Complex Roots And Oscillatory AR(2) Dependence

Why Complex Roots Matter

An AR(2) process can have dependence that alternates between relatively high and low values rather than fading smoothly in one direction. This pattern arises when the recurrence has a pair of complex conjugate roots. The roots determine both the pace of the oscillation and how quickly its influence decays.

For an actuarial application, imagine a hypothetical quarterly index of deviations in a portfolio’s claim counts from its long-run mean. An AR(2) model for that index could describe dependence that rises and falls over several quarters. The root calculations characterize that pattern; they do not, by themselves, establish a cause for it.

Definition: Write a centered AR(2) process as \(Y_t=\phi_1Y_{t-1}+\phi_2Y_{t-2}+\varepsilon_t\), where \(Y_t=X_t-\mu\), \(\mu\) is the process mean, and \(\varepsilon_t\) is a zero-mean white-noise innovation with variance \(\sigma^2\). The recurrence roots are the roots \(\lambda\) of \(\lambda^2-\phi_1\lambda-\phi_2=0\). When the roots are a nonreal conjugate pair, write them as \(\lambda=\rho e^{i\theta}\) and \(\bar{\lambda}=\rho e^{-i\theta}\), with \(\rho>0\) and \(0<\theta<\pi\).

Here, white noise means the innovations have constant finite variance and are uncorrelated at distinct times. For the covariance calculations below, assume also that each innovation is uncorrelated with earlier process values. No normality assumption is needed. The complex roots describe an oscillatory stationary process when their modulus satisfies \(\rho<1\).

From the Roots to the AR(2) Coefficients

Expanding the product of the two root factors gives a direct link between the roots and the model coefficients.

(\lambda-\rho e^{i\theta})(\lambda-\rho e^{-i\theta}) =\lambda^2-2\rho\cos(\theta)\lambda+\rho^2 =\lambda^2-\phi_1\lambda-\phi_2.

Matching coefficients yields \(\phi_1=2\rho\cos(\theta)\) and \(\phi_2=-\rho^2\). Thus, for a complex pair, \(\phi_2\) is negative, the modulus is \(\sqrt{-\phi_2}\), and the angle can be recovered using \(\cos(\theta)=\phi_1/(2\rho)\). These relationships are specific to the complex-root case; a real-root model does not have an oscillation angle.

A complex pair occurs when the discriminant \(\phi_1^2+4\phi_2\) is negative. The recurrence roots are then \(\phi_1/2 \pm i\sqrt{-(\phi_1^2+4\phi_2)}/2\). A negative discriminant tells us the roots are complex, but stationarity still depends on their modulus.

How the Roots Produce an Oscillating Autocorrelation

Let \(\gamma_k=\operatorname{Cov}(Y_t,Y_{t-k})\) and \(r_k=\gamma_k/\gamma_0\), where \(\gamma_0>0\). For lags \(k\geq2\), the innovation is uncorrelated with the earlier process value, so taking covariances in the AR(2) equation gives \(\gamma_k=\phi_1\gamma_{k-1}+\phi_2\gamma_{k-2}\). This is the same recurrence as the one whose characteristic equation defines the roots.

For complex roots, the real-valued solution to that recurrence has the form \(r_k=\rho^k(C\cos(k\theta)+D\sin(k\theta))\), for constants \(C\) and \(D\) set by the initial autocorrelations. This explains the two distinct features: \(\theta\) controls the oscillation, while \(\rho^k\) contributes geometric decay.

Result: The nominal oscillation period is \(2\pi/\theta\) time steps, and the angular frequency is \(\theta\) radians per time step. The recurrence roots are both inside the unit circle exactly when \(\rho<1\); this is the stationary case for a causal AR(2) process. If \(\rho=1\), the oscillations do not decay, and if \(\rho>1\), the recurrence solutions grow rather than decay.

The autocorrelation is not generally just \(\rho^k\cos(k\theta)\): its phase depends on the initial values. In particular, the lag-one autocorrelation follows from taking the covariance of the model equation with \(Y_{t-1}\). The innovation is uncorrelated with \(Y_{t-1}\), so \(\gamma_1=\phi_1\gamma_0+\phi_2\gamma_1\), which gives \(r_1=\phi_1/(1-\phi_2)\). In the complex-root case, these initial values give the more specific expression below.

r_k=\rho^k\left(\cos(k\theta)+ \frac{\cos(\theta)(1-\rho^2)} {(1+\rho^2)\sin(\theta)}\sin(k\theta)\right),\qquad k\geq0.

