Equivalent ARMA Forms and Common-Factor Cancellation

Why Equivalent ARMA Forms Matter

An ARMA model is specified by two lag polynomials: an autoregressive (AR) polynomial and a moving-average (MA) polynomial. Their degrees give the stated AR and MA orders, but those degrees do not always describe the simplest representation of the process. If the two polynomials share a factor, cancelling it leaves the same rational filter from innovations to observations, while reducing both orders.

This matters for identification. Data can reveal the process’s reduced ARMA representation, but an unreduced pair of polynomials may assign unnecessarily high orders to it. To interpret a proposed model correctly, check for common factors before treating its stated orders as the process’s minimal orders.

Definition: Let \(B\) denote the backshift operator, so \(BX_t=X_{t-1}\). A centered ARMA model can be written as \(\phi(B)Z_t=\theta(B)\varepsilon_t\), where \(Z_t=X_t-\mu\), \(\phi(B)=1-\phi_1B-\cdots-\phi_pB^p\), and \(\theta(B)=1+\theta_1B+\cdots+\theta_qB^q\). The innovation sequence \(\varepsilon_t\) has mean zero and variance \(\sigma^2\), and is uncorrelated across time. The AR order is the degree of \(\phi\); the MA order is the degree of \(\theta\). An ARMA representation is reduced if its AR and MA polynomials have no nonconstant common factor.

Recognizing and Cancelling a Common Factor

A common factor is a nonconstant polynomial that divides both \(\phi(B)\) and \(\theta(B)\). If \(\phi(B)=C(B)\phi^*(B)\) and \(\theta(B)=C(B)\theta^*(B)\), then their rational ratio simplifies: \(\theta(B)/\phi(B)=\theta^*(B)/\phi^*(B)\). The reduced orders are the degrees of \(\phi^*\) and \(\theta^*\), not the degrees before cancellation.

For example, a factor \(1-aB\) has a root at \(B=1/a\). When comparing roots, use the root of the polynomial, not the coefficient \(a\) itself. Cancelling a factor means cancelling the entire polynomial factor from both sides of the ARMA equation, not deleting a coefficient that happens to look similar.

Worked Example: Reducing a Second-Order Pair

A hypothetical centered monthly risk index is specified by \(\phi(B)Z_t=\theta(B)\varepsilon_t\), with \(\operatorname{Var}(\varepsilon_t)=4\), \(\phi(B)=(1-0.5B)(1-0.4B)\), and \(\theta(B)=(1-0.5B)(1+0.2B)\). Find the reduced model and check its AR stability and MA invertibility.

Method. Factor each polynomial, identify the shared factor, and cancel it. Then check the roots of the reduced polynomials against the usual root conditions.

First expand to see the stated degrees: \(\phi(B)=1-0.9B+0.2B^2\), since \((1-0.5B)(1-0.4B)=1-0.9B+0.2B^2\). Also, \(\theta(B)=1-0.3B-0.1B^2\), since \((1-0.5B)(1+0.2B)=1-0.3B-0.1B^2\).

Both contain \(1-0.5B\). Cancelling gives \[ (1-0.4B)Z_t=(1+0.2B)\varepsilon_t. \] The reduced model is ARMA(1,1), not ARMA(2,2). Its innovation variance remains \(4\); cancellation has not rescaled the innovations.

The reduced AR root solves \(1-0.4B=0\), so \(B=2.5\), outside the unit circle. The reduced MA root solves \(1+0.2B=0\), so \(B=-5\), also outside the unit circle. Thus the reduced model meets the usual causal-stationarity condition for the AR polynomial and the invertibility condition for the MA polynomial.

Check. Multiplying the reduced polynomials by \(1-0.5B\) recovers the expanded original polynomials exactly. This verifies the cancellation and the computed coefficients.

Why the Cancellation Works—and When to Be Careful

In the causal stationary setting, cancellation is more than a formal simplification. Suppose \(\phi(B)=C(B)\phi^*(B)\) and \(\theta(B)=C(B)\theta^*(B)\), and suppose the original AR polynomial has no roots on or inside the unit circle. Any factor \(C(B)\) of that polynomial also has all its roots outside the unit circle. Its inverse therefore defines a stable causal filter, so it can be applied to both sides of the model equation.

Factor both polynomials.
Write \(\phi(B)=C(B)\phi^*(B)\) and \(\theta(B)=C(B)\theta^*(B)\) for their shared factor \(C(B)\).

