Why the Operators Matter
An ARIMA model can be written as a compact equation involving lag polynomials, or as an equation involving differences of the observed series. Both forms describe the same relationship, but the placement of each operator matters: the autoregressive and differencing operators act on the observed series, while the moving-average operator acts on the innovations. Confusing these sides changes the model.
The goal here is to translate between the forms reliably. The examples use models without a constant or drift term, so that the operator algebra is visible without an additional deterministic component. Let \(X_t\) denote the observed series and let \(\varepsilon_t\) denote a white-noise innovation with mean zero and variance \(\sigma_\varepsilon^2\). The algebraic translations themselves do not depend on the particular value of that variance.
The Two Equivalent ARIMA Forms
Under the sign convention used here, the nonseasonal ARIMA\((p,d,q)\) model is written
where \(\phi(B)=1-\phi_1B-\cdots-\phi_pB^p\) and \(\theta(B)=1+\theta_1B+\cdots+\theta_qB^q\). The differencing order is \(d\); the autoregressive polynomial has degree \(p\), and the moving-average polynomial has degree \(q\). With \(W_t=(1-B)^dX_t\), the same equation is
The second form is an ARMA equation for the differenced series \(W_t\). The first form makes explicit how that series is obtained from the observations. In both forms, \(\phi(B)\) acts on \(X_t\) or \(W_t\), not on \(\varepsilon_t\); \(\theta(B)\) acts on \(\varepsilon_t\), not on the observed series.
The innovations are assumed to be white noise: they have mean zero, constant variance \(\sigma_\varepsilon^2\), and zero covariance at distinct times. These assumptions describe the innovation sequence; they do not mean that the observed \(X_t\) is white noise. Usual stationarity and invertibility conditions concern the roots of the AR and MA polynomials, respectively, but are not needed merely to expand or translate the equation.
Why Polynomial Multiplication Gives the Level Equation
The translation is justified directly by how \(B\) acts. For any series \(Z_t\), applying \(aB^j\) multiplies the observation \(Z_{t-j}\) by \(a\). Thus, multiplying two lag polynomials combines their coefficients just as ordinary polynomial multiplication does, with each resulting power \(B^k\) identifying the lag \(k\).
For example, for \(d=1\) and \(\phi(B)=1-\phi_1B\), the left-hand operator is
Applying this polynomial to \(X_t\) gives \(X_t-(1+\phi_1)X_{t-1}+\phi_1X_{t-2}\). Equivalently, applying \(\phi(B)\) to \(\Delta X_t=X_t-X_{t-1}\) gives \(\Delta X_t-\phi_1\Delta X_{t-1}\). Expanding that expression gives the same level equation, which verifies the translation.
Write the model as \(\phi(B)(1-B)^dX_t=\theta(B)\varepsilon_t\), keeping the observed series and innovations on their respective sides.
Set \(W_t=(1-B)^dX_t\), so the equation becomes \(\phi(B)W_t=\theta(B)\varepsilon_t\).
Multiply the left-hand polynomials to express the model in lags of \(X_t\); expand the right-hand polynomial separately to express it in lags of \(\varepsilon_t\).
The left side can reach lag \(p+d\), while the innovation side reaches lag \(q\). Confirm the constant term and every product coefficient.
Worked Example: Expand an ARIMA(1,1,1) Model
A hypothetical monthly claim index, after centering, is modeled by \((1-0.6B)(1-B)X_t=(1+0.4B)\varepsilon_t\). Write the differenced-series form and the expanded equation in levels. Then check both sides using \(X_t=120\), \(X_{t-1}=110\), \(X_{t-2}=105\), \(\varepsilon_t=5\), and \(\varepsilon_{t-1}=5\). The innovation process is assumed to be white noise; the listed values are a numerical equation check, not a specification of its distribution.
Method. First define \(\Delta X_t=(1-B)X_t\). The differenced-series equation is
Next multiply the left-hand polynomials: \((1-0.6B)(1-B)=1-B-0.6B+0.6B^2=1-1.6B+0.6B^2\). The level equation is therefore
Check in differences. \(\Delta X_t=120-110=10\), and \(\Delta X_{t-1}=110-105=5\). The left side is \(10-0.6(5)=10-3=7\). The right side is \(5+0.4(5)=5+2=7\).
Check in levels. The left side is \(120-1.6(110)+0.6(105)=120-176+63=7\). The right side remains \(7\). Both forms agree. The common mistake avoided is applying the moving-average factor to \(X_t\) rather than to the innovations.
Expanding Higher-Order Operators
When \(p\) or \(d\) is greater than one, the same multiplication rule applies. Each coefficient in a product is the sum of the products that contribute to the corresponding power of \(B\). It is useful to expand in a written line rather than infer coefficients by inspection. In particular, \((1-B)^2=1-2B+B^2\), not \(1-2B\): the final term is essential.
Worked Example: Expand an ARIMA(2,1,2) Model
A hypothetical weekly exposure index follows \((1-0.5B+0.2B^2)(1-B)X_t=(1+0.3B-0.1B^2)\varepsilon_t\). Write the differenced equation and fully expand the equation in levels. Check the two forms using \(X_t=200\), \(X_{t-1}=190\), \(X_{t-2}=177\), \(X_{t-3}=169\), and innovations \(\varepsilon_t=4\), \(\varepsilon_{t-1}=5\), \(\varepsilon_{t-2}=4\).
