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Statistical method in economics

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Introduction to statistical method in economics Problem 1 Suppose that E and Y represent coordinates on the Cartesian plane, where E and Y both have a standard normal distribution and are indepen- dent. Find the joint pdf of the polar coordinates v and 8. (Remember that vX = E X ‡ Y Xand tan 8 = Y .) Are vE and 8 independent?

Problem 2 Suppose that n random variables Es are independent and identically dis- tributed as Normal: Es ~ N(µ = fiOO, aX = 22†), s = fi, 2, ..., n. a.

Consider the case where n = 4. What is the probability that none fi Σ n ofb.the four random variables Define the average of (E the as: Ethan = fifi†? f i , nErandom X , E 3 , Evariables 4 ) is greater n s=fi E s . n What is the probability that E 4 is less than fifi†? c.

What is the sma11est. sntegev sampleΣsize nO that one would need in . n 0 — µ ≤. † ≤ O.9†? order to ensure t hat P E Problem 3 Assume the coin flips below are independent flips of a fair coin. a.Calculate the exagt probability that the number of heads in ten flips is less than or equal to 4. b.Use the novma1 appvoxsmatson to the binomial distribution to ap- proximate the probability you found in part a. and compare your answers in terms of accuracy. c.Now assume that you flip the coin fiOO times. The exact probability (up to three digits after the decimal point) that the number of heads is less than or equal to 4O is O.O28. What is the approximated probability (using

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2

the novma1 distribution) that the number of heads is less than or equal to 4O? d.

Suppose we now begin to use a different coin such that the flips are still independent but the probability of heads in each toss is now p = f i .

Suppose you flip this coin n = fiOO times. Compute the exact probability that H = 6 and compare it with an appvoxsmatson based on the Possson dsstvsbutson.

XO

Problem 4 Let E f i and EX be normally distributed random variables, so that (x—µ s )2 fi e 2a 2 ƒ E s(x) = , 2uas for s = fi, 2, and suppose that E f i and EX are independent. a. b.

(Hint: c.

Find the joint pdf of E f i and E X .

Define Y = E f i ‡ E X . Find the pdf of Y using the fi-step method use YX = E f i — EX as an "ancilliary variable").

Find the mean and variance of Y .

Problem 5 Suppose that E is distributed chi-squared with p degrees of freedom, where p is a positive integer. a.

Write down the pdf of E. Show that E can also be characterized as having a gamma distribution. What are the parameters of the gamma distribution? b.

Now suppose that Y is distributed normally with mean µ and .vari- Σ Y —µ X ‡ E . What ance aX, and that Y and E are independent. Let Z = a is the distribution of Z? c.

Suppose that p = 4. Use the appropriate table to find the value A such that Pr (Z > A) = O.O†

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The joint pdf of X and Y will be equal to the product of the marginal pdfs, since X and Y are independent. f X , Y (x, y) = f X (x) f Y (y) 1 −1 22x 1 −1 2y √ e 2π 2πe 2 2 1 −2 1 (x +y ) = e 2π The transformation into polar coordinates is 2 2 r 2 = X Y+ Y =

with inverse transformations

√

tan θ = X Y

= =

2

X r cos θ r sin θ

This yeilds the following matrix of partial derivatives. ∙ ¸ cos θ −r sin θ sin θ = rcos cosθ θ(r cos θ) − sin θ (−r sin θ) 2 =matrix, r cos + r Jacobian, sin2 θ ¢ is The determinant of this ¡ θthe 2 = r cos θ + sin θ =2r The transformations are unique, so we can use the 1-step method without modification.

1 −1 2r2 f rθ (r, θ) = r e 2π where r lies within [0, ∞] and θ lies within [0, 2π]. Because the ranges are not dependent and the joint pdf is separable, r and θ are also independent.

Problem 2 ³ X i− μ ´ μ a.

For a single random variable: P ( X ≤ 115) i= P ≤115− . σ σ X i− μ Notice that Z = i is distributed standard normal (Z ∼ N (0, 1)) so: i σ J

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³ ´ 115−100 √ P (X ≤ = P (Z ≤i 1). Using the Table you can i 115) = P Z ≤ i 225 find that this probability is approximately equal to: 0.8413. By indepen- dence: P (X 1 ≤ 115, X 2 ≤ 115, X 3 ≤ 115, X 4 ≤ 115) = 4 = 0.500 96.

P (X 1 ≤ 115) P (X ³ P115) P (X 4 ≤ ¡=´¢0.8413 ¡ P 2 ≤ 115) P (X 3 ≤ P 115) 2 n 225 ¢ = n n 1 b.

X = 1 X i∼ N 1 μ i, σ2 = N 100, n n i=1 i=1 i =1 i n n n µ ³ 15´ 2 ¶ ³ ¡ 15¢ 2 ´ X 4−100 N 100, , so: X ∼ N 4100, . Thus: Z = is a stan2 √ 2 n ¶ ( 15 ) = X 4 − 120 0 115−100 2 < 15 15 ( ) ( ) P (Z < 2) = 0.9772. à ! ³ − √2n ´ √ ¡¯ ¢ µ n ´ ¯ ³ ³ ´ X − μ c.

