ Z X

[1 − F(t + y − z)]dFk (z).

k=1 0

In the exponential case: t−x

−λ(t+y)

=e

+

∞ Z X

e−λ(t+y−z)

k=1 0 −λ(t+y)

=e

−λ(t+y)

=e 1.2

λk zk−1 −λz e dz (k − 1)!

Zt−x + e−λ(t+y−z) λeλz e−λz dz 0 h −λ(t+y) λ(t−x)

+e

e

i − 1 = e−λ(x+y) .

Pr{N(t) = k} = Pr {Wk ≤ t, Wk + Xk+1 > t} Z1 =

[1 − F(t − x)]dFk (λ), and in the exponential case

0

=λ e

k −λt

Zt

(λt)k e−λt xk−1 dx = , k! k!

0

1.3 E [γt ] = E WN(t)+1 − t = E [X1 ] [M(t) + 1] − t.

1 − F(x + y) 1.4 Pr γt > y δt = x = 1 − F(x) Z∞

1 − F(x + y) E γt δ t = x = dy.

1 − F(x) 0

An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

k = 0, 1, . . .

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53

2.1 K

4 + 5M(K − 1) K 2(K)

1 2 3 4 5

4.00 2.25 2.183 2.1138 2.1231

Block Period Cost =

*Replace on failure: 2 = 1.923, is best 2.2 We are asked to verify that M(n), as given, satisfies M(n) = 1 + (1 − q)M(n − 1) + qM(n − 2), n = 2.

1=1 (1 − q)(n − 1) (1 − q)q2 (1 − q)qn+1 − (1 − q)M(n − 1) = − 1+q (1 + q)2 (1 + q)2 qM(n − 2) =

qqn q(n − 2) qq2 + − 2 1+q (1 + q) (1 + q)2

?

M(n) = 1 − = 1− =

(1 − q)(n − 1) + q(n − 2) (1 − q)qn+1 − qn+1 q2 + − 2 1+q 1+q (1 + q)2 1 − q + 2q n q2 qqn+1 + − + 1+q (1 + q)2 1 + q (1 + q)2

n qn+2 q2 + = M(n) − 2 1 + q (1 + q) (1 + q)2

2.3 M(1) = p1 = β M(2) = p1 + p2 + p1 M(1) = β + β(1 − β) + β 2 = 2β M(3) = p1 + p2 + p3 + p1 M(2) + p2 M(1) = β + β(1 − β) + β(1 − β)2 + β(2β) + β(1 − β)β = β + β − β 2 + β − 2β 2 + β 3 + 2β 2 + β 2 − β 3 = 3β In general, M(n) = nβ.

∞ ∞ (λt)j

P 1 (λt)k e−λt P 1 1 1 3.1 E = = e−λt = e−λt eλt − 1 .

N(t) + 1 k! λt λt j=1 j! k=0 k + 1 Using the independence established in Exercise 3.3,

E

1

WN(t)+1 1 1 1 1 = E WN(t)+1 E = t+ 1 − e−λt = 1− 1 − e−λt .

N(t) + 1 N(t) + 1 λ λt λ λt

3.2 Because WN(t)+1 = t + γt and E[γt ] = λ1 , E[WN(t)+1 ] = t + λ1 . On the other hand E[X1 ] = λ1 and M(t) = λt, whence E [X1 ] {M(t) + 1} =

1 1 (λt + 1) = t + = E WN(t)+1 . λ λ

3.3 For t > τ, p(t) = Pr{No arrivals in (t − τ, t]} = e−λτ . For t ≤ τ, p(t) = Pr{No arrivals in (0, t]} = e−λt . Thus p(t) = e−λmin(τ,t) .

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Instructor Solutions Manual

3.4 Let l(x, y) =

x 1−x

if 0 ≤ y ≤ x ≤ 1 if 0 ≤ x < y ≤ 1.

Z1 Z1

E[L] = E[l(X, Y)] = Z1 =

l(xy)dx dy

 x 0 0 1  1  Z Z Z Z1 Z1 2 x  dy dx + (1 − x)  dydx = x2 dx+ (1 − x)2 dx = .

3

0

0

x

0

0

0

−2λx e 3.5 Pr[D(t) > x] = Pr{No birds in (t − x, t + x]} = −λ(x+t) e

for 0 < x < t for 0 < t ≤ x.

Z∞ Zt Z∞

1 Pr{D(t) > x}dx = e−2λx dx + e−λ(x+t) dx = 1 + e−2λt .

2λ t 0 0

d 2λe−2λx for 0 < x < t fT (t) = − Pr{D(t) > x} = λe−λ(x+t) for 0 < t ≤ x. dx

E[D(t)] =

4.1

1 1 M(t) → implies µ = 1 t µ

4.2 µ =

M(t) −

t σ 2 − µ2 → =1 µ 2µ2

implies σ 2 = 3

1 1 + ; α β

1 αβ = ; µ α+β

1 1 + α2 β 2

M(t) ≈

σ2 =

αβt σ 2 − µ2 αβ αβt + + . = 2 α+β α + β (α + β)2 2µ

Ry 1 2 4.3 1 − F(y) = exp − 0 θx dy = e− 2 θ y , y = 0 µ=

r Z∞ Z∞ 1 2 π [1 − F(y)]dy = e− 2 θ y dy = .

2θ 0

0

σ 2 + µ2 =

Z∞

y2 dF(y) =

0

Limiting mean age

σ 2 +µ2 2µ

=

q

2 . θ

2 πθ .

