ARMA Models in Lag-Operator Form

Why Use the Lag Operator?

An ARMA equation can contain several lagged observations and several lagged innovations. Writing each term separately is clear, but becomes cumbersome as the model orders increase. The backshift operator packages those lags into a polynomial expression, making the AR and MA parts easy to identify and manipulate.

This notation is compact rather than mysterious: each power of the operator simply shifts a time index backward. The main exam risks are losing a sign when moving terms across the equation and confusing an AR coefficient with the sign of its polynomial coefficient.

Definition: For a time series \(X_t\), the backshift operator \(B\) is defined by \(BX_t=X_{t-1}\). Repeated application gives \(B^jX_t=X_{t-j}\) for any nonnegative integer \(j\). The same rule applies to an innovation sequence \(\varepsilon_t\): \(B^j\varepsilon_t=\varepsilon_{t-j}\).

For this tutorial, use the following sign convention. An AR polynomial has alternating subtraction in its displayed coefficients, while an MA polynomial is written with plus signs for its displayed coefficients. The symbols \(\phi_j\) and \(\theta_j\) denote the model coefficients, not necessarily the signs that appear in the polynomials.

\phi(B)=1-\phi_1B-\phi_2B^2-\cdots-\phi_pB^p, \qquad \theta(B)=1+\theta_1B+\theta_2B^2+\cdots+\theta_qB^q.

An ARMA\((p,q)\) model in lag-operator form is

\phi(B)X_t=c+\theta(B)\varepsilon_t.

Here \(c\) is a constant, and \(\varepsilon_t\) is a white-noise sequence with mean zero, variance \(\sigma^2\), and zero covariance between distinct time points. The equation itself does not require us to impose stationarity in order to translate between the two notations. A model is described as ARMA\((p,q)\) when its AR and MA polynomials have orders \(p\) and \(q\), respectively; in particular, the coefficients at those highest orders are nonzero.

Reading and Expanding the Polynomials

The \(j\)th power of \(B\) shifts a variable back by \(j\) time periods. Thus, for example, \(B^2X_t=X_{t-2}\). Applying each polynomial to its series gives

\phi(B)X_t=X_t-\phi_1X_{t-1}-\cdots-\phi_pX_{t-p}, \qquad \theta(B)\varepsilon_t=\varepsilon_t+\theta_1\varepsilon_{t-1}+\cdots+\theta_q\varepsilon_{t-q}.

Consequently, the time-domain form of the model is

X_t-\phi_1X_{t-1}-\cdots-\phi_pX_{t-p} = c+\varepsilon_t+\theta_1\varepsilon_{t-1}+\cdots+\theta_q\varepsilon_{t-q}.

Solving this equation for \(X_t\) moves the lagged \(X\)-terms to the right-hand side. It does not change the signs of the innovation terms already on that side:

X_t=c+\phi_1X_{t-1}+\cdots+\phi_pX_{t-p} +\varepsilon_t+\theta_1\varepsilon_{t-1}+\cdots+\theta_q\varepsilon_{t-q}.

Key takeaway: In the convention used here, the AR polynomial coefficient on \(B^j\) is \(-\phi_j\), while the time-domain AR coefficient on \(X_{t-j}\) is \(+\phi_j\). The MA polynomial coefficient on \(B^j\) is \(+\theta_j\), matching the time-domain coefficient on \(\varepsilon_{t-j}\).

Worked Translations

Worked Example: Expand an ARMA(2,1) Operator Equation

A hypothetical monthly claim-count series is specified by \((1-0.6B+0.2B^2)X_t=4+(1+0.5B)\varepsilon_t\). Write its time-domain equation, identify the polynomial coefficients, and verify the signs in the solved form.

Method. Expand each polynomial by applying every power of \(B\) to the series immediately following it. The common mistake to avoid is reading the \(+0.2B^2\) term as a positive time-domain AR coefficient.

Step 1: Expand the AR side. Since \(BX_t=X_{t-1}\) and \(B^2X_t=X_{t-2}\), \[ (1-0.6B+0.2B^2)X_t =X_t-0.6X_{t-1}+0.2X_{t-2}. \]

Step 2: Expand the MA side. Since \(B\varepsilon_t=\varepsilon_{t-1}\), \[ 4+(1+0.5B)\varepsilon_t=4+\varepsilon_t+0.5\varepsilon_{t-1}. \]

Therefore, the equation before solving for \(X_t\) is \[ X_t-0.6X_{t-1}+0.2X_{t-2} =4+\varepsilon_t+0.5\varepsilon_{t-1}. \]

Step 3: Isolate the current observation. Add \(0.6X_{t-1}\) and subtract \(0.2X_{t-2}\) on both sides: \[ X_t=4+0.6X_{t-1}-0.2X_{t-2} +\varepsilon_t+0.5\varepsilon_{t-1}. \]

The model has AR order \(p=2\) and MA order \(q=1\). Comparing with \(\phi(B)=1-\phi_1B-\phi_2B^2\), we read \(\phi_1=0.6\) and \(\phi_2=-0.2\). Comparing with \(\theta(B)=1+\theta_1B\), we read \(\theta_1=0.5\). The negative time-domain coefficient on \(X_{t-2}\) agrees with \(\phi_2=-0.2\). The sign check confirms the expansion.

