Backward Shift Operator

Study guide: Exam MAS-II

The backward shift operator $B$ (also called the backshift or lag operator) maps a value of a Time Series to the value one period earlier, $Bx_t = x_{t-1}$ . Writing a model as polynomials in $B$ turns the autoregressive and moving average relationships between current and past values into algebra that can be expanded, factored and solved.

$B^{k}x_t = x_{t-k}$

$\nabla x_t = x_t - x_{t-1} = (1 - B)\,x_t$

$\phi(B)\,(1 - B)^{d}\,x_t = \theta(B)\,\varepsilon_t$

$\phi(B) = 1 - \phi_1 B - \cdots - \phi_p B^p$ is the AR polynomial and $\theta(B) = 1 + \theta_1 B + \cdots + \theta_q B^q$ the MA polynomial. $\varepsilon_t$ is white noise and $d$ the number of differences, which together make the ARIMA $(p, d, q)$ model. Cowpertwait writes the same model as $\theta_p(B)x_t = \phi_q(B)w_t$ , with $\alpha_i$ for the AR coefficients and $\beta_j$ for the MA ones. The letters are swapped relative to this vault's Box–Jenkins notation, so check which polynomial sits on which side.
Rules. $B$ is linear, leaves a constant unchanged ( $Bc = c$ ), and its powers add: $B^jB^k = B^{j+k}$ . Operator polynomials therefore multiply like ordinary polynomials, e.g. $(1 - 0.5B)(1 - B) = 1 - 1.5B + 0.5B^2$ .
Differencing. Differencing is $\nabla = 1 - B$ , repeated as $\nabla^d = (1 - B)^d$ . The seasonal difference is $\nabla_s = 1 - B^s$ , with $s = 4$ for quarterly and $s = 12$ for monthly data. A factor $(1 - B)$ inside an AR polynomial is a unit root: factor it out and it becomes the "I" in ARIMA.
Stationarity and invertibility. Treat $B$ as a number and solve $\phi(B) = 0$ . The model is stationary when every root exceeds $1$ in absolute value. The same test on $\theta(B) = 0$ checks invertibility. A root at exactly $B = 1$ is the unit root of a Random Walk.
Inverting an operator. For $\lvert\phi\rvert < 1$ , $(1 - \phi B)^{-1} = 1 + \phi B + \phi^2B^2 + \cdots$ , so an AR(1) is an MA( $\infty$ ) with weights $\psi_j = \phi^j$ . These $\psi$ weights set the width of a prediction interval.

Identifying an ARIMA Model from an Expanded Equation A quarterly claim-severity index is modelled as $x_t = 1.5x_{t-1} - 0.5x_{t-2} + \varepsilon_t - 0.3\varepsilon_{t-1}$ Identify $p$ , $d$ and $q$ , and check the remaining ARMA part for stationarity and invertibility.
Answer Collect the $x$ terms on the left and write both sides in $B$ :
 $\begin{align*} (1 - 1.5B + 0.5B^{2})\,x_t &= (1 - 0.3B)\,\varepsilon_t \\ (1 - 0.5B)(1 - B)\,x_t &= (1 - 0.3B)\,\varepsilon_t \end{align*}$ 
The AR polynomial has roots $B = 1$ and $B = 2$ . The root at $1$ is a unit root, so difference once: $(1 - 0.5B)\nabla x_t = (1 - 0.3B)\varepsilon_t$ .

The remaining AR root $B = 2$ exceeds $1$ , so the differenced series is stationary. The MA root $B = 1/0.3 = 3.33$ exceeds $1$ , so it is invertible.

ARIMA(1, 1, 1) with $\phi_1 = 0.5$ and $\theta_1 = -0.3$ . The index itself wanders, but its quarterly changes follow a stationary ARMA(1, 1).

One-Step Forecast from Operator Form Monthly reported claim counts follow $(1 - 0.6B)(1 - B)\,x_t = (1 + 0.3B)\,\varepsilon_t$ . The latest values are $x_t = 120$ and $x_{t-1} = 115$ , and the latest residual is $\hat\varepsilon_t = 2$ . Forecast $x_{t+1}$ .
Answer Expand the left side, $(1 - 0.6B)(1 - B) = 1 - 1.6B + 0.6B^2$ , and solve for $x_t$ : $x_t = 1.6x_{t-1} - 0.6x_{t-2} + \varepsilon_t + 0.3\varepsilon_{t-1}$ Shift forward one period and set the unknown future shock $\varepsilon_{t+1}$ to $0$ :
 $\begin{align*} \hat{x}_{t+1} &= 1.6(120) - 0.6(115) + 0.3(2) \\ &= 192 - 69 + 0.6 \\ &= 123.6 \end{align*}$ 
Read on the differenced scale: last month's change of $+5$ carries forward at $\phi = 0.6$ (giving $+3$ ), plus $0.3 \times 2 = 0.6$ from the MA term. That is $120 + 3.6 = 123.6$ claims.