4.4 List the children, family after family, to form a renewal process.

Children Family

MF #1

MMMF #2

F #3

F #4

MF #5

µ = 2E[#children in a family]. 1 female per family, Long run fraction female = µ1 = 12 .

4.5 m0 = 1 + .3m0

⇒

m2 = 1 + .5m2

⇒

1 10 = .7 7 m2 = 2 m0 =

93 30 28 Successive visits to state 1 form renewal instants for which µ = 1 + .6m0 + .4m2 = 35 . And π1 = 35 π = , π = 0 2 93 93 93 .

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55

βA βB ; Pr{B down} = . α A + βA αB + βB

βA βB Pr{System down} = . αA + βA α B + βB

5.1 (a) Pr{A down} =

1 (b) System leaves the failed state upon first component repair. E[Sojourn System down] = αA +α .

B

E[Sojourn System Down] so E[Cycle]

1 (αA + αB )

.

E[Cycle] = βA βB + β + β A B αA αB

(c) Pr[System down] =

(d) E[System Sojourn Up] = E[Cycle] − E[Sojourn down]

1 (αA + βA ) (αB + βB ) = −1 αA + αB βA βB 1 − F(y) , 0<x<y 1 − F(x)

Z∞ Z∞ 1 − F(y) E[X|X > x] = Pr{X > y|X > x}dy = x + dy 1 − F(x)

5.2 Pr{X > y|X > x} =

0

x

5.3 If successive fees are Y1 , Y2 , . . . then W(t) =

PN(t)+1 k=1

Yk when N(t) is the number of customers arriving in (0, t].

R∞ [1 − G(y)]dy E[W(t)] E[Y] lim = = R0∞ t→∞ t E[X] 0 [1 − F(x)]dx 5.4 The duration of the cycle in the obvious renewal process is the maximum of the two lives. The cycle duration when both are lit is the minimum. Let the bulb lives be ξ1 , ξ2 . The fraction time half lit = 1−

E [min {ξ1 , ξ2 }] E [max {ξ1 , ξ2 }]

1 1 (a) E [min {ξ1 , ξ2 }] = 2λ , E[max] = 2λ + λ1 ; Pr[Half lit] = 23 1 2 (b) E[min] = 3 , E[max] = 3 ; Pr[Half lit] = 12 .

6.1 (a) u0 = 1 u1 = p1 u0 = α u2 = p2 u0 + p1 u1 = α(1 − α) + α 2 = α u3 = p3 u0 + p2 u1 + p1 u2 = α(1 − α)2 + α[α + α(1 − α)] = α We guess un = α for all n so

?

α = α(1 − α)n−1 + α(p1 + p2 + · · · + (pn−1 )

= α(1 − α)n−1 + α 2 1 + (1 − α) + · · · + (1 − α)n−2

= α(1 − α)n−1 + α 1 − (1 − α)n−1 = α

(b) For excess life bn = pn+m = α(1 − α)n+m−1 = pm (1 − α)n Vn =

n X k=0

bn−k uk = pm (1 − α)n α

n X k=1

(1 − α)n−k pm = pm (1 − α)n + 1 − (1 − α)n = pm .

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6.2 We want Pr{γn = 3} for n = 10, p1 = p2 = · · · = p6 = 16 f0 =

1 6

= .16666

f1 =

1 1 + f0 6 6

= .19444

f2 =

1 1 + ( f0 + f1 ) 6 6

= .22685

f3 =

1 1 + ( f0 + f1 + f2 ) 6 6

= .26466

f4 =

1 ( f0 + f1 + f2 + f3 ) 6

= .1421

f5 =

1 ( f0 + f1 + f2 + f3 + f4 ) 6

= .16578

f6 =

1 ( f0 + f1 + f2 + f3 + f4 + f5 ) = .1934 6

f7 =

1 ( f1 + f2 + f3 + f4 + f5 + f6 ) = .197878 6

1 ( f2 + f3 + f4 + f5 + f6 + f7 ) = .198450 6 1 f9 = ( f3 + f4 + f5 + f6 + f7 + f8 ) = .1937166 6 1 f10 = ( f4 + f5 + f6 + f7 + f8 + f9 ) = .181892637 6

Pr X1 = 3 4 2/3 = = .190476 lim Pr {γn = 3} = = n→∞ E [X1 ] 7/2 21 f8 =

(Compare with problem 5.4 in III. The renewal theory gives us the limit) 6.3 That delaying the age of first birth, even P at constant family size, would lower population growth rates, was an important 1 v 2 3 conclusion in early work with this model. ∞ v=0 mv s = 2s + 2s = 1 solves to gives s = .5652 whence λ = s 1.77 compared to λ = 2.732 in the example.

6.4 (a)

P∞

v v=0 mv s = a

X

(b)

v

P∞

mv s = 1

v v=2 s =

as2 , |s| < 1.

1−s

solve to give

s=

−1 ±

√ 1 + 4a 2a , λ1 = √ 2a 1 + 4a − 1

q √ mv sv = as2 = 1 solves to give s = 1a and λ2 = a. Compare λ1 = √ 4 = 2 when a = 2 versus λ2 = 2 when a = 4.

9−1 That is 4 offspring at age 2 is equivalent to 2 offspring repeatedly.

