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!

  • $\begingroup$ $E[Y^a], E[Y^b], Var(Y^a)$ and $Var(Y^b)$ are of course all known too in terms of $\lambda^a, \lambda^b, \mu^a$ and $\mu^b$. $\endgroup$
    – M Tegling
    Commented Dec 5, 2014 at 14:49
  • $\begingroup$ There seems to be a typo. You've got the distribution of $N^a$ listed twice. $\endgroup$
    – Bill
    Commented Dec 5, 2014 at 16:11
  • $\begingroup$ Yes, indeed. The second one should of course be $N^b$. I'll edit that. $\endgroup$
    – M Tegling
    Commented Dec 5, 2014 at 17:31

2 Answers 2


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.

  • 1
    $\begingroup$ Good +1. Yet I think that each of the three minus signs at the end must be replaced by plus. $\endgroup$
    – Yves
    Commented Dec 5, 2014 at 17:21
  • $\begingroup$ @Yves Yes, you are correct. There should be plus signs. $\endgroup$
    – M Tegling
    Commented Dec 5, 2014 at 23:25
  • $\begingroup$ @Yves I second plus signs. +1 $\endgroup$
    – lanenok
    Commented Dec 6, 2014 at 15:22

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.

  • $\begingroup$ Thank you! Yes, it turned out to be a little easier than I though. I started out with a more complex problem and using the tower property, etc got really messy. I was using all sorts of substitutions and trying recursive moment generations and other methods. After stripping everything down again to the simple basic problem I guess I was too confused and fed up with it to be able to attack it rationally again. Thank you for your help! $\endgroup$
    – M Tegling
    Commented Dec 5, 2014 at 17:29

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