 Solution

The displayed equation has nominal orders $p=1,q=2$, withThe AR root is $2$, and the MA roots are $2$ and $10/9$, all strictly outside the unit disc. Thus its stationary solution is causal and invertible. There is also common factor cancellation in an ARMA model: the convergent inverse of $1-0.5B$ cancels the shared factor, givingConsequently the displayed representation is ARMA(1,2), but its minimal order is ARMA(0,1). This also follows without formal operator cancellation: the difference between any solution and the right-hand side satisfies $d_t=0.5d_{t-1}$. A stationary finite-variance difference must have zero variance, since stationarity would give $\mathrm{Var}(d_t)=0.25\mathrm{Var}(d_t)$; its constant mean must also vanish. A transient chosen by an arbitrary initial condition is not the stationary process whose ACF is requested.

For this moving-average model, independence of the noise giveswhile $\gamma (h)=0$ for $|h|\ge 2$, since the two noise sets are disjoint. Therefore its autocorrelation function isThe reduced moving-average inverse is $w_t={\sum }_{j\ge 0}0.9^j{x}_{t-j}$, whose coefficients are absolutely summable, confirming invertibility directly.