Invertibility Through MA Polynomial Roots

Why Polynomial Roots Determine Invertibility

For a moving-average model, the observed series is built from current and past innovations. Invertibility asks whether those innovations can, in turn, be recovered stably from current and past observations. The root test converts this question into an algebraic one: find the zeros of the model's MA polynomial and compare their distances from the origin with one.

This tutorial uses the convention that the coefficients appear with plus signs in the model. If a problem instead writes minus signs, preserve the signs as written when forming the polynomial. Changing that convention without changing the coefficients can reverse a conclusion.

Definition: Let \(X_t\) be a centered MA(q) process \(X_t=\varepsilon_t+\theta_1\varepsilon_{t-1}+\cdots+\theta_q\varepsilon_{t-q}\), where \(\varepsilon_t\) is white noise with mean zero and variance \(\sigma^2>0\), and \(\theta_q\ne 0\). Its MA polynomial is \(\Theta(z)=1+\theta_1z+\cdots+\theta_qz^q\). The process is invertible when the innovations can be represented by a convergent stable linear filter of present and past observations. Under this convention, it is invertible exactly when every root \(z\) of \(\Theta(z)=0\) satisfies \(|z|>1\).

The condition is strict. A root on the unit circle, where \(|z|=1\), does not meet the criterion. A root inside the unit circle also means the model is not invertible. Roots may be real or complex; for a polynomial with real coefficients, any non-real roots occur in conjugate pairs.

Why the Unit-Circle Criterion Works

Suppose the roots of the degree-\(q\) polynomial are \(\lambda_1,\ldots,\lambda_q\), including multiplicities. Since \(\Theta(0)=1\), its factorization can be written \(\Theta(z)=\prod_{j=1}^q(1-z/\lambda_j)\). Formally, recovering the innovations requires applying the reciprocal polynomial to the observations: \(\varepsilon_t=\Theta(B)^{-1}X_t\), where \(B\) denotes the lag operator, so \(BX_t=X_{t-1}\).

Each factor of the reciprocal has a geometric-series expansion when its root is outside the unit circle. For \(|\lambda_j|>1\), the ratio \(1/\lambda_j\) has absolute value less than one, and \((1-B/\lambda_j)^{-1}=\sum_{k=0}^{\infty}B^k/\lambda_j^k\). The coefficients decay geometrically. Multiplying the finitely many such expansions gives a convergent filter involving present and past \(X_t\) values. If a root is inside or on the unit circle, this expansion does not give a stable inverse filter: the corresponding reciprocal series fails to have summable, decaying coefficients. That is why the required condition concerns roots strictly outside the circle.

\Theta(z)=\prod_{j=1}^{q}\left(1-\frac{z}{\lambda_j}\right), \qquad |\lambda_j|>1 \ \text{for every }j \quad\Longleftrightarrow\quad \text{invertibility}

This derivation also explains a common source of error. The root \(\lambda_j\) is compared with one, but the geometric-series ratio is its reciprocal \(1/\lambda_j\). Thus the roots must be outside the unit circle—not inside it.

Applying the Root Test

A reliable calculation has two directions. If the model gives coefficients, solve \(\Theta(z)=0\) and check every root. If the problem gives roots, first form the polynomial with constant term one, expand it, and then read off the coefficients. For specified roots \(\lambda_1,\ldots,\lambda_q\), the appropriate polynomial is \(\prod_j(1-z/\lambda_j)\), not \(\prod_j(z-\lambda_j)\) without normalization.

Write the polynomial.
Copy the model's lag coefficients and signs into \(\Theta(z)=1+\theta_1z+\cdots+\theta_qz^q\).

Find all roots.
Factor the polynomial or use the quadratic formula when the degree is two. Include complex roots and multiplicities.

Compare root moduli with one.
Invertibility requires every root to have modulus strictly greater than one.

State the conclusion and check the convention.
Identify any inside or boundary root, and confirm that the polynomial used the model's actual coefficient signs.

Worked Example: Recovering Coefficients From Two Outside Roots

A hypothetical quarterly index is modeled as an MA(2) process whose MA polynomial has roots \(2\) and \(3\). Find its lag coefficients and determine whether the model is invertible.

Method. Form the constant-one polynomial from the roots, expand it, and apply the unit-circle criterion. The common mistake is to use \((z-2)(z-3)\) directly as the MA polynomial, which gives the wrong constant term and coefficient scale.

The normalized factors are \((1-z/2)\) and \((1-z/3)\). Therefore, \(\Theta(z)=(1-z/2)(1-z/3)\). Multiplying term by term gives \(\Theta(z)=1-z/3-z/2+z^2/6=1-(5/6)z+(1/6)z^2\).

Comparing with \(1+\theta_1z+\theta_2z^2\), the coefficients are \(\theta_1=-5/6\) and \(\theta_2=1/6\). The roots have moduli \(|2|=2\) and \(|3|=3\), both greater than one. The model is invertible. As a check, substituting \(z=2\) or \(z=3\) into the factored polynomial makes one factor zero.

Worked Example: A Root Inside the Unit Circle

A hypothetical monthly claim-count index has MA polynomial \(\Theta(z)=1-(7/3)z+(2/3)z^2\). Determine its roots and invertibility.