P

Chapter 8

1.1 Let

Tn = min t = 0; Bn (t) 5 −a or Bn (t) = b

√ √ = min t = 0; S[nt] 5 −a n or S[nt] = b n

√ √ 1 = min k = 0; Sk 5 −a n or Sk = b n n 1h √ Y √ i E [Tn ] = a n b n (III, Section 5.3) n → ab as n → ∞.

1.2 Begin with Z+∞ 1= −∞

√ 2 1 2 1 1 √ e 2 (x−λ t) dx = e− 2 λ t 2π

Z+∞ √ 1 2 1 eλ tx √ e− 2 x dx 2π

−∞

Z+∞ h i h i 1 2 1 2 1 2 1 2 1 = e− 2 λ t eλy √ e− 2 y /t dy = e 2 λ t E eλB(t) so E eλB(t) = e 2 λ t.

2π −∞

1.3 Pr

n

|B(t)| t >ε

o

√ = Pr {|B(t)| > εt} = 2 {1 − 8t (εt)} = 2 1 − 8 ε t

|B(t)| >ε →0 t

|B(t)| Pr >ε →1 t Pr

(1.8)

√ as t → ∞ because 8 ε t → 1 √ 1 as t → 0 because 8 ε t → .

2

1.4 Every linear combination of jointly normal random variables is normally distributed. " n # X X E αi B (ti ) = αi E [B (ti )] = 0 i=1

Var

" n X

# αi B (ti ) = E

nX

αi B (ti )

o2

 ( ) X X  = E αi B (ti ) α j B tj    i

i=1

=

n X n X

αi αj E B (ti ) B tj =

i=1 j=1

n X n X

j

αi αj min ti , tj .

i=1 j=1

1 (Using III, 5.3).

1.5 (a) Pr {Mτ = 0} = Pr {Sn drops to − a < 0 before rising 1 unit} = 1+α (b) In order that Mτ = 2, we must first have Mτ = 1 followed by moving ahead to 2, starting afresh from S0 = 1, before dropping a units below 1. The spatial homogeneity gives this second move the same probability as Pr{M(τ ) = 1}. Thus

a 2 a k Pr{Mτ = 2} = [Pr{Mτ = 1}]2 = 1+a . The argument repeats to give Pr{Mτ = k} = 1+a .

1 (c) Divide the state space into increments of length n and observe the Brownian motion as it crosses lines nk . That is, τ1 = n o n o min t = 0; B(t) = 1n or B(t) = − 1n , and if, for example, B(τ1 ) = 1n , then let τ2 = min t = τ1 ; B(t) = n2 or B(t) = 0n ,

An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

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Instructor Solutions Manual

etc. B(τ1 ), B(τ2 ), . . . has the same distribution as 1n S1 , 1n S2 . . . We apply part (b) to this approximating Brownian motion to see

nx na Pr{Mn (τ ) > x} = → e−x/a 1 + na As n → ∞, the partially observed Brownian gets closer to the Brownian motion. We conclude that M(τ ) is exponentially distributed with mean a (Rate parameter a1 ).

1.6 If Sn (e) is the compressive force on n balloons at a strain of e, we have E[Sn (e)] = nKe[1 − q(e)] × [1 − F(e)]. Because 1 log E[Sn (e)] = log nK + log e − log [1−q(e)][1−F(e)] , a plot of log E[Sn (e)] versus log e would have an intercept log nK reflecting material volume and modulus. Any departures from linearity would reflect the failure strain distribution plus departures from Hooke’s law.

1.7 (a) E[B(n + 1)|B(0), . . . , B(n)] = E[B(n + 1) − B(n)|B(0), . . . , B(n)] + B(n) = B(n). (b) E[B(n + 1)2 − (n + 1)|B(n)2 − n] = E[{B(n + 1)2 − B(n)2 − 1} + B(n)2 − n|B(n)2 − n] = B(n)2 − n + E[B(n + 1)2 − B(n)2 ] − 1 = B(n)2 − n.

1.8 The suggested approximations probably do well in mimicking Brownian motion in some tests, and √poorly in others. Because each BN (t) is differentiable, 1BN (t) will be on the order of 1t, where as 1B(t) is on the order of 1t.

Pr{B(u) 6= 0, t < u < t + b} 1 − ξ(t, t + b) = Pr{B(u) 6= 0, t < u < t + a} 1 − ξ(t, t + a) √ arc sin t/(t + b) = 0 < a < b. √ arc sin t/(t + a)

2.1 Pr{B(u) 6= 0, t < u < t + b | B(u) 6= 0, t < u < t + a} =

r √ a arc sin t/(t + b) = 0 < a < b.

2.2 lim √ t→0 arc sin t/t(t + a) b

2.3 Pr{M(t) > a} = 2{1 − 8t (a)} = Pr{B(t) > a}. But the joint distributions clearly differ (For 0 < s < t, it must be M(t) = M(s)).

2.4 The appropriate picture is The two paths are equally likely.

∂ 2z − x 1 2z − x ∂ ∂ 2z − x 1−8 =√ ϕ − 1−8 √ √ √ ∂x ∂z ∂x t t t t

1 ∂ 2z − x ϕ =√ √ t ∂z t

2 2z − x 2z − x = ϕ √ √ t t t 2.5 The Jacobian is one, whence 2 fM(t),Y(t) (z, y) = fM(t),B(t) (z, z − y) = t

z+y z+y ϕ √ √ t t

Z∞ Z∞ 2 y 2 2.6 fY(t) (y) = fM(t),Y(t) (z, y)dz = √ uϕ(u)du = √ ϕ √ = f|B(t)| (y). t √ t t 0

y/ t

Z Z R(t) 3.1 Pr √ > z = t √

1 − 1 (x2 +y2 ) e 2 dxdy = 2π

Z∞Z2π

1 − 1 r2 re 2 dθdr = − 2π

Z∞ n o 1 2 1 2 d e− 2 r = e− 2 z .

z 0

x2 +y2 >z

r r

Z∞ Z∞ R(t) R(t) π πt − 12 z2 E √ = Pr √ > z dz = e = and E[R(t)] = .

