Expected value of conditional poisson distribution

Question

Given two independent Poisson random variables $X$ and $Y$ with parameters $\lambda_{x}$ and $\lambda_{y}$ what is the $E[X|X+Y]$

I know that $X+Y$ is Poisson distributed with parameter $\lambda_{x}+\lambda_{y}$

I also know that $P[X=x|X+Y=k]$ is binomial with parameters $n=x+y$ and $$p= \frac{\lambda_{x}}{\lambda_{x}+\lambda_{y}}$$

So can I directly use that to find the expected value?

Answer 1

Notice that $E[X|X+Y]$ is a random variable which satisfices that $E[E[X|X+Y]]=E[X]$. Then, $E[X|X+Y]$ has as support $(X+Y)$'s and its law is given by: $$E[X|X+Y=k]=\sum_{x\in \mathcal{R}_x}x\mathbb{P}(X=x|X+Y=k)$$ I don't know if some of this answer your question. I wait for comments.

Answer 2

I believe you answered your own question with the binomial. Binomial with $(n,p)$ has mean $np$. Your question gave $p=\frac{\lambda_{x}}{\lambda_{x}+\lambda_{y}}$ and $n=x+y$, which I believe is $n=k$. Hence the answer.