# Spectral densities and their corresponding covariance functions.

Hey guys, I'm currently doing a course in stochastic processes and have come across something that has been wrecking my mind for a while.

So, let's say that I have some even, symmetric function $C(\tau)$, which is a covariance function of some wide-sense stationary random process.

Now, we know that there exists a Fourier transform of $C(\tau)$ which is defined by.. $S_X(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}C_X(\tau)e^{-i\omega\tau}\mathrm{d}\tau$

Finally, for my question (I think that the answer is correct, but I can't find it in my notes from the course, so I thought that I would double check).

As $C(\tau)$ is symmetric, we know that $C(\tau) = C(-\tau)$, and as a result... $S_X(\omega) = 2*\frac{1}{2\pi}\int_{0}^{\infty}C_X(\tau)e^{-i\omega\tau}\mathrm{d}\tau$

But, as the function is even, we know that the $i\sin$ term will always be $0$, due to the fact that $C(\tau)$ is even and $i\sin(blah)$ is odd.

So, as a result of this, can the integral be written as... $S_X(\omega) = \frac{1}{\pi}\int_{0}^{\infty}C_X(\tau)\cos(\omega\tau)\mathrm{d}\tau$

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Your first formula for $S_X$ is incorrect: $C_X$ is even but $\exp(-i\omega t)$ is not, so the integrand is not even. The second formula (with $\cos$) is correct. –  Alexandre Eremenko Apr 16 '13 at 11:49
In the case that $C_X$ was not even, would it be correct? What I really mean is, because $C_X$ is even, can I replace $e^(-i\omega\tau)$ by $\cos(\omega\tau)$, due to the cancellation of the $\sin$ term? –  PudgyPotato Apr 16 '13 at 12:13
certainly: the Fourier transform of an even function is a cosine transform –  Carlo Beenakker Apr 17 '13 at 8:44