Use the stable inverse.
Under the stated causal-stationarity condition, \(1/C(B)\) is a stable causal filter. Applying it to \(C(B)\phi^*(B)Z_t=C(B)\theta^*(B)\varepsilon_t\) removes \(C(B)\) from both sides.

Read the reduced orders.
The resulting equation is \(\phi^*(B)Z_t=\theta^*(B)\varepsilon_t\). Its AR and MA orders are the degrees of the reduced polynomials.

The key condition in this justification is that the common factor can be inverted as a stable filter. Formal polynomial cancellation still simplifies the ratio \(\theta(B)/\phi(B)\), but it does not by itself establish that two stochastic equations have identical solutions under every possible root condition or initialization.

For instance, if the shared factor has a root on the unit circle, cancellation can change which solutions satisfy the equation. With \(C(B)=1-B\), the equation \((1-B)Z_t=(1-B)\varepsilon_t\) is satisfied by \(Z_t=\varepsilon_t+K\) for any time-constant random variable \(K\), because \((1-B)K=0\). Cancelling \(1-B\) would leave \(Z_t=\varepsilon_t\), which omits those other solutions. The original AR polynomial has a unit root, so the usual stationary causal ARMA conditions do not hold. Root conditions therefore matter when interpreting cancellation as a statement about a process, not just a rational expression.

Theorem: For a causal stationary ARMA representation whose AR polynomial has no roots on or inside the unit circle, any common nonconstant factor of the AR and MA polynomials can be cancelled to give the same causal process and the same innovation sequence. The reduced representation has the same rational filter \(\theta(B)/\phi(B)\) and lower or equal orders. Without conditions that justify inverting the common factor, polynomial cancellation alone does not guarantee equivalence of all process solutions.

Reduced Orders and Identification

The transfer function from innovations to the centered process is the rational expression \(\theta(B)/\phi(B)\). A shared factor does not change that expression after cancellation. Consequently, an unreduced ARMA(p,q) specification can describe a process with lower minimal orders. For a reduced representation, the AR and MA polynomials have no common root, or equivalently, no nonconstant common polynomial factor.

The practical identification lesson is to distinguish a model’s written orders from the orders of its reduced representation. If the polynomials share a factor, simply reading off their degrees overstates the minimal orders. An unreduced specification can make the model look more complicated than the process’s rational filter requires.

Worked Example: Finding the Minimal Orders After Expansion

A hypothetical quarterly claim-count index has \(\phi(B)=(1-0.25B)(1+0.5B)\) and \(\theta(B)=(1-0.25B)(1-0.3B)\), with innovation variance \(9\). Expand the polynomials, find the minimal orders, and check the reduced root conditions.

Method. Identify the factorization before reading the orders from the expanded expressions.

Expanding the AR polynomial gives \(\phi(B)=1+0.25B-0.125B^2\): the coefficient of \(B\) is \(0.5-0.25=0.25\), and the coefficient of \(B^2\) is \((0.25)(-0.5)=-0.125\).

Expanding the MA polynomial gives \(\theta(B)=1-0.55B+0.075B^2\): the coefficient of \(B\) is \(-0.3-0.25=-0.55\), and the coefficient of \(B^2\) is \((-0.25)(-0.3)=0.075\).

Both polynomials contain \(1-0.25B\). Cancelling it leaves \[ (1+0.5B)Z_t=(1-0.3B)\varepsilon_t. \] The unreduced expressions have degrees two and two, but the minimal orders are ARMA(1,1). The innovation variance is still \(9\).

The reduced AR root is \(B=-2\), since \(1+0.5B=0\). The reduced MA root is \(B=1/0.3=10/3\), since \(1-0.3B=0\). Both roots are outside the unit circle. Multiplication of both reduced polynomials by the cancelled factor reproduces the original factorization, checking the result.

Common mistake avoided. Calling this an ARMA(2,2) process solely because the expanded polynomials have degree two confuses the stated representation with the minimal one.

Worked Example: Cancelling a Factor in an ARMA(1,2) Form

A hypothetical monthly portfolio indicator has \(\phi(B)=(1-0.6B)(1+0.2B)\) and \(\theta(B)=(1-0.6B)(1-0.5B+0.25B^2)\). Its innovation variance is \(1\). Reduce the model and check the roots of the reduced polynomials.

Method. Cancel the shared factor, then compute the roots of the remaining AR and MA polynomials. This also checks that polynomial degrees are counted after, not before, cancellation.

