variance in the time domain versus variance in frequency domain 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.
Variance is never defined as power. For a wide-sense stationary random process $X(t)$ with zero mean

$$\mu_X=E\{X(t)\}=0\tag{1}$$

the variance of $X(t)$ equals its power.

The autocorrelation of $X(t)$ is defined by

$$R_X(\tau)=E\{X^*(t)X(t+\tau)\}\tag{2}$$

The power of $X(t)$ is

$$P_X=E\{|X(t)|^2\}=R_X(0)\tag{3}$$

The variance of $X(t)$ is

$$\sigma^2_X=E\{|X(t)-\mu_X|^2\}=E\{|X(t)|^2\}-|\mu_X|^2=R_X(0)-|\mu_X|^2\tag{4}$$

Clearly, if $\mu_X=0$, we have $\sigma^2_X=R_X(0)=P_X$.

Note that the autocovariance of $X(t)$ is defined by

$$C_X(\tau)=E\{[X(t)-\mu_X]^*[X(t+\tau)-\mu_X]\}=R_X(\tau)-|\mu_X|^2\tag{5}$$

Consequently,

$$\sigma^2_X=C_X(0)\tag{6}$$

But since the power spectral density $S_X(\omega)$ is the Fourier transform of $R_X(\tau)$, the integral over $S_X(\omega)$ equals the power of $X(t)$ and not its variance, unless $\mu_X=0$:

$$P_X=R_X(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_X(\omega)d\omega\tag{7}$$

Also take a look at this related answer.

For the difference between the definition of autocorrelation in statistics and in signal processing look here.