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  "body": "\n## Definition\n\nLet $x_1, \\dots, x_n$ be observations of a real [[time series|EA3MF4VTEH]].\nThe *periodogram* of the data is the function\n$$I_n(\\omega) = \\frac{1}{n} \\abs{\\sum_{t=1}^{n} x_t e^{-i t \\omega}}^2, \\qquad \\omega \\in [-\\pi, \\pi].$$\nIt is usually evaluated at the *Fourier frequencies* $\\omega_j = 2\\pi j / n$ with $-\\pi \u003c \\omega_j \\le \\pi$, where it equals $\\frac1n \\abs{\\hat x_j}^2$ for the [[discrete Fourier transform|TST0GH66A5]] $\\hat x_j = \\sum_{t=1}^n x_t e^{-2\\pi i j t / n}$.\n\n## Remarks\n\nSome 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.\nAt every Fourier frequency $\\omega_j \\ne 0$, the periodogram equals $\\sum_{\\abs{h} \u003c n} \\hat\\gamma(h) e^{-i h \\omega_j}$, where $\\hat\\gamma$ is the sample autocovariance from the [[autocorrelation function|YCZHYY8VE5]].\nSo the periodogram is the Fourier transform of the sample autocovariance.\n\nExample: let $n = 12$ and $x_t = \\cos(2\\pi t / 4)$, a cycle of period $4$, so the frequency is $\\omega_3 = 2\\pi \\cdot 3/12$.\nWriting $\\cos u = (e^{iu} + e^{-iu})/2$ and using $\\sum_{t=1}^{12} e^{2\\pi i k t / 12} = 0$ for $k \\not\\equiv 0 \\pmod{12}$, one gets $\\sum_t x_t e^{-i t \\omega_3} = 12/2 = 6$.\nHence $I_{12}(\\omega_3) = 36/12 = 3$, while $I_{12}(\\omega_j) = 0$ at all other Fourier frequencies in $[0, \\pi]$.\n\nNon-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.\n\nFor a stationary process with summable autocovariances, $E[I_n(\\omega)] \\to 2\\pi f(\\omega)$ for $\\omega \\ne 0$.\nBut the variance of $I_n(\\omega)$ does not tend to $0$, so the periodogram is not a consistent estimator of the spectral density.\n",
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