Backshift Notation for AR and MA Equations

Why Use Backshift Notation?

An autoregressive or moving-average equation can contain several lagged terms. For example, an equation may refer to the current observation, the observation one time step earlier, and the observation two time steps earlier. Writing every time index is clear, but repeated expressions become cumbersome as the number of lags grows. Backshift notation replaces each lagged series value with an operator, so the equation can be written as a polynomial in that operator.

The notation is a compact way to organize the same terms, not a different model. To interpret a model or check its coefficients, a candidate should be able to move between the operator form and the time-indexed form without changing signs or shifting the wrong series. This tutorial focuses on that translation; it does not use the notation to derive stationarity or dependence properties.

Definition: The backshift operator, written $B$, shifts a time-indexed series back by one time step. For a series $x_t$, $Bx_t=x_{t-1}$. More generally, $B^k x_t=x_{t-k}$ for any nonnegative integer $k$, and $B^0x_t=x_t$. The operator acts on the series immediately to its right.

For example, $B^2x_t=x_{t-2}$: the exponent counts how many time steps to move backward. It does not mean that the value $x_t$ is squared. The same operator can act on an innovation series $\varepsilon_t$, so $B\varepsilon_t=\varepsilon_{t-1}$.

Lag Polynomials

A lag polynomial is an ordinary polynomial whose variable is the operator $B$. For instance, $(1-\phi_1B-\phi_2B^2)x_t$ means that each term in the polynomial acts on $x_t$. Distributing the operator gives $x_t-\phi_1x_{t-1}-\phi_2x_{t-2}$. The polynomial is a compact instruction for combining current and lagged values.

The signs in the polynomial come from the equation being rearranged. In the convention used here, an autoregressive equation with lag coefficients $\phi_j$ is written with those terms on the right in time-indexed form. Moving them to the left produces minus signs in the lag polynomial.

Definition: Under the convention used here, an autoregressive equation of order $p$, with constant $c$, is $$ x_t=c+\phi_1x_{t-1}+\cdots+\phi_px_{t-p}+\varepsilon_t. $$ Define $\phi(B)=1-\phi_1B-\cdots-\phi_pB^p$. The same equation is $\phi(B)x_t=c+\varepsilon_t$. The innovation sequence $\varepsilon_t$ represents the new disturbance at each time; when it is specified as white noise, it has constant mean and variance and zero covariance at nonzero lags.

A moving-average equation instead combines the current innovation with its lagged values. In the convention used here, the coefficients of lagged innovations appear with plus signs in the time-indexed equation, and therefore with plus signs in its lag polynomial. Other presentations may define moving-average coefficients with the opposite sign; always identify the convention in use before interpreting a polynomial.

Definition: Under the convention used here, a moving-average equation of order $q$, with mean $\mu$, is $$ x_t=\mu+\varepsilon_t+\theta_1\varepsilon_{t-1}+\cdots+\theta_q\varepsilon_{t-q}. $$ Define $\theta(B)=1+\theta_1B+\cdots+\theta_qB^q$. Then $x_t=\mu+\theta(B)\varepsilon_t$.

Expanding an AR Equation

To expand a lag-polynomial AR equation, apply each power of $B$ to $x_t$, distribute its coefficient, and then rearrange if the desired form has $x_t$ alone on the left. These steps are ordinary algebra once the shift rule is applied. A constant is not lagged in the polynomial $\phi(B)x_t$; it remains the constant on the right-hand side.

Worked Example: Writing an AR(2) Equation Compactly

Suppose a series satisfies $x_t=2+0.75x_{t-1}-0.20x_{t-2}+\varepsilon_t$, where $\varepsilon_t$ is white noise with mean $0$ and variance $1$. Write the equation using a lag polynomial, then expand the result to check it. The method is to move the lagged $x$-terms to the left and replace each lag by a power of $B$; this avoids carrying a time-index sign incorrectly.

Step 1: Move the autoregressive terms. Subtract $0.75x_{t-1}$ from both sides and add $0.20x_{t-2}$ to both sides:

x_t-0.75x_{t-1}+0.20x_{t-2}=2+\varepsilon_t.

Step 2: Replace lagged values by operator terms. Since $Bx_t=x_{t-1}$ and $B^2x_t=x_{t-2}$, the left side is $(1-0.75B+0.20B^2)x_t$. Thus the compact form is

(1-0.75B+0.20B^2)x_t=2+\varepsilon_t.

Step 3: Check by expanding. Applying the polynomial gives $x_t-0.75x_{t-1}+0.20x_{t-2}=2+\varepsilon_t$. Rearranging returns $x_t=2+0.75x_{t-1}-0.20x_{t-2}+\varepsilon_t$, exactly the original equation. The negative lag coefficient in the original equation correctly produced a positive $B^2$ term after being moved to the left.

Expanding an MA Equation

For a moving-average equation, the polynomial acts on the innovation sequence rather than on the observed series. Thus $\theta(B)\varepsilon_t$ expands into $\varepsilon_t$ and lagged innovations. The response $x_t$ is not itself shifted by this polynomial. Keeping track of which series follows $B$ is essential.

Worked Example: Translating an MA(2) Polynomial

Suppose $y_t=5+(1+0.40B-0.30B^2)u_t$, where $u_t$ is white noise with mean $0$ and variance $4$. Expand the equation into time-indexed form, and then identify the order and coefficients. The method is to apply each operator power to $u_t$, not to $y_t$; that distinction avoids a common series-substitution error.