Method. Define \(\Delta X_t=(1-B)X_t\). The differenced equation is
For the level equation, multiply term by term:
Thus the expanded level equation is
Check in differences. The differences are \(10\), \(13\), and \(8\), since \(200-190=10\), \(190-177=13\), and \(177-169=8\). The left side is \(10-0.5(13)+0.2(8)=10-6.5+1.6=5.1\). The right side is \(4+0.3(5)-0.1(4)=4+1.5-0.4=5.1\).
Check in levels. The left side is \(200-1.5(190)+0.7(177)-0.2(169)=200-285+123.9-33.8=5.1\), agreeing with the right side. The common mistake avoided is dropping the \(B^2\) contribution when multiplying the AR polynomial by \(1-B\).
Reading Orders and Keeping the Sides Straight
To identify orders from a polynomial equation, inspect the factors before expanding. The degree of \(\phi(B)\) gives \(p\), the exponent on \(1-B\) gives \(d\), and the degree of \(\theta(B)\) gives \(q\), provided the stated polynomials have nonzero highest-order coefficients. Expanding the left side can make its largest lag look like an AR order, but that lag includes differencing and is not, by itself, the value of \(p\).
Worked Example: Identify Orders and Expand a Second Difference
A hypothetical quarterly service-volume series is represented by \((1-0.4B+0.12B^2)(1-B)^2X_t=(1+0.5B)(1-0.2B)\varepsilon_t\). Determine \(p,d,q\), write the differenced-series equation, and expand both sides into lags.
Method. The AR polynomial has degree 2, the differencing factor is squared, and the MA side is a product of two degree-1 polynomials. Therefore \(p=2\), \(d=2\), and \(q=2\). Define \(\Delta^2X_t=(1-B)^2X_t\). The differenced-series form is
because \((1+0.5B)(1-0.2B)=1-0.2B+0.5B-0.1B^2=1+0.3B-0.1B^2\).
For the level equation, first use \((1-B)^2=1-2B+B^2\), then multiply:
The fully expanded equation is
The left side reaches lag 4 because the degree-2 AR polynomial is multiplied by a degree-2 differencing polynomial. The innovation side reaches lag 2 because the MA product has degree 2. These lag ranges confirm that the multiplication has been carried out consistently. The common mistake avoided is calling the expanded left-side maximum lag, 4, the AR order; the AR order is 2, and the other two lags come from differencing.
Common Translation Traps
The compact equation is often safer than an expanded equation until a specific lag representation is requested. After expansion, check each side independently and keep the original factorization available to verify the orders. These checks are particularly useful when the equation includes more than one factor on the left.
- Putting an operator on the wrong series. \(\theta(B)\) filters innovations; it does not multiply the observed series in the ARIMA representation.
- Forgetting that differencing acts on observations. The factor \((1-B)^d\) belongs with \(X_t\), not with \(\varepsilon_t\).
- Reading the expanded maximum lag as \(p\). The degree of the complete left-hand polynomial can be \(p+d\), even though the AR order is \(p\).
- Changing a sign convention partway through. Here the AR polynomial uses minus signs and the MA polynomial is written with its stated coefficients, which may be positive or negative.
- Expanding only one part of a product. Terms of matching powers must be combined, and every power in the product must be retained.
For example, in the first worked example, writing \((1-0.6B)X_t=(1+0.4B)\varepsilon_t\) is wrong: it omits the differencing factor. The correct equation is \((1-0.6B)(1-B)X_t=(1+0.4B)\varepsilon_t\), or, with \(W_t=\Delta X_t\), \((1-0.6B)W_t=(1+0.4B)\varepsilon_t\). The omitted factor changes the model from one for a differenced series to one that applies the AR polynomial directly to the observed series.
Check Your Understanding
For each problem, keep the observation-side and innovation-side operators separate before expanding.
- A centered hypothetical index follows \((1-0.25B)(1-B)X_t=(1+0.6B)\varepsilon_t\). Write its differenced-series equation and its expanded level equation. (Answer: \(\Delta X_t-0.25\Delta X_{t-1}=\varepsilon_t+0.6\varepsilon_{t-1}\); \(X_t-1.25X_{t-1}+0.25X_{t-2}=\varepsilon_t+0.6\varepsilon_{t-1}\).)
- A model is \((1+0.3B)(1-B)^2X_t=(1-0.4B)\varepsilon_t\). State \((p,d,q)\) and write the differenced-series equation. (Answer: \((p,d,q)=(1,2,1)\); \(\Delta^2X_t+0.3\Delta^2X_{t-1}=\varepsilon_t-0.4\varepsilon_{t-1}\).)
- Expand \((1-0.2B+0.1B^2)(1-B)X_t=(1+0.2B+0.05B^2)\varepsilon_t\). State the ARIMA orders and the level equation. (Answer: \((p,d,q)=(2,1,2)\); \(X_t-1.2X_{t-1}+0.3X_{t-2}-0.1X_{t-3}=\varepsilon_t+0.2\varepsilon_{t-1}+0.05\varepsilon_{t-2}\).)
- A candidate rewrites \((1-0.5B)(1-B)X_t=(1+0.2B)\varepsilon_t\) as \((1-0.5B)X_t=(1+0.2B)\varepsilon_t\). Identify the error and give the corrected differenced-series form. (Answer: The candidate omitted the differencing operator on the observed series; the correct form is \((1-0.5B)\Delta X_t=\varepsilon_t+0.2\varepsilon_{t-1}\).)
- For \((1-0.4B+0.1B^2)(1-B)^2X_t=(1+0.3B)(1-0.2B)\varepsilon_t\), state \((p,d,q)\) and the maximum lag on each expanded side. (Answer: \((p,d,q)=(2,2,2)\); the expanded observation-side maximum lag is 4 and the innovation-side maximum lag is 2.)