Calculate $\mathbb{E}[N(E) e^{-rE}]$ for Compound Poisson $(N(t))_t$ and exponential $E$.

Question

Let $(\xi_i)_i$ be an i.i.d. sequence of random variables, $(n(t))_t$ a Poisson process (independent of the $\xi_i$) with intensity $\lambda$ and let: $$ N(t) = \sum_{i=1}^{n(t)} \xi_i $$ be the associated compound Poisson process. Further, let $E$ be an exponentially distributed random variable, independent of $n(t)$ and $\xi_i$. We would now like to calculate the expectations: $$ \mathbb{E}[e^{-rE} N(E)] $$

Answer 1

Hints:

1. Calculate $\mathbb{E}(N_t)$.
2. Since $E$ and $(N_t)_{t \geq 0}$ are independent it holds that $$\mathbb{E}(e^{-rE} N(E)) = \mathbb{E} \left( \int_0^{\infty} e^{-rE} N(t) \, d\mathbb{P}_E(t) \right)$$ where $\mathbb{P}_E$ denotes the distribution of $E$. As $E$ is exponentially distributed, say with parameter $\mu>0$, this means that $$\mathbb{E}(e^{-rE} N(E)) = \mu \int_0^{\infty} e^{-rt} \mathbb{E}(N(t)) e^{-\mu t} \, dt.$$ Combining this with step 1, this allows you to calculate $\mathbb{E}(e^{-rE} N(E))$.