Conditional Expectation of Poisson variables Let $X \sim exp(\lambda)$ be a random variable. For a given value of $X$, let $Y,Z \sim \operatorname{Poisson}(X)$ be two Poisson random variables. I try to calculate $E(Y\mid Z)$. I tried first to find $E(Y\mid X),E(X\mid Z)$. The first term is trivial, but I'm not sure how to find out $E(X\mid Z)$ (actually I'm not even sure that these calculations give the right answer at the end. Does total expectation law simply solve it?)
 A: Let $d\mu(t)=\lambda e^{-\lambda t}dt$ $$P(Z=n|X)=e^{-X}\frac {X^{n}} {n!}.$$ So $$P(Z=n, X \leq x)= {\int_0^{x} {e^{-t} t^{n}}/n!} d\mu(t).$$ This gives $$P( X\leq x|Z=n)=\frac {{\int_0^{x} {e^{-t} t^{n}}/n!} d\mu(t)t} {{\int_0^{\infty} {e^{-t} t^{n}}/n!} d\mu(t)}.$$ Finally $$E(X|Z=n)=\frac {{\int_0^{\infty } t{e^{-t} t^{n}}/n!} d\mu(t)} {{\int_0^{\infty} {e^{-t} t^{n}}/n!} d\mu(t)}.$$
$E(Y|Z)$ cannot be calculated with the given information. However $E(Y|Z)=EY=EX=\lambda$ if you assume that $Y$ and $Z$ are independent.
A: This gives details on my comments. We have for each $z \in \{0, 1, 2, ...\}$:
\begin{align}
E[Y|Z=z] &= \sum_{y=0}^{\infty} y P[Y=y|Z=z]\\
&=\sum_{y=0}^{\infty} y \frac{P[Y=y,Z=z]}{P[Z=z]}
\end{align}
Now using the law of total probability in integral form (conditioning on $X=x$):
\begin{align}
P[Z=z] &= \int_{0}^{\infty} P[Z=z|X=x]f_X(x)dx\\
P[Y=y,Z=z] &= \int_0^{\infty} P[Y=y,Z=z|X=x]f_X(x)dx
\end{align}

Notes:

*

*We can compute $P[Z=z|X=x]$ from the information given in the question.


*We cannot compute $P[Y=y, Z=z|X=x]$ from the information given in the question. However, if we additionally assume $Y, Z$ are conditionally independent given $X$ (which was stated in later comments), then we can compute it.


*Note that if $Y, Z$ are conditionally independent given $X$, that does not mean $Y$ and $Z$ are independent.
A: Note, that by the tower property we can get
$$E[Y|Z]=E[E[Y|X]|Z]$$ $E[Y|X]$ is a function of $X$, which simply equals $X$, since $Y\sim Poisson(X)$, and so $E[Y|Z]=E[X|Z]$. From here we can use Murthy's answer to calculate it (I'm not sure, but I think that Michael's answer gives the same result, still it's more complicated to calculate).
Thank you everyone