2 2 t t

0

0

z

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59

3.2 The specified conditional process has the same properties as bt + B◦ (t) where B◦ (t) is the Brownian bridge. Whence E[B(t)|B(1) = b] = bt and Var[B(t)|B(1) = b] = t(1 − t) for 0 < t < 1.

3.3 We need only show that B(1) and B(u) − uB(1) are uncorrelated (Why?)

E[B(1){B(u) − uB(1)}] = E[B(1)B(u)] − uE B(1)2 = u − u = 0. (a) B(t) = {B(t) − tB(1)} + tB(1) The conditional distribution of B(t) − tB(1), given B(1), is the same as the unconditional distribution, by independence, where as tB(1), given B(1) = 0, is zero. (b) E[B◦ (s)B◦ (t)] = E[{B(s) − sB(1)}{B(t) − tB(1)}]

= E[B(s)B(t)] − sE[B(1)B(t)] − tE[B(s)B(1)] + stE B(1)2 = s − st − st + st = s(1 − t) for 0 < s < t < 1.

s t s t 3.4 E [W ◦ (s)W ◦ (t)] = E (1 − s)B (1 − t)B = (1 − s)(1 − t)min , 1−s 1−t 1−s 1−t 0 < s < t < 1.

= s(1 − t),

Z∞ Z+∞ 3.5 y[φt (y − x) − ϕt (y + x)]dy = yϕt (y − x)dy = x. −∞

0

3.6 M > z if and only if a standard Brownian motion reaches z before 0, starting from x.

3.7 The same calculation as in Problem 3.5 showes that A(t) is a martingale. For the (continuous time) martingale A(t), the maximal inequality is an equality.

3.8 Since ϕt (y − x) satisfies the diffusion equation, so will [ϕt (y − x) − ϕt (y + x)] and [ϕt (y − x) − ϕt (y − x) + ϕt (y + x)] 3.9 (a) E[B◦(F(s))B◦(F(t))] = F(s)[1 − F(t)] (b) The approximation is

for s < t.

1 FN (t) ≈ F(t) + √ B◦ (F(t)).

N .

4.1 Pr{max[B(t) − bt] > a} = e n o −2a(a−b) .

4.2 Pr max b+B(t) 1+t > a = Pr {max B(t) − at > a − b} = e

t 4.3 Pr max0≤u≤1 B◦(u) > a = Pr max (1 + t)B◦ − a(1 + t) > 0 t>0 1+t

2 = Pr max B(t) − at > a = e−2a . −2ab

t>0

4.4 If the drift of X(t) is µ0 , then the drift of X 0 (t) = X(t) − 12 (µ0 + µ1 ) is − 12 δ where δ = µ1 − µ0 > 0. If the drift of X(t) is µ1 , then the drift of X 0 (t) = X(t) − 21 (µ0 + µ1 ) is 12 δ. Set δ = µ1 − µ0 and apply the procedure in the text to X 0 (t). n o e−2µx/σ 2 − 1 A 4.5 Pr max X (t) > B = .

2 e−2µB/σ − 1 4.6 Pr{Double your money} = 12 .

4.7 Pr {Z(τ ) > a|Z(0) = z} = Pr {log Z(τ ) > log a|log Z(0) = log z}

  log az a − 21 σ 2 τ . = 1 − 8 √ σ τ 4.8 The Black-Scholes value where σ = .35 is $5.21.

60

Instructor Solutions Manual √

2

√

d 2λx + c e− 2λx . The constants are evalu4.9 The differential equation is dx 2 2 w(x) = 2λw(x) for x > 0. The solution is w(x) = c1 e √

ated via the boundary conditions w(0) = 1, limx→∞ w(x) = 0 whence w(x) = e−1 2λx . There are many other ways to evaluate w(x), including martingale methods.

4.10 The relevant computation is E[Z(t)e−rt |Z(0) = z] = zert e−rt = z.

5.1 (a) The formula follows easily from iteration. It is a discrete analog of (5.22). (b) Sum 1Vn = Vn − Vn−1 = βVn−1 + ξn to get the given formula. It is a discrete analog to (5.24).   t  1 Z1 Z  i 2Z h σ 5.2 E[S(t)V(t)] = E  V(s)ds V(t) = E[V(s)V(t)]ds = e−β(t−s) − e−β(t+s) ds   2β 0

=

σ 2 −βt e 2β 2

0

Zt

0

σ2 βeβs − βe−βs ds = 2 eβt + e−βt − 2 2β

0

2 σ = [cosh βt − 1] β 5.3 This is merely the result of Exercise 4.6 applied to the position process.

1 1 1 5.4 1V = VN t + − VN (t) = √ X[Nt]+1 − X[Nt] = √ 1X, 1X = X[Nt]+1 − X[Nt].

N N N √

n √ o 1 v N 1 Pr 1V = ± √ |VN (t) = v = Pr 1X = ±1 X[Nt] = v N = ∓ .

2 2N N √ ! √ !

