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I am working on the following problem arising in time series analysis.

Let us assume that $\sum_{h \in \mathbb{Z}} |\gamma(h)|<\infty$. I would like to prove that \begin{equation*} 1) \; \; \; \lim_{n\to +\infty} \sum_{h >n } \gamma(h) = 0 \end{equation*} and the stronger result \begin{equation*} 2) \; \; \; \lim_{n\to +\infty} \sqrt{n}\sum_{h >n } \gamma(h) = 0 \end{equation*}

I was able to prove 1) using the limit of the partial sum $S_n = \sum_{h=-\infty}^n\gamma(h)$, and then writing $\sum_{h > n} \gamma(h) = \sum_{h \in \mathbb{Z}} \gamma(h) - \sum_{h = -\infty}^{n} \gamma(h) \ $ and taking the limit. However, for the second question I think I need additional assumptions on the sequence $(\gamma(h))$ in order to prove it.

Davide Giraudo
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givo
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    What have you tried so far? – Avatrin Aug 01 '23 at 12:59
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    What’s your attempts? – user1176409 Aug 01 '23 at 13:12
  • I was able to prove 1) using the limit of the partial sum $S_n = \sum_{h=-\infty}^n\gamma(h)$, and then writing $\sum_{h > n} \gamma(h) = \sum_{h \in \mathbb{Z}} \gamma(h) - \sum_{h = -\infty}^{n} \gamma(h)
    $ and taking the limit. However, for the second question I think I need additional assumptions on the sequence $\gamma(h)$ in order to prove it.     – givo Aug 01 '23 at 13:20
  • please edit the post adding your effort – Sine of the Time Aug 01 '23 at 13:23
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    Hint: 2) is wrong without further conditions. – Ingix Aug 01 '23 at 13:31
  • Consider a decreasing sequence $a_n \to 0$, and define $\gamma(\pm n) = a_{n-1}-a_n$. Then $\sum_{h > n} \gamma(h) = a_n \to 0$. (2) would imply that $\lim_n a_n\sqrt n = 0$ also. Can you think of a decreasing sequence ${a_n}$ converging to $0$ such that $a_n\sqrt n$ does not converge to $0$? – Paul Sinclair Aug 02 '23 at 12:58

1 Answers1

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The example given by Paul Sinclair give a good path, but one has to be sure that $a_{n-1}-a_n$ gives indeed the covariance function of a stationary sequence.

To do so, take $X_t=\sum_{i\in\mathbb Z}a_i\varepsilon_{t-i}$, where $(\varepsilon_k)$ is i.i.d., centered and has unit variance, $a_i> 0$, $\sum_{i\in\mathbb Z}a_i<\infty$ and $\sqrt{N}\sum_{i>N}a_i\geqslant 1$. Then $$ \gamma(h)=\mathbb E[X_0X_h]=\sum_{i\in\mathbb Z}a_{h+i}a_i $$ hence $\sum_{h\in\mathbb Z}\gamma(h)<\infty$ and $\gamma(h)\geqslant a_0a_h$.

Davide Giraudo
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