Why Check the Roots?
An ARMA model can be written with lag polynomials, but its coefficients alone do not always make its behavior easy to judge. The roots of those polynomials provide a systematic test: the AR polynomial determines whether the process has a causal representation in current and past innovations, while the MA polynomial determines whether the innovations can be recovered from current and past observations. The location of the roots—not simply whether an individual coefficient is less than one—decides the result.
The key comparison is with the unit circle in the complex plane. For both tests in this tutorial, the relevant roots must be strictly outside that circle. A root on the circle does not pass the test. This root-location rule is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
The displayed sign convention matters. In this tutorial, the MA polynomial has plus signs in its coefficient definition. For instance, if \(\theta(B)=1-0.4B\), then \(\theta_1=-0.4\), and the polynomial to solve is \(1-0.4z=0\). Always form the polynomial actually written in the model before solving for its roots.
Why the Outside-Root Condition Works
The AR(1) case shows the basic reason. Suppose \(X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t\), with white-noise innovations. Its AR polynomial is \(\phi(z)=1-\phi z\), whose root is \(z=1/\phi\), provided \(\phi\ne0\). If \(|\phi|<1\), then the root has modulus greater than 1, and repeated substitution gives a convergent representation in current and past innovations.
For higher-order polynomials, factoring the polynomial gives the corresponding geometric-series reasoning for each factor. The root condition ensures that the inverse AR filter has coefficients that decay sufficiently for the causal representation. Similarly, invertibility requires the inverse MA filter to have a convergent representation in current and past observations. For an MA(1) polynomial \(1+\theta_1z\), the root is \(-1/\theta_1\); it lies outside the unit circle exactly when \(|\theta_1|<1\). Higher-order MA polynomials require checking all roots, not just one coefficient.
This derivation also explains why the root is the reciprocal of a familiar coefficient in the first-order case. The coefficient itself is not the root. The reciprocal relationship is a frequent source of reversed conclusions.
A Reliable Root-Test Procedure
Use the coefficients and signs in the stated ARMA model to form \(\phi(z)\) and \(\theta(z)\).
Find every root of \(\phi(z)=0\) for causality and every root of \(\theta(z)=0\) for invertibility.
For a complex root \(a+bi\), its modulus is \(\sqrt{a^2+b^2}\). A conjugate pair has the same modulus, so checking one root in the pair is enough once both have been identified.
All AR roots outside means causal; all MA roots outside means invertible. A root on or inside the unit circle fails its respective test.
For a real-coefficient polynomial, nonreal roots occur in conjugate pairs. If the roots are \(a+bi\) and \(a-bi\), both have modulus \(\sqrt{a^2+b^2}\). For a quadratic, the product of the roots can sometimes make the modulus check especially quick, but it does not replace solving the polynomial or identifying whether the roots are real or complex.
Worked Example: Testing a First-Order ARMA Model
Consider the hypothetical centered model \( (1-0.7B)(X_t-\mu)=(1+0.4B)\varepsilon_t \), where \(\varepsilon_t\) is white noise with mean zero and variance \(\sigma^2=1\). Determine whether the model is causal and invertible. The value of \(\mu\) does not affect either root test.
Method. Form each polynomial and solve for its root. Compare the modulus of the AR root with 1 for causality, and the modulus of the MA root with 1 for invertibility.
AR polynomial. We have \(\phi(z)=1-0.7z\). Solving \(1-0.7z=0\) gives \(0.7z=1\), hence \(z=1/0.7=10/7\). Its modulus is \(10/7\), which is greater than 1. The model is causal.
MA polynomial. We have \(\theta(z)=1+0.4z\). Solving \(1+0.4z=0\) gives \(z=-1/0.4=-2.5\). Its modulus is \(2.5\), which is greater than 1. The model is invertible.
Check. The AR coefficient has magnitude \(0.7<1\), consistent with its root being outside; the MA coefficient in the stated plus-sign convention has magnitude \(0.4<1\), also consistent with its root being outside. Thus this model is both causal and invertible. The common mistake to avoid is calling \(-2.5\) an inside root because it is negative: the test uses its modulus, not its sign.
Quadratics, Complex Roots, and Boundary Cases
For a quadratic polynomial, factorization is useful when available. Otherwise, use the quadratic formula and calculate the modulus of each resulting root. A negative discriminant produces complex roots; it does not prevent the test. A root with modulus exactly 1 lies on the unit circle and fails the strict condition, even if every other root is outside.
Worked Example: A Causal Model with a Boundary MA Root
Consider \( (1-1.3B+0.4B^2)(X_t-\mu)=(1-1.5B+0.5B^2)\varepsilon_t \), where \(\varepsilon_t\) is white noise with mean zero and variance \(\sigma^2=4\). Determine causality and invertibility.
Method. Factor the two polynomials, solve each factor equation, and check every root against the strict modulus condition.
