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Since Fourier transform of a random stationary process in time (in the case of existence) is not necessarily real, my question is what is the relation between the covariance of real and imaginary parts of the transformed random variable.

More specifically, according to Wiener-Khinchin theorem for time function $x(t)$ with some conditions the Fourier transform $X(\omega)$ exists. The question is can one recover $\mathbb{E}\left[\Re\{X(\omega)\},\Re\{X(\omega')\}\right]-\mathbb{E}\left[\Re\{X(\omega)\}\right]\mathbb{E}\left[\Re\{X(\omega')\}\right]$ and $\mathbb{E}\left[\Im\{X(\omega)\},\Im\{X(\omega')\}\right]-\mathbb{E}\left[\Im\{X(\omega)\}\right]\mathbb{E}\left[\Im\{X(\omega')\}\right]$

from $\mathbb{E}\left[\bar{x}(t),x(s)\right]-\mathbb{E}\left[\bar{x}(t)\right]\mathbb{E}\left[x(s)\right]$?

I would be happy to be guided to a reference as well. This is the most general case but the simpler case where the input signal is real is the main concern for me. I would be happy to be guided to a reference as well.

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    $\begingroup$ boy, i hope i didn't mangle the meaning by trying to fix notation. i know that some mathematicians use a different notation for the Fourier Transform (with $\hat{x}$), but can you stick with what is conventional for electrical engineers? also, can you define what you mean by "$\bar{x}(t)$"? is it complex conjugate? $\endgroup$ – robert bristow-johnson Jul 7 '14 at 18:25
  • $\begingroup$ Thank you very much for your edit. I'll do my best, although I am not from a signal processing background :) Its exactly complex conjugate. $\endgroup$ – Cupitor Jul 8 '14 at 12:27
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    $\begingroup$ I'm not sure I understand the question. Is this what you want: dsp.stackexchange.com/questions/13346/… ? $\endgroup$ – DanielSank Jul 9 '14 at 2:41
  • $\begingroup$ @DanielSank, yes that is very much related. But I don't understand what the answerer means by sampling enough samples. I am talking about a stochastic process in a complete theoritical framework. I don't need any empirical statistic. I believe actually the answer is that we get a diagonal covariance matrix for $X(\omega)$ and $\mathbb{E}\left[\Im\{X(\omega)\},\Im\{X(\omega')\}\right]-\mathbb{E}\left[\Im\{X(\omega)\}\right]\mathbb{E}\left[\Im\{X(\omega')\}\right]$ should be zero for nonequal frequencies and should be equal to the covariance for real part when $\omega=\omega'$ $\endgroup$ – Cupitor Jul 9 '14 at 16:58
  • $\begingroup$ I think the basic answer to your question is that the real and imaginary parts are jointly Gaussian distributed. Does that make sense / answer your question? $\endgroup$ – DanielSank Jul 9 '14 at 18:40
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Note that in general the Fourier transform of a stationary process $x(t)$ does not exist. The Wiener-Khinchin theorem only states that under certain conditions the power spectral density of $x(t)$ exists, and it can be computed as the Fourier transform of the autocorrelation function of $x(t)$.

Having said that, if for some reason one assumes that the Fourier transform of $x(t)$ exists, then you can do the math and see if you get a useful result. So, let's see. I assume that $x(t)$ is real-valued. This is not at all necessary, but just simplifies things a bit.

Since $\text{Re}\{X(\omega)\}=\frac12[X(\omega)+X^*(\omega)]$ and $\text{Im}\{X(\omega)\}=\frac{1}{2j}[X(\omega)-X^*(\omega)]$, we can compute the desired expectations from $E\{X(\omega)X(\omega')\}$ and $E\{X^*(\omega)X(\omega')\}$. For the first of these expectations we have

$$E\{X(\omega)X(\omega')\}=E\left\{\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \int_{-\infty}^{\infty}x(t')e^{-j\omega' t'}dt' \right\}=\\ =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}E\{x(t)x(t')\}e^{-j\omega t}e^{-j\omega't'}dtdt'$$

With the substution $\tau=t'-t$ and with the autocorrelation function $R_x(\tau)=E\{x(t)x(t+\tau)\}$ (assuming wide-sense stationarity of $x(t)$) we obtain

$$E\{X(\omega)X(\omega')\}=\int_{-\infty}^{\infty}R_x(\tau)e^{-j\omega'\tau}d\tau\int_{-\infty}^{\infty}e^{-j(\omega+\omega')t}dt$$

