# Connection between autocovariances and Fourier series of a continous function.

Let $f(w)$ be a continuous function of period $2 \pi$ then it's Fourier series is $$f(w) = \sum_{k = 0}^j \left(a_k \cos(kw) + b_k \sin(kw)\right)$$

I wrote that the autocovariances $\gamma(k)$ (of a stationary random process $X_t$) are equal to $a_k \pi$.

I can't recall why $\gamma(k) = a_k \pi$, could I be reminded of the connection?

• Thank you so much for providing the theorem. Could you help me a bit with showing $\gamma(k) = a_k \pi$ since it is not immediate from the proof. I get $\gamma(k) = \sum |a_k|^2 \cos(kw) + |b_k|^2 \sin(kw)$ by replacing the Fourier transforms with Fourier series. Aug 22 '15 at 18:50
• Did you apply the actual autocovar formula? $\Sigma f(\tau)f^*(t-\tau)$? Aug 22 '15 at 19:28
• so $\Sigma f(\tau)f^*(t-\tau) ) = \sum |f(v)|^2 e^{-2 \pi i v t}$ right? but then... Aug 22 '15 at 19:41
• I know that the spectral density is $g(\lambda) = (1/ \pi ) \sum \gamma(j) \cos( \lambda j)$ before your answer I was guessing this had something to do with it. But I can't seem to find $\gamma(k) = a_k \pi$. Aug 22 '15 at 19:48
• The problem is that In the proof I have $\gamma(t) = \int |f(v)|^2 e^{-2 \pi i v t} dv$ but I need $f(v) = \int \gamma \dots$ to match up the $a$ with the Fourier series of $f$. Could you explain to me how to proceed? Aug 23 '15 at 16:11