How to show that $var(\hat{\mu}) < var(\bar{X}) $for a stationary process ${X_t}$, where $X_t = \mu + Z_t + Z_{t-1} $? If ${X_t}$ is a stationary time series with mean $\mu$ then the usual estimator for $\mu$ is the sample mean  $\bar{X} = \frac{X_1+...+X_n}{n}$.
Assume we have $X_t = \mu + Z_t + Z_{t-1}$, where ${Z_t}$ is a white noise $WN(0,\sigma^2)$.
Can we find an unbiased estimator for $\mu$ of the form $\hat{\mu} = a_1X_1+...+a_nX_n$ such that $Var(\hat{\mu}) < Var(\bar{X})$?
 A: Given the MA(1) model $x_t=\mu + z_t + z_{t-1}$, you can think of your vector of observations $\mathbf{x}=(x_1,x_2,\dots,x_n)$ as coming from an intercept-only regression model
$$
\mathbf{x} = \mathbf{1}\mu+\boldsymbol\epsilon
$$
where $\mathbf{1}=(1,1,\dots,1)^T$, $\boldsymbol\epsilon \sim N(\mathbf{0},\mathbf{\Sigma})$ and $\mathbf\Sigma$ is tridiagonal with $2\sigma^2$ on the diagonal and $\sigma^2$ on the two off-diagonals.
The general least square solution of $\mu$ is then unbiased and efficient and  is given by
$$
\hat\mu =(\mathbf{1}^T \mathbf{\Sigma}^{-1} \mathbf{1})^{-1} \mathbf{1}^T \mathbf{\Sigma}^{-1} \mathbf{x}.
$$
This is simply a weighted average of the observations with weights proportional to the sum of each column of $\mathbf{\Sigma}^{-1}$.  This MLE is also computed by the R arima function (by numerically maximising the likelihood computed via the Kalman filter) when fitting an MA(1) model including an intercept.
A closed form solution can be derived however. The entries of $\mathbf{\Sigma}^{-1}$ are given by
$$
(\mathbf{\Sigma}^{-1})_{ij}=\frac{\sigma^{-2}(-1)^{i-j}}{2(n+1)}
\begin{cases}
(n-i+1)j &\text{for} & j<i \\
(n-j+1)i &\text{for} & j\ge i,
\end{cases}
$$
see Huang & McColl (1997), eqs. (22) and (23).
From this, for odd $n$, one can show that the even columns of $\mathbf\Sigma^{-1}$ sum to zero whereas the sums of the odd columns are all equal so the MLE of $\mu$ is
$$
\hat\mu = \frac2{n+1}(x_1 + x_3 + \dots + x_n).
$$
This estimator has variance $\operatorname{Var}\hat\mu=\frac{4\sigma^2}{n+1}$, whereas the sample average has variance $\operatorname{Var}\bar x=\frac{4\sigma^2}n (1-\frac1{2n})$.
For even $n$ the weights of the MLE have a certain regular oscillatory behaviour. For example, for $n=12$, the MLE is
$$
\hat\mu=\frac1{42}(6 x_1 + 1 x_2 + 5 x_3 + 2 x_4 + 4 x_5 + 3x_6 + 3x_7+ 4x_8 +2x_9+ 5x_{10}+ 1x_{11}+ 6x_{12}).
$$
