Wiener-Khinchin theorem problem

Question

It is stated by the Wiener-Khinchin theorem that you can obtain the spectral density of a stochastic process $X_t$ merely by taking the Fourier-transform of its auto-correlation function \begin{equation} S(\omega) = \int_{-\infty}^{\infty} \mathbb{E}[X(t) X^*(t+\tau)] e^{-i\omega\tau}d\tau \end{equation} This works when $\mathbb{E}[X(t) X^*(t')]$ contains a term such as $|t-t'|$ where both $t$s cancel out when you replace $t' = t+\tau$, leaving you with $\tau$ only. However, how can you perform the integral when you get both $t$ and $\tau$ in the same function?

Answer 1

You are correct that there is a problem in the case when $E[X(t)X(t')]$ is not a function of $|t-t'|$. This is because the Wiener-Khinchin Theorem only applies when the process is wide sense stationary. One of the properties of a wide-sense stationary process is that that $E[X(t)X(t')]$ depends only on the difference between the two times $|t-t'|$. If this is not the case, then the process is not stationary, so Wiener-Khinchin does not apply, so you will get problems when trying to apply it.