= Solution
Every $X_n$ is integrable: induction gives $|X_n|\leq a^n|x|+\sum_{j=0}^{n-1}a^j$. Since the innovation has <expectation> zero, the recursion implies
$$
\mathbb EX_{n+1}=a\mathbb EX_n+\mathbb E\theta_{n+1}=a\mathbb EX_n.
$$
Starting from $\mathbb EX_0=x$, induction therefore gives
$$
\boxed{\mathbb EX_n=a^nx\qquad(n\geq0).}
$$
This holds for every $a>0$, including $a=1$. The <independence> of the signs also identifies this as an <autoregressive process of order one>, although zero mean of each sign already suffices for the displayed <expectation> recursion.
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