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Probability and Statistics for Eng...

9th Edition
Jay L. Devore
ISBN: 9781305251809

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Chapter
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BuyFindarrow_forward

Probability and Statistics for Eng...

9th Edition
Jay L. Devore
ISBN: 9781305251809
Textbook Problem

a. Let X1 have a chi-squared distribution with parameter v1 (see Section 4.4), and let X2 be independent of X1 and have a chi-squared distribution with parameter v2. Use the technique of Example 5.22 to show that X1 + X2 has a chi-squared distribution with parameter v1 + v2.

b. In Exercise 71 of Chapter 4, you were asked to show that if Z is a standard normal rv, then Z2 has a chi-squared distribution with v = 1. Let Z1, Z2, . . . , Zn  be n independent standard normal rv’s. What is the distribution of Z12 +…+ Zn2 ? Justify your answer.

c. Let X1, … , Xn be a random sample from a normal distribution with mean µ and variance σ2. What is the distribution of the sum Y = i = 1 n [ ( X i μ ) / σ ] 2 ? Justify your answer.

a.

To determine

Prove that X1+X2 has a chi-squared distribution with parameter ν1+ν2.

Explanation

Given info:

Reference- Example 5.22: The random variables X1 and X2 are independently distributed with chi-squared distribution. The parameter of X1 is ν1 and that of X2 is ν2.

Calculation:

The probability density function (pdf) of a random variable X having a chi-squared distribution with parameter ν is given as:

fX(x)={12ν2Γ(ν2)xν21ex2, x00,                         otherwise.

Denote To=X1+X2. Thus, using the above pdf and the technique in Example 5.22, the cdf of To is obtained as follows:

FTo(t)=P(X1+X2t)=0t0tx1[12ν12Γ(ν12)x1ν121ex12][12ν22Γ(ν22)x2ν221ex22]dx2dx1 (as x1+x2=t)=1(2ν12Γ(ν12))(2ν22Γ(ν22))0t0tx1x1ν121x2ν221ex1+x22dx2dx1=1(2ν12Γ(ν12))(2ν22Γ(ν22))0t0tx1x1ν121x2ν221et2dx2dx1

Make the transformations:

x1=tp and x2=t(1p), so that x1+x2=t is true and 0p1.

The Jacobian determinant of the transformation is:

|J(t,p)|=|x1t x1px2t x2p|=|p             t(1p) t|=|tpt+tp|=t

b.

To determine

Find the distribution of Z12+Z22+...+Zn2.

c.

To determine

Find the distribution of the sum Y=i=1n(Xiμσ)2.

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