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

  • 1
    $\begingroup$ What have you tried so far? $\endgroup$
    – πr8
    Apr 18, 2021 at 10:02

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.