2
$\begingroup$

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 $\gamma(h) \equiv Cov(X_{t+h}, X_t)$.

Improve this question
$\endgroup$
5
  • $\begingroup$ 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
    Commented May 25, 2015 at 15:46
  • $\begingroup$ That was a typo. $\endgroup$
    – Monolite
    Commented May 25, 2015 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
    Commented May 26, 2015 at 2:44
  • $\begingroup$ LHS = left-hand side. But I think it should be just the other way round! $\endgroup$
    – Christoph Hanck
    Commented May 26, 2015 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
    Commented May 26, 2015 at 11:05

1 Answer 1

Reset to default
1
$\begingroup$

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, $$\lim_{n \rightarrow \infty} n^{-1} \sum_{|h| <n} |\gamma(h)| = \lim_{n \rightarrow \infty} 2|\gamma(n)| =0$$

Improve this answer
$\endgroup$

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.