Expected value of a product of two compound Poisson processes I'm working on my master thesis now and I've been struggling with a problem for some while now and no one seems to be able to help me or point me in any direction. So now I reach out to see if someone here can help me.
Basically I have two (dependent) compound Poisson processes with no parameters in common. I want to find an analytical expression of the covariance of those processes or at least a useful approximation. What remains to compute is this the following:

Searched:
$E[Y^a Y^b]$, where $Y^a = \sum_{i=1}^{N^a} X_i^a$ and $Y^b = \sum_{i=1}^{N^b} X_i^b$
Known distributions:
$N^a \sim Po(\lambda^a)$, $N^b \sim Po(\lambda^b)$, $X^a \sim Exp(1/\mu^a)$, $X^b \sim Exp(1/\mu^b)$
Known parameters:
$\lambda^a, \lambda^b, \mu^a, \mu^b$
Correlations:
$Cov(N^a,N^b)$ and $Cov(X^a,X^b)$ are non-zero and can be found. Otherwise independent ($N^a \perp X^a$, etc).

How can I solve this? Does any analytically tractable solution exist? Can I approximate the solution in any way?
If anything is unclear please let me know and I'll explain further!
Thanks in advance!
 A: Use the tower porperty of conditional expectations. 
\begin{eqnarray}
\mathbb{E}\left[Y^a Y^b\right] 
&=& \mathbb{E}\left[\mathbb{E}\left[Y^a Y^b\ |\ N_a, N_b \right]\right]
\\
&=& \mathbb{E}\left[\mathbb{E}\left[\left(\sum_{i=1}^{N_a}X_i^a\right)\left(\sum_{i=1}^{N_b}X_i^b\right) \Bigg|\ N_a, N_b \right]\right]
\\
&=& \mathbb{E}\left[\mathbb{E}\left[\sum_{i=1}^{N_a}\sum_{j=1}^{N_b}X_i^aX_j^b\ \Bigg|\ N_a, N_b \right]\right]
\\
&\overset{1}{=}& \mathbb{E}\left[\sum_{i=1}^{N_a}\sum_{j=1}^{N_b} \mathbb{E}\left[ X_i^aX_j^b\ \right]\right]
\\
&\overset{2}{=}& \mathbb{E}\left[ N_aN_b\left(\text{Cov}(X^a, X^b) - \mathbb{E}\left[X^a\right]\mathbb{E}[X^b]\right)\right]
\\
&=&\left(\text{Cov}(N^a, N^b) - \mathbb{E}\left[N^a\right]\mathbb{E}[N^b]\right)\left(\text{Cov}(X^a, X^b) - \mathbb{E}\left[X^a\right]\mathbb{E}[X^b]\right)
\end{eqnarray}
In step 1 the sums are moved out of the integral (finite sums) and the conditioning can be removed. In step 2 we use that we can express the expectation of the product with known stuff, and we sum.
A: I guess we start like this:
\begin{align}
Y^aY^b &= \left(\sum_{i=1}^{N^a}X^a_i\right)\left(\sum_{i=1}^{N^b}X^b_i\right)\\
       &= \sum_{i=1}^{N^a}\sum_{j=1}^{N^b}X^a_iX^b_j
\end{align}
The expectations of the terms of the sum are all the same at $E\{X^aX^b\}$, which you 
say we may assume known (or estimated?).  Also, the terms of the sum are independent of
how many terms there are (that is $X^aX^b$ is independent of $N^aN^b$).  So, the
expectation of the sum (by an argument you have probably seen many times if you
work with stochastic processes) is $E\{X^aX^b\}E\{N^aN^b\}$.  Now, we are done.  If you
know $Cov(N^a,N^b)$, then you know $E\{N^aN^b\}$.
Did I misunderstand the question somehow?  Or maybe I have made an error?  That seemed too easy.
