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The autoregressive polynomial is $\Phi (z)=1-{\sum }_{j=1}^p{\varphi }_j{z}^j$. Substituting the backshift operator gives the autoregressive filter. Zeros outside the unit disk give the causality root criterion for an autoregressive model; zeros on the unit circle obstruct a nondegenerate stationary innovation-driven solution.

The autoregressive polynomial factors as $(1-0.4z)(1+0.7z)$, with zerosBoth have modulus greater than one. The causality root criterion for an autoregressive model therefore gives a causal stationary solution. The moving-average polynomial has its only zero at $-2$, also outside the unit circle, so the invertibility of a moving-average model holds. There is no common root to cancel.

To use unit-variance white noise, put $W_t=Z_t/2$. One suitable pair isThen $\tilde{\varphi }(B)X_t=\tilde{\theta }(B)W_t$ with $W\sim \mathrm{WN}(0,1)$. The factor two changes the innovation scale, not the zero of the moving-average polynomial.

For a stationary linear autoregression, causal time series means that the observation uses only current and past driving noise:This is a convergent in the sense of mean-square convergence infinite moving-average representation. The stronger usual stable-filter definition requires absolute summability; the argument below also handles the square-summable definition. Let $\Phi (z)=1-{\sum }_{k=1}^p{\varphi }_k{z}^k$ and $\Psi (z)={\sum }_{j\ge 0}{\psi }_j{z}^j$. Substituting the filter into the recurrence and comparing coefficients of the orthogonal noise gives ${\psi }_0=1$ and ${\psi }_j={\sum }_{k=1}^{\min (p,j)}{\varphi }_k{\psi }_{j-k}$. HenceThe Cauchy-Schwarz inequality ensures that $\Psi$ is an analytic function in this disk. Thus $\Phi$ has no zero strictly inside it. A boundary zero is also impossible: a zero of multiplicity $m\ge 1$ at $z_0=e^{i{\omega }_0}$ makes $|1/\Phi (r{e}^{i\omega }){|}^2$ at least a constant times $[(1-r{)}^2+(\omega -{\omega }_0{)}^2{]}^{-m}$ near that point. Its integral over $\omega$ diverges as $r\uparrow 1$. On the other hand, orthogonality of complex exponentials givesa contradiction. Therefore the causality root criterion for an autoregressive model isUnder absolute summability, the shorter boundary argument is continuity of $\Psi$ on the closed disk and the identity $\Phi \Psi =1$ there.

By the causality root criterion for an autoregressive model, choose $r>1$ smaller than the modulus of every root of $\Phi$. The function $\Psi =1/\Phi$ is an analytic function on and inside $|z|=r$. Writing $M={\max }_{|z|=r}|1/\Phi (z)|$, the Cauchy estimate gives $|{\psi }_j|\le M{r}^{-j}$. For $h\ge 0$, independence of the noise in the infinite moving-average representation givesThe autocovariance is symmetric in the lag. Thus exponential decay holds withThis is exponential autocovariance decay of a causal autoregression. If $\Phi$ is constant, there are no roots and any $r>1$ works.