Hi All: I'm trying to better understand the connection between variance of a time series and the integral of the spectral density over all frequencies. Rather than going through all of the relations, if one looks at the link below,
https://en.wikipedia.org/wiki/Spectral_density, the relations are shown there in the power spectral density section.
To summarize what is said there:
The power, $P$, is defined as the limit as $P=\lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2$.
$S_{xx}(\omega)$ is defined as the limit as $S_{xx}(\omega)=\lim_{T \to \infty} E|\hat{x}_T(\omega)|^2$. So, I think of it as the expected instantaneous power at frequency $\omega$. I hope that's the way to think of it.
$\hat{x}_{T}(\omega)$ is defined as $\frac{1}{\sqrt T} \int_0^{T} x(t) e^{-i\omega t} dt$. So the fourier transform of $x_t$ divided by $\sqrt T$.
Finally, $\gamma_{0}$ is defined as the power contained in the total frequency band so $\int_{\omega} S_{xx} (\omega)$. My confusion is the following:
In the time domain-econometrics world, $\gamma_{0}$ is defined as $\mathrm{Var}(X_{t}) = E(X_{t} - \mu)^2$ where $\mu$ is the mean of $X_{t}$.
So, my confusion stems from the fact that when the variance is defined as power, there is no mean subtracted from $X_{t}$. Does that mean that $X_{t}$ has already been de-meaned when it's written as $X_t$. If not, then how can $\gamma_{0}$ be viewed as the variance when statistical variance is the (expected value of the deviation of a random variable from it's mean) squared yet the view of variance in the DSP world is a weighted (by frequency) combination of $E(X_{t}^2)$ over all frequencies. My only conclusion is that the mean is already being subtracted out or is assumed to be zero ? I basically don't understand how variance is defined in the DSP framework. Is it viewed as expected power rather than expected variability ? Thanks.