Expected value of $Y=\prod_{j=1}^{N} X_{j}$, where $N\sim\operatorname{Poisson}(\lambda)$.

Question

Let $$Y=\prod_{j=1}^{N} X_{j},$$ $X_{j}$, for $j=1,2, ...$, are identically and independently distributed with mean $\lambda$ and variance $\sigma^{2}$ and $N\sim \operatorname{Poisson}(\lambda).$

How would I find $E(Y)$?

$E(Y)=E(E(\prod_{j=1}^{N} X_{j} | N))$

$E(\prod_{j=1}^{N} X_{j} | N)=E(\prod_{j=1}^{n} X_{j} | N=n)=E(\prod_{j=1}^{n} X_{j})=E(X_{1}^{n})$?

Answer 1

By independence $$E\left[\prod_{k = 1}^n X_k \,\Bigg|\, N = n\right] = \prod_{k = 1}^nE[X_k\mid N = n] = \lambda^n$$

So $$E\left\{E\left[\prod_{k = 1}^nX_k\,\Bigg|\,N = n\right]\right\} = E[\lambda^N] = \sum_{k = 0}^\infty \lambda^k\cdot e^{-\lambda}\frac{\lambda^k}{k!} = e^{\lambda^2-\lambda}.$$

Answer 2

First we make use of the obvious: the Law of Total/Iterated Expectation and the Independence and Identicallity of $\{X_k\}_\infty$ (all of which are also independent of $N$).   Then we use $\mathsf M_N(t)=\mathsf e^{\lambda(\mathsf e^t-1)}$ which is the moment generating function of our Poisson distributed random variable.   (That is: $\mathsf E(\mathsf e^{tN})$

$$\begin{align}\mathsf E(Y) =&~ \mathsf E(\mathsf E(\prod_{k=1}^N X_k\mid N)) & \text{by LoTE}\\[1ex]=&~ \mathsf E(\mathsf E(X_\star)^N) & \text{because iid}\\[1ex]=&~ \mathsf E(\mathsf e^{N\ln \lambda}) & \text{...} \\[1ex]=&~ \mathsf M_N(\ln \lambda) & \text{recognise mgf} \\[1ex] =&~ \mathsf e^{\lambda (\mathsf e^{\ln \lambda}-1)} & \text{use mgf}\\[1ex] =&~ \mathsf e^{\lambda^2-\lambda}\end{align}$$