To see why the expression holds, the recurrence has solutions formed from \(\rho^k e^{ik\theta}\) and its conjugate, so its real solutions are combinations of \(\rho^k\cos(k\theta)\) and \(\rho^k\sin(k\theta)\). Setting \(r_0=1\) fixes the cosine coefficient at 1. The lag-one relation gives \(r_1=2\rho\cos(\theta)/(1+\rho^2)\). Substituting \(k=1\) into the proposed expression then fixes the sine coefficient as \(\cos(\theta)(1-\rho^2)/((1+\rho^2)\sin(\theta))\). The recurrence and these two initial values determine all subsequent \(r_k\), proving the expression.

The oscillation’s peaks need not occur at exact multiples of the nominal period because of that phase. The period is a useful description of the root-driven cycle, not a guarantee that every observed peak is separated by exactly that many time steps. A smaller \(\rho\) means faster decay; a \(\rho\) near 1 means the oscillatory pattern can persist over many lags.

Find the recurrence roots.
Solve \(\lambda^2-\phi_1\lambda-\phi_2=0\), and check that the discriminant is negative before interpreting the roots as a complex pair.

Calculate the modulus and angle.
Write one root as \(a+ib\). Then \(\rho=\sqrt{a^2+b^2}\) and \(\theta\) is the angle in \((0,\pi)\) whose cosine is \(a/\rho\).

Interpret the dependence.
If \(\rho<1\), the process is stationary; report the period \(2\pi/\theta\) in the data’s time steps and the decay factor \(\rho\) per step.

Worked Examples

Worked Example: Finding the Cycle in a Quarterly AR(2) Model

A hypothetical quarterly portfolio index, after subtracting its mean, follows \(Y_t=0.6Y_{t-1}-0.16Y_{t-2}+\varepsilon_t\). The innovations have mean zero, variance \(1\), are uncorrelated over time, and are uncorrelated with earlier process values. Find the roots, stationarity conclusion, and nominal period. Calculate the first four autocorrelations to show the pattern. The method is the recurrence-root calculation; it avoids confusing the roots’ angle with a period measured in quarters.

The recurrence equation is \(\lambda^2-0.6\lambda+0.16=0\). Its discriminant is \(0.6^2-4(0.16)=0.36-0.64=-0.28\), so the roots are complex: \(\lambda=0.3\pm i\sqrt{0.28}/2=0.3\pm0.264575i\).

Their modulus is \(\rho=\sqrt{0.3^2+0.264575^2}=\sqrt{0.09+0.07}=0.4\), and \(\cos(\theta)=0.3/0.4=0.75\). Therefore \(\theta=\arccos(0.75)\approx0.722734\) radians per quarter, and the nominal period is \(2\pi/0.722734\approx8.694\) quarters. Since \(0.4<1\), the process is stationary and the oscillation decays geometrically.

For the autocorrelations, \(r_0=1\), and \(r_1=0.6/(1-(-0.16))=0.6/1.16\approx0.517241\). Applying \(r_k=0.6r_{k-1}-0.16r_{k-2}\) gives \(r_2=0.6(0.517241)-0.16(1)=0.150345\), \(r_3=0.6(0.150345)-0.16(0.517241)=0.007448\), and \(r_4=0.6(0.007448)-0.16(0.150345)=-0.019586\). Rounded to six decimals, the values are \(r_1=0.517241\), \(r_2=0.150345\), \(r_3=0.007448\), and \(r_4=-0.019586\). The sign change is consistent with an oscillatory pattern. The modulus, angle, period, and stationarity conclusion agree with the roots and coefficients.

Worked Example: Interpreting a Slowly Decaying Oscillation

A hypothetical quarterly claims-deviation process has model \(Y_t=1.5Y_{t-1}-0.81Y_{t-2}+\varepsilon_t\). Its innovations are zero-mean white noise with variance \(4\), uncorrelated with earlier process values. Find the recurrence-root modulus, angle, and nominal period, then calculate the geometric decay factors at lags 5 and 10. The method is to solve for the roots and interpret their polar form; it avoids treating a long period as evidence that the process is nonstationary.

The recurrence equation is \(\lambda^2-1.5\lambda+0.81=0\). Its discriminant is \(1.5^2-4(0.81)=2.25-3.24=-0.99\), so \(\lambda=0.75\pm i\sqrt{0.99}/2\approx0.75\pm0.497494i\). The modulus is \(\rho=\sqrt{0.75^2+0.497494^2}=\sqrt{0.5625+0.2475}=0.9\). Thus \(\cos(\theta)=0.75/0.9=5/6\), and \(\theta=\arccos(5/6)\approx0.585686\) radians per quarter.

The nominal period is \(2\pi/0.585686\approx10.728\) quarters. Since \(0.9<1\), the roots are inside the unit circle and the process is stationary, although the decay is relatively slow. The geometric factors are \(0.9^5=0.59049\) and \(0.9^{10}=0.3486784401\). These are envelope factors, not exact autocorrelation values: the trigonometric part also affects each lag. The calculations check because \(\phi_1=2(0.9)(5/6)=1.5\) and \(\phi_2=-(0.9)^2=-0.81\).

