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.
Best Answer
The term you are missing is calculated the same way you did with the second one: $$E\left[ N(t)\sum_{i=1}^{N(t)}(X_i) \right] = E \left [E\left( N(t) \sum_{i=1}^{N(t)} X_i \mid N(t)\right)\right]$$ $$= E\left[N(t)E\left(\sum_{i=1}^{N(t)} X_i \mid N(t)\right)\right] = E[N^2(t)\mu] = \mu[\lambda t+(\lambda t)^2].$$