Characteristic Roots And AR(p) Stationarity

Why Characteristic Roots Matter

An autoregressive model describes a series partly through its own past values. The coefficients alone do not always make it obvious whether a shock to the series will fade, persist, or grow. The roots of the model’s characteristic polynomial provide a systematic way to answer that question: for a causal AR model, stationarity requires every root of the lag polynomial to lie outside the unit circle.

The same model can be written using two different root conventions. Roots of the lag polynomial must be outside the unit circle, while roots of the recurrence equation must be inside it. These statements agree because the two sets of roots are reciprocals. Stating which polynomial is being solved prevents a frequent sign and direction error.

Definition: An AR(p) model with mean $\mu$ is $$ X_t-\mu=\phi_1(X_{t-1}-\mu)+\cdots+\phi_p(X_{t-p}-\mu)+\varepsilon_t, $$ where $\phi_p\ne 0$, and $\varepsilon_t$ is white noise with mean zero and finite variance $\sigma_\varepsilon^2>0$. The lag polynomial is $$ \phi(B)=1-\phi_1B-\phi_2B^2-\cdots-\phi_pB^p, $$ where $B X_t=X_{t-1}$. Its characteristic roots are the complex solutions $z$ of $\phi(z)=0$. The causal stationarity criterion is that every such root satisfy $|z|>1$.

Here, “causal” means that the current value can be represented using current and past innovations, not future ones. A weakly stationary process has a constant finite mean, a finite variance that does not depend on time, and autocovariances that depend only on the time separation. For the standard causal AR(p) model with nondegenerate white-noise innovations, the outside-the-unit-circle criterion is the condition for a stationary causal solution.

How the Root Criterion Works

The AR equation can be written $\phi(B)(X_t-\mu)=\varepsilon_t$. If it can be inverted as a stable causal filter, the centered series is a weighted sum of present and past innovations. Those weights must decay sufficiently for the resulting process to have a finite, time-invariant variance.

Write the lag polynomial.
Put the model in centered form and collect its lag terms as $\phi(B)=1-\phi_1B-\cdots-\phi_pB^p$.

Solve the polynomial equation.
Find all complex values $z$ for which $\phi(z)=0$, counting repeated roots according to their multiplicity.

Compare each root’s modulus with one.
The model has a causal stationary solution only if every root has modulus strictly greater than one.

Check the convention.
If instead using roots of the recurrence equation, take reciprocals: those roots must have modulus strictly less than one.

Why Roots Outside the Circle Give Decaying Effects

A short derivation makes the direction of the criterion clear. Suppose the lag polynomial factors over the complex numbers as $\phi(B)=\prod_{j=1}^p(1-B/z_j)$, with roots $z_j$ counted with multiplicity. When every $|z_j|>1$, each factor has a convergent geometric-series inverse for lags on or inside the unit circle. For a single factor,

\frac{1}{1-B/z_j}=\sum_{k=0}^{\infty}\frac{B^k}{z_j^k},\qquad |z_j|>1.

The coefficient on lag $k$ therefore decays geometrically in magnitude. Multiplying the inverse factors gives a causal filter $\psi(B)=1/\phi(B)=\sum_{k=0}^{\infty}\psi_kB^k$. With distinct roots outside the circle, its coefficients decay at a geometric rate; repeated roots can add polynomial factors in $k$, but the coefficients still decay fast enough to be square-summable. Thus, for white noise with finite variance,

X_t-\mu=\sum_{k=0}^{\infty}\psi_k\varepsilon_{t-k}, \qquad \operatorname{Var}(X_t)=\sigma_\varepsilon^2\sum_{k=0}^{\infty}\psi_k^2<\infty.

The weights depend on the lag $k$, not on calendar time $t$, so this representation gives a constant mean and a covariance structure that depends only on lag. Conversely, if a lag-polynomial root is inside the unit circle, the corresponding causal response includes a growing component rather than a decaying one; a root on the unit circle prevents the response from decaying geometrically. Neither case yields a stable causal inverse with finite-variance shock effects. This is why the condition is strict: $|z_j|=1$ does not pass.

