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A causal time series satisfying $\Phi (B)X_t={\epsilon }_t$ with nondegenerate white noise has a square-summable present-and-past filter. The identity $\Phi (z)\Psi (z)=1$ excludes roots inside the unit disk; a pole of $1/\Phi$ on its boundary also prevents square summability. Thus all roots of $\Phi$ lie strictly outside the unit circle.

A weakly stationary process satisfying an autoregressive model can depend on future driving white noise. For $X_t=\varphi X_{t-1}+{\epsilon }_t$ with $|\varphi |>1$, the convergent in the sense of mean-square convergence solution is $X_t=-{\sum }_{j\ge 1}{\varphi }^{-j}{\epsilon }_{t+j}$. Its autocovariance is ${\sigma }^2{\varphi }^{-|h|}/({\varphi }^2-1)$. This is different from a causal time series.

A weakly stationary process has finite second moments, a mean $m=E{X}_t$ independent of $t$, and a covariance depending only on the separation of the observations. Its autocovariance function and autocorrelation function arewhere the autocorrelation requires $\gamma (0)>0$.

Write the autoregressive polynomial as $A(z)=1-\alpha z+\alpha z^2$, so the autoregressive model is $A(B)X_t={ϵ}_t$, with $B$ the backshift operator. It is important to distinguish a weakly stationary process from a causal time series: weak stationarity alone does not require $X_t$ to depend only on present and past white noise.

For a two-sided weakly stationary process, the two-sided stationary inverse of an autoregressive polynomial exists and is unique if $A$ has no root on the unit circle. To find the excluded parameters, set $z=e^{i\lambda }$. The imaginary part of $A(z)=0$ givesThe case $\alpha =0$ is harmless. The possible frequencies are $0,\pi ,\pm \pi /3$. Since $A(1)=1$, $A(-1)=1+2\alpha$ and $A(e^{\pm i\pi /3})=1-\alpha$, the exceptional parameters are $-1/2$ and $1$. Thus, with nondegenerate white noise, the literal two-sided answer isFor completeness, away from those values $1/A(z)$ has an absolutely summable Laurent series on an annulus containing the unit circle. Its bilateral coefficients define an $L^2$-convergent linear filter of a stationary time series applied to ${ϵ}_t$, giving the required solution. Applying the same inverse to the equation gives uniqueness. At an exceptional value, the spectral measure of a stationary time series would have to satisfy $|A(e^{i\lambda }){|}^2{\mu }_X(d\lambda )={\sigma }^{2}d\lambda /(2\pi )$. A zero of $A$ on the unit circle makes the resulting density nonintegrable near that zero, contradicting finite variance.

If the usual additional causal time series convention is intended, the causality root criterion for an autoregressive model requires the roots of $A$ to lie outside the closed unit disk. Equivalently, the roots of $u^2-\alpha u+\alpha$ lie strictly inside it. The real quadratic stability conditions reduce to $|\alpha |<1$ and $1+2\alpha >0$, giving the narrower answerThe printed question does not explicitly impose this extra convention. Both answers have therefore been distinguished. If ${\sigma }^2=0$, the exceptional parameters permit nonunique stationary homogeneous solutions rather than the nonexistence conclusion above.

At $\alpha =-1/12$,Consequently the Wold representation isThe coefficients are absolutely summable, and ${\psi }_0=1$. The past of $X$ lies in the closed linear span of the past white noise; conversely ${ϵ}_t=A(B)X_t$ recovers that noise from present and past $X$. Thus ${ϵ}_t$ is orthogonal to the past of $X$ and is its one-step linear prediction error, as required for the Wold decomposition. There is no deterministic component.

Using orthogonality of the white noise, put $H=|h|$ and calculate the autocovariance by geometric sums:HenceIn particular, $\gamma (0)=66{\sigma }^2/65$ and $\gamma (1)=-6{\sigma }^2/65$; these also satisfy the Yule-Walker equations for this autoregressive model.

A stationary causal time series exists. The infinite moving-average representationconverges in the sense of mean-square convergence because the driving variables are independent random variables and ${\sum }_j{4}^{-j}<\infty$. Shifting the series gives $X_t=\frac{1}{2}{X}_{t-1}+{\epsilon }_t$. Its expected value is zero and its autocovariance isThus it is a weakly stationary process; since the driving variables have a normal distribution, it is also a Gaussian process and a strictly stationary process. Throughout the stationarity discussion, take ${\sigma }^2>0$; degenerate zero noise permits trivial constant solutions.

A unit-root autoregressive process retains shocks permanently, whereas a causal time series with $|\varphi |<1$ reverts towards its mean. Testing the unit root determines whether stationary autoregressive analysis is appropriate or differencing is needed. For the model without an intercept or trend, use the Dickey–Fuller test against the lower-sided alternative $\varphi <1$ near the null. PutThis is the ordinary regression statistic for a zero coefficient when $\Delta X_t$ is regressed on $X_{t-1}$, but its null probability distribution is not the usual Student law. Under the standard unit-root initialization $X_1=o_P(\sqrt{n})$ and innovations independent of the starting value,where $W$ is standard Brownian motion. If $d_{\alpha }$ is the lower $\alpha$-quantile of $D$, the asymptotic level-$\alpha$ critical region isThe deterministic terms and null initialization must match the critical-value table. For exact finite-sample size of a statistical test, calibrate the statistic from its Gaussian random walk null with the specified initial condition and noise scale; for a zero starting value its distribution is scale-free. The printed two-sided recurrence alone specifies neither an initial law nor a universal finite-sample critical value. Ordinary normal quantiles do not give the intended size of a statistical test.

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.

A zero-mean autoregressive process of order one satisfieswhere ${\epsilon }_t$ is white noise with variance ${\sigma }_{\epsilon }^2>0$. A causal time series uses present and past innovations only. For $|\alpha |<1$ the serieshas mean-square convergence because ${\sum }_j{\alpha }^{2j}<\infty$, and it supplies the causal weakly stationary process. Its variance is ${\sigma }_{\epsilon }^2/(1-{\alpha }^2)$. For $h\ge 0$, only the common innovations contribute to the autocovariance:Evenness then givesWhen $\alpha <0$, the signs alternate; when $\alpha =0$, the process is white noise. With nondegenerate innovations, $|\alpha |\ge 1$ does not give a causal finite-variance stationary solution. These covariance calculations need uncorrelated innovations, not Gaussianity; Gaussian innovations additionally yield a Gaussian process.