Calculate $E[X + Y\mid X + Y + Z = n]$ where $X,Y,Z$ are i.i.d Poisson

Question

Let $X$, $Y$, and $Z$ be independent Poisson random variables with $ \lambda = 1$. Calculate $E[X + Y\mid X + Y + Z = n]$.

I believe this is it: $$ E[X + Y\mid X + Y + Z = n] = \frac{E[X+Y]E[X+Y+Z=n]}{E[X+Y+Z=n]}$$ and by the definition of expectation this gives: $$ \sum_k \frac{P(X + Y = k)P(X + Y + Z=n)}{P(X + Z = n - Y)} \implies$$ $$ \sum_k \frac{ e^{-(\lambda_X + \lambda_Y)} \times \frac{(\lambda_X + \lambda_Y)^k}{k!} \times [e^{-(\lambda_X + \lambda_Z)} \times \frac{(\lambda_X + \lambda_Z)^{n - Y}}{(n - Y)!}]} {[e^{-(\lambda_X + \lambda_Z)} \times \frac{(\lambda_X + \lambda_Z)^{n - Y}}{(n - Y)!}]} \implies $$ $$ \frac{ \lambda_Y^k(\lambda_X+\lambda_Z)^{n-k} } { (\lambda_X + \lambda_Y + \lambda_Z)^{n} } \times { \frac{n!}{k!(n-k)!} } $$ $$= {n \choose k}\left(\frac{\lambda_X}{\lambda_X + \lambda_Y + \lambda_Z}\right)^k \times \left(\frac{\lambda_X + \lambda_Z}{\lambda_X + \lambda_Y + \lambda_Z}\right)^{n-k}$$

Answer 1

All expectations in this answer are conditional on $X+Y+Z=n$; I'm omitting that for conciseness.

By symmetry, $E[X]=E[Y]=E[Z]$. On the other hand, $E[X]+E[Y]+E[Z]$ $=E[X+Y+Z]=n$. Thus $E[X]=E[Y]=E[Z]=\frac n3$, and thus $E[X+Y]=E[X]+E[Y]=\frac{2n}3$.

Note that I didn't use the specific distribution of the variables or even their independence; this follows by symmetry alone.

Answer 2

On several occasions some form of the following question has appeared here:

Suppose $U\sim\operatorname{Poisson}(\lambda)$ and $V\sim\operatorname{Poisson}(\mu)$ are independent. How do you prove that the conditional distribution of $U$ given the event $\big[U+V=w\big]$ is $\operatorname{Binomial}(w, \lambda/(\lambda+\mu))\text{?}$

Recall that the assumptions stated in the question entail that $X+Y\sim\operatorname{Poisson}(2)$ and $X+Y,$ $Z$ are independent. Thus with $U=X+Y$ and $V=Z,$ we get the conditional distribution of $X+Y$ given $X+Y+Z=n$ is $\operatorname{Binomial}(n,2/3).$