Physical meaning of average values of random signals This question might be a bit stupid, anyway, i'll risk it, since i want to get better understanding of this subject.

Let's consider random signal x(t), and let's say that we know that it is ergodic by mean, and by autocorrelation (this implies that it is wide sense stationary).

This means that time averaged mean of this signal is equal to it's statistic averaged mean, same goes for autocorrelation.

So we have  the following:

$\mu_X=E(x)=\frac{1}{T} \int_T x(t)dt=\langle x(t)\rangle  \\ R_{xx}(\tau)=E[x(t)x(t+\tau)]= \lim_{T-&gt;\infty} \frac{1}{T} \int_T x(t)x(t+\tau)dt=\langle x(t)x(t+\tau)\rangle$

Then, as i found in one of the books i have, it says following:

$\langle x(t) \rangle$- corresponds to DC level of given signal

$\langle x(t) \rangle ^2$ - corresponds to normalized power of DC component

$\langle x^2(t) \rangle$ - corresponds to total average normalized power

$\overline{\sigma_x^2}= \langle x^2(t) \rangle - \langle x(t) \rangle^2$-corresponds to average normalized power in AC component of the signal

$\overline{\sigma_x}$-corresponds to rms of AC component

My question is, is there any way to prove this, or to gain intuitive understanding why this is true. Any help appreciated!
Dilip: I'm confident that what you said is correct but the weirdo sample path that you refer to ( the one with probability zero ) can be used to estimate the autocorrelation at different lags, namely $\hat{R}_{xx}(\tau)$. This is true even though it's only one realization and this is the case only because of the ergodicity assumption. So, to a person coming in new, my opinion is that your answer doesn't give enough significance to the one realization. It's a weirdo but a useful weirdo for sure  !!!!!