# Variance of Weakly Stationary Time Series

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 $\gamma(h)= \mathrm{Cov}\left(X_t, X_{t + h}\right) = \mathrm{E}\left[X_tX_{t + h}\right]$

Show that: $$Var\left( \frac{X_1 + X_2 +\ldots+ X_n}{n}\right) = \frac{\gamma(0)}{n} + \frac{2}{n}\sum_u \left( 1−\frac{u}{n}\right)\gamma(u)$$

Edit: So far, I've gotten this... $Var(X\bar) = \frac{1}{n^2} \sum_{i=1}^{n} \sum_{j=1}^{n} Cov(X_i,X_j) = \frac{1}{n^2} \sum_{i-j=-n}^{n} (n-|i-j)\gamma(i-j) = \frac{1}{n} \sum_{u=-n}^{n} (1- \frac{|u|}{n})\gamma (u)$

How am I supposed to come up with the $\frac{\gamma(0)}{n} + \frac{2}{n}$?

• Problems like this you can typically solve/prove by mathematical induction: (1) first show the formula holds for $n=1$ (2) Second show that if the formula holds for $n=k-1$, that it then holds for $n=k$. Oct 9 '16 at 17:08
• However you solve it, you almost certainly need to apply the formula $\mathrm{Var}\left( X + Y \right) = \mathrm{Var}(X) + 2 \mathrm{Cov}(X,Y) + \mathrm{Var}(Y)$. More generally, $\mathrm{Var}(\sum_i^n X_i) = \sum_{i=1}^n\sum_{j=1}^n \mathrm{Cov}(X_i, X_j)$. You can probably just brute force, figure it out that way. Oct 9 '16 at 17:16
• Possible duplicate of Proof of variance of stationary time series
– Ben
Oct 18 '19 at 4:56

$$\text{Var}\left( \frac{X_1 + X_2 +\ldots+ X_n}{n}\right) = \frac 1{n^2}E[(X_1 + X_2 +\ldots+ X_n)^2]$$
Decompose $(X_1 + X_2 +\ldots+ X_n)^2$, manipulate, and you will be able to arrive at the right-hand side.