Causality and Invertibility from ARMA Roots

Why Roots Matter

An ARMA equation is a compact description of how a series depends on its past and on shocks. But the equation alone does not always tell us whether the series can be built from present and past shocks, or whether those shocks can be reconstructed from present and past observations. These properties are causality and invertibility, and both are determined by the roots of the model’s lag polynomials.

The root conditions are more than technical restrictions. They distinguish a stable one-sided representation from one that requires future shocks, and they identify a canonical form in which the innovations are recoverable. Polynomial factorization also reveals when a written ARMA specification contains canceling factors or when a noninvertible moving-average specification has an alternative invertible representation.

Polynomials and Root Conditions

Write a real ARMA(\(p,q\)) model for the centered series \(Y_t=X_t-\mu\) as

A(B)Y_t=M(B)\varepsilon_t, \qquad A(z)=1-\phi_1z-\cdots-\phi_pz^p, \qquad M(z)=1+\theta_1z+\cdots+\theta_qz^q.

Here \(B\) is the lag operator, so \(BY_t=Y_{t-1}\); \(\varepsilon_t\) is white noise with mean zero and variance \(\sigma_\varepsilon^2\). The variable \(z\) is used to find polynomial roots: a root \(r\) of \(A\), for example, satisfies \(A(r)=0\). The signs in the definitions above fix the coefficient convention used in the examples.

Definition: The AR polynomial satisfies the causality condition if \(A(z)\ne 0\) for every \(z\) with \(|z|\leq 1\). The MA polynomial satisfies the invertibility condition if \(M(z)\ne 0\) for every \(z\) with \(|z|\leq 1\). Equivalently, every root of each polynomial must lie strictly outside the unit circle. Causality is a condition on \(A\); invertibility is a condition on \(M\).

The strict inequality matters: a root on the unit circle is not allowed. If a polynomial has degree zero, it has no roots and its corresponding condition is satisfied. A root \(r\) can be checked directly by solving the polynomial equation, or by factoring the polynomial into linear factors over the complex numbers. For real-coefficient polynomials, nonreal roots occur in conjugate pairs.

Why the AR Roots Govern Causality

If the AR condition holds, division by \(A(B)\) gives the formal transfer function from shocks to observations, \(M(z)/A(z)\). The causal representation is a one-sided filter,

Y_t=\sum_{j=0}^{\infty}\psi_j\varepsilon_{t-j}.

The AR root condition ensures that the reciprocal \(1/A(z)\) can be expanded as a power series around zero whose coefficients decay sufficiently fast. In fact, because the closest AR root is strictly outside the unit circle, this power series converges on a disk larger than the unit disk. Multiplying by the finite polynomial \(M(z)\) gives the coefficients \(\psi_j\) of the causal filter. The shocks entering \(Y_t\) are therefore current and past shocks, not future ones.

A first-order example shows why the location of the root is the key. If \(A(z)=1-\phi z\), then its root is \(r=1/\phi\) when \(\phi\ne0\). The condition \(|r|>1\) is equivalent to \(|\phi|<1\). In that case, \(1/A(B)\) has a convergent geometric expansion. When the root is inside the unit circle, that expansion does not give a stable causal filter; a stationary solution, if one is considered, cannot be represented as a stable one-sided filter of the specified shocks.

Worked Example: Checking an AR(1) Root

Suppose \(Y_t=0.6Y_{t-1}+\varepsilon_t\), where \(\varepsilon_t\) is white noise with variance \(4\). Its AR polynomial is \(A(z)=1-0.6z\). Find the root and assess causality.

Root. Solve \(1-0.6z=0\):

z=\frac{1}{0.6}=1.667.

Since \(|1.667|>1\), the root is strictly outside the unit circle, so the AR polynomial meets the causality condition. To see the resulting filter, solve formally for \(Y_t\) by expanding:

Y_t=(1-0.6B)^{-1}\varepsilon_t =\sum_{j=0}^{\infty}0.6^j\varepsilon_{t-j} =\varepsilon_t+0.6\varepsilon_{t-1}+0.36\varepsilon_{t-2}+\cdots.