AR roots. The polynomial factors as \(1-1.3z+0.4z^2=(1-0.8z)(1-0.5z)\), since the product of the factors has middle coefficient \(-0.8-0.5=-1.3\) and quadratic coefficient \(0.8(0.5)=0.4\). The roots are \(z=1/0.8=1.25\) and \(z=1/0.5=2\). Both moduli exceed 1, so the model is causal.
MA roots. The polynomial factors as \(1-1.5z+0.5z^2=(1-z)(1-0.5z)\). The roots are \(z=1\) and \(z=2\). The root at 1 has modulus exactly 1, while the other has modulus 2. Since not all roots are strictly outside the unit circle, the model is not invertible.
Check. The AR test passes for both AR roots, and the MA test fails at its boundary root. The variance of the white noise does not change either conclusion. The common mistake is to treat “on the unit circle” as equivalent to “outside”; the invertibility condition is strict.
Worked Example: Comparing Moduli for Complex Roots
Consider the hypothetical model \( (1-0.5B+0.5B^2)(X_t-\mu)=(1+2.4B+2B^2)\varepsilon_t \), where \(\varepsilon_t\) is white noise with mean zero and variance \(\sigma^2=9\). Determine whether it is causal and invertible.
Method. Solve each quadratic, then calculate the modulus using the real and imaginary parts.
AR roots. The equation \(1-0.5z+0.5z^2=0\), multiplied by 2, is \(z^2-z+2=0\). The quadratic formula gives \(z=(1\pm\sqrt{1-8})/2=(1\pm i\sqrt{7})/2\). Each root has real part \(1/2\) and imaginary part \(\pm\sqrt{7}/2\). Its modulus is \(\sqrt{(1/2)^2+(\sqrt{7}/2)^2}=\sqrt{(1+7)/4}=\sqrt{2}\), which exceeds 1. The model is causal.
MA roots. Solve \(1+2.4z+2z^2=0\). Dividing the equation by 2 gives \(z^2+1.2z+0.5=0\). The discriminant is \(1.2^2-4(0.5)=1.44-2=-0.56\), so the roots are \(z=(-1.2\pm i\sqrt{0.56})/2=-0.6\pm i\sqrt{0.56}/2\). Their common modulus is \(\sqrt{0.6^2+(\sqrt{0.56}/2)^2}=\sqrt{0.36+0.14}=\sqrt{0.5}\), which is less than 1. The model is not invertible.
Check. The AR roots are both outside, while the conjugate MA roots are both inside. Therefore, the model is causal but not invertible. The common mistake is to use only the real part of a complex root; the required comparison is the modulus, which incorporates both parts.
What the Root Test Does—and Does Not—Say
Causality and invertibility describe different representations. Causality concerns expressing the centered process using present and past innovations. Invertibility concerns expressing the innovations using present and past observations. Passing one condition does not imply passing the other, as the complex-root example demonstrates.
The conclusions also depend on checking the correct polynomial. Do not apply the AR root condition to the MA polynomial or vice versa. Nor should a candidate substitute a coefficient-size rule for a polynomial-root test: for orders greater than one, several coefficients jointly determine the roots. A model may have a coefficient whose magnitude exceeds 1 and still pass a relevant root test, or coefficients that individually appear small while the polynomial has a root that fails it.
A useful final report names the roots or their moduli and states the two conclusions independently. For example: “All AR roots have modulus greater than 1, so the model is causal; one MA root has modulus 1, so it is not invertible.” This makes the reasoning transparent and prevents a correct calculation from being paired with the wrong interpretation.
Check Your Understanding
For each model, form the relevant roots and apply the strict unit-circle conditions. Keep causality and invertibility separate.
- A centered ARMA model has \(\phi(z)=1-0.8z\) and \(\theta(z)=1+0.25z\). Find both roots and state whether the model is causal and invertible. (Answer: AR root \(1.25\), MA root \(-4\); causal and invertible.)
- A model has \(\phi(z)=1+0.6z\) and \(\theta(z)=1-1.4z\). Find the roots and identify which property fails. (Answer: AR root \(-1.6667\), MA root \(0.7143\); causal but not invertible.)
- For \(\phi(z)=1-1.1z+0.28z^2\), factor the polynomial, find both roots, and test causality. (Answer: \((1-0.7z)(1-0.4z)\), roots \(1.4286\) and \(2.5\); causal.)
- For \(\theta(z)=1+z+z^2\), solve for the roots and test invertibility. (Answer: roots \((-1+i\sqrt{3})/2\) and \((-1-i\sqrt{3})/2\), each with modulus 1; not invertible.)
- A model has \(\phi(z)=1-0.4z\) and \(\theta(z)=1+1.5z+z^2\). The MA roots are complex. Find their modulus and state both conclusions. (Answer: the AR root is \(2.5\), so the model is causal; the MA roots are \(-0.75\pm0.6614i\), each with modulus 1, so it is not invertible.)