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Statistics / Mathematical Statistics with Applications 7 / Chapter 5 / Problem 39E

Mathematical Statistics with Applications | 7th Edition | ISBN: 9780495110811 | Authors: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer

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Suppose that Y1 and Y2 are independent Poisson distributed

Chapter 5, Problem 39E

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QUESTION:

Problem 39E

Suppose that Y1 and Y2 are independent Poisson distributed random variables with means λ1 and λ2, respectively. Let W = Y1 + Y2. In Chapter 6 you will show that W has a Poisson distribution with mean λ1 + λ2. Use this result to show that the conditional distribution of Y1, given that W = w, is a binomial distribution with n = w and p = λ1/(λ1 + λ2).1

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QUESTION:

Problem 39E

Suppose that Y1 and Y2 are independent Poisson distributed random variables with means λ1 and λ2, respectively. Let W = Y1 + Y2. In Chapter 6 you will show that W has a Poisson distribution with mean λ1 + λ2. Use this result to show that the conditional distribution of Y1, given that W = w, is a binomial distribution with n = w and p = λ1/(λ1 + λ2).1

ANSWER:

Solution

Step 1 of 2

We have to show that W has a poisson distribution with mean $\lambda{}{\ }_{1}+\lambda{}{\ }_{2}$

Given that conditional distribution of Y1 given that w is a binomial distribution with n=w and $p=\lambda{}{\ }_{1}(\lambda{}{\ }_{1}+\lambda{}{\ }_{2})$

Let Y1 and Y2 are two independent poisson random variables with mean $\lambda{}{\ }_{1}and\ \lambda{}{\ }_{2}$

And $W=y{\ }_{1}+y{\ }_{2}$

Here $Y{\ }_{1}\sim{}Poisson(\lambda{}{\ }_{1})\ \ \ and\ Y{\ }_{2}\sim{}Poisson(\lambda{}{\ }_{2})\$

Then $P(Y{\ }_{1}=k)=\frac{e{\ }^{-\lambda{}{\ }_{1}}\ \lambda{}{\ }_{1}^{k}}{k!}$

$P(Y{\ }_{2}=k)=\frac{e{\ }^{-\lambda{}{\ }_{2}}\ \lambda{}{\ }_{2}^{k}}{k!}$

Then $P(Y{\ }_{1}+y{\ }_{2}=k)=\frac{e{\ }^{-(\lambda{}{\ }_{1}+\lambda{}{\ }_{2})}\ (\lambda{}{\ }_{1}+\lambda{}{\ }_{2}){\ }^{k}}{k!}$

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