Expanding the Multiplicative SARIMA Operator

Why Expand a Multiplicative SARIMA Operator?

A seasonal ARIMA specification is often written compactly as a product of ordinary and seasonal lag polynomials. That notation keeps the model organized, but it can hide terms that arise when the factors are multiplied. Expanding the product makes every lag and coefficient explicit, which is useful for checking a specification or writing the model as an equation in observations and innovations.

The key algebraic point is straightforward: multiplying an ordinary term at lag \(i\) by a seasonal term at lag \(js\) produces a term at lag \(i+js\). Its coefficient is the product of the two coefficients, including their signs. This tutorial focuses on carrying out that multiplication correctly. An expansion by itself does not establish that a model is stationary or invertible.

Definition: The backshift operator \(B\) acts on a time series by \(By_t=y_{t-1}\), and hence \(B^k y_t=y_{t-k}\). For seasonal period \(s\), the seasonal backshift is \(B^s\), so \(B^{js}y_t=y_{t-js}\). A lag polynomial is a finite polynomial in \(B\) applied to a series.

Ordinary and Seasonal Polynomial Conventions

For an ordinary autoregressive polynomial of order \(p\), use \(\phi(B)=1-\phi_1B-\cdots-\phi_pB^p\). Its seasonal counterpart of order \(P\) and period \(s\) is \(\Phi(B^s)=1-\Phi_1B^s-\cdots-\Phi_PB^{Ps}\). The minus signs are part of the conventional AR-polynomial definition; the values of the parameters themselves may be positive or negative.

For moving-average terms, use \(\theta(B)=1+\theta_1B+\cdots+\theta_qB^q\) and \(\Theta(B^s)=1+\Theta_1B^s+\cdots+\Theta_QB^{Qs}\). The plus signs are conventional here, too. Thus a negative moving-average parameter appears as a negative coefficient in the polynomial. Keeping these conventions visible prevents a common error: copying an AR sign convention into an MA factor, or changing a parameter’s sign twice.

A multiplicative SARIMA model without a constant can be written \(\Phi(B^s)\phi(B)(1-B)^d(1-B^s)^D y_t =\Theta(B^s)\theta(B)\varepsilon_t\). Here \(d\) is the ordinary differencing order, \(D\) is the seasonal differencing order, and \(\varepsilon_t\) is white noise with mean zero and variance \(\sigma^2\). White noise means that the innovations have zero covariance at distinct times; independent, identically distributed innovations are a stronger assumption sometimes imposed. The finite polynomial expansions below are algebraic identities and do not depend on a stationarity or invertibility condition. Those conditions concern the stochastic behavior of the specified model, not whether the multiplication is valid.

Key result: If \(A(B)=\sum_i a_iB^i\) and \(C(B^s)=\sum_j c_jB^{js}\), then \(A(B)C(B^s)=\sum_i\sum_j a_i c_j B^{i+js}\). At any particular lag \(k\), add the products \(a_i c_j\) for all pairs satisfying \(i+js=k\). If several pairs produce the same power, their coefficients must be combined.

Why the Cross-Product Has Lag \(i+js\)

The rule follows directly from the definition of the backshift. Applying the two operators in succession gives \(B^i(B^{js}y_t)=B^i y_{t-js}=y_{t-js-i}\). This is the observation at lag \(i+js\), so the operator product is \(B^{i+js}\). Its coefficient is the product of the two original coefficients because ordinary polynomial multiplication multiplies coefficients when multiplying terms.

More explicitly, distribute every term of \(A(B)\) across every term of \(C(B^s)\): \[ (a_iB^i)(c_jB^{js})=a_ic_jB^{i+js}. \] Then collect terms with equal powers of \(B\). This last step matters when the seasonal and ordinary factors have terms that can produce the same lag. Do not assume the resulting coefficient list contains every integer lag, or that each lag appears only once.

Write each factor with its signs.
Use the stated AR or MA convention and substitute the parameter values exactly once.

Distribute the factors.
Multiply each term in one polynomial by every term in the other, including the constant terms.

Track powers and coefficients.
Add exponents of \(B\), multiply signed coefficients, and combine any repeated powers.

Apply the result to the series.
Replace each \(B^k y_t\) by \(y_{t-k}\), or each \(B^k\varepsilon_t\) by \(\varepsilon_{t-k}\).