Worked Example: Recover Polynomials from a Time-Domain Equation

A hypothetical index series satisfies \[ X_t=2+0.7X_{t-1}-0.1X_{t-2} +\varepsilon_t-0.4\varepsilon_{t-1}+0.3\varepsilon_{t-2}. \] Write its lag-operator form and identify \(p\), \(q\), and all polynomial coefficients.

Method. First move all lagged \(X\)-terms to the left. Then collect the current and lagged innovations on the right. The mistake to avoid is copying the time-domain coefficient on \(X_{t-2}\) directly into the AR polynomial without changing its sign.

Step 1: Move the lagged observations. Subtract \(0.7X_{t-1}\) from both sides and add \(0.1X_{t-2}\) to both sides: \[ X_t-0.7X_{t-1}+0.1X_{t-2} =2+\varepsilon_t-0.4\varepsilon_{t-1}+0.3\varepsilon_{t-2}. \]

Step 2: Express the left side as an AR polynomial. The coefficients on \(B\) and \(B^2\) are \(-0.7\) and \(+0.1\), respectively: \[ \phi(B)=1-0.7B+0.1B^2. \] In the convention \(\phi(B)=1-\phi_1B-\phi_2B^2\), this means \(\phi_1=0.7\) and \(\phi_2=-0.1\).

Step 3: Express the innovation terms as an MA polynomial. The right side has coefficients \(1\), \(-0.4\), and \(0.3\) on \(\varepsilon_t\), \(\varepsilon_{t-1}\), and \(\varepsilon_{t-2}\): \[ \theta(B)=1-0.4B+0.3B^2. \] Thus \(\theta_1=-0.4\) and \(\theta_2=0.3\).

The lag-operator equation is \((1-0.7B+0.1B^2)X_t=2+(1-0.4B+0.3B^2)\varepsilon_t\), with \(p=2\) and \(q=2\). Expanding it reproduces the given equation after isolating \(X_t\), so the translation is checked.

Why the Operator Form Is Equivalent

The equivalence follows directly from the definition of \(B\), not from a special property of an ARMA process. For example, distribute the AR polynomial over \(X_t\):

\begin{aligned} (1-\phi_1B-\phi_2B^2)X_t &=1X_t-\phi_1(BX_t)-\phi_2(B^2X_t)\\ &=X_t-\phi_1X_{t-1}-\phi_2X_{t-2}. \end{aligned}

Each term in the expansion is determined by substituting \(B^jX_t=X_{t-j}\). For the MA polynomial, the same distribution gives

\begin{aligned} (1+\theta_1B+\theta_2B^2)\varepsilon_t &=\varepsilon_t+\theta_1(B\varepsilon_t)+\theta_2(B^2\varepsilon_t)\\ &=\varepsilon_t+\theta_1\varepsilon_{t-1}+\theta_2\varepsilon_{t-2}. \end{aligned}

Substitution of these two expansions into \(\phi(B)X_t=c+\theta(B)\varepsilon_t\) yields the time-domain equation term by term. Conversely, grouping the lagged observations and innovations in a time-domain equation produces exactly the corresponding polynomials. This justification also shows the scope of the notation: the operator is acting on a series, and the polynomial is a compact way to list its lagged terms.

Worked Example: Evaluate Both Sides at One Time Point

A hypothetical quarterly reserve index follows \[ (1-0.4B)X_t=1.7+(1+0.2B-0.1B^2)\varepsilon_t. \] At time \(t=5\), suppose \(X_5=5\), \(X_4=5\), \(\varepsilon_5=1\), \(\varepsilon_4=2\), and \(\varepsilon_3=1\). Expand the equation and check it numerically at \(t=5\).

Method. Translate the operator equation first, then substitute the values at the matching time indices. A frequent trap is using \(\varepsilon_2\) for the \(B^2\varepsilon_5\) term; two shifts from time 5 give time 3.

