Prove variance of mean estimator of an AR(1) model I need to solve the proof as given in the screenshot below. I had tried to do something (under My solution) but I don't know how to proceed with the proof. I'm not sure whether I did something wrong or mistake. Please help me to continue with the proof. Any help will be greatly appreciated! Thanks!


Compute directly (not the best homework question, if that's what this is):
\begin{align*}
Var(\bar{Y}) &= Var(\frac{1}{n} \sum_{t=1}^n Y_t) \\
             &= \frac{1}{n^2} Var( \sum_{t=1}^n Y_t) \\
             &= \frac{1}{n} \gamma(0) + 2 \frac{1}{n} \sum_{h = 1}^{n-h} \frac{n-h}{n} \gamma(h),
\end{align*}
where $\gamma(h) = \frac{\beta^h}{1-\beta^2}$ is the autocovariance function at lag $h$.
Substituting $\gamma(0) = \frac{1}{1-\beta^2}$ gives the first term in your sum
$$ 
\frac{1}{n} \gamma(0) = \frac{1}{n} \frac{1}{1-\beta^2}.
$$
Similarly,
\begin{align*}
2 \frac{1}{n} \sum_{h = 1}^{n-h} \frac{n-h}{n} \gamma(h) &= \frac{2}{n^2(1-\beta^2)}  \sum_{h=1}^{n-h} (n-h) \beta^h \\
&=  \frac{2}{n^2(1-\beta^2)}  \sum_{h=1}^{n-h} \sum_{j=1}^{n-h} \beta^j \\ 
&=  \frac{2}{n^2(1-\beta^2)}  \sum_{h=1}^{n-h} ( \frac{1-\beta^{n+1-h}}{1-\beta}) \\
&=  \frac{2}{n^2(1-\beta^2)(1-\beta)}  \sum_{h=1}^{n-h} ( 1-\beta^{n+1-h} )\\
&= \cdots.
\end{align*}
