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Question: Let $X$ and $Y$ be independent random variables, each $\mathrm{Poi}(\lambda)$ distributed, and $Z = X+Y$.

Given $Z=n$, what is the expectation of $X - Y$? Explain it.

My Work: From solving a previous part of this question, I believe that $Z$ is distributed: $Z\sim\mathrm{Poi}(2\lambda)$, and that the $P(X=k \mid Z=n)$ is the pdf of a binomial distribution.

Thus, the expectation of $X$, given $Z=n$, is $n/2$.

I'm confused, however, on how to prove the expectation of $X - Y$. My guess is that $E(X-Y)$ given $Z=n$ should be something like $0$? (Since $E(X \mid Z=n)$ is $n/2$ and $E(Y | Z=n)$ should also be $n/2$?)

Could someone explain how to prove something like this for $E(X-Y | Z=n)$ though?

Thanks!

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    If you can show both $E(X|Z=n)$ and $E(Y|Z=n)$ are $n/2$ (or just that they're both the same), then you're done (conditional expectation is linear) – 2017-01-12
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    The distribution of $X$ and $Y$ does not matter (as long as it is the same). In view of symmetry, $E(X\mid Z) = E(Y\mid Z) = Z/2$, so the answer is always $0$. – 2017-01-12
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    Of course the symmetry argument is (perfectly rigorous and) the neatest one. But what prevents you to compute the joint distribution of $(X,Y)$ conditionally on $Z$, that is, the value of $$P(X=k,Y=\ell\mid Z=n)$$ for every $(k,\ell)$ and to deduce the desired quantity $$E(X-Y\mid Z=n)=\sum_{k,\ell}(k-\ell)P(X=k,Y=\ell\mid Z=n)\ ?$$ – 2017-01-12

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