# Question on Power Spectral Density and Wiener-Khinchin theorem

I tried asking this question on stackexchange and I also extensively researched it online without results so I will ask here.

In my textbook the Wiener-Khinchin theorem is used to connect the auto-correlation definition of PSD with an "intuitive interpretation" of power spectral density for deterministic signals. It says:

$S_{xx}(\omega) = \lim_{T\rightarrow\infty} \frac{1}{2T}\mathbb E\left[|FT\{X(t)*I_{[-T,T]}\}|^2\right]$

But such a theorem assumes you can take the fourier transform of a(truncated) realization of the random process, which unless I am missing something - may not be the case. Such a realization may not be integrable.

So what is going on? Is there a different definition of integral which allows you to do this?

• Do we assume that $\mathbb E |X(0)| < \infty$? If yes, then, by Fubini, $\mathbb{E} \int_{-T}^T |X(t)| dt = \int_{-T}^T \mathbb{E} |X(t)| dt = 2 T \mathbb{E} |X(0)|$ is finite, and hence $\int_{-T}^T |X(t)| dt$ is finite a.s. Apr 14, 2018 at 21:54

Not every stationary stochastic process has a power spectral density, in general it has a power spectral measure. But in any case $X*1_{[-T,+T]}$ is itself a stationary process, whose realisations (smooth or not) are never integrable on $\mathbb R$ (except if $X\equiv 0$). They have a Fourier transform only in the sense of tempered distributions: its values at a point (frequency) $\omega$ are not defined, and taking the expectation of its square doesn't make sense.