1 1 v N 1 1 v N 1 E [1V |VN (t) = v ] = + √ − −√ + = −v = −v1t.

2N 2N N N 2 N 2 h i 1 2 1 = = 1t.

E 1V 2 |VN (t) = v = √ N N

Chapter 9

1.1 Let X(t) be the number of trucks at the loader at time t. Then X(t) is birth and death process for which λ0 = λ1 = λ and µ1 = µ2 = µ · (λ2 = µ0 = 0) . Then θ0 = 1, θ1 = ñ = µλ , and θ2 = ñ2 π0 = Long run fraction of time of no trucks are at the ñ+ñ2 . Since trucks load at a rate of µ per unit time, long 1 loader = 2 . Fraction of time trucks are loading = 1 − π0 = 2 1+ñ+ñ

1+ñ+ñ ñ +ñ2 1+ñ run loads per unit time = µ (1 − π0 ) = µ =λ .

1+ñ+ñ2 1+ñ+ñ2 2.1 For the M/M/2 system, θk = 2ñk for k = 1.

1 + ñ 2 −1 = .

1 − ñ 1 − ñ 2 (1 − ñ) k 1 − ñ ; πk = π0 = ñ , k = 1 1 + ñ 1 + ñ 2ñ3 2ñ L0 = , L= .

1 − ñ2 1 − ñ2 L0 , L → ∞ as ñ → 1.

2 (See solution to 2.1) 2.2 W = λ1 L = µ1 2 1−ñ When X

θk = 2

X

ñk − 1 =

λ = 2, µ = 1.2 then ñ =

λ 1 = 2µ 1.2

and  W=

1   2

 2

1−

   = 3.27 1 2

1.2

1 For a single server queue W = µ−λ = 21 = 5. This helps explain why many banks have a single line feeding several tellers (M/M/s) rather than a separate line for each teller (s parallel M/M/1 systems).

ñ , π = 2 1−ñ ñk , k = 1.

2.3 For the M/M/2 system θ0 = 1, θk = 2ñk for k = 1, π0 = 1− k 1+ñ 1+ñ

∞ 1 − ñ X k 2ñ kñ = .

1 + ñ 1 − ñ2 k=1

1 1 1 1 2 1 W = π0 + π1 + + π2 + + π2 + · · · µ µ µ 2µ µ 2µ ∞ 1 1 X = (π0 + π1 + π2 + · · ·) + (k − 1) πk µ 2µ k=2

∞

1 1 1 − ñ X 1 1 k W= + (k − 1) ñ = µ µ 1 + ñ µ 1 − ñ2 L=

X

kπk = 2

k=2

and L = λW. An Introduction to Stochastic Modeling, Instructor Solutions Manual c 2011 Elsevier Inc. All rights reserved.

62

Instructor Solutions Manual

2.4 (a) λ0 = λ1 = λ2 = λ, µ1 = µ2 = µ3 = µ, µ0 = λ3 = 0 (b) θ0 = 1, θk = ñk , k = 1, 2, 3, ñ = µλ .

1 − ñ4 1 − ñ . . π0 = 1 − ñ 1 − ñ4

(c) Fraction of customers lost = π3 = 1−ñ4 ñ3 .

1−ñ X

θk = 1 + ñ + ñ2 + ñ3 =

2.5 Observe that µπk = λπk−1 . Following the hint, we get for j = 1 π0 P00j (t) = λπ0 P1j (t) − λπ0 P0j (t) = µπ1 P1j (t) − λπ0 P0j (t) and πk P0 kj (t) = µπk Pk−1,j−1 (t) + λπk Pk+1,j (t) − (λ + µ)πk Pk,j (t) = λπk−1 Pk−1,j−1 (t) + µπk+1 Pk+1,j (t) − (λ + µ)πk Pk,j (t).

Upon summing the above, the µ terms drop out and we get P0j (t) = λPj−1 (t) − λPj (t),

j=1

Similar analysis yields P00 (t) = −λP0 (t).

Set Qn (t) = eλt Pn (t). Then Qn (t) = Qn−1 (t) The solution (using Q0 (0) = P0 (0) = 1) is 0

(λt)n , n = 0, 1, 2, . . . hence n! (λt)n e−λt Pn (t) = n!

Qn (t) =

Compare with Theorem 5.1.

2.6 We want the smallest c for which

∞ P k=c+1

log 1000 log 1/ñ

log 1000 c∗ = log v/ñ

1 πk = ñc+1 5 1000 . Want smallest c for which

c+1 =

[x] = Integer part of x.

2.7 Since X(0) = 0 we set Pj (t) = P0j (t). The forward equations in this case are P0j (t) = λPj−1 (t) + µ( j + 1)Pj+1 (t) − (λ + µj)Pj (t). Multiply by j h i jP0j (t) = λ( j − 1 + 1)Pj−1 (t) + µ ( j + 1)2 − ( j + 1) Pj+1 (t) − λjPj(t) − µj2 Pj (t), Sum: M 0 (t) = λM(t) + λ − µM(t) + µP1 (t) − µP1 (t) − λM(t) = λ − µM(t), M(0) = 0. Let Q(t) = eµt M(t).

Q0 (t) = eµt M 0 (t) + eµt µM(t) = eµt M 0 (t) + µM(t) = λeµt , t = 0 λ Q(t) = µ

Zt

µeµt dt =

0

M(t) = e−µt Q(t) =

λ µt e −1 µ

λ 1 − e−µt . µ

t=0

j = 0.

