0
$\begingroup$

In my time series class, we got taught, that an MA($\infty $) process with $$ \sum_i^\infty \theta_i^2 < \infty $$ is weakly stationary.

Obviously the mean exists and is time independent, but why does the second moment, so the covariance, also exist? From the assumption I can only derive, that the variance, but not the covariance with some lag $h$ $$ Cov(x_t, x_{t-h}) = \sum_i^\infty \theta_i \theta_{i+h} $$ exist, which is, as far as I know, required for weak stationarity.

$\endgroup$
  • $\begingroup$ When you say existent do you mean exists? Weak stationarity means that the mean and variance are finite and don't change over time and the autocovariance only depends on the lag. $\endgroup$ – Michael R. Chernick Nov 25 '17 at 17:57
  • $\begingroup$ So the autocovariance does not have to exist? $\endgroup$ – Luca Thiede Nov 25 '17 at 17:58
  • $\begingroup$ I didn't say that. Mainly I wanted to understand what you meant because your English was not correct. $\endgroup$ – Michael R. Chernick Nov 25 '17 at 18:14
  • $\begingroup$ Sorry, I edited it. $\endgroup$ – Luca Thiede Nov 25 '17 at 18:16
4
$\begingroup$

For any pair of random variables $x_t$ and $x_{t-h}$, the squared-covariance obeys the upper bound:

$$\mathbb{Cov}(x_t, x_{t-h})^2 = \mathbb{Corr}(x_t, x_{t-h})^2 \mathbb{V}(x_t) \mathbb{V}(x_{t-h}) \leqslant \mathbb{V}(x_t) \mathbb{V}(x_{t-h}).$$

Hence, if the variance values are both finite, the squared-covariance must also be finite, which means that the covariance is also finite. This is a general property that works in any context, so you don't need to use the specific form for the covariance that occurs in an MA model.

Alternatively, if you particularly want to establish this result just for the particular case of the MA($\infty$) process, you could do this as follows. Using the Cauchy-Schwartz inequality combined with your finite variance condition you have:

$$\begin{equation} \begin{aligned} \mathbb{Cov}(x_t, x_{t-h})^2 &= \Bigg( \sum_{i=0}^\infty \theta_i \theta_{i+h} \Bigg)^2 \\[6pt] &\leqslant \Bigg( \sum_{i=0}^\infty \theta_i^2 \Bigg) \Bigg( \sum_{i=0}^\infty \theta_{i+h}^2 \Bigg) \\[6pt] &\leqslant \Bigg( \sum_{i=0}^\infty \theta_i^2 \Bigg)^2 < \infty. \end{aligned} \end{equation}$$

$\endgroup$

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.