Identify $E(X\mid X+Y)$ for $X$ and $Y$ independent and Poisson I am currently trying to understand how does an instance of the EM algorithm is derived when there are "hidden variables" in this old paper: http://www.ncbi.nlm.nih.gov/pubmed/18238264.
I don't really understand how the expectation step is derived, and what bothers me the most, is that I cannot prove the following statement (right before 2.11):
$$
\mathbb{E}[X|X+Y] = \frac{(X+Y)\lambda_X}{\lambda_X+\lambda_Y}
$$
where $X \sim Poisson\{\lambda_X\}$ and $Y \sim Poisson\{\lambda_Y\}$
To begin, I wrote the discrete expectation as follows:
$$
\mathbb{E}[X|X+Y] = \sum_{k=1}^{X+Y} k P(X=k|X+Y)\\
= \sum_{k=0}^{X+Y} k \frac{P(X+Y|X=k)P(x)}{P(X+Y)}
$$
Where I can replace
$$ P(X+Y=k) = \frac{(\lambda_X+\lambda_Y)^ke^{-(\lambda_X+\lambda_Y)}}{k!}$$
$$ P(k=x) = \frac{\lambda_X^xe^{-\lambda_X}}{x!}$$
$$ P(X+Y=k|x) = P(Y=x-k) = \frac{\lambda_Y^{x-k}e^{-\lambda_Y}}{(x-k)!}$$
but I dont really know how to mix that together, because I don't really see what does the event $X+Y$ stands for, and how does it rely on the x in the integral ?
Thank you very much for you help
 A: Here is a hint:  Assuming $X$ and $Y$ are Poisson and independent, the conditional distribution of $X$ given that $X+Y = n$ is binomial with parameters $n$ and $p = \lambda_X/(\lambda_X + \lambda_Y)$.

Let us prove the above statement (which is stronger than simply doing the expectation calculation).  Consider the random variable $X + Y = N$.  Then 
$$\Pr[X = x \mid N = n] = \frac{\Pr[(X = x) \cap (N = n)]}{\Pr[N = n]}= \frac{\Pr[(X = x) \cap (Y = n-x)]}{\Pr[N = n]}.
$$  The numerator is straightforward.  The denominator is $$\begin{align*} \Pr[N = n] &= \sum_{x=0}^n \Pr[(X = x) \cap (Y = n-x)] \\
&= \sum_{x=0}^n e^{-\lambda_X} \frac{\lambda_X^x}{x!} e^{-\lambda_Y} \frac{\lambda_Y^{n-x}}{(n-x)!} \\
&= \frac{e^{-(\lambda_X+\lambda_Y)}}{n!} \sum_{x=0}^n \binom{n}{x} \lambda_X^x \lambda_Y^{n-x} \\ 
&= \frac{e^{-(\lambda_X + \lambda_Y)}}{n!} (\lambda_X + \lambda_Y)^n, \end{align*}$$ which of course proves that the sum of independent Poisson variables is Poisson with rate equal to the sum of the individual rates.  Then $$\Pr[X = x \mid N = n] = e^{-\lambda_X} \frac{\lambda_X^x}{x!} e^{-\lambda_Y} \frac{\lambda_Y^{n-x}}{(n-x)!} e^{\lambda_X + \lambda_Y} \frac{n!}{(\lambda_X + \lambda_Y)^n} = \binom{n}{x} \left(\frac{\lambda_X}{\lambda_X + \lambda_Y}\right)^x \left(\frac{\lambda_Y}{\lambda_X + \lambda_Y}\right)^{n-x},$$ which of course is binomial with the claimed parameters.  The distributional properties immediately follow.
