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.


1 Answer 1


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:

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


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