Let $X$ be an infinite moving average time series, i.e.
$$
X(t) = \sum\_{k = -\infty}^\infty a\_j \varepsilon\_{t-j}, \quad t \in \mathbb{Z},
$$
where $\varepsilon\_{j}$ are uncorrelated zero mean, finite variance and identically distributed random variables.


In my opinion, $\mathbb{E}X(t) = 0$ for all $t \in \mathbb{Z}$ is easy to prove with if $\sum\_j \vert a\_j \vert < \infty$. This is often referred to as short memory of the moving average, cf. Section 4.2.4 in [1].


But what happens if we only assume $\sum\_j \vert a\_j \vert^2 < \infty$? This is a common assumption, e.g. in Wold's decomposition, and can include both long and short memory moving averages. Is there some alternative way to prove $\mathbb{E}X(t) = 0$ for all $t \in \mathbb{Z}$ under these conditions?


References
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[1] J. Beran, Y. Feng, S. Ghosh, and R. Kulik. Long-memory processes. Springer, 2016.



Expected value of long memory moving average
This is a slightly more elementary version of Anthony Quas's answer, without appealing to martingales.


By the condition $\sum\_j|a\_j|^2<\infty$, the sequence of random variables $X\_n:=\sum\_{j=-n}^n a\_j\epsilon\_{-j}$ is [Cauchy convergent](https://en.wikipedia.org/wiki/Cauchy%27s_convergence_test) in $L^2$ and hence in $L^1$. So, $X\_n$ converges to a limit $X(0)$ in $L^1$. Since $EX\_n=0$ for all $n$, we have $EX(0)=0$.


(In view of [Kolmogorov's maximal inequality](https://en.wikipedia.org/wiki/Kolmogorov%27s_inequality), $X\_n\to X(0)$ almost surely as well.)


