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Example 3.3 of Ross's book on Introduction to Probability models is If X and Y are independent Poisson random variables with respective means λ1 and λ2, calculate the conditional expected value of X given that X +Y = n. And the solution is quite straight forward. Now, what if instead we had $P\{X|X+Y \ne n\}$? Can I just replace $P\{Y=\}$ with $1-P$?

$ P\{X=k|X+Y=n\} = \frac{P\{X=k \ , \ X+Y=n\}}{P\{X+Y = n\}} = \frac{P\{X=k \ , \ Y=n-k\}}{P\{X+Y = n\}} = \frac{P\{X=k\} P\{Y=n - k\}}{P\{X+Y = n\}} $

$ P\{X=k|X+Y \ne n\} = \frac{P\{X=k \ , \ X+Y \ne n\}}{P\{X+Y \ne n\}} = \frac{P\{X=k \ , \ Y \ne n-k\}}{P\{X+Y \ne n\}} = \frac{P\{X=k\} P\{Y \ne n - k\}}{P\{X+Y \ne n\}} = \frac{P\{X=k\}(1-P\{Y = n-k\})}{1-P\{X+Y=n\}} $

Does that work? Thanks

Ram Zi
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  • Yes that seems to work, though why you might want to do that is another matter – Henry Oct 16 '21 at 17:37
  • Thanks @Henry. The problem I'm facing now is that in the original question, when you write everything down and most of the terms cancel out each other, we will end up with a binomial distribution, and so the expected value of that distribution can be calculated. but, when I do mine, terms do not cancel out, and I do not seem to be able to get a distribution out of it. I'm not sure if I'm doing it right or not. – Ram Zi Oct 16 '21 at 17:43

1 Answers1

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Yes that seems to work.

You have a conditional distribution and since everything is Poisson distributed you can write the terms down.

For $0 \le k \le n$ it gives $${n \choose k}\frac{\lambda_1^k \left((n-k)!e^{\lambda_2}-\lambda_2^{n-k} \right) }{n!e^{(\lambda_1+\lambda_2)}-(\lambda_1+\lambda_2)^n }$$

For $k \gt n$ it gives $$\frac{\frac{1}{k!}\lambda_1^k }{e^{\lambda_1}-\frac{ 1}{n!}e^{-\lambda_2}(\lambda_1+\lambda_2)^n }$$

Summing over $k$ will give $1$. But it does not have a well known distribution.

Henry
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