A compound Poisson random variable $S$ is defined as: $S=\displaystyle\sum^N_{i=1}X_i,$ where $N$ is a random draw from a Poisson distribution with intensity parameter $\lambda$, and $X_i$ are independent identically distributed random variables.

What is $\operatorname{Cov}(S,N)$?

$N$ is a Poisson random variable with mean $\lambda $. $S$ is compound Poisson random variable with Poisson parameter $\lambda $ and component distribution $F$.

$X_i$ where i =1,2,...is a sequence of iid r. v.s having distribution $F$

  • $\begingroup$ What is N? You are interested in covariance with the number of variables? $\endgroup$
    – Tim
    Oct 17, 2016 at 16:52
  • 2
    $\begingroup$ I edited the question to make it self-contained. I think it's an interesting question. I don't recall seeing and answer to it before. $\endgroup$
    – Aksakal
    Oct 17, 2016 at 17:21
  • 3
    $\begingroup$ You cannot answer the question without defining the distribution of the $X_i$'s. $\endgroup$
    – Xi'an
    Oct 17, 2016 at 17:58
  • $\begingroup$ dxaston has already answered this question correctly. So Now no matter, if it is deleted or not. $\endgroup$ Oct 18, 2016 at 15:03

1 Answer 1


We'll denote $\mu_N = \text{E}(N)$, $\mu_X = \text{E}(X)$ and $\sigma_N^2 = \text{Var}(N)$. To find the covariance we can use the formula

\begin{align} \text{Cov}(S, N) &= \text{E}(SN) - \text{E}(S) \text{E}(N) \\ &= \text{E}(SN) - \mu_N^2 \mu_X \end{align}

where the second equality is found by taking an iterated expectation

\begin{align} \text{E}(S) &= \text{E} [ \text{E}(S \mid N) ] \\ &= \text{E} \left [ \text{E} \left (\sum_{i=1}^{N} X_i \mid N \right ) \right ] \\ &= \text{E} [ N \text{E}(X) ] \\ &= \mu_N \mu_X . \end{align}

To find $\text{E}(SN)$ we make a similar conditioning argument

\begin{align} \text{E}(SN) &= \text{E} \left [ \text{E}(SN \mid N) \right ] \\ &= \text{E} \left [ N^2 \text{E}(X) \right ] \\ &= \mu_X \left ( \sigma_N^2 + \mu_N^2 \right ) \end{align}

and so we get

\begin{align} \text{Cov}(S, N) &= \mu_X \sigma_N^2 . \end{align}

  • $\begingroup$ But E(S)=$\lambda$E(X) $\endgroup$ Oct 18, 2016 at 19:17
  • $\begingroup$ @Dhamnekar Yes, that's what I have. $\lambda = \mu_N$ when $N \sim$ Poisson$(\lambda)$. I gave the general solution where $N$ need not be Poisson distributed. $\endgroup$
    – dsaxton
    Oct 18, 2016 at 19:22

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