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3.1 X(t) and Y(t) are independent random variables, each Poisson distributed where E[X(t)] = λ

Zl [1 − G(y)]dy 0

E[Y(t)] = λ

Z1 G(y)dy.

0

Observe that Y(t) is the number of points (Wk , Vk ) in the triangle Bt = {(w, v) : 0 ≤ w ≤ t, 0 ≤ v ≤ t − w} and that At and Bt are disjoint. Apply Theorem V, 6.1.

3.2 Method A: v = .5, τ 2 = .2, λ = 1, ñ = λv = 5λ

λ τ 2 + v2 .45λ = .5 + = .95 when λ = 1.

W = v+ 2 (1 − ñ) 2−λ If λ → 2 then W → ∞.

Method B : v = .4, τ 2 = .9, λ = 1, ñ = λv = .4λ

λ τ 2 + v2 1.06λ = .4 + = 1.28 W = v+ 2 (1 − ñ) .8(2.5 − λ)

when λ = 1

Method A is preferred when λ = 1, but B is better when λ < .3 or λ < 1.7 . . ..

4.1 When the faster server is first D = 1400 and π(1,1) = .3977 When the slower is first D = 1600 and π(1,1) = .4082 Slightly more customers are lost when the slower server is first.

4.2 Slower customers have priority σ=

2 5

τ=

4 15

Faster customers have priority σ= Lp 2 3 4 11

4 2 τ= 15 5 Ln L 1.6

2.27

82 55

1.85

Slower have priority Faster have priority Lslow .67 1.49

Lfast L 1.60 2.27 .36 1.85

Total L reduced with faster customers having priority.

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4.3 A birth and death process with λn = λ and µn = µ + rn ∞ X

1 λk π0 = P πk = π0 . θk = 1 + µ1 µ2 · · · µk θk k=1 P Rate at which customers depart prior to service is πk rk .

X

λk , µ1 µ2 · · · µk

4.4 λn = λ, n = 0; µ0 = 0, µ1 = α, µk = β, k = 2.

k λ β λ β λ θ0 = 1 θ1 = = θk = , k ≥ 1. α α β α β

∞ X βX λ k β λ θk = 1 + = 1+ ; 0 < λ < β. α β α β −λ k=1 k β(β − λ) λ α(β − λ) , πk = , k=1 π0 = α(β − λ) + βλ α(β − λ) + βλ β 4.5 λ0 = λ1 = λ2 = λ, µ1 = µ, µ2 = µ3 = 2µ.

λ 1 λ 2 1 λ 3 θ0 = 1, θ1 = θ2 = θ3 = . µ 2 µ 4 µ 1 π0 = 2 πk = θk π0 1 λ 3 λ 1 λ + 1+ + µ 2 µ 4 µ k

10 10 , k=0 5.1 Pr {X3 = k} = 1 − 15 15 k 10 Pr {X3 > k} = 15 c 10 ≤ .01 Want c such that 15 2 such that c log ≤ log .01 3 log .01 such that c = log 32 c = 11.36 c∗ = 12.

6.1 Let x be the rate of feedback. Then x + λ go in to Server #1, and .6(x + λ) go in and out and Server #2. But x = .2 of the output of Server #2. Therefore x = (.2)(.6)(x + λ) = .12x + .12λ .88x = .12λ x =

12 3 λ= .

88 11

Server #2: Arrival rate

6 3 15 = .6(x + λ) = +2 = .

10 11 11 15/11 6 π20 = 1 − = .

3 11 Server #3: Arrival rate = .4(x + λ) = π30 = 1 −

10 11

10/11 6 = 2 11

Long run Pr {X2 = 0, X3 > 0} =

6 5 30 × = .

11 11 121

Chapter 10

10.1 (a) Since the rows of the matrix P(t) add to 1, we may represent P(t) in the form

P(t) =

φ(t) 1 − φ(t) 1 − ψ(t) ψ(t)

and set Q =

−a a b −b

where φ(0) = 1, φ 0 (0) = −a, ψ(0) = 1, ψ 0 (0) = −b. The (1,1) term of the matrix equation P0 = PQ is the first-order linear equation φ 0 (t) = −(a + b)φ(t) + b.

The general solution is the sum of a particular solution and a general solution of the homogeneous equation, obtained by a family of exponentials; in detail φ(t) =

b + ce−(a+b)t a+b

(1)

The constant c is obtained from the first derivative at t = 0 in (1); thus −a = φ 0 (0) = −c(a + b) =⇒ c =

a a+b

which gives the result φ(t) =

b a −(a+b)t + e , a+b a+b

1 − φ(t) =

a a −(a+b)t − e . a+b a+b

Similarly one obtains ψ(t) by interchanging a and b. (b) We can use the result of part (a) to obtain E[V(t)|V(0) = v1 ] = v1 P11 (t) + v2 P12 (t)

a + be−µt b − be−µt = v1 + v2 a+b a+b av1 + bv2 bv − bv 1 2 = + e−µt a+b a+b (c) Interchange a with b and v1 with v2 in (1b), to obtain the indcated result. (d) Replace v1 with v21 and v2 with v22 in the result of part (b). (e) Interchange a with b and replace v1 with v21 , v2 with v22 . in the result of part (d).

