Spectral density of product Let $X_t$ and $Y_t$ stationary, zero mean, independent processes with $\phi_x(\lambda)$ and $\phi_y(\lambda)$ spectral densities. 
How can I prove that the process $Z_t=X_tY_t$ has a spectral density:
$\phi_z(\lambda)=\int_{-\pi}^{\pi}\phi_x(\lambda-\omega)\cdot\phi_y(\omega)d\omega$
Thanks for the help!
 A: This question needs some ideas from random processes and some from
Fourier theory.
The autocorrelation function of a (continuous-time finite-variance) stationary random process
$\{X_t\colon -\infty < t < \infty\}$ is $R_X(t) = E[X_{\tau}X_{\tau+t}]$ and the
spectral density $S_X(\omega)$ is the Fourier transform of $R_x(t)$. For your
problem, the independence of the $\{X_t\}$ and $\{Y_t\}$ processes gives that
$$R_Z(t) = E[Z_{\tau}Z_{\tau+t}] = E[X_{\tau}Y_{\tau}X_{\tau+t}Y_{\tau+t}]
= E[X_{\tau}X_{\tau+t}]E[Y_{\tau}Y_{\tau+t}]= R_X(t)R_Y(t).$$
So much so for random processes. From Fourier transform theory, we have
that the transform of a product of two functions is the convolution of their
Fourier transforms.  If you are not familiar with this, 
see, for example, the last paragraph of Section 5.8 of the 
Wikipedia article on the Fourier transform.  Thus,
$$S_Z(\omega) = \int_{-\infty}^\infty S_X(\lambda)S_Y(\omega-\lambda)\,
\mathrm d\lambda.\tag{1}$$
For your particular application with a discrete-time random process
(a.k.a. time series), similar results apply but in $(1)$ the limits
work out to be $-\pi$ and $\pi$. (You will need to know about the discrete-time
Fourier transform to get to this). Note also
that you might be missing a $\frac{1}{2\pi}$
factor in the result you state as wanting to prove in your question.
