Conditional expectation of a random variable conditional on a function of the random variable

Question

I have a problem understanding the following about conditional expectations:

Say $X$ and $Y$ are two discrete random variables. Then the conditional expectation of $X$ conditional on $Y$ is $E[X|Y] = \psi(Y)$ where $\psi(y) = E[X|Y=y]$. $E[X|Y]$ is a random variable depending on $Y$, whereas $E[X|Y=y]$ is a function of $y$.

My problem is this: Assume for example $X$ and $Y$ being independent Poisson RVs of mean $\lambda_1$ and $\lambda_2$ respectively. I can compute the conditional pmf $P(X=k|X+Y=n)$ and the conditional expectation $E[X|X+Y=n]$, namely $$E[X|X+Y=n] = \frac{\lambda_1 n}{\lambda_1+\lambda_2}$$

But what is $E[X|X+Y]$? One can't just substitute and say $$E[X|X+Y=n] = \frac{\lambda_1 (X+Y)}{\lambda_1+\lambda_2}$$ since $E[X|X+Y]$ is not a function of $X$.

Some help in my lack of understanding is welcome!

Answer 1

Indeed $\mathsf E(X\mid X+Y)$ is not a function of (or rather, "measured against") $X$; nor $Y$.   It is a random variable measured over (the sigma algebra of) the random variable $(X+Y)$.

Let's give this a new name, say, $Z=X+Y$

Then do you have any problems interpreting the following? : $\quad\mathsf E(X\mid Z) = \dfrac{\lambda_1 Z}{\lambda_1+\lambda_2}$

---

The expected value of $X$, measured relative to the sum of $X+Y$, is a particular fraction of that sum.

Answer 2

You can compute this:

$$P(X,X-Y=z) = P(X,Y=z-X) = P(Y=z-X| X) P(X),$$

where the r.h.s. is a product of two Poissons. So:

$$P(X|X+Y) = P(X,X+Y) / P(X+Y).$$

The denominator is going to be Poisson distributed. Can you finish it from here?