# Why the sum of the sample autocovariance function of a stationary process is zero?

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|

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

$$\newcommand{\hgamma}{\hat{\gamma}}$$ 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*}