Worked Example: Expanding an Ordinary AR(1) and Seasonal AR(1)

A hypothetical monthly claims index \(w_t\) is modeled with ordinary AR parameter \(\phi_1=0.6\) and seasonal AR parameter \(\Phi_1=0.4\) at period \(s=3\). Expand the AR operator and write the resulting equation. Also identify how many terms come from the nonconstant seasonal term.

Method. Use the AR convention \(\phi(B)=1-\phi_1B\) and \(\Phi(B^3)=1-\Phi_1B^3\). Thus the factors are \(1-0.6B\) and \(1-0.4B^3\).

Multiply term by term. The constant \(1\) in the seasonal factor reproduces the ordinary polynomial \(1-0.6B\). The nonconstant term \(-0.4B^3\) multiplies both ordinary terms: \((-0.4B^3)(1)=-0.4B^3\), and \((-0.4B^3)(-0.6B)=+0.24B^4\). Therefore,

$$ (1-0.6B)(1-0.4B^3) =1-0.6B-0.4B^3+0.24B^4. $$

There are exactly two terms contributed by the nonconstant seasonal term because the ordinary factor has two terms. The full product has four terms: the two terms reproduced by the seasonal constant and those two cross-products.

Apply the operator. Setting the expanded operator applied to \(w_t\) equal to the innovation gives \(w_t-0.6w_{t-1}-0.4w_{t-3}+0.24w_{t-4}=\varepsilon_t\). The highest lag is \(1+3=4\), as predicted by multiplying the highest powers.

Check and common mistake avoided. The lag-4 coefficient is the product of two negative coefficients, so it is positive. Counting only the seasonal terms, or forgetting that the seasonal nonconstant term multiplies the ordinary \(B\) term as well as its constant, would omit \(+0.24B^4\). Direct distribution gives all four displayed terms.

Worked Example: Expanding an AR(2) with a Seasonal AR(1)

A hypothetical quarterly exposure series \(x_t\) has ordinary AR parameters \(\phi_1=0.5\), \(\phi_2=-0.2\), and seasonal AR parameter \(\Phi_1=0.3\) with period \(s=4\). Find the expanded operator, write the equation, and report the coefficient at lag 5.

Method. The ordinary polynomial is \(1-\phi_1B-\phi_2B^2=1-0.5B+0.2B^2\). The plus sign on \(B^2\) results from subtracting \(\phi_2=-0.2\). The seasonal polynomial is \(1-0.3B^4\).

Distribute the factors. The seasonal constant reproduces \(1-0.5B+0.2B^2\). Multiplying the nonconstant seasonal term by each ordinary term gives \((-0.3B^4)(1)=-0.3B^4\), \((-0.3B^4)(-0.5B)=+0.15B^5\), and \((-0.3B^4)(+0.2B^2)=-0.06B^6\). Consequently,

$$ (1-0.5B+0.2B^2)(1-0.3B^4) =1-0.5B+0.2B^2-0.3B^4+0.15B^5-0.06B^6. $$

The lag-5 coefficient is \(+0.15\), obtained as \((-0.5)(-0.3)=+0.15\). Applying the polynomial to the series yields \(x_t-0.5x_{t-1}+0.2x_{t-2}-0.3x_{t-4}+0.15x_{t-5}-0.06x_{t-6}=\varepsilon_t\). There is no lag-3 term because no pair of powers in the factors adds to 3. The largest lag is \(2+4=6\).

Check and common mistake avoided. The ordinary \(B^2\) coefficient is \(+0.2\), not \(-0.2\), because of the AR definition and the negative parameter. The cross-product at lag 6 is therefore negative: \((+0.2)(-0.3)=-0.06\). Substituting the parameter into the polynomial before distributing avoids losing that sign.

Moving-Average Products Use the Same Algebra

The multiplication rule is identical for MA factors; only the chosen polynomial convention differs. With \(\theta(B)=1+\theta_1B+\cdots\) and \(\Theta(B^s)=1+\Theta_1B^s+\cdots\), apply the product to the innovation series. The resulting terms are lagged innovations, not lagged observations. Keep the sign of each MA parameter as given.

Worked Example: Expanding Ordinary MA(2) and Seasonal MA(1)

A hypothetical quarterly payment-count series has an ordinary MA polynomial with \(\theta_1=0.5\), \(\theta_2=-0.2\), and a seasonal MA parameter \(\Theta_1=0.3\) at period \(s=4\). Expand the MA operator, write the innovation equation, and identify the coefficient at lag 5.