This argument concerns the causal solution. In models with roots inside the unit circle, a stationary solution may sometimes be constructed using future innovations, but that is not the standard causal AR model. The root test here should therefore be applied with the causal interpretation stated.

Worked Applications

Worked Example: Factoring an AR(3) Lag Polynomial

A hypothetical monthly claims-index deviation follows $X_t-10=\frac{1}{3}(X_{t-1}-10)+\frac{1}{4}(X_{t-2}-10)-\frac{1}{12}(X_{t-3}-10)+\varepsilon_t$. Assume $\varepsilon_t$ are independent $N(0,4)$ innovations. Determine whether the model has a causal stationary solution. Name the method and identify a likely mistake.

Method. Form $\phi(B)$, factor it, and compare the modulus of every lag-polynomial root with one. The mean is $\mu=10$; the innovation mean is zero and its variance is $4$.

Form the polynomial. The coefficients are $\phi_1=\frac{1}{3}$, $\phi_2=\frac{1}{4}$, and $\phi_3=-\frac{1}{12}$, so

\phi(B)=1-\frac{1}{3}B-\frac{1}{4}B^2+\frac{1}{12}B^3.

Factor and solve. Multiplication gives $(1-\frac{1}{2}B)(1-\frac{1}{3}B)(1+\frac{1}{2}B)$: the quadratic product of the first and third factors is $1-\frac{1}{4}B^2$, and multiplying by $1-\frac{1}{3}B$ gives $1-\frac{1}{3}B-\frac{1}{4}B^2+\frac{1}{12}B^3$, as required. Setting each factor to zero gives

B=2,\qquad B=3,\qquad B=-2.

Their moduli are $2,3,2$, each greater than one. The model therefore has a causal stationary solution. The method avoids confusing the lag-polynomial roots with recurrence roots, which would have reciprocal values inside the unit circle.

Check. Every factor has constant term one, and their product reproduces the original polynomial, including the positive cubic coefficient. All three roots pass the strict modulus test.

Worked Example: One Root Inside the Unit Circle

A hypothetical quarterly reserve-index deviation follows $X_t-6=2.5(X_{t-1}-6)-1.0(X_{t-2}-6)+\varepsilon_t$, with independent $N(0,1)$ innovations. Does it have a causal stationary solution? State the method and the mistake to avoid.

Method. Construct the lag polynomial and factor it before deciding. The mean is $6$, and the innovation distribution has mean zero and variance $1$.

The lag polynomial is $\phi(B)=1-2.5B+B^2$. Its factorization is

\phi(B)=(1-2B)(1-0.5B) =1-2.5B+B^2.

Setting the factors to zero gives $B=0.5$ and $B=2$. Their moduli are $0.5$ and $2$, so one root is inside the unit circle. The model fails the causal stationarity criterion.

The failure is also visible in the inverse filter: the factor $1/(1-2B)$ has coefficients $1,2,4,8,\ldots$, which grow rather than decay. The likely mistake is to pass the model because one of its roots is outside the circle. Every root must be outside, not merely at least one. The factorization reproduces all three coefficients of the polynomial, confirming the calculation.

Worked Example: Translating to Recurrence Roots

A hypothetical monthly portfolio-index deviation follows $X_t-4=0.6(X_{t-1}-4)-0.08(X_{t-2}-4)+\varepsilon_t$, where $\varepsilon_t$ are independent $N(0,9)$ innovations. Check stationarity using both root conventions.

Method. First find lag-polynomial roots, then form the recurrence equation to verify the reciprocal relationship. The mean is $4$; the innovations have mean zero and variance $9$.

The lag polynomial is $\phi(B)=1-0.6B+0.08B^2$. Factoring gives

\phi(B)=(1-0.2B)(1-0.4B), \qquad B=5,\;2.5.