10.2 Assuming that g, gt , gtt , gx , gxx are smooth functions, we have f (x, t) =

Z

ft (x, t) =

Z

g(µ, t)eiµx dµ

R

gt (µ, t)eiµx dµ

R

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ftt (x, t) =

Z

fxx (x, t) =

Z

gtt (µ, t)eiµx dµ

R

g(µ, t)(iµ)2 eiµx dµ

R

Z

gtt + 2gt + µ2 g eiµx dµ ftt + 2ft − fxx = R

By definition, f is a solution of the telegraph equation if and only if the left side is zero. But the Fourier transform is 1:1, so that we must have gtt + 2gt + µ2 g = 0 for almost all µ.

10.3 The ordinary differential equation for g is linear and homogeneous with constant coefficients. The form of the solution depends on whether p |µ| < 1, |µ| = 1 or |µ| > 1. In every case we have the quadratic equation r2 + 2r + µ2 = 0 whose solution is r = −1 ± 1 − µ2 which leads to the following three-fold system:

q q |µ| < 1 implies g(t) = e−t A cosh t 1 − µ2 + B sinh t 1 − µ2 |µ| = 1 implies g(t) = e−t (A + Bt)

q q |µ| > 1 implies g(t) = e−t A cos(t µ2 − 1) + B sin(t µ2 − 1) for suitable constants A, B. Exercises (10.4) and (10.5) can be paraphrased as follows: the idea is to charcterize those second-order differential operators which arise from random evolutions in one dimension. Consider the following set-up: let L be the set of second order differential operators of the form (10.32) for some v1 < v2 , q1 > 0, q2 > 0. Let M be the set of differential operators of the form (10.33) a20 = 1, a212 < 4a02 a20 , a10 > 0, v1 < a01 /a10 < v2 , a00 = 0

(2)

The revised exercises are written as follows: Exercise (10.4’). Let L ∈ L for some v1 < v2 , q1 > 0, q2 > 0. Then L ∈ M with a20 = 1, a11 = v1 + v2 , a02 = v1 v2 , a10 = q1 + q2 , a01 = q1 v2 + q2 v1 , a00 = 0.

Exercise (10.5’). Let L ∈ M for some (aij ) satisfying the above conditions (2). Then L ∈ L. Thus we have a necessary and sufficient condition for a second order operator to be associated with a random evolution process.

Solution of Exercises (10.4’), (10.5’): If L ∈ L, equation (2) holds and we can make the suitable identifications: a20 = 1, a11 = v1 + v2 , a02 = v1 v2 , a10 = q1 + q2 , a01 = v1 q2 + v2 q1 It is immediate that these satisfy the conditions (2) hence L ∈ M. Conversely, suppose that L ∈ M. Let v1 , v2 be the roots of the equation λ2 + a11 λ + a02 = 0. From the hypotheses, both roots are real and can be labeled so that v1 < v2 .

Next, we define q1 , q2 as the solution of the 2 × 2 system a10 = q1 + q2 ,

a01 = q1 v2 + q2 v1

The unique solution is q1 = (a01 − v1 a10 )/(v2 − v1 ),

q2 = (v2 a10 − a01 )/(v2 − v1 ).

In both formulas the denominator is positive; in addition we have the hypothesis v1 < a01 /a10 < v2 , which shows that the numerators are also positive. Hence the ratios are positive, so that, in particular, q1 + q2 > 0. Finally we need to verify

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the condition on a01 /a10 . But this follows by dividing the second equation by q1 + q2 to obtain v1 <

a01 < v2 a10

But a01 /a10 is a convex combination of v1 , v2 , hence it must lie on the segment from v1 to v2 . The proof is complete.• For completeness, we include the self-contained proof of (10.4) and (10.5): 10.4 In case N = 2 the determinant in eqn. (10.31) equals (−q1 − ∂t + v1 ∂x ) (−q2 − ∂t + v2 ∂x ) − q12 q21 = 0.

When we multiply out the diagonal terms there appear 9 terms for the product and another one from the off-diagonal terms.

The determinant is zero, hence we have only eight non-zero terms, namely ∂t2 once, ∂t ∂x twice, v1 v2 ∂x2 once, ∂t twice, ∂x twice When we add these eight terms we obtain (10.32), as required.

10.5 The polynomial equation is λ2 − λ(v1 + v2 ) + v1 v2 = 0. By inspection the roots are λ = v1 , v2 , where we may assume that v1 < v2 .

The coefficient a10 = q1 + q2 > 0. Finally a01 /a10 = (v1 q2 + v2 q1 )/(q1 + q2 ), a convex combination of v1 , v2 . Hence this fraction belongs to the interval (v1 , v2 ), as required.

10.6 If we are given v1 , v2 , q1 , q1 we can define a20 = 1, a11 = − (v1 + v2 ), a02 = v1 v2 , a10 = q1 + q2 , a01 = (v1 q2 + v2 q1 )(q1 + q2 ).

10.7 Since the roots are distinct, the associated cubic polynomial separates into three linear factors: λ3 + a21 λ2 + a12 λ + a03 = (λ − v1 )(λ − v2 )(λ − v3 ), where we label the v’s so that v1 < v2 < v3 , which proves i’).

To prove ii‘), note that a11 /a20 is a convex combination of the three quantities v1 + v2 , v2 + v3 , v1 + v3 . which are also in increasing order. This set can equally well be described as the open interval (v1 + v2 , v2 + v3 ), which was to be proved.

To prove the second part of ii’) note that a02 /a20 = (q1 v2 v3 + q2 v1 v3 + q3 v1 v2 )/(q1 + q2 + q3 ) is a convex combination of the three products v1 v2 , v1 v3 , v2 v3 . This convex combination belongs to the interval (m, M) where m = mini,j vi vj , M = maxi,j vi vj which proves the second half of ii‘).