Step 1: Expand the left side. \[ (1-0.4B)X_t=X_t-0.4X_{t-1}. \]

Step 2: Expand the right side. \[ 1.7+(1+0.2B-0.1B^2)\varepsilon_t =1.7+\varepsilon_t+0.2\varepsilon_{t-1}-0.1\varepsilon_{t-2}. \]

The complete time-domain equation is \[ X_t-0.4X_{t-1} =1.7+\varepsilon_t+0.2\varepsilon_{t-1}-0.1\varepsilon_{t-2}. \] Solving for \(X_t\) gives \[ X_t=1.7+0.4X_{t-1}+\varepsilon_t+0.2\varepsilon_{t-1}-0.1\varepsilon_{t-2}. \]

Step 3: Evaluate the left side at \(t=5\). \[ X_5-0.4X_4=5-0.4(5)=5-2=3. \]

Step 4: Evaluate the right side at \(t=5\). \[ 1.7+\varepsilon_5+0.2\varepsilon_4-0.1\varepsilon_3 =1.7+1+0.2(2)-0.1(1)=1.7+1+0.4-0.1=3. \] Both sides equal \(3\), so the numerical check passes. The example confirms that \(B^2\) shifts the index back two periods and that the negative MA coefficient remains negative in the time-domain equation.

Common Traps and a Reliable Check

When moving quickly between notations, it helps to separate three tasks: determine which series the polynomial acts on, apply the lag shifts, and only then interpret coefficients. In particular, an AR polynomial's displayed signs are not the same as the corresponding time-domain signs.

Identify the target series.
In \(\phi(B)X_t\), every power of \(B\) shifts \(X_t\). In \(\theta(B)\varepsilon_t\), every power shifts \(\varepsilon_t\).

Replace each power explicitly.
Write \(B^jX_t=X_{t-j}\) or \(B^j\varepsilon_t=\varepsilon_{t-j}\) before simplifying signs.

Move terms only after expansion.
When isolating \(X_t\), move the lagged \(X\)-terms across the equality and keep the innovation terms on their existing side.

Check the highest lags.
The largest power of \(B\) in \(\phi(B)\) gives \(p\), and the largest power in \(\theta(B)\) gives \(q\), provided the corresponding coefficients are nonzero.

A coefficient sign error can turn a different model into the apparent answer. For instance, if \(\phi(B)=1-0.3B+0.2B^2\), the correct expansion is \(X_t-0.3X_{t-1}+0.2X_{t-2}\). After solving for \(X_t\), the lagged-observation terms become \(+0.3X_{t-1}-0.2X_{t-2}\). Writing \(+0.2X_{t-2}\) in the solved equation would be inconsistent with the original polynomial.

The operator form is useful because it provides a compact, systematic representation of the AR and MA lag structures. It also makes it straightforward to communicate model order and coefficient conventions. For aligned work with the SOA Exam SRM (Statistics for Risk Modeling) syllabus, translate an expression term by term rather than relying on a memorized pattern.

Check Your Understanding

For each problem, expand or collect the terms as requested, and check coefficient signs against the stated convention.

Expand \((1+0.3B-0.2B^2)X_t=(1-0.6B)\varepsilon_t\) into a time-domain equation, then solve for \(X_t\). (Answer: \(X_t+0.3X_{t-1}-0.2X_{t-2}=\varepsilon_t-0.6\varepsilon_{t-1}\); hence \(X_t=-0.3X_{t-1}+0.2X_{t-2}+\varepsilon_t-0.6\varepsilon_{t-1}\).)
Write the lag-operator form and identify the orders for \(X_t=-1+0.5X_{t-1}+0.25X_{t-3}+\varepsilon_t+0.4\varepsilon_{t-2}\). (Answer: \((1-0.5B-0.25B^3)X_t=-1+(1+0.4B^2)\varepsilon_t\); \(p=3,\ q=2\).)
For \((1-0.2B+0.1B^2)X_t=3+(1-0.5B+0.2B^2)\varepsilon_t\), give the expanded equation and the values of \(\phi_1,\phi_2,\theta_1,\theta_2\). (Answer: \(X_t-0.2X_{t-1}+0.1X_{t-2}=3+\varepsilon_t-0.5\varepsilon_{t-1}+0.2\varepsilon_{t-2}\); \(\phi_1=0.2,\ \phi_2=-0.1,\ \theta_1=-0.5,\ \theta_2=0.2\).)
Suppose \((1+0.5B)X_t=2+(1-0.25B+0.1B^2)\varepsilon_t\), with \(X_6=4,\ X_5=2,\ \varepsilon_6=1,\ \varepsilon_5=4,\ \varepsilon_4=2\). Evaluate each side at \(t=6\) and determine whether the values are equal. (Answer: left side \(=4+0.5(2)=5\); right side \(=2+1-0.25(4)+0.1(2)=2.2\); they are not equal.)
A time-domain model is \(X_t=5-0.4X_{t-1}+0.2X_{t-2}+\varepsilon_t-0.3\varepsilon_{t-1}\). Move its lagged observations to the left, write both polynomials, and state the orders. (Answer: \((1+0.4B-0.2B^2)X_t=5+(1-0.3B)\varepsilon_t\); \(\phi(B)=1+0.4B-0.2B^2,\ \theta(B)=1-0.3B\), with \(p=2,\ q=1\).)

What You'll Learn