P X =P ≤ Z ≤ 3 n= 5 n 3 −μ ≤5 =P ¯ ¯ √ 15n ¯ ≤ n √15 0.95. From the table we know that: P (Z ≤ ¢1.96) ' 0.975 and using the sym¡ metry of the normal distribution this implies that P (−1.96 ≤ Z ≤ 1.96) ' dard normal random variable: P X √ 4 < 115 = P n 2 0.95, so = 34. 574. We want the smallest 3= 1.96 ⇒ n = (1.96 · 3) integer and it is n 0 = 35.

(H) in 10 independent flips of a fair coin is disPa. r o b l e The m 3 number of heads 10 k 10− k 1 ¡ . P (0 ¢ ¡ ¢4 P tributed Binomial 10, ≤ H ≤ 4) = (0.5) (0.5) = 2 ¡10 ¢ P 4 k=0 k 10 (0.5) = £¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢¤ k=0 k 101024 = = (0.5) 10 100 +101 +102 + 3 +104 = 0.376 95 386 b. Since H is binomial we can calculate its mean and variance: E [ H ] = 10 · (0.5) = 5,V ar [H] = 10 · (0.5) (1 − 0.5) = 2.5. The approximation re- lies on the E [H assumption is] distributed a normal random 5 'similar √H −H variable, so: that = H√ −2.5 Z ∼ Nto(0, 1). Therefore: P (0 ≤ H ≤ 4) = V ar[H] µ ¶ ³ ´ [H] − E [H ] √− 5 √− 1 ' P (−3.162 ≤ Z ≤ − √0 −VEar[H] √4 − E [H]' P P ≤ V√Har[H] ≤ V ar[H] 0.632) = 2.5 ≤ Z ≤ 2.5 P (Z ≤ 3.162) − P (Z ≤ 0.632) ' 0.999 − 0.736 = 0.263 . Thus the approximation is not very accurate for n = 10. µ ¶ ³ − 50 ´ [H] H −√ E[H] 4 0−E[H] − 10 √ √ √0 −E √ c.

Now P (0 ≤ H ≤ 40) = P ≤ ≤ ' P ≤ Z ≤ = 25 25 V ar[H] V ar[H] V ar[H] P (−10 ≤ Z ≤ −2) = P (Z ≤ 10) − P (Z ≤ 2) ' 1 − 0.977 = 0.023, which is quite close to the exact probability. ¡ ¢¡ ¢ ¡ 6 ¢100− 6 100 1 d.

Exact calculation: P (H = 6) = 120 − 1 = 0.15.

6 20 Approximation: as n → ∞ , p → 0, (np) → λ the binomial distribution converges to the Poisson distribution with parameter λ. Since here np =

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5 we can approximate the distribution with a Poisson distribution where e− λ λ6 λ = 5: P (H = 40) ' = 0.146 . Clearly, this is a good 6! = e −5 5 6 6! approximation.

Problem 4

(x − μ i )2

√ 1 2σ 2i − First of all, to have valid pdfs, we must use f ( x )=X i . As 2πσ i e always, sorry about the typo (I omitted the negative sign). a. Because X 1 and X 2 are independent, the joint pdf is again the product of the marginal pdfs:

fX 1

X2

(x 1 , x 2 ) = f X 1 (x 1 ) f X 2 (x 2 ) 2 − ( x −1 μ )1 − 1 1 1 2σ 2 √ 2πσ 2e = √ 2πσe1 µ³ x 1− μ 1 ´ 2 ³ x 2− μ + 1 − 21 σ1 σ2 = e 2πσ 1 σ 2

b.

We will use Y1 for Y . We start with the transformations Y1

= X 1 + X 2 Y2

= X1

− X2 which will yeild the following inverse transformations: Y 1 + Y2 X 1 = Y1 −2Y2 X 2 = 2 Then the matrix of partial derivatives is ∙1 2 1 2

So the Jacobian is

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1¸ 21

−

¯¡ 12¢ ¡ −12¢ − ¡ 12¢¡ ¢12

2

¯=

1 2

(x − 2 μ )2 2σ 22 2

´ 2¶

2

The transformation is unique, so we can use the 1-step method without modification.

