# The Expectation of a function of independent random variables

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.

• Henry, some people take a very grim view on multiple edits. We have had very bad cases of that in the past, so doing it presses people's buttons. They may even think that you are trying to hog more attention to your question, and give it negative attention instead. When many edits are done, the site software generates messages to us moderators so that we can intervene. Jan 22, 2015 at 17:12
• As a relatively new user your chances of knowing all this were somewhere between slim and none. Your goal of leaving a polished question for others to think about is fine. You are not the first user with a need to spend some time proofreading. To that end something called a sandbox has been created. My advice is that you edit your question there. Doing that in meta is not nearly as disruptive, and we have MathJax support there. So when you are satisfied, simply copy/paste the question back here. Jan 22, 2015 at 17:19
• But, I hope you enjoy the site. IMHO we need more users at graduate student level so, in spite of the above, welcome to Math.SE from me. Jan 22, 2015 at 17:20
• Are you looking for an exact evaluation ? I don't see one, but if you look for an asymptotic as $n\to\infty$, then surely you can approximate $1/(1-sG(\omega)\gamma_i^{-1})$ by $e^{-sG(w)\gamma_i^{-1}}$ up to some error, since $\gamma_i\to\infty$. Then you get something tractable. I'm not sure what hypotheses you need.
– Sary
Feb 3, 2015 at 2:33
• thanks. i am looking to see if the steps are correct. and please can you right what you proposed as an answer. thank you very mu ch c Feb 3, 2015 at 3:30