2
$\begingroup$

I'm having trouble to understand a passage in the mean square error of an estimator.

Let $\{X_t\}$ be a stationary process of a time series with mean $\mu$, thus the sample mean estimator is $$\overline{X}_n=\frac{1}{n}(X_1+X_2+\dots +X_n)$$ The mean squared error of this estimator is $$E[\overline{X}_n-\mu]^2=Var(\overline{X}_n)$$ since that $\overline{X}_n$ is a unbiased estimator of $\mu$. Then

$$Var(\overline{X}_n)=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n Cov(X_i,X_j)\qquad (1)$$

$$=\frac{1}{n^2}\sum_{i-j=-n}^n (n-|i-j|)\gamma(i-j)\qquad (2)$$

$$=\frac{1}{n}\sum_{h=-n}^n \Big(1-\frac{|h|}{n}\Big)\gamma (h)$$

where $\gamma (i-j)=Cov(X_{t+(i-j)},X_t)$ and $\gamma(h)=Cov(X_{t+h},X_t)$ (autocovariance function).

I can't figure out what they make from (1) to (2).

I make a test with $n=1$ and it works, but I don't understand what they did, since in (1) I have a sum with $n^2$ terms and in (2) I have just $2n+1$ terms.

$\endgroup$
1
$\begingroup$

As you mentioned $Cov(X_i,X_j)=\gamma(i-j)$. See in the total sum $\sum_{i=1}^n\sum_{j=1}^n Cov(X_i,X_j)$ how many times $\gamma(k)$ appears. This is the number of solutions of $i-j=k$ for $1\leq i,j\leq n$.

Of course $i=j+k$ and since $1\leq i,j \leq n$, then $k+1\leq i \leq n+k$ which means that $i\in\{1,\dots,n\}\cap\{k+1,n+k\}$. It is not difficult to see that there are $n-|k|$ possible $i$ for which $j$ is uniquely determined and hence it has $k$ solutions. So: $$ \frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n Cov(X_i,X_j)=\frac{1}{n^2}\sum_{k=-n}^n (n-|k|)\gamma(k). $$ To number of terms on the right side is indeed $\sum_{k=-n}^n (n-|k|)=n^2$.

$\endgroup$
  • $\begingroup$ What you did with $\frac{1}{n}$ in the right side of the expression? $\endgroup$ – Roland Sep 29 '16 at 23:12
  • $\begingroup$ Just a typo.... $\endgroup$ – Arash Sep 29 '16 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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