1

f Y 1 Y 2 (y1 , y2 ) =

1 2

−

e

⎛ Ã ⎝

12

2πσ σ

1

=

e

4π σ 1 σ 2

=

e

− −

σ1

µ

1 2 σ 21σ 22 1 2 2 2σ1 σ2

!2

y 1+y 2 2

σ2

³

2

µ

σ 2 2 4

(

1 2! 2 ⎞ 2 −μ2 σ2

y −y

+

y1 +y 2 2

−μ

´ 1

2

−

2y 1 y −22 μ y 1− 2

1

=

(−

e

12

4πσ2y σy1 −22 μ − e

2 1

Z

)

∞

−∞

y 1− 2 2 μ y +2μ 1 1 y −2 1 2 2 2 8σ 1

1

2 μ y +2μ −

μ y +2μ 121 2 1y − 2 2 2 μ y +2μ − 2 8σ 1 2− 2 σ 2+σ (y (μ + 1 μ ))2 1 1 2 8 σ 21σ 22

2 1

2 μ y +2μ −

y1 −2y 2

1

´

−μ

2

2

2

)

2μ2 2−

σ 2 1 4 1+

(

2 2− y12−2y y1+y

2

4μ y 1+4μ y +4μ 2 2

2

2

( μ − μ )) 2

× −2y 1 y 2 +2μ y 1 +2μ 2 y 1− 1

2

− μ1 )) 2

2 12 μ y 2+2μ 2 y −2μ − 2 8σ 2

2 1

2μ2 2

) ×

2μ2 2−

e

2 1

1 µ −¶ ¶

4μ1 y2 +4μ 2)

− 4 μ y1 −1

2 ( μ + μ )) +(y − 1 2 2 8 σ 21σ 22

To get the pdf of Y , we must integrate over Y2.

2 σ 2+ σ 2 2)( ( y − 1 ( μ1+ μ 2)) +(y − (2μ Z ∞ − ( 1 1 8 σ 21σ 22 e f Y (y1) = − ∞ 4πσ 1 σ 2 e

³

⎠

2

2

1 2

2 2

2

+σ

y 12+2y y +y

1 ( σ 21+σ 22)( (y − 1 − 1 4πσ 1 σ 2 e 2y y1 −22 μ y 1− 2 μ y +2μ 121 2 1y − −σ σ 4π 2 8σ 1 e 1 2

=

Ã

−μ1

−

−2y 1 y 2 +2μ y 1 +2μ 2 y 1− 1

( σ 21+σ (y22)−

2

2 12 μ y 2+2μ 2 y −2μ − 2 8σ 2 2

( μ1− μ 2))

2μ2 2

2 1

2μ 2 2

2 1

2μ 2 2

dy2

×

8σ 21σ 22

2 μ2 − 2 y y +2μ y +2μ y − 1 2 1 2 1 1 2−

2 1

2 1

2 μ y +2μ y − 2 2 μ2 − 2

8σ 2

dy2

which has general. By other methods, it can be proved ¡ no closed form, ¢ in 2 that Y ˜N μ +1 μ , σ2 + σ2 .

We can show this here if we let σ 1 = σ 2 = σ.

2 1 1 4πσ

f Y (y1) =

− ( 2e

2 − 2 σ (y 1

)

2y 1 y −22 μ y 1− 2

e

= =

−

√

(y −

e

−

π2σ 1 e− √ π 2σ

R because ∞ integrate to 1.− ∞

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1

√ 1 π2σ

1

( μ + μ )) 1 2 8σ 4

μ y +2μ 121 2 1y − 8σ 2 ( μ1+ μ 2)) 4σ2

(y −1 ( μ +1μ )) 2 4σ 2

e−

2

2

2

Z

2 2

Z

∞ −∞

( 2σ 2 )(y2 − ( μ 1 − μ 2 ) ) 2 8σ 4

∞ −∞

e−

( 2σ 2 ) ( y 2 − ( μ 1 − μ 2 ) ) 2

2 μ y +2μ −

√

1 π2σ

e

2 1

(

×

8σ 4 2 μ 2 − 2 y y 1+2μ 2 y +2μ 1 2 y −1 1 2− 2 σ−2 ( y 2 − ( μ 1 − μ 2 )8σ )2 4

)

2 12 μ y 2+2μ y − 2 μ 2− 8σ 2

dy2

dy2 is a standard normal, and must

dy

2

2 2

)

¶

c. E (Y ) = E (X 1 + X 2 ) = μ 1 + μ 2 , since X 1 and X 2 are independent, normally distributed randome variables. Similarly, V (Y ) = V (X 1 + X 2 ) = σ 21+ σ 22 (since X 1 and X 2 are independent, their covariance is zero).

Problem 5 a.

X is distributed χ 2 with p degrees of freedom, so its pdf is p x 1 2 − 1e − p f ( x )= x 2 Γ p2 ¡ ¢2 2 A gamma distribution for a random variable Y is of the form y 1 f (y)= y α − 1e − β Γ (α) β α You can see that if we let α = p and2β = 2, X has a gamma distribution. b.

We learned in class that the square of a standard normal random has a χ 2 distribution with one degree of freedom.

Thus ¡ y− μ ¢variable 2 2 ˜χ(1) . In addition, we learned that the sum of two independent σ χ 2 variables will also have a χ 2 distribution, with degrees of freedom equal to the sum of the degrees of freedom of initial random variables. Because Y and X are independent, Y 2 and X will also independent, and we can ¡ y−be μ¢ 2 2 apply this property to conclude that + X ˜χ σ (p+1) . c. If p = 4, we use the fourth row of the table given in class, and look for the column corresponding to α = 0.05. We can see that A = 9.488.

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