Method. Under the MA convention, the factors are \(\theta(B)=1+0.5B-0.2B^2\) and \(\Theta(B^4)=1+0.3B^4\).

Multiply term by term. The constant seasonal term reproduces \(1+0.5B-0.2B^2\). The seasonal term \(+0.3B^4\) contributes \((0.3B^4)(1)=0.3B^4\), \((0.3B^4)(0.5B)=0.15B^5\), and \((0.3B^4)(-0.2B^2)=-0.06B^6\). Thus,

$$ (1+0.5B-0.2B^2)(1+0.3B^4) =1+0.5B-0.2B^2+0.3B^4+0.15B^5-0.06B^6. $$

If \(x_t=\theta(B)\Theta(B^4)\varepsilon_t\), then \(x_t=\varepsilon_t+0.5\varepsilon_{t-1}-0.2\varepsilon_{t-2}+0.3\varepsilon_{t-4}+0.15\varepsilon_{t-5}-0.06\varepsilon_{t-6}\). The lag-5 coefficient is \(0.15\), and the highest lag is 6.

Check and common mistake avoided. The lag-6 coefficient is negative because it multiplies \(+0.3\) by \(-0.2\). A frequent error is to change the sign of the given MA parameter because AR polynomials use minus signs. Here the negative \(\theta_2\) is already represented by the coefficient \(-0.2B^2\); no additional sign change is made.

Expansion Checks and Limits

Several quick checks help catch algebra errors without replacing the full calculation. First, the constant term of each normalized factor is 1, so the product’s constant term is also 1. Second, the largest power is the sum of the largest powers in the two factors, provided their leading coefficients are nonzero. Third, each coefficient should be traceable to a pair of terms in the original factors. If multiple pairs lead to one power, add their signed products rather than listing duplicate powers.

For example, if ordinary and seasonal terms happen to generate the same lag, the coefficient at that lag is the sum of their contributions. The coefficient rule \(\sum_{i+js=k}a_i c_j\) accounts for this automatically. Do not assume every intermediate lag is present: a seasonal factor with period 4 only has powers that are multiples of 4, and its products with ordinary terms create shifts of those lags.

Expansion also does not alter the model’s assumptions or prove a property of the time series. It is an exact rearrangement of finite polynomials. Stationarity, invertibility, differencing choices, and the interpretation of the resulting ACF or PACF require separate reasoning. For an integrated specification, ordinary and seasonal differencing operators can be multiplied in the same way, but they should not be confused with the AR and MA factors.

Key takeaway: Preserve the stated AR and MA sign conventions, distribute every term, add lag exponents, multiply signed coefficients, and combine equal powers. Then apply each resulting \(B^k\) to the appropriate series. These steps make the multiplicative specification explicit without making claims beyond the algebra.

Check Your Understanding

Expand each product carefully. For equation questions, replace each lag power by the corresponding lagged series or innovation.

Expand \((1-0.25B)(1-0.6B^4)\). State the highest lag. (Answer: \(1-0.25B-0.6B^4+0.15B^5\); highest lag 5.)
Expand the MA product \((1+0.3B-0.1B^2)(1-0.5B^3)\). (Answer: \(1+0.3B-0.1B^2-0.5B^3-0.15B^4+0.05B^5\).)
For period \(s=2\), expand the seasonal AR polynomial \((1-0.4B^2)(1+0.2B^4)\). (Answer: \(1-0.4B^2+0.2B^4-0.08B^6\).)
For a hypothetical MA specification \(x_t=(1-0.2B+0.1B^2)(1+0.5B^3)\varepsilon_t\), write the expanded equation in innovations. (Answer: \(x_t=\varepsilon_t-0.2\varepsilon_{t-1}+0.1\varepsilon_{t-2}+0.5\varepsilon_{t-3}-0.1\varepsilon_{t-4}+0.05\varepsilon_{t-5}\).)
Expand \((1-B)(1-0.3B)(1-0.5B^4)\), and identify the coefficient at lag 5. (Answer: \(1-1.3B+0.3B^2-0.5B^4+0.65B^5-0.15B^6\); the lag-5 coefficient is \(0.65\).)

What You'll Learn