In the framework of time series analysis

Why does $\lim_{n \rightarrow \infty} n^{-1} \sum_{|h| <n} |\gamma(h)| = \lim_{n \rightarrow \infty} 2|\gamma(n)| $?

The LHS (left hand side) sequence of functions we are taking the limit of is $$\{ |\gamma(1)|, \frac{1}{2}(|\gamma(1)| + |\gamma(2)|), \frac{1}{3}(|\gamma(1)| + |\gamma(2)| + |\gamma(3)|), \dots \} $$

The RHS (right hand side) sequence we are taking the limit of is $$\{2 |\gamma(1)|, 2 |\gamma(2)| , 2 |\gamma(3)|, \dots \}$$

Adding some steps in-between might help me greatly.

Where $\gamma(h)$ is the auto-covariance function defines as $\gamma(h) \equiv Cov(X_{t+h}, X_t)$.

  • $\begingroup$ Is there any relationship between the $k$ on the left side and $n$ in your $n^{-1} \sum_{|h| <k} |\gamma(h)| = 2|\gamma(n)|$? $k$ does not appear on the right side at all, and there presumably is a reason why it disappears. $\endgroup$ – Dilip Sarwate May 25 '15 at 15:46
  • $\begingroup$ That was a typo. $\endgroup$ – Monolite May 25 '15 at 16:03
  • $\begingroup$ RHS? LHS? What do these acronyms mean? and, a more personal question if I may, are you left-handed? Also, what happened to the absolute value signs? $\endgroup$ – Dilip Sarwate May 26 '15 at 2:44
  • $\begingroup$ LHS = left-hand side. But I think it should be just the other way round! $\endgroup$ – Christoph Hanck May 26 '15 at 4:39
  • $\begingroup$ @DilipSarwate I made so many mistakes I feel like I should just delete the question. It's disrespectful to people trying to answer, I will edit in the absolute values and swap the acronyms. Please tell me if I should just delete it, thanks for the patience up to now. $\endgroup$ – Monolite May 26 '15 at 11:05

The left hand side sequence you are taking the limit of is $$\left\{ \frac{1}{n}[|\gamma(-n)| + |\gamma(-(n-1))| + \ldots+|\gamma(0)|+|\gamma(1)| + |\gamma(2)| + \ldots+|\gamma(n)|] \right\}, $$ as you sum over $|h|$, not $h$. By stationarity (a condition which should be mentioned somewhere in the result you are referring to), $\gamma(-n)=\gamma(n)$.

Now, as $n\to\infty$, we (again that should be mentioned somewhere) have that $\gamma(n)\to0$, as the memory of a stationary series does not extend into the infinite past. In that case, $$\lim_{n \rightarrow \infty} n^{-1} \sum_{|h| <n} |\gamma(h)| = \lim_{n \rightarrow \infty} 2|\gamma(n)| =0$$


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