Both lag-polynomial roots have modulus greater than one, so the model is causal stationary. For the recurrence convention, substitute a trial deviation $m_t=\lambda^t$ into the homogeneous equation $m_t=0.6m_{t-1}-0.08m_{t-2}$. Dividing by $\lambda^{t-2}$ gives

\lambda^2-0.6\lambda+0.08=0 =(\lambda-0.2)(\lambda-0.4).

The recurrence roots are $0.2$ and $0.4$, the reciprocals of $5$ and $2.5$. Both are inside the unit circle, giving the same conclusion under the alternative convention. The common mistake is to apply the outside-circle rule to these recurrence roots. The rule applies outside to roots of $\phi(B)$, and inside to recurrence roots.

Boundary Cases and Exam Traps

The modulus comparison is about distance from zero in the complex plane: for a complex root $z=a+bi$, its modulus is $|z|=\sqrt{a^2+b^2}$. A nonreal root and its conjugate have the same modulus, so both pass or fail together. A root of modulus exactly one is on the unit circle and fails the causal stationarity criterion.

A useful second convention begins with the homogeneous recurrence for deviations from the mean. Substituting $m_t=\lambda^t$ into the AR(p) recurrence gives

\lambda^p-\phi_1\lambda^{p-1}-\cdots-\phi_p=0.

If $z\ne0$ solves $\phi(z)=0$, then $\lambda=1/z$ solves this recurrence equation. Thus $|z|>1$ is equivalent to $|\lambda|<1$. Always identify the polynomial before deciding which inequality to use.

Using the wrong direction. Outside applies to lag-polynomial roots; inside applies to recurrence roots.
Checking only one root. Every root, including complex and repeated roots, must satisfy the condition.
Treating the boundary as acceptable. A root with modulus exactly one does not produce a decaying causal response.
Ignoring the model convention. The result is the standard causal AR stationarity test; do not silently interpret an unstable causal model as a stationary process driven only by past innovations.

The root test is useful because it converts the long-run behavior of an AR model into a finite algebraic check. It does not, by itself, estimate coefficients or diagnose whether a fitted model is appropriate; it answers the specific question of whether the specified causal dynamics are stable. This root-location skill is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

For each model, write the lag polynomial, find its roots, and apply the causal stationarity criterion. Where relevant, distinguish lag-polynomial roots from recurrence roots.

A hypothetical monthly index has $X_t-\mu=0.4(X_{t-1}-\mu)+\varepsilon_t$, where $\varepsilon_t$ are independent $N(0,1)$. Find the lag-polynomial root and classify the model. (Answer: $\phi(B)=1-0.4B$, root $2.5$; causal stationary.)
A hypothetical quarterly payment index has $X_t-\mu=1.2(X_{t-1}-\mu)+\varepsilon_t$, where $\varepsilon_t$ are independent $N(0,2)$. Find the root and explain the result. (Answer: $\phi(B)=1-1.2B$, root $5/6$; not causal stationary.)
A hypothetical annual index has $X_t-\mu=-0.5(X_{t-1}-\mu)+\varepsilon_t$, with independent $N(0,1)$ innovations. Find the root, including its sign, and classify the model. (Answer: $\phi(B)=1+0.5B$, root $-2$, modulus $2$; causal stationary.)
A hypothetical quarterly claims index has $X_t-\mu=0.7(X_{t-1}-\mu)-0.1(X_{t-2}-\mu)+\varepsilon_t$, where $\varepsilon_t$ are independent $N(0,4)$. Factor its lag polynomial and classify it. (Answer: $\phi(B)=1-0.7B+0.1B^2=(1-0.2B)(1-0.5B)$; roots $5$ and $2$; causal stationary.)
A hypothetical yearly reserve index has $X_t-\mu=1.25(X_{t-1}-\mu)-0.25(X_{t-2}-\mu)+\varepsilon_t$, with independent $N(0,1)$ innovations. Factor the lag polynomial and explain why a root on the boundary matters. (Answer: $\phi(B)=1-1.25B+0.25B^2=(1-B)(1-0.25B)$; roots $1$ and $4$; not causal stationary because the root at $1$ lies on the unit circle.)

What You'll Learn