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For a weakly stationary process with mean $m$ and nonzero variance, put $\gamma (h)=\mathrm{Cov}(X_{t+h},X_t)$ and $\rho (h)=\gamma (h)/\gamma (0)$. The function $\rho$ is its autocorrelation function. For observations $x_1,\dots ,x_N$, one common sample autocorrelation function isA correlogram plots these estimates against lag; the population analogue plots $\rho (h)$.

The lag-$h$ partial autocorrelation function measures the correlation between $X_t$ and $X_{t-h}$ after removing their linear projections on the intervening observations. Its sample counterpart can be calculated by the Yule-Walker equations: solveand take $\hat{\alpha }(h)={\hat{\varphi }}_{h,h}$, assuming the matrix is nonsingular. This defines a standard sample partial autocorrelation function.

For a minimal causal autoregressive model of order $p$, the autocorrelation typically decays geometrically, possibly with damped oscillations, whereas the partial autocorrelation function is zero beyond lag $p$: after those $p$ preceding values have been projected out, the innovation is orthogonal to older values. For a moving-average model of order $q$, nonoverlapping sets of driving white noise give $\gamma (h)=0$ for $|h|>q$; its partial autocorrelation function generally tails off when the representation is invertible. Thus an ACF cutoff suggests an MA order; a PACF cutoff suggests an AR order. An ARMA model generally has neither cutoff. These are population properties; sample noise makes cutoffs approximate, so fitted residuals and uncertainty should also be inspected. Under independent white noise, fixed-lag sample correlations have approximate standard error $N^{-1/2}$, giving a rough diagnostic band rather than a universal simultaneous test.

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.

For a causal autoregressive model $X_t={\sum }_{j=1}^p{\varphi }_j{X}_{t-j}+{ϵ}_t$ with centered white noise of variance ${\sigma }^2$, multiplying by past values and taking expectations yields $\gamma (h)={\sum }_j{\varphi }_j\gamma (h-j)$ for $h\ge 1$, and $\gamma (0)={\sum }_j{\varphi }_j\gamma (j)+{\sigma }^2$. These Yule-Walker equations determine the autocovariance from the coefficients and noise variance, or estimate coefficients from observed autocovariances. Causality supplies the needed orthogonality of the current noise to past observations.