## fourier transform of a finite-energy random process

Hi all,
I've got the following problem: I have a random process in time-domain, call it $\displaystyle h(t)$; assuming that it's a finite-energy process (e.g. it's mean value $\displaystyle E\{h(t)\}$ is non-zero only for $\displaystyle 0<t<t_0$), I can consider the Fourier transform:
$\displaystyle H(w)=\int_{-\infty}^{\infty} h(t)e^{-jwt}\, dt$
Now, if $\displaystyle h(t)$ has a joint PDF of N-th order $\displaystyle f_h(h_1,h_2,...,h_N)$, what's the joint PDF of N-th order of the process $\displaystyle H(w)$?

I tried to tackle the problem in this way: considering a fixed value for frequency $\displaystyle \bar{w}$, I can extract a r.v. $\displaystyle H(\bar{w})$ from the random process $\displaystyle H(w)$:
$\displaystyle H(\bar{w})=\int_{-\infty}^{\infty} h(t)e^{-j\bar{w}t}\, dt$
If I approximate the integral as:
$\displaystyle H(\bar{w})\simeq\sum_{i=0}^{t_0/{\Delta}\tau} h(i{\Delta}\tau)e^{-j\bar{w}i{\Delta}\tau}{\Delta}\tau$
I can consider $\displaystyle H(\bar{w})$ as a function of the finite set of random variables $\displaystyle h(i{\Delta}\tau)$. This means that the PDF of $\displaystyle H(\bar{w})$ is a sort of convolution of the PDFs of the r.v. $\displaystyle h(i{\Delta}\tau)$ (scaled by the exponentials)... does this make sense?
Is there a better solution to my original problem?

Thanks indeed!