10.8 From the formula for σ 2 , we have

Z∞ 1 −r/a σ = π r2 dr e 2π a 2

0

1 = 2a

Z∞ r2 e−r/a dr 0

Z∞ a2 = ρ 2 e−ρ dρ 2 0

=a

2

Chapter 11

√ R 2 11.1 The moment of order k of the normal distribution is obtained through the formula 2πmk = R xk e−x /2 dx. If k is odd the . integrand is an odd function, hence mk is zero if k is odd. If k is even, we can write k = 2n for some n = 1, 2, ... Next, we integrate by parts: ∞

Z √ 2 2πm2n = x2n−1 xe−x dx −∞

Z∞ =− −∞ Z∞

=+

2 x2n−1 d e−x /2

(2n − 1)x2n−2 e−x /2 dx 2

−∞

=

√

2π(2n − 1)m2n−2

For example m0 = 1, m2 = m0 , m4 = 3m2 = 3m0 and in general m2n = (2n − 1)(2n − 3) · · · m0 =

(2n)!

2n n!

11.2 The difference quotient is

φ(t + h) − φ(t) = E ei(t+h)X − eitX .

The integrand satisfies the conditions that for each t

lim ei(t+h)X − eitX = 0,

h→0

|ei(t+h)X − eitX | ≤ 2

which implies, by the bounded convergence theorem, that the integral tends to zero: limh→0 E ei(t+h)X − eitX = 0 in other words limh→0 (φ(t + h) − φ(t)) = 0, as required.

11.3 Write φX (t) = f (t) = u(t) + iv(t) where u, v are continuous functions with u(t)2 + v(t)2 ≤ 1.

Then f (t) = u(t) + iv(t) Z Z f (t) dt = u(t) + i v(t) dt Z = Reiθ , where R = f (t)dt Z Z Z −iθ e f (t) dt = cos θ u(t)dt + sin θ v(t)dt Z

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Now use Cauchy-Schwarz as follows: Z Z Z [cos θ u(t) + sin θ v(t)] dt cos θ u(t)dt + sin θ v(t) dt = Z p ≤ u(t)2 + v(t)2 dt Z = |f (t)| dt ≤ 1

which was to be proved.

11.4 We can extend theorem (11.3) to intervals of the type: (a, b), (a, b], [a, b). We work with the functions ψ ± which satisfy ψ − ≤ 1(a,b) ≤ ψ + ,

lim ψ − (x) = 1(a,b) (x),

lim ψ + (x) = 1[a,b] (x)

→0

→0

Now take the expectation, let n → ∞ and then → 0: Eψ − (Xn ) ≤ P[a < Xn < b] ≤ Eψ + (Xn )

(3)

Eψ − (X) ≤ lim inf P[a < Xn < b] ≤ lim sup P[a < Xn < b] ≤ Eψ + (X) n n P[a < X < b] ≤ lim inf P[a < Xn < b] ≤ lim sup P[a < Xn < b] ≤ P[a ≤ X ≤ b] n

n

If P[X = a] and P[X = b] are both zero, then the extreme members in (3) are equal and we have lim P(a < Xn < b) = P(a < X < b) = P(a ≤ X ≤ b) n

which was to be proved.

The proof for (a, b] and [a, b) is entirely similar. In the first case we have Eψ − (Xn ) ≤ P[a < Xn ≤ b] ≤ ψ + (Xn )

(4)

Eψ (X) ≤ lim inf P[a < Xn ≤ b] ≤ lim sup P[a < Xn ≤ b] ≤ Eψ (X) −

+

n

n

P[a < X ≤ b] ≤ lim inf P[a < Xn ≤ b] ≤ lim sup P[a < Xn ≤ b] ≤ P[a < X ≤ b] n

n

If P[X = a] and P[X = b] are both zero, then the extreme members in (4) are equal and we have lim P(a < Xn ≤ b) = P(a < X ≤ b) = P(a ≤ X ≤ b) n

11.5 The complex exponential is defined by the power series eiz =

∞ X (iz)n /n!. n=0

Moving the first three terms to the left side, we have eiz − 1 − iz − (iz)2 /2 =

∞ X (iz)n /n! n=3 ∞

eiz − 1 − iz − (iz)2 /2 X n−2 = (iz) /n! z2 n=3

The term on the right is bounded by |z|e|z| . Hence eiz − 1 − iz 1 + ≤ |z|e|z| → 0, 2 z2

z→0

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11.6 If z = α + iβ is an arbitrary complex number, the multiplicative property of the exponential implies that α+iβ

e

α iβ

= e e

r

= e |e | = e |cos β + i sin β| == e cos2 β + sin2 β = eα . α

iβ

α

α

11.7 For 0 ≤ θ ≤ π/2, cos θ ≤ 1, so that sin θ =

Zθ

Zθ cos t dt ≤

0

dt = θ.

0

Meanwile θ → sin θ is a convex function so that it always lies above its chord with respect to any two points, esp. on the interval [0, π/2] where the chord is the line with equation y = (2/π)θ. Combining these two estimates, we have the two-sided estimate 2 θ ≤ sin θ ≤ θ, π Z 11.8

e R2

−(x2 +y2 ) 2

Z∞Z2π 2 dxdy = e−r /2 r dr dθ 0 0

= 1 · 2π

0≤θ ≤

π 2