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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!

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@did Thank you for the hint! You're right, $\hat{\mu}$ is also a probability measure. However could you give me a hint how to calculate $E[(\frac{\eta}{\lambda}h(Y_1))^{N_t}]$, this is what I actually have to calculate, right? –  user20869 Oct 2 '12 at 17:56
    
Actually, I'm done, if $E[h(Y_j)]=1$, but I do not see this. Of course $E[h]=1$, but why the composition? –  user20869 Oct 2 '12 at 18:21
    
Thank you for your help –  user20869 Oct 2 '12 at 19:32
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1 Answer

up vote 1 down vote accepted

Since $N_t$ is a random variable, what you computed so far is $\mathbb E(Z_t\mid N_t)$, not $\mathbb E(Z_t)$. To get the value of $\mathbb E(Z_t)$, consider $\mathbb E(\mathbb E(Z_t\mid N_t))$. You will probably end up being forced to use a hypothesis which is missing from your post, namely that $\hat\mu$ is a probability measure.

By definition, if $\hat\mu$ is a probability measure, then $\mathbb E(h(Y))=\displaystyle\int h(y)d\mu(y)=\int d\hat\mu(y)=1$.

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