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I am cross-posting a question I asked on cross-validated here. It is a mathematical doubt on the application of the dominated convergence theorem in the time series setting.

I leave the auto-covariance function $\gamma()$ undefined as I think it is not relevant to the step I am having problems on. Here is the question (for completeness I upload the whole statement and proof):

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focusing on the second proposition the proof states:

enter image description here

And If $ \sum_{h = -\infty}^{\infty} |\gamma(h)| < \infty$ then the dominated convergence theorem gives:

$$\lim_{n \rightarrow \infty} n Var(\bar{X_n}) = \lim_{n \rightarrow \infty} \sum_{|h| < n} \Big( 1 - \frac{|h|}{n} \Big) \gamma(h) = \sum_{h = -\infty}^{\infty} \gamma(h) $$

I understand the proof up until the dominated convergence theorem is used, how is it applied?

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  • $\begingroup$ Is there a typo in the last line? Should it read $\lim_n \sum_{|h|<n} \dots = \sum_{h=-\infty}^{\infty} \gamma(h)$? $\endgroup$ – saz May 27 '15 at 4:33
  • $\begingroup$ @saz Yes you are right I just checked! edited. Could you explain to me how the dominated convergence theorem is applied? $\endgroup$ – Monolite May 27 '15 at 7:38
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Consider $(\mathbb{Z},\mathcal{P}(\mathbb{Z}))$ endowed with the counting measure

$$\mu(B) := \sum_{h \in \mathbb{Z}} \delta_h(B).$$

Then for any integrable function $f \in L^1$, we have

$$\int f \, d\mu = \sum_{h \in \mathbb{Z}} f(h). \tag{1}$$

Define

$$f_n(h) := \begin{cases} \left(1- \frac{|h|}{n} \right) \gamma(h), & |h| < n, \\ 0, & |h| \geq n \end{cases}$$

Then $|f_n| \leq |\gamma| \in L^1$ and $f_n(h) \to \gamma(h)$ for all $h \in \mathbb{Z}$. Therefore, by the dominated convergence theorem

$$ \lim_{n \to \infty} \int f_n \, d\mu = \int f \, d\mu.$$

By $(1)$ and the definition of $f_n$, this is equivalent to

$$\lim_{n \to \infty} \sum_{|h|<n} \left(1- \frac{|h|}{n} \right) \gamma(h) = \sum_{h \in \mathbb{Z}} \gamma(h).$$

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  • $\begingroup$ $f_n(h) \rightarrow \gamma(h)$ for all $h \in Z$? does it not go to $0$ for the $|h| \ge n$? Aside from this minor doubt I understood the rest, thanks a lot! $\endgroup$ – Monolite May 27 '15 at 15:24
  • $\begingroup$ @Monolite For any (fixed) $h \in \mathbb{Z}$, we have $|h| <n$ for $n$ sufficiently large. Consequently, $$f_n(h) = \left(1- \frac{|h|}{n} \right) \gamma(h)$$ for $n$ sufficiently large and therefore $f_n(h) \to \gamma(h)$ as $n \to \infty$. $\endgroup$ – saz May 27 '15 at 15:25
  • $\begingroup$ Got it, it was evident. Thanks for clarifying anyway. $\endgroup$ – Monolite May 27 '15 at 15:39

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