= Solution
Apply $1-\phi B$ to the observed process and call the result $V_t$. Then
$$
V_t=Z_t+W_t-\phi W_{t-1}.
$$
Its only nonzero <covariance> lags are
$$
A=\gamma_V(0)=\sigma_z^2+(1+\phi^2)\sigma_w^2,\qquad C=\gamma_V(1)=-\phi\sigma_w^2.
$$
Seek an invertible moving-average factor $V_t=\varepsilon_t+\vartheta\varepsilon_{t-1}$ with $\operatorname{Var}(\varepsilon_t)=\nu$. Matching these covariances requires $\nu(1+\vartheta^2)=A$ and $\nu\vartheta=C$. Solving gives
$$
\boxed{\nu=\frac{A+\sqrt{A^2-4\phi^2\sigma_w^4}}2,\qquad
\vartheta=-\frac{\phi\sigma_w^2}{\nu},\qquad\phi_{\rm AR}=\phi.}
$$
These are the three requested parameters in the positive-sign moving-average convention. For nonzero total noise <variance>,
$$
A-2|C|=\sigma_z^2+(1-|\phi|)^2\sigma_w^2>0,
$$
so $|\vartheta|<1$ and the larger quadratic root gives the invertible factor.
<Covariance> matching alone would not identify arbitrary processes in distribution. To obtain an actual representation on the given space, define
$$
\varepsilon_t=(1+\vartheta B)^{-1}V_t=\sum_{j\geq0}(-\vartheta)^jV_{t-j}.
$$
The series converges in L2. The spectrum of $V$ is $(A+2C\cos\omega)/(2\pi)=\nu|1+\vartheta e^{-i\omega}|^2/(2\pi)$, so this filtered process has constant spectrum $\nu/(2\pi)$ and is <white noise>. Thus
$$
\boxed{(1-\phi B)Y_t=(1+\vartheta B)\varepsilon_t,\qquad\varepsilon\sim\operatorname{WN}(0,\nu).}
$$
This is the <invertible ARMA factorization of an AR(1)-plus-noise process>. If $\sigma_w^2=0$, it reduces to AR(1); if $\phi=0$, it reduces to <white noise>. If $\sigma_z^2=0$ and $\sigma_w^2>0$, then $\vartheta=-\phi$, the common factor cancels and $Y=W$. Therefore the orders are at most (1,1); no unnecessarily minimal-order claim is made in those degenerate cases.
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