expected value of exponential compound poisson process

Question

Let $Z(t)=\sum_{i=1}^{N(t)} X_i$ and let $N(t)$ be a Poisson process with parameter $\lambda$ and $X_1,X_2,\dots$ positive iid random variables with density function $f_X(x)$, independent of $N(t)$. How can I calculate $E[e^{\lambda t-\mu Z(t)}]$?

Answer 1

Since $N_t$ and all the $X_i$ are independent and the latter are identically distributed we have \begin{align} \mathbb E\left[e^{-\mu Z_t}\right]&=\mathbb E\left[\prod_{i=1}^{N_t}e^{-\mu X_i}\right]=\sum_{n=0}^\infty\prod_{i=1}^n\mathbb E\left[e^{-\mu X_i}\right] \frac{(\lambda t)^n}{n!}e^{-\lambda t}=\sum_{n=0}^\infty \left(\mathbb E\left[e^{-\mu X_1}\right]\right)^n\frac{(\lambda t)^n}{n!}e^{-\lambda t}\\ &=\exp\Big((\mathbb E\left[e^{-\mu X_1}\right]-1)\lambda t\Big)\,. \end{align}