= Exponential limit of a one-sided periodogram
{title2=$I_T^+(\omega)/s(\omega)\xrightarrow{d}\operatorname{Exp}(1)$}
Let $X_t$ be a centered real stationary <Gaussian> <time series> with absolutely summable <autocovariances> and two-sided angular-frequency <spectral density> $f$. The parent <periodogram> uses $I_T(\omega)=(2\pi T)^{-1}|\sum_{t=1}^T X_te^{-it\omega}|^2$. At an interior positive frequency define the one-sided quantities
$$
I_T^+(\omega)=2I_T(\omega),\qquad s(\omega)=2f(\omega),\qquad 0<\omega<\pi.
$$
Thus the doubling applies to both the statistic and its target <spectral density>.
For Fourier-grid frequencies $\omega_T=2\pi k_T/T\to\omega\in(0,\pi)$ with $f(\omega)>0$, <set>
$$
A_T=\frac1{\sqrt{\pi T}}\sum_{t=1}^TX_t\cos(t\omega_T),\qquad
B_T=\frac1{\sqrt{\pi T}}\sum_{t=1}^TX_t\sin(t\omega_T).
$$
Then $I_T^+(\omega_T)=A_T^2+B_T^2$. Summability makes their <variances> tend to $f(\omega)=s(\omega)/2$ and their <covariance> tend to zero. Joint Gaussianity therefore gives
$$
\frac{I_T^+(\omega_T)}{s(\omega)}\xrightarrow{d}\frac{\chi_2^2}{2}
\sim\operatorname{Exp}(1).
$$
Under these Gaussian assumptions the fourth <moments> also converge, so $\operatorname{var}(I_T^+(\omega_T))\to s(\omega)^2$. An unsmoothed <periodogram> is therefore not a <consistent estimator> of a positive <spectral density>. At zero and the <Nyquist frequency> $\pi$, the sine coordinate vanishes and the real-only calibration differs; the interior doubling and two-coordinate argument do not apply there.<Introduction to Time Series Analysis, Lecture 20 by Peter Bartlett>[Slides 2–9; converted from cycles per observation to angular frequency]{ref}
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