Step 1: Apply the lag powers. The constant term $1$ gives $u_t$, $0.40B$ gives $0.40u_{t-1}$, and $-0.30B^2$ gives $-0.30u_{t-2}$. Therefore,

y_t=5+u_t+0.40u_{t-1}-0.30u_{t-2}.

Step 2: Read the order and coefficients. The largest innovation lag is 2, so this is an MA(2) equation. In the stated convention, $\mu=5$, $\theta_1=0.40$, and $\theta_2=-0.30$.

Step 3: Verify the polynomial. Applying $1+0.40B-0.30B^2$ to $u_t$ gives $u_t+0.40u_{t-1}-0.30u_{t-2}$, which is the innovation part of the expanded equation. The variance of $u_t$ is not needed for this algebraic translation.

Translating in Both Directions

An equation may already be written as a lag polynomial, or it may be given in time-indexed form. In either case, the reliable method is to label each operator by the series it acts on, replace $B^k$ by a lag of $k$, and preserve the coefficient signs. If a polynomial is a product of simpler polynomials, multiply the polynomials first or distribute them across the series; both approaches must give the same coefficients.

Worked Example: Multiplying Lag Polynomials Before Expanding

Suppose $(1-0.50B)(1+0.25B)x_t=6+\varepsilon_t$, where $\varepsilon_t$ is white noise with mean $0$ and variance $9$. Write the equation as a time-indexed AR equation. The method is to multiply the polynomials, then apply the resulting terms to $x_t$; this avoids dropping the cross-product when multiplying lag polynomials.

Step 1: Multiply the polynomials. The constant product is $1$. The coefficient of $B$ is $0.25-0.50=-0.25$, and the coefficient of $B^2$ is $(-0.50)(0.25)=-0.125$. Hence

$$ (1-0.50B)(1+0.25B)=1-0.25B-0.125B^2. $$

Step 2: Apply the polynomial. The given equation becomes $x_t-0.25x_{t-1}-0.125x_{t-2}=6+\varepsilon_t$.

Step 3: Isolate the current observation. Add $0.25x_{t-1}$ and $0.125x_{t-2}$ to both sides:

x_t=6+0.25x_{t-1}+0.125x_{t-2}+\varepsilon_t.

Step 4: Check. Moving the two lag terms back to the left gives coefficients $-0.25$ and $-0.125$, matching the multiplied polynomial. The constant remains 6, and the innovation term remains $\varepsilon_t$.

A Short Justification of the Polynomial Form

The compact notation works because the backshift operator is linear: for constants $a$ and $b$, $B(ax_t+by_t)=aBx_t+bBy_t$. Its repeated application gives $B^kx_t=x_{t-k}$. Consequently, applying the AR polynomial term by term gives

(1-\phi_1B-\cdots-\phi_pB^p)x_t =x_t-\phi_1x_{t-1}-\cdots-\phi_px_{t-p}.

Starting with the time-indexed AR equation and subtracting every lagged $x$-term from both sides produces precisely this expression, equal to $c+\varepsilon_t$. This establishes the compact form directly; it is algebraic and does not require a distributional assumption or stationarity. Similarly, applying $1+\theta_1B+\cdots+\theta_qB^q$ to $\varepsilon_t$ gives the current innovation plus the stated lagged innovations, which justifies the compact MA form.

Common Exam Traps

Reading $B^2x_t$ as a square. It is $x_{t-2}$, not $(x_t)^2$.
Applying the polynomial to the wrong series. In an MA equation, $\theta(B)$ acts on $\varepsilon_t$; in the AR polynomial form, $\phi(B)$ acts on $x_t$.
Changing signs without moving terms. In an AR polynomial, the signs reflect moving lagged $x$-terms to the left. Expand the polynomial and compare it with the rearranged equation rather than guessing the signs.
Changing the convention for MA coefficients silently. Here the time-indexed MA form uses plus signs for its coefficients. A different convention must be identified explicitly.
Lagging the intercept. The constant $c$ in $\phi(B)x_t=c+\varepsilon_t$ remains $c$; it is not replaced by a time-indexed series term.

Key takeaway: $B^k$ shifts the series immediately to its right back $k$ time steps. Expand each polynomial term against that series, preserve signs, and verify the result by rearranging it into the original equation.

Check Your Understanding

For each question, translate the given equation as requested. Use the convention stated in the question.

Write $(1-0.40B)z_t=3+e_t$ in time-indexed form. (Answer: $z_t=3+0.40z_{t-1}+e_t$)
Expand $w_t=2+(1-0.60B+0.10B^2)a_t$ into time-indexed form. (Answer: $w_t=2+a_t-0.60a_{t-1}+0.10a_{t-2}$)
Write $u_t+0.20u_{t-1}+0.05u_{t-2}=e_t$ in lag-polynomial form. (Answer: $(1+0.20B+0.05B^2)u_t=e_t$)
Expand $(1-0.70B)(1+0.10B)y_t=4+e_t$, then isolate $y_t$. (Answer: $y_t=4+0.60y_{t-1}+0.07y_{t-2}+e_t$)
Write $x_t-0.30x_{t-1}=1+e_t+0.80e_{t-1}$ in lag-polynomial form, leaving the innovation polynomial factored if desired. (Answer: $(1-0.30B)x_t=1+(1+0.80B)e_t$)

What You'll Learn