MGF of sum of $N(t)$ iid random variables where $N(t)$ is a Poisson process

Question

Given $(N(t),t\geq0)$ is a Poisson process with constant rate $\lambda\in\mathbb{R}$ (perhaps positive if required).

Let $X_i$ be iid random variables that are also independent of the Poisson process.

Define $X(t) = \sum_{i=1}^{N(t)} X_i$ as the sum of the $X_i$'s at time $t$.

I want to find the MGF of $X(t)$. Is it a simple application of the fact that the MGF of $Y = Y_1 + Y_2 +... Y_n$ for $Y_i$ iid is the product of their MGFs?

So far I've computed $$ M_{X(t)}(x) = \mathbb{E}\bigg( e^{x[X_1 + X_2 +...+ X_{N(t)}]} \bigg) = \mathbb{E}\bigg(\prod_{i=1}^{N(t)} e^{xX_i} \bigg) $$ and I think I might be overcomplicating the problem.

Answer 1

Yes, we need $\lambda \ge 0.$

It is not a simple matter of taking the product of the MGFs since there is a random number of terms. Although the variables are independent conditionally on $N(t),$ they are not unconditionally independent.

Your first couple steps are right. The key to proceeding is to condition on $N(t)$ and use the law of total probability.

$$ E\left(\prod_{i=1}^{N(t)}e^{x X_i}\right) = \sum_{n=0}^\infty E\left(\prod_{i=1}^{N(t)}e^{x X_i} \bigg \vert \;N(t)= n\right)P(N(t)=n) $$ Now that the $N(t)$ is conditioned on, we can treat it as a constant, and now, we can treat the $X_i$ as i.i.d. and of a fixed number. So this conditional MGF factors and we have $$ E\left(\prod_{i=1}^{N(t)}e^{x X_i}\right) = \sum_{n=0}^\infty (M_X(t))^n P(N(t)=n).$$

I'll let you take it from here.

Answer 2

$Ee^{X(t)}=Ee^{\sum_{i=1}^{N(t)} X_i} =E(Ee^{\sum_{i=1}^{N(t)} X_i}|N(t))$. Now $E(e^{\sum_{i=1}^{N(t)} X_i}|N(t))$ can be evaluated by treating $N(t)$ as a fixed integer and its value is $M_o^{N(t)}$ where $M_0$ is the MGF of $X_i$. Hence $Ee^{X(t)}=\sum_{k=0}^{\infty} P\{N(t)=k\} M_o^{k}=\sum_{k=0}^{\infty} M_o^{k} e^{-\lambda t}\frac {(\lambda t)^{k}} {k!}$. The sum of this series is $e^{-\lambda t}e^{\lambda t M_o}$.