Periodogram

Remarks

Some authors divide by $2\pi n$ instead of $n$, so that the periodogram estimates the spectral density $f(\omega) = \frac{1}{2\pi} \sum_h \gamma(h) e^{-i h \omega}$ directly. At every Fourier frequency $\omega_j \ne 0$, the periodogram equals $\sum_{\abs{h} < n} \hat\gamma(h) e^{-i h \omega_j}$, where $\hat\gamma$ is the sample autocovariance from the autocorrelation function. So the periodogram is the Fourier transform of the sample autocovariance.

Example: let $n = 12$ and $x_t = \cos(2\pi t / 4)$, a cycle of period $4$, so the frequency is $\omega_3 = 2\pi \cdot 3/12$. Writing $\cos u = (e^{iu} + e^{-iu})/2$ and using $\sum_{t=1}^{12} e^{2\pi i k t / 12} = 0$ for $k \not\equiv 0 \pmod{12}$, one gets $\sum_t x_t e^{-i t \omega_3} = 12/2 = 6$. Hence $I_{12}(\omega_3) = 36/12 = 3$, while $I_{12}(\omega_j) = 0$ at all other Fourier frequencies in $[0, \pi]$.

Non-example: the function $\omega \mapsto \abs{\sum_t x_t e^{-it\omega}}$ without the square is not the periodogram; it is the amplitude spectrum of the data, and its square divided by $n$ is the periodogram.

For a stationary process with summable autocovariances, $E[I_n(\omega)] \to 2\pi f(\omega)$ for $\omega \ne 0$. But the variance of $I_n(\omega)$ does not tend to $0$, so the periodogram is not a consistent estimator of the spectral density.