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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.

Interpret stationarity in the usual second-order time-series sense and assume nondegenerate noise, ${\sigma }_z^2>0$. For a two-sided autoregressive equation, the missing existence condition isIt is important to separate this from causality. If $|\varphi |<1$, the unique stationary solution is $X_t={\sum }_{j\ge 0}{\varphi }^j{Z}_{t-j}$. If $|\varphi |>1$, there is still a stationary solution, but it is anticausal:Both expansions converge in L2 because their coefficients are square summable. Substitution verifies the equation. Their means are zero and their covariance functions depend only on lag. Uniqueness follows by iterating the equation backward in the first case and forward in the second: the remainders ${\varphi }^n{X}_{t-n}$ or ${\varphi }^{-n}{X}_{t+n}$ tend to zero in L2 for any stationary finite-variance solution. This is the stationary versus causal solution of a two-sided AR(1) equation.

For $\varphi =\pm 1$, iteration givesThe variance of the right side is $n{\sigma }_z^2$. The variance of the left side is at most $4\mathrm{Var}(X_t)$ by stationarity and Cauchy-Schwarz inequality. These are incompatible as $n\to \infty$. Thus no weakly stationary finite-variance solution exists at those unit roots.

If the intended claim includes a causal innovation representation, its condition is instead $|\varphi |<1$, as in the next part. The stated white noise equation alone does not say that $Z_t$ is orthogonal to the past of $X$. If zero innovation variance is allowed, the unit-root exclusion has degenerate exceptions, such as random constant solutions when $\varphi =1$; the nondegenerate convention is necessary for the asserted nonexistence.