Interpreting the second integral as a distribution, and with the power spectrum $S_x(\omega)=\mathcal{F}\{R_x(\tau)\}$ we finally get

$$E\{X(\omega)X(\omega')\}=2\pi\delta(\omega+\omega')S_x(\omega')= 2\pi\delta(\omega+\omega')S_x(-\omega)=2\pi\delta(\omega+\omega')S_x(\omega)\tag{1}$$

because $S_x(\omega)$ is an even function (we assumed $x(t)$ to be real-valued). The other expectation $E\{X^*(\omega)X(\omega')\}$ can be derived in a completely analogous manner. The result is

$$E\{X^*(\omega)X(\omega')\}=2\pi\delta(\omega-\omega')S_x(\omega)\tag{2}$$

From (1) and (2) we can obtain the desired expectations

$$E\{\text{Re}\{X(\omega)\}\text{Re}\{X(\omega')\}\}= \pi S_x(\omega)[\delta(\omega-\omega')+\delta(\omega+\omega')]\\ E\{\text{Im}\{X(\omega)\}\text{Im}\{X(\omega')\}\}= \pi S_x(\omega)[\delta(\omega-\omega')-\delta(\omega+\omega')]$$

It remains to compute the quantities $E\{\text{Re}\{X(\omega)\}$ and $E\{\text{Im}\{X(\omega)\}$. We can easily derive them from the expectations $E\{X(\omega)\}$ and $E\{X^*(\omega)\}$:

$$E\{X(\omega)\}=E\left\{ \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \right\}= \int_{-\infty}^{\infty}E\{x(t)\}e^{-j\omega t}dt= \mu_x\int_{-\infty}^{\infty}e^{-j\omega t}dt=2\pi\mu_x\delta(\omega)$$

with $\mu_x=E\{x(t)\}$, where we have again assumed wide-sense stationarity of $x(t)$. Obviously, we get the same result for $E\{X^*(\omega)\}$:

$$E\{X^*(\omega)\}=2\pi\mu_x\delta(\omega)$$

This results in

$$E\{\text{Re}\{X(\omega)\}\}=2\pi\mu_x\delta(\omega)\quad\text{and}\quad E\{\text{Im}\{X(\omega)\}\}=0$$

So we have

$$E\{\text{Re}\{X(\omega)\}\}\cdot E\{\text{Re}\{X(\omega')\}\}=4\pi^2\mu_x^2\delta(\omega)\delta(\omega')\\ E\{\text{Im}\{X(\omega)\}\}\cdot E\{\text{Im}\{X(\omega')\}\}=0\tag{3}$$

Combining (3) with (1) and (2) you obtain the desired result. From the result you can see that the covariances of the real parts of $X(\omega)$ and $X(\omega')$ vanish everywhere except for $\omega=\pm\omega'$. The same is true for the covariances of the imaginary parts. The difference is an additional term for $\omega=\omega'=0$ for the covariance of the real parts.

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    $\begingroup$ @Cupitor: You're welcome. Unfortunately, the only reference I know about is a paper about the D(T)FT of windowed noise. It also doesn't consider the covariances that you're interest in. You can find the paper here: users.ece.gatech.edu/mrichard/DFT%20of%20Noise.pdf $\endgroup$ – Matt L. Jul 10 '14 at 10:45
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    $\begingroup$ @jojek: Oh well, I felt like shuffling around some integrals ... :) $\endgroup$ – Matt L. Jul 10 '14 at 10:46
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    $\begingroup$ @Cupitor: Yes, but the techniques are pretty standard. I learned them when I studied DSP and probability theory. So I didn't really invent anything new here, unfortunately ;) $\endgroup$ – Matt L. Jul 10 '14 at 10:48
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    $\begingroup$ @Cupitor: Yes, you just miss a factor of $\pi$ in the first equation. $\endgroup$ – Matt L. Jul 10 '14 at 11:09
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    $\begingroup$ @Cupitor: It means that $\text{Re}\{X(\omega)\}$ and $\text{Re}\{X(\omega')\}$ are orthogonal for $\omega\neq\pm\omega'$ (and the same is true for the imaginary part), and that the power of $\text{Re}\{X(\omega)\}$ and $\text{Im}\{X(\omega)\}$ is infinite. If this makes a lot of sense is another question because normally $X(\omega)$ doesn't even exist. The $\delta$ impulses (and, consequently, the infinite powers) occur because you consider infinitely long signals. If you were to do the same thing with windowed signals, then you would get finite values for the powers. $\endgroup$ – Matt L. Jul 12 '14 at 15:36

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