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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 $\frac{\gamma(0)}{n} + \frac{2}{n}$?

FBeller
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    Problems like this you can typically solve/prove by [mathematical induction](https://en.wikipedia.org/wiki/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 $n=k$. – Matthew Gunn Oct 09 '16 at 17:08
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    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. – Matthew Gunn Oct 09 '16 at 17:16
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    Possible duplicate of [Proof of variance of stationary time series](https://stats.stackexchange.com/questions/240010/proof-of-variance-of-stationary-time-series) – Ben Oct 18 '19 at 04:56

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Since the mean is zero, we have that

$$\text{Var}\left( \frac{X_1 + X_2 +\ldots+ X_n}{n}\right) = \frac 1{n^2}E[(X_1 + X_2 +\ldots+ X_n)^2]$$

Decompose $(X_1 + X_2 +\ldots+ X_n)^2 $, manipulate, and you will be able to arrive at the right-hand side.

Alecos Papadopoulos
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