Why Vanishing Autocovariance Does Not Imply the Stated Average Limit

Summary

The document asks about a step in a proof concerning estimation of the mean and autocovariance function. The stated condition is that autocovariance tends to zero at large lags, and the question is why the average of its absolute values over lags should have a particular limit.

The included answer argues that autocovariance is symmetric around zero and then attempts to simplify the sum by pairing negative and positive lags. However, it does not establish the claimed limit: a sequence tending to zero need not have its averages equal to twice its endpoint value, and the sum depends on all lags in the range. Thus, the response appears mathematically incorrect or incomplete. The document is useful chiefly as a prompt to check the proof carefully; it does not provide a valid derivation or evidence resolving the question.

Key ideas

• Autocovariance for a stationary process is symmetric across positive and negative lags.
• A sum across lags includes all terms in the interval, not only the endpoint values.
• Convergence of autocovariance to zero alone does not justify replacing its average with a multiple of its endpoint value.
• The supplied answer does not adequately prove the claimed limit.

Tags

# Misunderstanding of time series autocovariance


# Misunderstanding of time series autocovariance












I'm reading the "Time Series: Theory and Methods (2nd ed.)" by P.J.Brockwell and R.A.Davis. I've stopped at the one moment at pp.218-219 (Chapter 7 "Estimation of the mean and the Autocovariance function"). In the proof of theorem 7.1.1 if

$\gamma(n) \rightarrow 0$ as $n \rightarrow +\infty$

then $$lim_{n \rightarrow +\infty} n^{-1} \sum_{|h| < n} \left(|\gamma(h)| \right) = 2 \lim_{n \rightarrow +\infty} \left( |\gamma(n)| \right) = 0$$.

Could anyone explain me the first equality in this part of the proof, pls? I spend much time, but suppose, I'm not so intelligent for self-understanding...((((

## Answer by numerairX (score 1)

https://quant.stackexchange.com/a/45490

### this answer is on hold

first it used the fact that your function $y$ is symmetric around 0 (proof) can be found here, so i don't need to type everything.

then just expanding the summation $$lim_{n \rightarrow +\infty} n^{-1} \sum_{|h| < n} \left(|\gamma(h)| \right) = lim_{n \rightarrow +\infty} n^{-1} * \frac{(y(-n)+y(n))*2n}{2} $$ because h is from -n to n. Then it's pretty self explanatory given that y is symmetric around 0.

$$orig = lim_{n \rightarrow +\infty}y(-n)+y(n) = lim_{n \rightarrow +\infty} 2 y(n)$$

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