# 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 $$Var(\hat{\mu}) < Var(\bar{X})$$?

• What have you tried so far?
– πr8
Apr 18, 2021 at 10:02

## 1 Answer

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 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 $$\hat\mu=\frac1{42}(6 x_1 + 1 x_2 + 5 x_3 + 2 x_4 + 4 x_5 + 3x_6 + 3x_7+ 4x_8 +2x_9+ 5x_{10}+ 1x_{11}+ 6x_{12}).$$