The coefficients decay because \(|0.6|<1\). For example, the variance of this causal series is \(4\sum_{j=0}^{\infty}0.6^{2j}=4/(1-0.36)=6.25\). Thus the root check and the explicit filter agree.

Why the MA Roots Govern Invertibility

Invertibility asks whether the shocks can be expressed as a stable one-sided filter of the observations. If \(M(z)\) has no roots in or on the unit circle, its reciprocal has a convergent power series around zero. Rearranging the model then yields

\varepsilon_t=\sum_{j=0}^{\infty}\pi_jY_{t-j}.

This representation makes the innovations recoverable from current and past observations. It also gives a precise sense in which the shocks are the new information in the series: after applying the inverse filter, the remaining sequence is white noise. If the MA root condition fails, formal inversion may instead require future observations or coefficients that do not decay, so it does not provide a stable one-sided recovery.

Worked Example: Inverting a First-Order MA Model

Let \(Y_t=\varepsilon_t+0.4\varepsilon_{t-1}\), with white-noise innovations of variance \(1\). The MA polynomial is \(M(z)=1+0.4z\). Find its root and recover the innovations.

Root check. Solving \(1+0.4z=0\) gives \(z=-2.5\), whose modulus is greater than \(1\). The model is invertible.

Since \(M(B)^{-1}=(1+0.4B)^{-1}\), expand the inverse as a geometric series:

\varepsilon_t=(1+0.4B)^{-1}Y_t =\sum_{j=0}^{\infty}(-0.4)^jY_{t-j} =Y_t-0.4Y_{t-1}+0.16Y_{t-2}-0.064Y_{t-3}+\cdots.

The coefficients shrink in magnitude because \(0.4^j\) tends to zero. The formula recovers each innovation from current and past observations; it does not use future values of \(Y_t\).

Factorization, Cancellation, and Equivalent MA Forms

Factoring the polynomials makes both root checks and representation issues visible. A common factor in \(A(z)\) and \(M(z)\) cancels from their ratio, the transfer function \(M(z)/A(z)\). Such a specification is nonminimal: the same transfer function can be written with lower-degree polynomials. Root conditions should therefore be assessed on the reduced polynomials when discussing the actual transfer function. An uncanceled common factor can obscure the effective dynamics or make a nominal polynomial degree misleading.

A different issue arises when an MA polynomial has roots inside the unit circle. For real coefficients, reflecting such a root \(r\) to \(1/\bar r\) produces a root outside the circle. The resulting normalized MA polynomial has the same frequency-dependent magnitude up to a constant factor. Adjusting the white-noise variance by that factor preserves the spectrum. Thus a noninvertible MA specification can have an invertible representation with the same second-order behavior. Without Gaussian innovations, matching spectra establishes equality of second-order properties, not necessarily equality of the full distributions.

For a real MA(1), the reflection is especially transparent. If \(M(z)=1+\theta z\), its root is \(-1/\theta\). When \(|\theta|>1\), that root is inside the unit circle. Replacing \(\theta\) with \(1/\theta\) moves the root outside; multiplying the innovation variance by \(\theta^2\) compensates for the change in the polynomial’s magnitude on the unit circle.

Worked Example: Reflecting an MA Root

Consider \(Y_t=\varepsilon_t+1.5\varepsilon_{t-1}\), where the white-noise variance is \(4\). Find an invertible MA(1) representation with the same spectrum.

The original polynomial is \(M(z)=1+1.5z\), with root \(z=-1/1.5=-0.6667\). Since its modulus is less than \(1\), the given representation is not invertible. Reflecting the root gives \(-1/0.6667=-1.5\), corresponding to the normalized polynomial \(M^*(z)=1+\frac{2}{3}z\). The new root has modulus \(1.5>1\), as required.

For \(|z|=1\), the original polynomial factors in magnitude as

|1+1.5z|=1.5\left|1+\frac{2}{3}z\right|.

