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Suppose that $X_1$ and $X_2$ are independent Poisson distributed random variables with means $\lambda_1$ and $\lambda_2$. Given is that $W = X_1 + X_2$ is also Poisson distributed with mean $\lambda_1 + \lambda_2$. Use this result to show that the conditional distribution of $X_1$, given that $W=w$, is a binomial distribution with $n=w$ and $p = \lambda_1 /(\lambda_1 + \lambda_2)$.

I looked up the PDFs for the Poisson distribution and the binomial distribution. I also looked at the formula for conditional probabilities. However, I cant seem to find the clue where to start. Could anyone please help me out?

Ashot
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dreamer
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1 Answers1

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$$P(X_1 = x \mid W = w) = \dfrac{P(X_1 = x \text{ and } W = w)}{P(W=w)} $$ Note that if $X_1 = x$ and $W=w$ you must have $X_2 = w - x$. Now evaluate these probabilities using the Poisson distribution.

Michael Hardy
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Robert Israel
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    If I understood correctly what you mean, then I get the following: [(lambda2)^(w-x) * e^-(lambda2)/(w-x)!] / [{(lambda1 + lambda2)^w * e^-(lambda1 + lambda2)} / w!]. Is this correct? And if so, how could I proceed from here? Thank you a lot for your help, it is highly appreciated. – dreamer Mar 22 '13 at 19:44
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    You left out the $P(X_1 = x)$ part from the numerator. – Robert Israel Mar 22 '13 at 19:55
  • Got it! Thanks a lot for your explanations! – dreamer Mar 24 '13 at 12:05