TITLE: Stochastic integral with a Poisson process
QUESTION [7 upvotes]: I have a Poisson process $X_t$ for $t\ge0$. How I can find a process $b_t$ such that 
$$\exp ({\alpha X_t})=1+\int_0^t b_{s^{-}}dX_s$$ where $\alpha\in\mathbb{R}$
and  what would be the expectation of $\exp ({\alpha X_t})$. The last question is how i can find expectation and variance of this process $\int_0^t \exp(\alpha X_{s^-})dX_s.$
Thanks.

REPLY [6 votes]: I thinks this is an application of Ito's Lemma with noncontinuous semimartingales. Of course, in the case that $X$ is a Poisson process, it simpliefies  to
$$f(X_t)-f(X_0)=\sum_{0< s\le t} \Delta [f(X_{s})],$$
where $f(x)=e^{\alpha x}$.
Since the Poisson process has a jump size 1 a.s., 
$$\sum_{0<s\le t}\Delta f(X_s)=\int_0^t \Delta f(X_s)dX_s=\int_0^t f(X_{s-}+1)-f(X_{s-})dX_s.$$
So the process you are looking for is 
$$b_s=f(X_s+1)-f(X_s)=\exp(\alpha X_s)(e^\alpha-1).$$
With that, the last question is easy to answer, since
$$\int_0^t \exp(\alpha X_{s^-})dX_s=\frac1{e^{\alpha}-1}[\exp(\alpha X_t)-1],$$
and you can use the moment generating function for $X_t$
$$\mathbb{E} \exp(\alpha X_t)=\exp\{\lambda t (e^{\alpha}-1)\}$$
to figure out expectation and variance of $\int_0^t \exp(\alpha X_{s^-})dX_s.$