Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
Question
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Chapter 5.4, Problem 65E

a.

To determine

Prove that the marginal distribution of Y1 is exponential with mean 1.

a.

Expert Solution
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Explanation of Solution

Calculation:

Consider that Y1 and Y2 are two continuous real valued random variables with joint probability density function of f(y1,y2).

Then, the marginal probability functions of Y1 and Y2 are defined as,

f1(y1)=f(y1,y2)dy2 and f2(y2)=f(y1,y2)dy1.

Exponential distribution:

A random variable Y is said to follow the exponential distribution with mean 1, if and only if the probability density function of Y is f(y)=ey,y>0.

Consider that C1=12ey1.

Thus, using the joint probability density functions of Y1 and Y2, the marginal probability of Y1 is calculated below:

f1(y1)=0[1α{C1(12ey2)}]ey1y2dy2=ey10ey2dy2αC1ey10(12ey2)ey2dy2=ey1[ey2]0αC1ey1[0ey2dy220e2y2dy2]=ey1[0+1]αC1ey1[[ey2]0[e2y2]0]=ey1αC1ey1[[0+1][0+1]]=ey1αC1ey1[0]=ey1

Thus, it is proved that the marginal distribution of Y1 is exponential with mean 1.

b.

To determine

Find the marginal distribution of Y2.

b.

Expert Solution
Check Mark

Answer to Problem 65E

The marginal distribution of Y2 is Exponential with mean 1.

Explanation of Solution

Calculation:

Consider that C2=12ey2.

Thus, using the joint probability density functions of Y1 and Y2, the marginal probability of Y2 is calculated below:

f2(y2)=0[1α{(12ey1)C2}]ey1y2dy1=ey20ey1dy1αC2ey20(12ey1)ey1dy1=ey2[ey1]0αC2ey2[0ey1dy120e2y1dy1]=ey2[10]αC2ey2[[ey1]0[e2y1]0]=ey2αC2ey2[[0+1][0+1]]=ey2αC2ey2[0]=ey2

Thus, the marginal distribution of Y2 is Exponential with mean 1.

c.

To determine

Prove that Y1 and Y2 are independent if and only if α=0.

c.

Expert Solution
Check Mark

Explanation of Solution

The two continuous random variables Y1 and Y2 are said to be independent if and only if f(y1,y2)=f1(y1)f2(y2), where f(y1,y2) is the joint probability density function of Y1 and Y2 and f1(y1),f2(y2) are the marginal probability density functions of Y1 and Y2, respectively.

Substitute α=0 in the joint probability distribution.

That is,

f(y1,y2)=[1(0){(12ey1)(12ey2)}]ey1y2=[10]ey1y2=ey1y2

From Part (a) and Part (b), the marginal distribution of Y1 is Exponential with mean 1 and the marginal distribution of Y2 is Exponential with mean 1.

Here, the ranges of Y1 and Y2 are not dependent on each other and according to the rule of independence the probability of joint distribution is equal to the product of two marginal distributions if and only if α=0.

That is,

f1(y1)f2(y2)=(ey1)(ey2)=ey1y2=f(y1,y2)

Thus, it is proved that Y1 and Y2 are independent if and only if α=0.

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Chapter 5 Solutions

Mathematical Statistics with Applications

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