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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.