Method. Factor or solve the polynomial and assess each root separately. Do not conclude that a model is invertible merely because one of its roots is outside the circle.

Set the polynomial equal to zero and multiply by \(3\): \(3-7z+2z^2=0\), or \(2z^2-7z+3=0\). Factor: \((2z-1)(z-3)=0\). Thus \(z=1/2\) or \(z=3\).

The root moduli are \(1/2=0.5\) and \(3\). Since \(0.5<1\), one root lies inside the unit circle; the model is not invertible. The root at \(3\) does not compensate for the root at \(1/2\). As a check, the polynomial factors as \((1-2z)(1-z/3)\), which expands to \(1-(7/3)z+(2/3)z^2\), the stated polynomial.

Worked Example: Complex Roots and Their Moduli

A hypothetical quarterly reserve-change series has MA polynomial \(\Theta(z)=1-(1/2)z+(1/4)z^2\). Determine whether its complex roots satisfy the invertibility criterion.

Method. Solve for the roots and compute their moduli. For a complex root \(a+bi\), the modulus is \(\sqrt{a^2+b^2}\); checking only the real part is not sufficient.

Multiply the equation by \(4\): \(z^2-2z+4=0\). The quadratic formula gives \(z=(2\pm\sqrt{(-2)^2-4(1)(4)})/2\). The discriminant is \(4-16=-12\), so \(z=(2\pm\sqrt{-12})/2=1\pm i\sqrt{3}\).

For either root, the modulus is \(\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2\). Both roots therefore lie outside the unit circle, and the model is invertible. The conjugate pair is expected because the polynomial has real coefficients. Substitution is also consistent: the sum of the roots is \(2\), and their product is \(4\), matching the roots of \(z^2-2z+4\).

Worked Example: The Boundary Case

A hypothetical MA(1) model has polynomial \(\Theta(z)=1-z\). Determine its root and explain why the equality case matters.

Method. Solve the polynomial equation and use the strict root condition. The common mistake is to treat “at least one” as equivalent to “strictly greater than one.”

Solving \(1-z=0\) gives \(z=1\), with modulus \(|1|=1\). The root is on the unit circle, not outside it. Consequently, the model is not invertible under the strict criterion. This remains true even though the root is not inside the circle: a boundary root fails the requirement \(|z|>1\).

As a direct check, the reciprocal polynomial would require expanding \((1-B)^{-1}\) as \(1+B+B^2+\cdots\). These coefficients do not decay, so the expansion is not a stable inverse filter.

Common Traps and Interpretation

The root test is short, but several errors can change its result. Read the polynomial and the question's convention before solving; a root calculation is only as reliable as the polynomial being tested.

Confusing roots with reciprocal roots. If the root is \(\lambda\), the geometric-series ratio is \(1/\lambda\). Invertibility requires \(|\lambda|>1\), not \(|\lambda|<1\).
Using the wrong coefficient signs. For \(X_t=\varepsilon_t-\alpha\varepsilon_{t-1}\), the polynomial is \(1-\alpha z\), not \(1+\alpha z\). The stated model determines the signs.
Checking only real roots. Complex roots count too. Compute their modulus; the real part alone does not determine whether a root lies outside the circle.
Ignoring equality. A root with modulus exactly one fails the strict condition. Do not round a modulus near one before deciding which side it lies on.
Checking only one root. Every root must be outside. One inside or boundary root is enough to make the model noninvertible.

A root-based conclusion is about invertibility of the specified MA polynomial. It does not by itself establish other properties of a time-series model, such as stationarity conditions for an autoregressive polynomial. Keep the polynomial being tested and the property being tested distinct.

Key takeaway: For \(\Theta(z)=1+\theta_1z+\cdots+\theta_qz^q\), find every solution of \(\Theta(z)=0\). The MA(q) model is invertible if and only if every root has modulus strictly greater than one. When roots are supplied instead, construct \(\Theta(z)=\prod_j(1-z/\lambda_j)\) before reading off the lag coefficients.

Check Your Understanding

For each problem, use the polynomial as written, find or assess every root, and apply the strict unit-circle criterion.

A hypothetical MA(1) model has \(\Theta(z)=1+0.4z\). Find the root and decide invertibility. (Answer: root \(-2.5\); invertible.)
A hypothetical MA(1) model has \(\Theta(z)=1-1.2z\). Find the root and decide invertibility. (Answer: root \(5/6\); not invertible.)
A hypothetical MA(2) model has \(\Theta(z)=1-(3/4)z+(1/8)z^2\). Find the roots and decide invertibility. (Answer: roots \(2\) and \(4\); invertible.)
A hypothetical MA(2) polynomial is specified to have roots \(0.8\) and \(2\). Expand its constant-one polynomial and decide invertibility. (Answer: \(\Theta(z)=1-(7/4)z+(5/8)z^2\); not invertible because \(0.8<1\).)
A hypothetical MA(2) model has \(\Theta(z)=1-(1/3)z+(1/9)z^2\). Find its roots and decide invertibility. (Answer: roots \(3e^{i\pi/3}\) and \(3e^{-i\pi/3}\), each with modulus \(3\); invertible.)

What You'll Learn