Real Roots And AR(2) Dependence Patterns

Why Real Roots Matter

An AR(2) process can have dependence that fades at two different rates. When its decay roots are distinct and real, its autocorrelation function is a weighted sum of two exponential sequences. The root with the larger absolute value decays more slowly and often determines the pattern at long lags. Knowing both roots helps distinguish a gradual tail from a faster decline, and it prevents a common mistake: treating the roots of the AR polynomial as the decay rates themselves.

This tutorial considers a stationary AR(2) model with distinct real decay roots. Complex roots produce a different, oscillatory pattern and are outside the scope here.

Definition: Write a centered AR(2) process as \(Z_t=X_t-\mu\), where

Z_t=\phi_1 Z_{t-1}+\phi_2 Z_{t-2}+\varepsilon_t.

Here \(\mu=E[X_t]\), and \(\varepsilon_t\) is a zero-mean white-noise innovation with finite variance that is uncorrelated with past values of the process. The decay roots are the solutions \(r\) of

$$ r^2-\phi_1 r-\phi_2=0. $$ For a stationary causal AR(2), these roots satisfy \(|r|<1\). They are reciprocals of the roots of the AR polynomial \(1-\phi_1 z-\phi_2 z^2\).

The distinction between the two kinds of roots is important. The decay roots solve the recurrence for autocorrelations. The roots of the AR polynomial are commonly used to state the stationarity condition, and must lie outside the unit circle for a causal stationary process. The two conventions describe the same condition because the roots are reciprocals. The earlier material on AR(2) stationarity establishes when the model is stationary; here the focus is how its distinct real decay roots shape dependence.

From AR(2) Coefficients to Autocorrelation

Let \(\gamma_k=\operatorname{Cov}(Z_t,Z_{t-k})\), and let \(\rho_k=\gamma_k/\gamma_0\) be the autocorrelation at lag \(k\). For a stationary process, \(\rho_0=1\). Since innovations are uncorrelated with past process values, taking covariance with \(Z_{t-k}\) gives, for \(k\geq 2\),

\gamma_k=\phi_1\gamma_{k-1}+\phi_2\gamma_{k-2}, \qquad \rho_k=\phi_1\rho_{k-1}+\phi_2\rho_{k-2}.

At lag 1, covariance with \(Z_{t-1}\) gives \(\gamma_1=\phi_1\gamma_0+\phi_2\gamma_1\). Rearranging and dividing by \(\gamma_0\) yields

\rho_1=\frac{\phi_1}{1-\phi_2},

provided \(1-\phi_2\ne 0\), as holds for a stationary AR(2). This initial value, together with the recurrence and \(\rho_0=1\), determines the autocorrelation sequence.

Theorem: If the decay roots \(r_1\) and \(r_2\) are distinct and real, then the AR(2) autocorrelation function has the form

\rho_k=c_1 r_1^k+c_2 r_2^k,\qquad k\geq 0,

where

c_1=\frac{\rho_1-r_2}{r_1-r_2}, \qquad c_2=\frac{r_1-\rho_1}{r_1-r_2}, \qquad \rho_1=\frac{\phi_1}{1-\phi_2}.

The coefficients satisfy \(c_1+c_2=1\), so the formula gives \(\rho_0=1\).

Why the Two-Exponential Formula Holds

The recurrence determines each later autocorrelation from the two preceding values. To see why powers of the decay roots solve it, try a sequence of the form \(\rho_k=r^k\). Substitution into the recurrence gives \(r^k=\phi_1r^{k-1}+\phi_2r^{k-2}\). Dividing by \(r^{k-2}\), when \(r\ne 0\), gives \(r^2-\phi_1r-\phi_2=0\). Thus each root generates a sequence satisfying the recurrence, and any linear combination of those sequences also satisfies it.

For distinct roots, choose the coefficients to match the two initial conditions. From \(c_1+c_2=1\) and \(c_1r_1+c_2r_2=\rho_1\), subtracting \(r_2(c_1+c_2)=r_2\) from the second equation gives \(c_1(r_1-r_2)=\rho_1-r_2\). This gives the stated expression for \(c_1\); subtracting from the sum gives \(c_2\). The recurrence and the initial values are therefore satisfied, so the resulting sequence is the AR(2) autocorrelation function.

The coefficients are not necessarily both positive. Therefore “weighted sum” means a linear combination, not necessarily a probability mixture or a convex average. A negative coefficient can affect the early-lag shape even when both roots are positive.

