Let $(N_t)_t$ be a homogeneous Poisson process with intensity $\lambda>0$. How can I prove that $Y_t=\exp(rN_t-\lambda t(e^r-1))$ is a Martingal w.r.t. the canonical filtration, $r\in \mathbb{R}$, $t\geq > 0$.
My attempt
- $Y_t$ is adapted to $\mathcal{F}_t$ since $N_t$ is adapted to $\mathcal{F}_t$ and $Y_t=f(N_t)$ is a measurable function of $N_t$.
- Integrability: $E(|Y_t|)<\infty$ ( I don't really see at first glance why that is the case)
For $0\leq s\leq t$, one has to show:$$E(\exp(rN_t-\lambda t(e^r-1))|\mathcal{F}_s)=\cdots = \exp(rN_s-\lambda s(e^r-1))$$
However, I don't know how to deal with the $\exp$ in $E(\cdot)$. It might be noteworthy that I know that $N_t-\lambda t$ (i. e. the compensated poisson process) and $(N_t-\lambda t)^2-\lambda t$ are martingales.
Can someone help?
Hints: $$N_t - N_s \sim \text{Poi}(\lambda(s-t))$$
And if $X$ is a Poisson random variable the MGF of $X$ is $$\mathbb E\left[e^{uX}\right] = e^{\lambda (e^u-1)}$$