Let $t\in [0,a]$, $X_t:=\sum_{i=1}^{N_t}Y_i$ be a compounded Poisson process, i.e. $N_t$ a Poisson process with parameter $\lambda>0$ and $Y_i$ are iiid with distribution $\mu$. Now let $S_t:=\exp{(\sum_{i=1}^{N_t}f(Y_i)+(\lambda-\eta)t)}$, where $\eta>0$ and $f(x):=\log{(\frac{\eta}{\lambda}h(x))}$. We assume that there is a absolutely continuous measure $\hat{\mu}$ with respect to $\mu$ such that $h$ is the Radon-Nikodym derivative, i.e. $h(x):=\frac{d\hat{\mu}}{d\mu}(x)$. I want to prove $E[Z_t]=1$. What I did so far:
$E[Z_t]=E[\exp{(f(Y_1))}\cdots\exp{(f(Y_{N_t}))}\exp{((\lambda-\eta)t)}]=\exp{((\lambda-\eta)t)}E[\exp{(f(Y_1))}\cdots\exp{(f(Y_{N_t}))}]$
Now $\exp{(f(Y_j))}=\exp{(\log(\frac{\eta}{\lambda}h(Y_j)))}=\frac{\eta}{\lambda}h(Y_j)$, so first since the $Y_j$ are identical distributed we have
$E[\exp{(f(Y_1))}\cdots\exp{(f(Y_{N_t}))}]=(\frac{\eta}{\lambda})^{N_t}E[h(Y_1)^{N_t}]$
and then by independence
$(\frac{\eta}{\lambda})^{N_t}E[h(Y_1)^{N_t}]=(\frac{\eta}{\lambda})^{N_t}E[h(Y_1)]^{N_t}=(\frac{\eta}{\lambda}E[h(Y_1)])^{N_t}$
We end up with $E[Z_t]=\exp{((\lambda-\eta)t)}(\frac{\eta}{\lambda}E[h(Y_1)])^{N_t}$. If this should be one, we must have
$(\frac{\eta}{\lambda}E[h(Y_1)])^{N_t}=\exp{(-(\lambda-\eta)t)}$
Here I'm stuck. I guess I have to use, that $h$ is the Radon-Nikodym derivative and the distribution of $N_t$. Or did I a mistake so far? Some help would be appreciated!