Worked Examples

Worked Example: Two Positive Decay Rates

A centered stationary AR(2) model has \(\phi_1=0.80\) and \(\phi_2=-0.15\). Find its decay roots and autocorrelation at lags 1 and 2, and identify its long-lag decay rate.

Method and stationarity check. Solve the decay-root equation \(r^2-0.80r+0.15=0\). It factors as \((r-0.50)(r-0.30)=0\), so the distinct roots are \(r_1=0.50\) and \(r_2=0.30\). Both have absolute value less than 1, consistent with stationarity.

Initial autocorrelation. Using the AR(2) lag-1 equation, \(\rho_1=0.80/[1-(-0.15)]=0.80/1.15=16/23\), or approximately \(0.695652\).

Coefficients. Assign \(r_1=0.50\) and \(r_2=0.30\). Then \(c_1=(16/23-0.30)/(0.50-0.30)=91/46\), and \(c_2=(0.50-16/23)/(0.50-0.30)=-45/46\). Their sum is \(46/46=1\), as required.

Lag 2 and check. The two-root formula gives \(\rho_2=(91/46)(0.50)^2-(45/46)(0.30)^2=(22.75-4.05)/46=0.406522\) (to six decimals). The recurrence independently gives \(\rho_2=0.80(16/23)-0.15=0.406522\), agreeing. The slower component is \(0.50^k\), because \(0.50\) has the larger absolute value. The negative coefficient on the \(0.30^k\) component is a reminder that the coefficients need not be positive.

Common mistake avoided: Assuming the two coefficients must be positive because both roots are positive. The computed \(c_2=-45/46\) shows why the initial autocorrelation must be used to find the coefficients rather than inferred from the root signs.

Worked Example: A Negative Root and an Alternating Component

Consider a centered stationary AR(2) model with \(\phi_1=0.50\) and \(\phi_2=0.24\). Find its decay roots, write its autocorrelation function, and calculate \(\rho_2\).

Find and check the roots. The equation is \(r^2-0.50r-0.24=0\), which factors as \((r-0.80)(r+0.30)=0\). Thus \(r_1=0.80\) and \(r_2=-0.30\). Both absolute values are below 1.

Find the initial value and coefficients. The lag-1 autocorrelation is \(\rho_1=0.50/(1-0.24)=0.50/0.76=25/38\). Therefore \(c_1=(25/38+0.30)/(0.80+0.30)=182/209\), and \(c_2=(0.80-25/38)/(0.80+0.30)=27/209\). The coefficients sum to \(209/209=1\).

Write the sequence and calculate lag 2. For \(k\geq0\), \(\rho_k=(182/209)(0.80)^k+(27/209)(-0.30)^k\). At lag 2, \(\rho_2=(182/209)(0.64)+(27/209)(0.09)=118.91/209=0.568947\) (to six decimals). The recurrence check gives \(\rho_2=0.50(25/38)+0.24=0.568947\).

The negative root contributes an alternating term: its sign changes with each increase in \(k\), while its magnitude shrinks by a factor of \(0.30\) at each step. The \(0.80^k\) component decays more slowly and dominates at long lags. This example illustrates an alternating component without requiring the overall autocorrelation to alternate in sign.

Common mistake avoided: Concluding that any negative decay root forces the full autocorrelation function to alternate in sign. Here the positive, slower-decaying component is larger, and \(\rho_1\) and \(\rho_2\) are both positive.

Worked Example: Comparing the Dominant Decay Rate

A second centered stationary AR(2) model has \(\phi_1=1.10\) and \(\phi_2=-0.28\). Find its two-root representation and compare its long-lag decay with the first worked example.

Roots and stationarity. The decay equation \(r^2-1.10r+0.28=0\) factors as \((r-0.70)(r-0.40)=0\). The roots are \(0.70\) and \(0.40\), both strictly within the unit circle.

Initial autocorrelation. \(\rho_1=1.10/[1-(-0.28)]=1.10/1.28=55/64=0.859375\).

Coefficients and lag 2. With \(r_1=0.70\) and \(r_2=0.40\), \(c_1=(55/64-0.40)/(0.70-0.40)=49/32\) and \(c_2=(0.70-55/64)/(0.70-0.40)=-17/32\). The coefficients sum to \(32/32=1\). Thus \(\rho_k=(49/32)(0.70)^k-(17/32)(0.40)^k\). At lag 2 this is \((49/32)(0.49)-(17/32)(0.16)=0.6653125\). The recurrence check is \(1.10(0.859375)-0.28=0.6653125\).

