 Solution

Take the driving signs $a_t=\alpha (t)$ to be independent across times, with $\mathbb{E}{a}_t=0$ and $a_t^2=1$. Temporal independence of the drivers is an assumption needed in addition to their stated equal one-time probabilities. For $X_t=\delta y(t)=a_{t-2}{a}_{t-1}+a_t$,As a useful scale check, $X_t$ takes values $-2,0,2$ with probabilities $1/4,1/2,1/4$. Its variance is therefore $\boxed{\mathbb{E}{X}_t^2=2}$, rather than the unit variance of a single driving sign.