Therefore the original spectrum, proportional to \(4|1+1.5z|^2\), equals the spectrum from \(M^*(z)\) when the new innovation variance is \(4(1.5)^2=9\). The invertible representation is consequently \(Y_t=\eta_t+\frac{2}{3}\eta_{t-1}\), where \(\eta_t\) is white noise with variance \(9\). Its shocks are recoverable through the stable inverse filter with coefficients \((-2/3)^j\). This is an equivalent second-order representation; it does not assert that the original \(\varepsilon_t\) and the recovered \(\eta_t\) are the same shocks.

Checking a Full ARMA Representation

For an ARMA model, causality and invertibility are separate checks. An AR root does not determine invertibility, and an MA root does not determine causality. The checks can be summarized as follows.

Property	Polynomial	Root requirement	Consequence
Causality	\(A(z)\)	All roots have modulus greater than \(1\)	A stable filter of current and past innovations represents \(Y_t\)
Invertibility	\(M(z)\)	All roots have modulus greater than \(1\)	A stable filter of current and past observations recovers \(\varepsilon_t\)

A compact way to apply the checks is to factor each polynomial, list its roots, and compare each root’s modulus with \(1\). If the AR and MA polynomials share a factor, first identify the reduced transfer function and state whether the root check concerns the unreduced specification or the minimal representation.

Worked Example: Checking an ARMA Model and Recovering Its Innovations

Consider the hypothetical model

(1-0.5B)Y_t=(1+0.3B)\varepsilon_t,

where \(\varepsilon_t\) is white noise with variance \(2\). Check both root conditions and write the innovation-recovery filter.

The AR root solves \(1-0.5z=0\), so \(z=2\), which is outside the unit circle. The MA root solves \(1+0.3z=0\), so \(z=-3.333\), also outside the unit circle. The representation is causal and invertible.

Rearrange the model to express the shocks in terms of observations:

\varepsilon_t=(1+0.3B)^{-1}(1-0.5B)Y_t =\left(\sum_{j=0}^{\infty}(-0.3)^jB^j\right)(1-0.5B)Y_t.

For the first coefficients, the inverse MA filter is \(1-0.3B+0.09B^2-0.027B^3+\cdots\). Multiplying by \(1-0.5B\) gives

\varepsilon_t =Y_t-0.8Y_{t-1}+0.24Y_{t-2}-0.072Y_{t-3}+\cdots.

For example, the coefficient on \(Y_{t-1}\) is \(-0.3-0.5=-0.8\), and that on \(Y_{t-2}\) is \(0.09+0.15=0.24\). The coefficients decay, as expected from the invertible root condition. The innovation variance does not affect either root check.

Common Pitfalls and Why the Conditions Matter

Checking the coefficient instead of the root. For \(1-\phi z\), the root is \(1/\phi\), not \(\phi\). The condition \(|\phi|<1\) follows from the root condition; it is not the root itself.
Using the AR condition for the MA polynomial. Causality checks roots of \(A(z)\); invertibility checks roots of \(M(z)\). Both must be checked independently.
Allowing a unit root. A root with modulus exactly \(1\) does not satisfy the strict condition. The associated inverse or causal expansion will not have the required stable decay.
Assuming an invertible representation recovers the original shocks from every specification. Invertibility recovers the innovations associated with that representation. A reflected, spectrally equivalent MA form generally has different innovations and may have a different innovation variance.
Ignoring common factors. A shared factor cancels from the transfer function. Work with the reduced representation when describing the effective dynamics and do not interpret canceling polynomial degrees as distinct dynamics.

These conditions matter for interpretation as well as algebra. Causality gives a stable account of the series in terms of past shocks, while invertibility makes the shocks estimable from the observed history. An invertible, reduced representation is therefore the natural basis for treating ARMA innovations as recoverable model quantities rather than unobserved labels.

Check Your Understanding

For each question, use the polynomial roots rather than relying only on the coefficient values.

For \(A(z)=1-0.25z\), find the root and determine whether the AR polynomial satisfies the causality condition.
For \(M(z)=1-0.8z\), find the root and determine whether the MA polynomial is invertible.
Explain why a root exactly on the unit circle fails the stated root condition.
An MA(1) has \(M(z)=1+2z\) and innovation variance \(3\). What reflected coefficient and innovation variance give an invertible representation with the same spectrum?
In one sentence each, distinguish what causality and invertibility allow us to represent as a stable one-sided filter.