Suppose that we have two stationary random processes $x(t)$ and $y(t)$ with probability density functions $f_{x}(x)$ and $f_{y}(y)$ respectively. Now suppose we form:
$z(t) = x(t) \ast y(t)$
What is the probability density function of $z$?
I understand that if:
$z(t) = x(t) + y(t)$
Then we have:
$f_{z} = f_{x} \ast f_{y}$
So, considering that if:
$h = xy \Rightarrow f_{h}(h) = \int f_{x}(x) f_{y}(\frac{h}{x}) \frac{1}{|x|} dx$
And that:
$z(t) = x(t) \ast y(t) = \int x(\tau)y(t-\tau)d\tau$
Would we then have:
$f_{z}(z) = \int f_{x}(x) f_{y}(\frac{z}{x}) \frac{1}{|x|} dx \ast \int f_{x}(k) f_{y}(\frac{z}{k}) \frac{1}{|k|} dk$
Or not? If not, what is the correct distribution and how can I find it?
The distribution of $z(t) = \displaystyle\int x(\tau)y(t-\tau)d\tau$ involves the joint distribution of the processes $(x(\tau))_{\tau}$ and $(y(\tau))_{\tau}$. Hence the distributions of $x(t)$ and $y(t)$ (even assuming stationarity) are quite unsufficient to determine the distribution of $z(t)$.