Compare rates. The first worked example has dominant root \(0.50\), whereas this model has dominant root \(0.70\). Since \(0.70^k\) declines more slowly than \(0.50^k\), this second model has the slower asymptotic decay, assuming the coefficient on its dominant component is nonzero. Here that coefficient is \(49/32\), so it is nonzero.

Common mistake avoided: Comparing only the lag-1 autocorrelations. A larger \(\rho_1\) may suggest stronger short-lag dependence, but the dominant decay root identifies the slowest exponential rate in the long-lag tail.

Interpreting the Dependence Pattern

For distinct real roots inside the unit circle, each component \(r_j^k\) tends to zero as \(k\) increases. The magnitude \(|r_j|\) controls how quickly that component fades: a root close to 1 in absolute value gives slow decay, while a root near 0 gives rapid decay. If the coefficient for the largest-magnitude root is nonzero, that root governs the asymptotic rate. If its coefficient is zero, the other component governs instead.

Two positive roots: both exponential components retain their signs across lags. The full autocorrelation need not be a simple monotone decline, because one coefficient can be negative.
A negative root: its component changes sign from one lag to the next. Whether the full autocorrelation changes sign depends on the sizes and signs of both components.
Two negative roots: both components alternate in sign; their relative sizes and coefficients still determine the resulting sequence.

A distinct-root formula does not apply when the decay roots coincide: the recurrence then has a different form involving a factor of \(k\). Nor should this real-root formula be used for complex roots. The useful diagnostic here is simple: first solve the decay-root equation, check that the roots are distinct and real, and then use the two initial conditions to calculate the coefficients.

A Root-Convention Trap

For \(\phi_1=0.80\) and \(\phi_2=-0.15\), the decay equation has roots \(0.50\) and \(0.30\). The AR polynomial is \(1-0.80z+0.15z^2\), whose roots are \(2\) and \(10/3\). These are reciprocals of the decay roots.

Wrong calculation: Use \(2^k\) and \((10/3)^k\) as the autocorrelation decay components. These terms grow rather than decay, contradicting stationarity.

Correction: Use the roots \(0.50\) and \(0.30\) of \(r^2-\phi_1r-\phi_2=0\) for the recurrence solution. Equivalently, take reciprocals of the AR-polynomial roots. The resulting components tend to zero, as a stationary autocorrelation tail should.

Key takeaway: In a stationary AR(2) with distinct real decay roots, the autocorrelation is a linear combination of two powers. The root magnitudes determine the component decay rates; the initial autocorrelation determines their coefficients. Keep decay roots separate from AR-polynomial roots, and do not assume the coefficients or the full autocorrelation must have a particular sign.

Check Your Understanding

For each problem, use the decay-root equation and the two initial conditions where needed.

A centered AR(2) has \(\phi_1=0.60\) and \(\phi_2=-0.08\). Find its decay roots and state whether both satisfy the stationary decay-root condition. (Answer: roots \(0.40\) and \(0.20\); both have absolute value less than 1.)
A centered AR(2) has \(\phi_1=0.50\) and \(\phi_2=0.24\). What is its lag-1 autocorrelation, and which root determines the slowest decay? (Answer: \(\rho_1=0.50/0.76=25/38\approx0.6579\); the root \(0.80\).)
For the model with \(\phi_1=0.80,\ \phi_2=-0.15\), calculate \(\rho_2\) using the recurrence. (Answer: \(\rho_2=0.80(16/23)-0.15=0.406522\), to six decimals.)
A stationary AR(2) has distinct real decay roots \(0.60\) and \(-0.20\), and \(\rho_1=0.50\). Find \(c_1,c_2\), assigning \(r_1=0.60\) and \(r_2=-0.20\). (Answer: \(c_1=(0.50+0.20)/0.80=0.875\), \(c_2=0.125\).)
Two stationary AR(2) models have distinct real decay roots \((0.45,0.25)\) and \((0.72,0.10)\). Assuming the coefficient on each largest-magnitude root is nonzero, which has slower long-lag decay, and why? (Answer: The model with roots \((0.72,0.10)\), because its dominant root has larger absolute value, \(0.72>0.45\).)

What You'll Learn