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):

enter image description here

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?

  • $\begingroup$ 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

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

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


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

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

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

  • $\begingroup$ $f_n(h) \rightarrow \gamma(h)$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.