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Question
Chapter 5, Problem 91SE

a.

To determine

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

a.

Expert Solution

Explanation of Solution

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.

In that case, the above integral becomes:

FTo(t)=1(2ν12Γ(ν12))(2ν22Γ(ν22))0t01(tp)ν121(t(1p))ν221et2|J(t,p)|dpdt=et22ν1+ν22Γ(ν1+ν22)β(ν12,ν22)0ttν121+ν22101pν121(1p)ν221tdpdt=et22ν1+ν22Γ(ν1+ν22)0ttν12+ν221[1β(ν12,ν22)01pν121(1p)ν221dp]dt.

Now, consider the quantity:

1β(ν12,ν22)01pν121(1p)ν221dp.

This is the integration of the probability density function (pdf) of the beta distribution (first kind) taken over the whole range (0,1). Hence,

1β(ν12,ν22)01pν121(1p)ν221dp=1.

Thus, the cdf reduces to:

FTo(t)=et22ν1+ν22Γ(ν1+ν22)0ttν12+ν221[1β(ν12,ν22)01pν121(1p)ν221dp]dt=et22ν1+ν22Γ(ν1+ν22)0ttν12+ν221dtFTo(t)=0tet22ν1+ν22Γ(ν1+ν22)tν12+ν221dt_.

Compare this with the pdf of the chi-squared distribution. It is clearly observed that the quantity within the integral, et22ν1+ν22Γ(ν1+ν22)tν12+ν221, is the pdf of a chi-squared distribution where the degrees of freedom is ν=ν1+ν22.

It is known that the cdf of a distribution is unique. Thus, using the uniqueness property of the cdf of a random variable, it is proved that X1+X2 has a chi-squared distribution with parameter ν1+ν2_.

b.

To determine

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

b.

Expert Solution

Answer to Problem 91SE

The distribution of Z12+Z22+...+Zn2 is a chi-squared distribution with parameter ν=n_.

Explanation of Solution

Given info:

Reference- Exercise 71: The random variables Z1,Z2,...,Zn are independently distributed standard normal variables.

Calculation:

The result in Exercise 71 shows that, for a standard normal random variable, Z, the variable Z2 has a chi-squared distribution with ν=1.

The result in part a shows that the addition of two variables having chi-squared distributions with parameters ν1 and ν2 produces another variable, which also a chi-squared distribution with parameter ν=ν1+ν2. This result can be generalized for the addition of n such variables with chi-squared distributions.

For the random variables Z1,Z2,...,Zn, each Zi2 has a chi-squared distribution with νi=1, for i=1,2,...,n.

Thus, by generalization of the result in part a and the result in Exercise 71, the random variable Z12+Z22+...+Zn2 will have a chi-squared distribution. The parameter of this distribution is:

1+1+...+1 (upto n terms)=i=1n(1)=n.

Hence, the distribution of Z12+Z22+...+Zn2 is a chi-squared distribution with parameter ν=n_.

c.

To determine

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

c.

Expert Solution

Answer to Problem 91SE

The distribution of the sum Y=i=1n(Xiμσ)2 is chi-squared distribution with parameter ν=n_.

Explanation of Solution

Given info:

The random variables X1,X2,...,Xn form a random sample taken from a normal distribution with mean μ and variance σ2.

Calculation:

For a normally distributed random variable, X, with mean μ and variance σ2, the standard normal variable, Z, obtained is:

Z=Xμσ.

Thus, for the normally distributed random variables Xi each having n μ and variance σ2, the corresponding standard normal variable is

Zi=Xiμσ, for i=1,2,...,n.

Now,

Y=i=1n(Xiμσ)2=i=1nZi2=Z12+Z22+...+Zn2.

Thus, Y is the sum of the squares of n standard normal variables.

The result in part b shows that the distribution of Z12+Z22+...+Zn2 is a chi-squared distribution with parameter ν=n.

Hence, the distribution of the sum Y=i=1n(Xiμσ)2 is chi-squared distribution with parameter ν=n_.

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Chapter 5 Solutions

Probability and Statistics for Engineering and the Sciences

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.2 - Prob. 22ECh. 5.2 - Prob. 23ECh. 5.2 - Prob. 24ECh. 5.2 - Prob. 25ECh. 5.2 - Prob. 26ECh. 5.2 - Prob. 27ECh. 5.2 - Prob. 28ECh. 5.2 - Prob. 29ECh. 5.2 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.3 - Prob. 37ECh. 5.3 - Prob. 38ECh. 5.3 - Prob. 39ECh. 5.3 - Prob. 40ECh. 5.3 - Prob. 41ECh. 5.3 - Prob. 42ECh. 5.3 - Prob. 43ECh. 5.4 - Prob. 46ECh. 5.4 - Prob. 47ECh. 5.4 - Prob. 48ECh. 5.4 - Prob. 49ECh. 5.4 - Prob. 50ECh. 5.4 - Prob. 51ECh. 5.4 - Prob. 52ECh. 5.4 - Prob. 53ECh. 5.4 - Prob. 54ECh. 5.4 - Prob. 55ECh. 5.4 - Prob. 56ECh. 5.4 - Prob. 57ECh. 5.5 - Prob. 58ECh. 5.5 - Prob. 59ECh. 5.5 - Prob. 60ECh. 5.5 - Prob. 61ECh. 5.5 - Prob. 62ECh. 5.5 - Prob. 63ECh. 5.5 - Prob. 64ECh. 5.5 - Prob. 65ECh. 5.5 - Prob. 66ECh. 5.5 - Prob. 67ECh. 5.5 - Prob. 68ECh. 5.5 - Prob. 69ECh. 5.5 - Prob. 70ECh. 5.5 - Prob. 71ECh. 5.5 - Prob. 72ECh. 5.5 - Prob. 73ECh. 5.5 - Prob. 74ECh. 5 - Prob. 75SECh. 5 - Prob. 76SECh. 5 - Prob. 77SECh. 5 - Prob. 78SECh. 5 - Prob. 79SECh. 5 - Prob. 80SECh. 5 - Prob. 81SECh. 5 - Prob. 82SECh. 5 - Prob. 83SECh. 5 - Prob. 84SECh. 5 - Prob. 85SECh. 5 - Prob. 86SECh. 5 - Prob. 87SECh. 5 - Prob. 88SECh. 5 - Prob. 89SECh. 5 - Prob. 90SECh. 5 - Prob. 91SECh. 5 - Prob. 92SECh. 5 - Prob. 93SECh. 5 - Prob. 94SECh. 5 - Prob. 95SECh. 5 - Prob. 96SECh. 5 - Prob. 97SECh. 5 - Prob. 98SE
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