ARMA Models in Lag-Polynomial Form

Why Use Lag-Polynomial Form?

An ARMA model can be written as a collection of lagged observations and innovations. When the orders $p$ and $q$ increase, that expanded equation can become cumbersome. Lag-polynomial notation packages the same terms into two polynomials: one describing the autoregressive part and one describing the moving-average part. This makes the model compact and makes the role of the intercept easier to track.

That last point matters because the intercept in an ARMA equation is generally not the unconditional mean of the series. The autoregressive coefficients determine how the intercept translates into the mean. The moving-average coefficients do not enter that relationship when the innovations have mean zero.

Lag Operators and the ARMA Definition

The lag operator $L$ shifts a time-indexed variable back by one period: $LX_t=X_{t-1}$. Repeated application gives $L^2X_t=X_{t-2}$, and, more generally, $L^jX_t=X_{t-j}$. A polynomial in $L$ therefore represents a weighted sum of current and lagged values.

Definition: Let $\{X_t\}$ be a covariance-stationary process with unconditional mean $\mu=E[X_t]$. An ARMA($p,q$) model in centered lag-polynomial form is $$ \phi(L)(X_t-\mu)=\theta(L)\varepsilon_t, $$ where $$ \phi(L)=1-\phi_1L-\cdots-\phi_pL^p,\qquad \theta(L)=1+\theta_1L+\cdots+\theta_qL^q. $$ The innovations $\{\varepsilon_t\}$ are assumed to have mean zero, constant finite variance $\sigma_\varepsilon^2$, and zero covariance across distinct times. The signs shown here are the convention used throughout this tutorial.

The corresponding expanded equation is $$ (X_t-\mu)-\phi_1(X_{t-1}-\mu)-\cdots-\phi_p(X_{t-p}-\mu) =\varepsilon_t+\theta_1\varepsilon_{t-1}+\cdots+\theta_q\varepsilon_{t-q}. $$ The autoregressive polynomial acts on deviations from the mean in this form; the moving-average polynomial acts on current and lagged innovations. A coefficient can be zero, so the displayed orders need not mean every lag has a nonzero coefficient.

An equivalent form uses an intercept $c$ and the uncentered series: $$ \phi(L)X_t=c+\theta(L)\varepsilon_t. $$ Here $c$ is the constant term in the equation. It is not automatically equal to $\mu$. The relation between them follows by taking expectations, as shown below.

Worked Example: Writing an ARMA(1,1) as Polynomials

Suppose a hypothetical stationary series, measured in index points, satisfies $$ X_t=2+0.6X_{t-1}+\varepsilon_t+0.4\varepsilon_{t-1}, $$ where the innovations have mean zero, constant finite variance, and zero covariance across distinct times. The coefficient $0.6$ has absolute value less than one, so this ARMA(1,1) has a stationary causal solution under the stated innovation assumptions.

Move the autoregressive lag term to the left and use $LX_t=X_{t-1}$: $$ (1-0.6L)X_t=2+(1+0.4L)\varepsilon_t. $$ Thus $\phi(L)=1-0.6L$, $\theta(L)=1+0.4L$, and $c=2$. To express the model in centered form, first compute its mean using the intercept-mean relationship derived below: $$ \mu=\frac{2}{1-0.6}=5\text{ index points}. $$ The equivalent centered equation is $$ (1-0.6L)(X_t-5)=(1+0.4L)\varepsilon_t. $$ The intercept is 2 index points, while the unconditional mean is 5 index points. They differ because the autoregressive coefficient feeds the mean back through time.

How the Intercept Determines the Mean

For a covariance-stationary process, $E[X_t]$ is the same value $\mu$ at every time. Take expectations in the intercept form, $\phi(L)X_t=c+\theta(L)\varepsilon_t$. Each lag of $X_t$ has expectation $\mu$, and each innovation lag has expectation zero. Consequently, the left side has expectation $\mu(1-\phi_1-\cdots-\phi_p)$, while the right side has expectation $c$.

Result: If $\phi(1)=1-\phi_1-\cdots-\phi_p\ne 0$, then the unconditional mean in the intercept form is $$ \mu=\frac{c}{\phi(1)} =\frac{c}{1-\phi_1-\cdots-\phi_p}. $$ Equivalently, $c=\mu\phi(1)$. The moving-average coefficients do not appear because all innovation terms have mean zero.

This result depends on the model and assumptions: the process has a finite, time-invariant mean, and the innovations have mean zero. It is not a formula for the mean of any arbitrary series. If $\phi(1)=0$, the equation for the mean reduces to $c=0$; it does not determine a unique mean from the intercept. The formula involving division by $\phi(1)$ applies only when that quantity is nonzero.