Worked Example: Building Coefficients From a Desired Cycle

A hypothetical monthly index is to be modeled with recurrence roots of modulus \(0.7\) and angle \(\pi/3\). The innovations are zero-mean white noise with variance \(2\), uncorrelated with earlier process values. Find the AR(2) coefficients, stationarity conclusion, nominal period, and lag-one autocorrelation. The method is to expand the conjugate-root product; it avoids estimating the angle from rounded decimal coefficients when the angle is given exactly.

The roots are \(0.7e^{i\pi/3}\) and \(0.7e^{-i\pi/3}\). From the coefficient relationships, \(\phi_1=2(0.7)\cos(\pi/3)=1.4(0.5)=0.7\), and \(\phi_2=-(0.7)^2=-0.49\). The resulting recurrence is \(Y_t=0.7Y_{t-1}-0.49Y_{t-2}+\varepsilon_t\).

The modulus is \(0.7<1\), so the process is stationary. Its nominal period is \(2\pi/(\pi/3)=6\) months. The lag-one autocorrelation is \(r_1=\phi_1/(1-\phi_2)=0.7/1.49=70/149\approx0.469799\). As a check, the roots have sum \(2(0.7)\cos(\pi/3)=0.7\) and product \(0.7^2=0.49\), matching the recurrence coefficients’ sum and product relationships.

Root Conventions and Common Traps

Some presentations use the roots of the AR characteristic polynomial \(A(z)=1-\phi_1z-\phi_2z^2\), rather than the recurrence roots used above. The two conventions give reciprocal roots. If \(\lambda\) solves the recurrence equation, then \(z=1/\lambda\) solves \(A(z)=0\). Thus, for a stationary AR(2) model, recurrence roots have modulus less than 1, while characteristic-polynomial roots have modulus greater than 1.

Wrong: For the first example, the recurrence roots have modulus \(0.4\), so the process is nonstationary because the modulus is below 1. Correction: That conclusion reverses the criterion. Recurrence roots inside the unit circle imply stationarity; it is the roots of \(A(z)\) that must be outside.
Wrong: Use \(2\pi\theta\) as the period. Correction: The angle is in radians per time step, so the period is \(2\pi/\theta\) time steps.
Wrong: Treat \(\rho^k\) as the exact autocorrelation at lag \(k\). Correction: \(\rho^k\) describes geometric decay; the cosine and sine terms determine the autocorrelation’s value and sign at each lag.
Wrong: Report a period without its time unit. Correction: The unit follows the observations: a period calculated from monthly data is in months, and one calculated from quarterly data is in quarters.

Key takeaway: For a complex recurrence-root pair \(\rho e^{\pm i\theta}\), the modulus controls geometric decay and the angle controls oscillation. Check stationarity with \(\rho<1\), calculate the nominal period as \(2\pi/\theta\), and keep recurrence-root and characteristic-polynomial conventions distinct.

Check Your Understanding

For each problem, identify the relevant root quantities and interpret them in the stated time unit.

A quarterly AR(2) model has \(\phi_1=0.8\) and \(\phi_2=-0.25\). Find the root modulus, angle, nominal period, and stationarity conclusion. (Answer: \(\rho=0.5\), \(\theta=\arccos(0.8)\approx0.6435\), period \(\approx9.764\) quarters; stationary.)
A monthly model is specified by recurrence roots \(0.6e^{\pm i\pi/4}\). Find \(\phi_1,\phi_2\), the period, and the stationarity conclusion. (Answer: \(\phi_1=0.6\sqrt{2}\approx0.8485\), \(\phi_2=-0.36\), period \(8\) months; stationary.)
A quarterly AR(2) model has \(\phi_1=1.2\) and \(\phi_2=-0.64\). Find its modulus, angle, and nominal period, and determine stationarity. (Answer: \(\rho=0.8\), \(\cos(\theta)=0.75\), \(\theta\approx0.7227\), period \(\approx8.694\) quarters; stationary.)
A monthly model has recurrence roots \(0.5e^{\pm i\pi/3}\). Find its coefficients, nominal period, and the geometric envelope factor at lag 4. (Answer: \(\phi_1=0.5\), \(\phi_2=-0.25\), period \(6\) months, envelope factor \(0.5^4=0.0625\); stationary.)
A quarterly AR(2) model has \(\phi_1=1.4\) and \(\phi_2=-0.64\). Find the root modulus, the cosine of the angle, the nominal period, and the stationarity conclusion. (Answer: \(\rho=0.8\), \(\cos(\theta)=7/8\), period \(2\pi/\arccos(7/8)\) or approximately \(12.431\) quarters; stationary.)

What You'll Learn