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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.

Let $\Phi (z)=1-{\varphi }_{1}z-\cdots -{\varphi }_p{z}^p$. The causality root criterion for an autoregressive model,is a sufficient condition for a weakly stationary process solution, and gives the causal solution $X_t={\sum }_{j\ge 0}{\psi }_j{\epsilon }_{t-j}$. Indeed $1/\Phi$ is analytic in a disk of radius greater than one, so its Taylor coefficients ${\psi }_j$ decay geometrically. The series therefore has mean-square convergence and defines a linear filter of a stationary time series; multiplying by $\Phi$ verifies the equation.

For the specified order-two model, put $\phi =(1+\sqrt{5})/2$. Since $\phi -{\phi }^{-1}=1$,When $\alpha \ne 0$, its roots are $1/(\alpha \phi )$ and $-\phi /\alpha$. Both have modulus greater than one exactly when $|\alpha |<1/\phi$. The case $\alpha =0$ is white noise and satisfies the same inequality. ConsequentlyThis is the causal stationarity condition selected above. Without causality, a two-sided autoregressive model can also have a noncausal stationary autoregression when its polynomial has no unit-circle zeros. Thus the inequality is not a necessary condition for all conceivable weak stationarity solutions; the causal interpretation is the one required by the subsequent innovations and forecasting questions.

To find the Wold representation, expand $1/\Phi (z)={\sum }_{j\ge 0}{\psi }_j{z}^j$. Matching coefficients gives ${\psi }_0=1$, ${\psi }_1=\alpha$ and ${\psi }_j=\alpha {\psi }_{j-1}+{\alpha }^2{\psi }_{j-2}$. HenceHere $F_1=F_2=1$ are the Fibonacci numbers. The coefficient formula also follows by partial fractions in the displayed factorization. Absolute summability follows from $|\alpha \phi |<1$ and $|\alpha /\phi |<1$. The past observation span equals the past noise span, because the causal series gives one inclusion and ${\epsilon }_t=X_t-\alpha X_{t-1}-{\alpha }^2{X}_{t-2}$ gives the reverse inclusion. Thus these noises are the linear innovations, and there is no additional deterministic component. This is the Fibonacci-coefficient autoregression.

The best linear prediction from an infinite past is the orthogonal projection onto the closed span of $X_T,X_{T-1},\dots$. Future white noise is orthogonal to that span. Substitution in the recursion therefore givesThe respective errors are ${\epsilon }_{T+1}$ and $\alpha {\epsilon }_{T+1}+{\epsilon }_{T+2}$, so their mean squared errors are ${\sigma }^2$ and $(1+{\alpha }^2){\sigma }^2$. No normal distribution assumption is needed for these linear forecasts.

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.

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.

If an autoregressive model has white noise of positive variance, a unique weakly stationary process solving its equation on all integer times exists precisely when its autoregressive polynomial has no root on the unit circle. In that case $1/A(z)$ has an absolutely summable Laurent series on an annulus containing the unit circle; its coefficients define a bilateral linear filter of a stationary time series. Applying that inverse also proves uniqueness among bounded second-moment solutions. A root on the unit circle forces the candidate spectral density of a stationary process to have a nonintegrable singularity. The stronger causality root criterion for an autoregressive model requires every root outside the closed unit disk.