The derivation can also be seen by writing the centered model in the uncentered form. Since $X_t=(X_t-\mu)+\mu$, $$ \phi(L)X_t=\phi(L)(X_t-\mu)+\mu\phi(L)1. $$ A constant sequence is unchanged by a lag, so $\phi(L)1=1-\phi_1-\cdots-\phi_p=\phi(1)$. Substitution of the centered model therefore gives $$ \phi(L)X_t=\theta(L)\varepsilon_t+\mu\phi(1). $$ Thus the intercept is $c=\mu\phi(1)$, as claimed. This also shows directly why the autoregressive coefficients, rather than the moving-average coefficients, control the conversion.

Worked Example: Recovering the Mean in an ARMA(2,1)

Consider a hypothetical process measured in monthly index points: $$ X_t=1.4+0.4X_{t-1}-0.1X_{t-2} +\varepsilon_t+0.3\varepsilon_{t-1}. $$ Assume zero-mean innovations with constant finite variance and zero covariance across distinct times. Here $|0.4|+|-0.1|=0.5<1$, which is a sufficient contraction condition for a stationary causal solution with this finite moving-average component.

Moving the autoregressive terms to the left gives $$ (1-0.4L+0.1L^2)X_t=1.4+(1+0.3L)\varepsilon_t. $$ The autoregressive polynomial evaluated at one is $$ \phi(1)=1-0.4+0.1=0.7. $$ Therefore the unconditional mean is $$ \mu=\frac{c}{\phi(1)}=\frac{1.4}{0.7}=2.00\text{ index points}. $$ The centered representation is $$ (1-0.4L+0.1L^2)(X_t-2) =(1+0.3L)\varepsilon_t. $$ Notice that the $0.3$ moving-average coefficient affects the short-run innovation terms, but not the mean calculation.

Translating Between the Two Forms

A reliable conversion has two steps: identify the autoregressive polynomial with the stated sign convention, then use $c=\mu\phi(1)$ or $\mu=c/\phi(1)$. The centered form is useful when the model is described in terms of deviations from its long-run level. The intercept form is often convenient for writing an equation directly in the observed series. Either form describes the same model when the conversion is done consistently.

Worked Example: Converting a Centered Model to an Intercept

Suppose a hypothetical quarterly output index has unconditional mean $\mu=12$ index points and satisfies $$ (1-0.5L-0.2L^2)(X_t-12) =(1+0.25L-0.1L^2)\varepsilon_t. $$ Assume zero-mean innovations with constant finite variance and zero covariance across distinct times. The autoregressive coefficients have absolute values summing to $0.7<1$, a sufficient condition for a stationary causal solution under these assumptions.

The autoregressive polynomial evaluated at one is $$ \phi(1)=1-0.5-0.2=0.3. $$ The intercept in the uncentered equation must therefore be $$ c=\mu\phi(1)=12(0.3)=3.60\text{ index points}. $$ So the equivalent intercept form is $$ (1-0.5L-0.2L^2)X_t =3.60+(1+0.25L-0.1L^2)\varepsilon_t. $$ For example, expanding the left side gives $X_t-0.5X_{t-1}-0.2X_{t-2}$. The intercept is 3.60 index points, not 12: 12 is the unconditional mean.

Common Mistake: Calling the Intercept the Mean

A frequent error is to read $c$ in $\phi(L)X_t=c+\theta(L)\varepsilon_t$ and report it as the mean. That is only correct in special cases, such as when the autoregressive coefficients sum to zero, so that $\phi(1)=1$. In an AR(1) model, for instance, $X_t=c+\phi_1X_{t-1}+\varepsilon_t$ has mean $c/(1-\phi_1)$, not $c$, provided the stationary mean exists.

The correction is to take expectations or evaluate the autoregressive polynomial at one before interpreting the constant. A large positive sum of autoregressive coefficients can make $\phi(1)$ small, so a relatively small intercept can correspond to a larger mean. If the sum is negative, $\phi(1)$ can exceed one, and the mean can be smaller in magnitude than the intercept. The moving-average coefficients do not alter this unconditional-mean calculation under zero-mean innovations.

Key takeaway: In the convention $\phi(L)=1-\sum_{i=1}^{p}\phi_iL^i$, the intercept and mean satisfy $c=\mu\phi(1)$. State which form of the ARMA model is being used, keep the polynomial signs consistent, and do not equate the intercept with the unconditional mean without checking $\phi(1)$.

Check Your Understanding

Use the sign convention and assumptions stated in this tutorial.

Write $X_t=1+0.3X_{t-1}+\varepsilon_t-0.2\varepsilon_{t-1}$ in lag-polynomial intercept form. Identify $\phi(L)$, $\theta(L)$, and $c$.
For an AR(1) intercept model with $c=1.2$ and $\phi_1=0.4$, calculate the unconditional mean, assuming a stationary process and zero-mean innovations.
In an ARMA(2,1) model, $\phi_1=0.2$, $\phi_2=0.3$, and $c=2.5$. Find $\phi(1)$ and the mean, assuming $\phi(1)\ne0$.
Explain why changing a moving-average coefficient does not change the unconditional mean under the zero-mean innovation assumption.
If $\phi(1)=0$, what restriction does the expectation equation place on $c$, and why can the usual formula for $\mu$ not be used?