# Stationarity of MA($\infty$) process

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.

For any pair of random variables $x_t$ and $x_{t-h}$, the squared-covariance obeys the upper bound:
$$\mathbb{Cov}(x_t, x_{t-h})^2 = \mathbb{Corr}(x_t, x_{t-h})^2 \mathbb{V}(x_t) \mathbb{V}(x_{t-h}) \leqslant \mathbb{V}(x_t) \mathbb{V}(x_{t-h}).$$
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{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}