Suppose that $\{X_t\}$ is a weakly stationary time series with mean
$\mu = 0$ and a covariance function $\gamma(h)$, $h \geq 0$, $\mathrm{E}[X_t] = \mu = 0$ and $\gamma(h)= \operatorname{Cov}\left(X_t, X_{t + h}\right) = \mathrm{E}\left[X_tX_{t + h}\right]$
Show that:
$$
\operatorname{Var}\left( \frac{X_1 + X_2 +\ldots+ X_n}{n}\right) = \frac{\gamma(0)}{n} + \frac{2}{n}\sum_{m=1}^{n-1} \left( 1−\frac{m}{n}\right)\gamma(m).
$$
So far, I've gotten this:
\begin{align}
\operatorname{Var}(\bar X ) &= \frac{1}{n^2} \sum_{i=1}^{n} \sum_{j=1}^{n} \operatorname{Cov}(X_i,X_j) \\[7pt]
&= \frac{1}{n^2} \sum_{i-j=-n}^{n} (n-|i-j|)\gamma(i-j) \\[7pt]
&= \frac{1}{n} \sum_{m=-n}^{n} \left(1- \frac{|m|}{n}\right)\gamma (m)
\end{align}
How am I supposed to come up with the $\frac{\gamma(0)}{n} + \frac{2}{n}$?
Best Answer
Since this post has attracted so many answers, it seems worthwhile to show the idea.
Here is a diagram of the covariance matrix $\Sigma = \operatorname{Cov}(X_1,X_2,\ldots, X_n).$ Values that are necessarily equal receive the same color. It has this diagonal striped pattern because the covariances depend only on the absolute lags--and the lags index the diagonals.
The variance of a sum of random variables $X_1+\cdots +X_n$ is the sum of all their variances and covariances, taken in all orders. This is a consequence of the multilinear property of covariance. It is easily demonstrated by observing $X_1+\cdots +X_n$ is the dot product of the random vector $\mathbf{X}=(X_1,\cdots,X_n)$ and the vector $\mathbf{1}=(1,1,\ldots 1)$ (with $n$ components). Therefore the variance of the sum is
$$\operatorname{Var}(X_1+\cdots+X_n) = \mathbf{1}^\prime \Sigma \mathbf{1},$$
which the rules of matrix multiplication tell us is the sum of all the entries of $\Sigma.$
The formula in the question sums the entries of $\Sigma$ by color:
There are $n$ copies of $\gamma_0$ (in red, on the diagonal).
There are $2(n-1)$ copies of $\gamma_1$ (in orange, on both sides of the diagonal: this is where the factor of $2$ comes from).
There are $2(n-2)$ copies of $\gamma_2$ (in yellow).
... and so on, up to $2$ copies of $\gamma_{n-1}$ (in blue).
Therefore, by merely looking at the figure, we obtain
$$\operatorname{Var}(X_1+\cdots+X_n) = n\gamma_0 + 2(n-1)\gamma_1 + 2(n-2)\gamma_2 + \cdots + 2\gamma_{n-1}.$$
The general pattern is
The question asks for the variance of $1/n$ times this sum. Again, according to the multilinear property of variance, we must multiply the variance of the sum by $1/n^2.$ Doing that to the preceding formula gives the answer,
Comparing this to the formula in the question helps us interpret the question's "$1/n$" factors as really being $1/n=n/n^2,$ $(1-1/n)/n= (n-1)/n^2,$ and so on down to $(1-(n-1)/n)/n = 1/n^2.$