# Stochastic integral with a Poisson process

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.

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What is the context of this question? Usually if you encounter stochastic integrals you would know a bit of stochastics... –  Jonas Teuwen Apr 18 '11 at 21:50

## 1 Answer

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 $\int_0^t \exp(\alpha X_{s^-})dX_s.$

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Your formula for $E\exp(\alpha X_t)$ is not correct. –  Did Apr 25 '11 at 20:53
@Didier Piau Thanks! You mean the exponent $-1$ right? –  GWu Apr 26 '11 at 1:11
Yup.    –  Did Apr 26 '11 at 5:07