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The sample autocovariance function for a sample $X_1, X_2, \dots, X_n$ from a stationary process is defined as:

$$\widehat{\gamma}(h) = n^{-1}\displaystyle\sum_{t=1}^{n-h}(x_{t+h}-\bar{x})(x_t-\bar{x})$$

with $\widehat{\gamma}(-h) = \widehat{\gamma}(h)\;$ for $\;h = 0,1, ..., n-1$. $\bar{x}$ is the mean.

Show that $\sum_{|h| <n}\widehat{\gamma}(h) = 0$

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  • $\begingroup$ The equality holds for any sequence $\{x_1, \ldots, x_n\}$. The stationary assumption is redundant. $\endgroup$
    – Zhanxiong
    Commented Mar 22, 2021 at 3:30

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$\newcommand{\hgamma}{\hat{\gamma}}$ This is Exercise $7.3$ from the classic text Time Series: Theory and Methods. We can prove it by a series of reductions of the original equality.

First note by $\hgamma(h) = \hgamma(-h)$, the equality to be proven is $$0 = \hgamma(0) + 2\sum_{h = 1}^{n - 1}\hgamma(h). $$

Next note by definition of $\hgamma$, without loss of generality we can assume $\bar{x} = 0$. Therefore, it suffices to prove that under the condition of $x_1 + \cdots + x_n = 0$, it holds that \begin{align*} 0 = \sum_{t = 1}^n x_t^2 + 2\sum_{h = 1}^{n - 1}\sum_{t = 1}^{n - h}x_{t + h}x_t. \end{align*}

But this is just the result of expanding the quadratic form \begin{align*} 0 = \bar{x}^2 = (x_1 + x_2 + \cdots + x_n)^2. \end{align*}

This completes the proof.

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