If we have a Poisson process $ \lbrace N(t), t > 0 \rbrace $ with rate $\lambda > 0 $ and if we have a random variable $S$ having a uniform distribution on the interval $[0,2]$. I was wondering how can I obtain the moment-generating function of the random variable $N(t+S)$.
I know the moment-generating function of $ N(t) $, however I can really see how to take into account the fact that $S$ is a random variable.
Thanks.
$\newcommand{\E}{\mathbb E}$ I will assume here that $S$ is independent of $N$. (Very often people writing about this neglect to state that assumption. And so it is here, if in fact that was intended, and if it wasn't then the nature of the dependence would need to be mentioned at least implicitly.)
You can use the law of total expectation: \begin{align} M_{N(t+S)}(y) & = \E(e^{uN(t+S)}) = \E(\E(e^{uN(t+S)}\mid S)) = \E(e^{\lambda (t+S)(e^u-1)}) \\[15pt] & = \int_0^2 e^{\lambda (t+s)(e^u-1)} \frac{ds}{2} = \frac{e^{\lambda t(e^u-1)}}{2} \int_0^2 e^{\lambda s(e^u-1)} \, ds. \end{align}
So you have $\displaystyle \int_0^2 e^{ks} \, ds$ for $k=\lambda(e^u-1)$.