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An autoregressive process of order one obeyswhere $\varphi$ is real and ${\epsilon }_t$ is centered white noise of variance ${\sigma }^2>0$. The zero-mean version is $X_t=\varphi X_{t-1}+{\epsilon }_t$; in the affine form $X_t=c+\varphi X_{t-1}+{\epsilon }_t$, the stationary mean satisfies $c=(1-\varphi )\mu$.

For $|\varphi |<1$, its unique weakly stationary process solution on all integer times is the causal time-series representationThe sum converges in the sense of mean-square convergence because ${\sum }_{j\ge 0}{\varphi }^{2j}=1/(1-{\varphi }^2)<\infty$. It is an infinite moving-average representation using only present and past driving noise. Weak white noise suffices for this second-order conclusion; strong white noise also gives strict stationarity. The autocovariance of an AR(1) process is ${\sigma }^2{\varphi }^{|h|}/(1-{\varphi }^2)$, so its variance is ${\sigma }^2/(1-{\varphi }^2)$ and its ACF is ${\varphi }^{|h|}$. See Bartlett, lecture 5: AR(1) representation and causality.

The appeal of an AR(1) is that one coefficient expresses how a deviation persists. For $0<\varphi <1$, disturbances propagate with weights $1,\varphi ,{\varphi }^2,\dots$. For $-1<\varphi <0$, these weights alternate sign. With $\mu =10$, $\varphi =0.7$, and $X_t=12$, the recursive forecasts of an autoregressive process give $11.4$ and $10.98$ one and two steps ahead. If ${\sigma }^2=1$, the stationary variance is $100/51\simeq 1.9608$, while the one-step forecast error variance is only $1$. Under weak white noise these are best linear forecasts; independent driving noise also makes them conditional expectations given the observed past.

The condition $|\varphi |<1$ describes a stationary causal solution, not every finite recursion with those coefficients. Starting from a prescribed value gives the transient in finite initialization of an AR(1) process. Nor is it a condition for all possible two-sided stationary solutions: when $|\varphi |>1$, noncausal stationary autoregression uses future driving noise. See stationary versus causal solution of a two-sided AR(1) equation for the boundary cases.

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 noncausal stationary autoregression exists. On the two-sided time axis defineThis series converges in the sense of mean-square convergence, and direct subtraction gives $X_t-2X_{t-1}={\epsilon }_t$. The expected value and autocovariance areConsequently this is a weakly stationary process and, by Gaussianity, a strictly stationary process. It is an example of noncausal stationary autoregression, since $X_t$ uses future noise. In particular, $X_{t-1}$ is correlated with ${\epsilon }_t$, so the usual causal variance recursion is inapplicable. If an additional assumption required the driving noise to be independent of past observations, this solution would be excluded; that assumption is not stated here.