Assume we have for some index $i>n$ ($n \in \mathbb{N} $) the following ${\it Independent \ Random \ Variables}$ $$h_i \sim \text {i.i.d }\ \ \mathcal{CN}(0,1) \ \ \text{ Complex Gaussian}$$ $$\Omega_i \sim \text {i.i.d with pdf }\ \ f_{\Omega_i}(\omega_i)$$ $$\gamma_i \in \Xi \ \ \text{ a Poisson Point Process with intensity } \lambda $$
I would like to find the ${\it Laplace}$ transform of the function $$Y=\sum\limits_{i \ >\ n} \gamma_{i}^{-1} |h_{i}|^2G(\Omega_{ i}) $$ where $G(.)$ is some function of $\Omega_{ i}$ (let us not go through the details of what $G$ is but assume we know it is defined from $-\infty$ to $\infty$.)
This is the way I proceed
\begin{align*} \nonumber \mathcal{L}_Y(s)&= \mathbb{E}\left(e^{-s Y}\right)\\ &=\mathbb{E}\Bigg[e^{-s \sum\limits_{i \ >\ n} \gamma_{i}^{-1} |h_{i}|^2G(\Omega_{ i})}\Bigg]\\\nonumber &=\mathbb{E}\Bigg[\prod_{i>n}\bigg(e^{-s \ \gamma_{i}^{-1}|h_{i}|^2G(\Omega_{ i}) }\bigg)\Bigg]\\\nonumber &\stackrel{(a)}{=}\mathbb{E}_{\{\Omega_{ i}\},\Xi}\Bigg[\prod_{i>n}\mathbb{E}_{|h|^2}\bigg(e^{-s \ \gamma_{i}^{-1}|h|^2 G(\Omega_{ i}) }\bigg)\Bigg]\\\nonumber &\stackrel{(b)}{=}\mathbb{E}_{\{\Omega_{ i}\},\Xi }\Bigg[\prod_{{i>n}}\Bigg(\frac{1}{1+s \ G(\Omega_{ i})\ \gamma_{i}^{-1} }\biggl)\Bigg]\\\nonumber &\stackrel{(c)}{=}\mathbb{E}_{\Xi }\Bigg[\prod_{{i>n}}\mathbb{E}_{\Omega}\Bigg(\frac{1}{1+s G(\Omega) \gamma_{i}^{-1} }\biggl)\Bigg]\\\nonumber &\stackrel{(d)}{=}\mathbb{E}_{\Xi }\Bigg[\prod_{{i>n}}\int_{-\infty}^{+\infty}\Bigg(\frac{1}{1+s G(\omega)\gamma_{i}^{-1} }\biggl) f_{\Omega(\omega})\ d\omega\Bigg]\\\nonumber &\stackrel{(e)}{=} ????? \end{align*}
where (a) follows from the fact that the $|h_i|^2$ are independent, so the expectation of the product is the product of the expectation, (b) follows from the moment generating function of an exponential random variable because $h_i$ is Gaussian so $|h_i|^2$ is exponential; (c)same reasoning as (a) but for $\Omega_i$;(d)from the taking the expectation of a function of the random variable $\Omega_i$.
Do you agree with the steps above ?
Part 2: Next I would like to utilize the {\it probability generating function for the Poisson Point Process} to continue part (e)
The formula is stated next, let $\Phi$ be a Poisson Point Process with intensity $\lambda$. Then
$$\mathbb{E}\left( \prod_{x\in \Phi} v(x)\right)= \text{exp} \left(-\int_{\mathbb{R}^d} [1-v(x)]\lambda dx\right)$$
How can I use this theorem to continue step (e)? any thoughts?
Thanks.