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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...((((

Hope, you'll help me.

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It's not obvious, so let's approach this from first principles.

We only know, from the definition of $\gamma(n)\to 0,$ that for every $\epsilon\gt 0$ there exists an $N(\epsilon)\ge 0$ for which $n\ge N(\epsilon)$ implies $|\gamma(n)|\lt \epsilon.$

Let's exploit this to estimate the sum. Pick any $\epsilon$ as in the definition, set $N=N(\epsilon/2)$, and for $m \gt N$ break the sum into two parts where every term in the latter part is bounded by $\epsilon/2,$ and (over)estimate the second sum:

$$s_m = \frac{1}{m}\sum_{h=0}^{m-1} |\gamma(h)|=\frac{1}{m}s_N + \frac{1}{m}\sum_{h=N}^{m-1} |\gamma(h)| \lt \frac{1}{m}s_N + \frac{1}{m}(m-N)\epsilon/2 \le \frac{1}{m}s_N + \epsilon/2.$$

By choosing $m$ to be greater than either $N$ or $2s_N/\epsilon$ you can guarantee $s_N/m \lt \epsilon/2.$ In such cases, $s_N/m + \epsilon_/2 \lt \epsilon.$ This shows that for any $\epsilon\gt 0,$ there is an $m$ for which $0 \le s_n \lt \epsilon$ for all $n\ge m.$ That's the very definition of having $0$ as a limit.


Notice that the quoted statement cannot be taken literally as a sequence of substitutions: it is not necessarily the case that for any $n,$ $s_n$ equals (or even is bounded above) by $2|\gamma(n)|.$ After all, it is possible that all the $\gamma(n)$ are eventually zero, but provided any single $\gamma(h)$ is nonzero, $s_n$ is always strictly positive. I suspect the authors are implicitly quoting some theorem about summing convergent series. Indeed, this is an elementary result in Cesàro summation--but the factor of $2$ is superfluous.

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  • $\begingroup$ Thank you! Looks well proved and simple, yes. Factor 2 here rises from an symmetric of $\gamma(h)$ - $\gamma(h) = \gamma(-h)$. Maybe authors wanted to say that partial weighted sums and $\gamma(n)$ sequence have just have an 0 limit, nothing more. $\endgroup$ – Dmitriy May 7 '19 at 13:41
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    $\begingroup$ I had initially guessed that the phrase "$|h|\lt n$" might refer to the range $-n+1,-n+2,\ldots,n-1,$ but then abandoned that interpretation after noticing the assumption was only about what happens as $n\to\infty$ rather than $n\to\pm\infty.$ For symmetric $\gamma$--and that's almost always the case in time series analysis--I can now see why the factor of $2$ appears. Nothing needs to change in the demonstration, though, because for every $h\gt 0$ you can just replace every appearance of "$\gamma(h)$" in the assertion with "$\gamma(h)+\gamma(-h)$." $\endgroup$ – whuber May 7 '19 at 13:46
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    $\begingroup$ Yeah, I understood. Thank you again! Sometimes proofs of simple facts stop us for a looooong time.(((( $\endgroup$ – Dmitriy May 7 '19 at 13:47

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