The expansions are \(\phi(B)=1-0.4B-0.12B^2\), because the \(B\) coefficient is \(-0.6+0.2=-0.4\) and the \(B^2\) coefficient is \((-0.6)(0.2)=-0.12\). For the MA polynomial, \(\theta(B)=1-1.1B+0.55B^2-0.15B^3\): the coefficients after the constant are \(-0.5-0.6=-1.1\), \(0.25+0.3=0.55\), and \((-0.6)(0.25)=-0.15\).

The shared factor is \(1-0.6B\). Cancelling it gives \[ (1+0.2B)Z_t=(1-0.5B+0.25B^2)\varepsilon_t. \] The reduced model is ARMA(1,2), with innovation variance \(1\).

The AR root is \(-5\). To check the MA roots, solve \(1-0.5B+0.25B^2=0\). The quadratic formula gives \(B=(0.5\pm\sqrt{0.25-1})/0.5=1\pm i\sqrt{3}\). Each root has modulus \(\sqrt{1^2+(\sqrt{3})^2}=2\). The reduced AR roots and MA roots are therefore all outside the unit circle. This verifies the usual causal and invertible root conditions for the reduced model.

Check. The common factor has root \(B=1/0.6=5/3\); it is a shared factor because it appears explicitly in each original polynomial. The expanded degrees were two and three, but cancellation leaves degrees one and two.

Common Exam Traps

Reading orders before reducing. An expanded AR polynomial of degree \(p\) and MA polynomial of degree \(q\) may not give the minimal orders. Factor both before concluding.
Confusing a coefficient with a root. The factor \(1-0.6B\) has root \(B=1/0.6\), not \(0.6\). Check the root by substituting it into the factor.
Cancelling only one side. A common factor must divide both polynomials. Removing it from only the AR or only the MA polynomial changes the rational filter.
Assuming every algebraic cancellation is safe for every process. A common unit root is not covered by the causal stationary justification above. State the relevant root conditions before claiming equivalence of stochastic solutions.
Changing the innovation variance without cause. Cancelling an identical factor from numerator and denominator leaves the transfer function unchanged; it does not itself change \(\sigma^2\).

Key takeaway: Factor the AR and MA polynomials, cancel any common nonconstant factor, and report the reduced orders. The cancellation preserves the rational filter; treating it as an equivalent causal stationary process also requires the appropriate root conditions.

Check Your Understanding

For each problem, reduce the polynomials before stating the minimal orders. Check roots when requested, and note when the usual cancellation conditions fail.

A centered series has \(\phi(B)=(1-0.4B)(1+0.3B)\) and \(\theta(B)=(1-0.4B)(1+0.2B)\), with innovation variance \(2\). Give the reduced model, its orders, and the common root. (Answer: \((1+0.3B)Z_t=(1+0.2B)\varepsilon_t\), ARMA(1,1), \(\sigma^2=2\), common root \(B=2.5\).)
Suppose \(\phi(B)=(1-0.2B)(1-0.7B)\) and \(\theta(B)=(1-0.2B)(1+0.1B)\). Find the reduced model and check the reduced AR and MA roots. (Answer: \((1-0.7B)Z_t=(1+0.1B)\varepsilon_t\), ARMA(1,1); roots \(10/7\) and \(-10\), both outside the unit circle.)
Suppose \(\phi(B)=(1+0.5B)(1-0.25B)\) and \(\theta(B)=(1+0.5B)(1-0.4B+0.1B^2)\). Find the reduced orders and the shared root. (Answer: reduced ARMA(1,2), with \(\phi^*(B)=1-0.25B\), \(\theta^*(B)=1-0.4B+0.1B^2\), and shared root \(B=-2\).)
For \(\phi(B)=1-B\) and \(\theta(B)=1-B\), the equation is \((1-B)Z_t=(1-B)\varepsilon_t\). Why does formal cancellation not establish that every solution is \(Z_t=\varepsilon_t\)? (Answer: \(Z_t=\varepsilon_t+K\), for a time-constant random variable \(K\), also satisfies the equation because \((1-B)K=0\); the shared factor has a unit root, so the stationary causal cancellation argument does not apply.)
A proposed ARMA(2,2) model has \(\phi(B)=(1-0.3B)(1+0.4B)\) and \(\theta(B)=(1-0.3B)(1-0.2B)\). State the minimal orders and check whether the reduced model meets the AR stability and MA invertibility root conditions. (Answer: reduced ARMA(1,1), \(\phi^*(B)=1+0.4B\), \(\theta^*(B)=1-0.2B\); roots \(-2.5\) and \(5\), both outside the unit circle.)

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