Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
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Chapter 7.2, Problem 34E

Suppose that W1 and W2 are independent χ2-distributed random variables with ν1 and ν2 df, respectively. According to Definition 7.3,

F = W 1 / v 1 W 2 / v 2

has an F distribution with ν1 and ν2 numerator and denominator degrees of freedom, respectively. Use the preceding structure of F, the independence of W1 and W2, and the result summarized in Exercise 7.30(b) to show

  1. a E(F) = ν2/(ν2 − 2), if ν2 > 2.
  2. b V ( F ) = [ 2 v 2 2 ( v 1 + v 2 2 ) ] / [ v 1 ( v 2 2 ) 2 ( v 2 4 ) ] , if ν2 > 4.

*7.30    Suppose that Z has a standard normal distribution and that Y is an independent χ2-distributed random variable with ν df. Then, according to Definition 7.2,

T = Z Y / v

has a t distribution with ν df.

  1. a If Z has a standard normal distribution, give E(Z) and E(Z2). [Hint: For any random variable, E(Z2) = V(Z) + (E(Z))2.]
  2. b According to the result derived in Exercise 4.112(a), if Y has a χ2 distribution with ν df, then

    E ( Y a ) = Γ ( [ v / 2 ] + a ) Γ ( v / 2 ) 2 a , if  v > 2 a .

    Use this result, the result from part (a), and the structure of T to show the following. [Hint: Recall the independence of Z and Y.]

    1.                                 i.            E(T) = 0, if ν > 1.
    2.                               ii.            V(T) = ν/(ν − 2), if ν > 2.

a.

Expert Solution
Check Mark
To determine

Prove that E(F)=ν2(ν22),ifν2>2.

Explanation of Solution

Let F=W1/ν1W2/ν2

Here, W1 has a χ2-distributed random variables with ν1 df and W2 has a χ2-distributed random variables with ν2 df. W1 and W2 are independent.

E(F)=E(W1/ν1W2/ν2)=ν2ν1E(W1W2)=ν2ν1E(W1)E(1W2)

From the exercise 7.30(b), if Y has a χ2 distribution with ν df, then E(Ya)=Γ([ν/2]+a)Γ(ν/2)2a,ifν>2a.

Here,

E(1W2)=E(W21)=Γ([ν2/2]1)Γ(ν2/2)21=Γ([ν2/2]1)(ν221)Γ(ν221)21=2×21ν22=1ν22

Then,

E(F)=ν2ν1×ν1×1ν22=ν2ν22,ν2>2

Hence proved.

b.

Expert Solution
Check Mark
To determine

Prove that V(F)=2ν22(ν1+ν22)ν2(ν22)2(ν24),ifν2>4.

Explanation of Solution

The variance of F is calculated as follows:

V(F)=E(F2){E(F)}2

E(1W22)=E(W22)=Γ([ν2/2]2)Γ(ν2/2)22=Γ([ν2/2]2)(ν221)(ν222)Γ(ν222)22=22(ν221)(ν222)=22×22(ν22)(ν24)=1(ν22)(ν24)

E(W12)=V(W1)+{E(W1)}2=2ν1+(ν1)2=ν1(ν1+2)

E(F2)=E(W1/ν1W2/ν2)2=ν22ν12E(W12W22)=ν22ν12E(W12)E(1W22)=ν22ν12×ν1(ν1+2)×1(ν22)(ν24)=ν22ν12ν1(ν1+2)(ν22)(ν24)

Then,

V(F)=ν22ν12ν1(ν1+2)(ν22)(ν24)(ν2ν22)2=ν22ν12ν1(ν1+2)(ν22)(ν24)ν22(ν22)2=ν22(ν22)[ν1(ν1+2)ν12(ν24)1(ν22)]=ν22(ν22)[ν1(ν1+2)(ν22)ν12(ν24)ν12(ν24)(ν22)]=ν22(ν22)[ν1(ν1+2)(ν22)ν12(ν24)ν12(ν24)(ν22)]=ν22(ν22)[ν1{(ν1+2)(ν22)ν1(ν24)}ν12(ν24)(ν22)]=ν22(ν22)[(ν1ν22ν1+2ν24)(ν1ν24ν1)ν1(ν24)(ν22)]=ν22(ν22)[(ν1ν22ν1+2ν24)ν1ν2+4ν1ν1(ν24)(ν22)]=ν22(ν22)[2ν1+2ν24ν1(ν24)(ν22)]=2ν22(ν1+ν22)ν1(ν22)2(ν24),ν2>4

Hence proved.

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Chapter 7 Solutions

Mathematical Statistics with Applications

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