Expected value of Poisson pdf $p(a\mid X)$, where $X$ is also Poisson distributed I had a hard time figuring out how to phrase this question, so please forgive the confusing title. 
To be concrete: the expected value of a function $g(X)$ of a random variable $X$, where $X$ has the pdf $f(x)$, can be computed as:
$$ E[g(X)] = \int_R \! g(x) f(x) \, \mathrm{d}x $$
Now in my case, I have
$$ f(x) = \mathrm{Poisson}(x\mid\lambda) = \frac{\lambda^x}{x!}{e^{-\lambda}} $$
and $g(x)$ is also Poisson, except its mean is $X$, so we have
$$ g(x) = \mathrm{Poisson}(a\mid x) = \frac{x^a}{a!}{e^{-x}} $$
and what I want is the integral
$$ E[\mathrm{Poisson}(a\mid X)] = \int_R \!  \mathrm{Poisson}(a\mid x) \mathrm{Poisson} (x\mid\lambda) \, \mathrm{d}x \\
 = \int_0^\infty \! \frac{x^a}{a!}{e^{-x}} \frac{\lambda^x}{x!}{e^{-\lambda}} \, \mathrm{d}x
$$
or the discrete version is fine as well:
$$ E[\mathrm{Poisson}(a\mid X)] = \sum_x \frac{x^a}{a!}{e^{-x}} \frac{\lambda^x}{x!}{e^{-\lambda}} $$
Do either of these have an analytic solution? What if $$ \mathrm{Poisson}(a\mid x) $$ was approximated by a normal distribution?
Physically this situation arises when e.g. a Poisson process produces some random number of particles, which then go on to trigger another cascade of particles, and we are interested in the pdf of the number of particles in the second cascade. So we sum over all the possible numbers of particles coming from the first process. This is straightforward to do numerically, but I just wonder if there isn't a way to do it analytically... I tried to compute it via characteristic functions but that just made it worse I think:
$$ E[\mathrm{Poisson}(a\mid X)] = \frac{1}{2\pi}\frac{1}{a!}\Gamma(a+1) \int_{-\infty}^{+\infty} (1-i \omega)^{-a-1}\cdot e^{\lambda\left(e^{i \omega} - 1 \right)} \, \mathrm{d} \omega$$
(assuming I didn't screw up getting that far)
 A: If you are looking for the expectation rather than the distribution, then you can use the law of total expectation
$$\mathbb{E}[a] = \mathbb{E}_X\left[\mathbb{E}_{a \mid X}[a \mid X]\right]$$
and in this example this is 
$$ =  \mathbb{E}_X\left[X\right] = \lambda$$
If you want further evidence, you could try simulation.  For example in R, 
set.seed(1)
cases <- 1000000
lambda <- 7
X <- rpois(cases, lambda)
a <- rpois(cases, X)

gives
> mean(X)
[1] 6.998076
> mean(a)
[1] 6.999449

with the final result closer to $\lambda$ than it deserves to be
A: The Poisson distribution is a discrete distribution.  Its support is the set $\{0,1,2,3,\ldots\}$.  If $X\sim\operatorname{Poisson}(\lambda),$ then
$$
\operatorname{E}(g(X)) = \sum_{x=0}^\infty g(x) \frac{e^{-\lambda} x^\lambda}{x!}.
$$
(Be careful about when I use capital $X$ and when I use lower-case $x$.)
A: $$X\sim\text{Po}(\lambda), Y\mid X\sim\text{Po}(X)\implies\mathbb{P}(Y=y)=\sum_x \mathbb{P}(Y=y\mid X=x)\mathbb{P}(X=x)$$
$$\implies\mathbb{P}(Y=y)=\sum_x \frac{x^y e^{-x}}{y!}\frac{\lambda^xe^{-\lambda}}{x!}=\frac{e^{-\lambda}}{y!}\sum_{x\ge0}\frac{x^y\left(\lambda e^{-1}\right)^x}{x!}=\frac{e^{-\lambda}}{y!} T_y(\lambda e^{-1})$$
where $T_y$ are Touchard polynomials - I'm not sure these simplify much more in the general case.
