# Covariance of a poisson distribution and a condition sum of variables

Let ${N(t),t ≥ 0}$ be a Poisson process with rate $λ$ that is independent of the sequence $X_1,X_2,\ldots$ of independent and identically distributed random variables with mean $μ$ and variance $σ^2$. Find $$\operatorname{Cov} \left(N(t),\sum_{i=1}^{N(t)} X_i \right)$$

I think the way to go is using the Covariance formula \begin{align} \operatorname{Cov}(X,Y) = \operatorname{E}[XY] - \operatorname{E}(X)\operatorname{E}(Y) \end{align}

I can find the second term relatively easily

\begin{align} \operatorname{E}[N(t)] & = \lambda t \\[10pt] \operatorname{E} \left[ \sum_{i=1}^{N(t)}(X_i)\right] & = \operatorname{E} \left[\operatorname{E}\sum_{i=1}^{N(t)}(X_i)\mid N(t) = N\right] \\[10pt] & = \operatorname{E}[N(t) \mu] \\[10pt] & = \lambda t \mu \end{align} Hence the second term gives \begin{align} \operatorname{E}(X)\operatorname{E}(Y) = (\lambda t)^2 \mu \end{align}

However, for the first term, I am not sure how to find the expectation of the product of the two distributions. Any advice on the route would be much appreciated.

• When working with compound Poisson processes it is very natural to condition on $N(t),$ thus: $$\operatorname{cov}(X,Y) = \operatorname{cov}(\operatorname{E}(X\mid Z),\operatorname{E}(Y\mid Z)) + \operatorname{E}(\operatorname{cov}(X,Y\mid Z)).$$ – Michael Hardy Aug 8 '17 at 20:29

The term you are missing is calculated the same way you did with the second one: $$E\left[ N(t)\sum_{i=1}^{N(t)}(X_i) \right] = E \left [E\left( N(t) \sum_{i=1}^{N(t)} X_i \mid N(t)\right)\right]$$ $$= E\left[N(t)E\left(\sum_{i=1}^{N(t)} X_i \mid N(t)\right)\right] = E[N^2(t)\mu] = \mu[\lambda t+(\lambda t)^2].$$
You could make use of the tower property as follows (similar to your own calculations) $$E \left[ N(t) \sum_{i=1}^{N(t)} X_i \right] = E \left\{ E \left[ N(t) \sum_{i=1}^{N(t)} X_i \mid N(t) \right] \right\} = E[(N(t))^2 EX_1] = (VN(t) + (EN(t))^2) \cdot EX_1.$$ Using this and your result, we have $$\text{Cov}(N(t), \sum_{i=1}^{N(t)} X_i) = EX \cdot VN(t).$$ This is a general solution, where $N(t)$ need not be Poisson distributed.
$\newcommand{\c}{\operatorname{cov}}\newcommand{\E}{\operatorname{E}}$One can use the identity $$\c(X,Y) = \c(\E(X\mid Z),\E(Y\mid Z)) + \E(\c(X,Y\mid Z)).$$
Thus we have: \begin{align} \c\left( N ,\sum_{i=1}^N X_i \right) & = \c\left( \E(N\mid N), \E\left( \sum_{i=1}^N X_i \mid N \right) \right) + \E\left( \c\left( N,\sum_{i=1}^N X_i \mid N \right) \right) \\[10pt] & = \c\left( N,N\mu \right) + 0 \\ & \qquad\qquad \text{(assuming the sequence $X_1,X_2,\ldots$ is independent of $N$)} \\[10pt] & = \mu\c(N,N) = \mu\lambda t. \end{align}