# What does A(B) mean in time series

So I have been reading some papers regarding time series, mainly from Granger and Engle. I am a bachelor econometrics student, but I have never seen such notation before. For example, A(B)(1-B)x(t) = -az(t-1) + b(t). I know that B is the backward shift operator. Could someone clarify this?

another example would be that time series x(t) = a(B)epsilon(t)

• Can you provide a link tho these papers? – Bob Jansen Jul 21 at 14:46
• I dont have direct links, I got them from my university. But the articles that I am referring too are 1- ''Co-integration and Error Correction: Representation, Estimation, and Testing '' by Engle and Granger 1987 and 2 - ''Some properties of time series data and their use in econometric model specification'' by Granger – cem Jul 21 at 15:16

As you said, $$B$$ is the lag or backward shift operator such that $$BX_t=X_{t-1}$$ and $$B^pX_t=X_{t-p}$$. Let $$A$$ now be polynomial, say $$A(x)=a_1 x + a_2x^2+...+a_px^p$$. Then,
\begin{align} A(B) X_t &= \left( a_1 B + a_2B^2+...+a_pB^p\right) X_t \\ &=a_1 X_{t-1} + a_2 X_{t-2} + ... + a_p X_{t-p} \end{align} and \begin{align} \big(1-A\big)(B) X_t &= \big(1-A(B)\big) X_t \\ &= \left( 1- a_1 B - a_2B^2-...-a_pB^p\right) X_t \\ &=X_t-a_1 X_{t-1} - a_2 X_{t-2} - ... - a_p X_{t-p}. \end{align} Thus, if you write $$\big(1-A\big)(B) X_t=c+\varepsilon_t$$, you get \begin{align} X_t&=c+a_1 X_{t-1} + a_2 X_{t-2} + ... + a_p X_{t-p}+\varepsilon_t \\ &= c + \sum_{i=1}^p a_iX_{t-i}+\varepsilon_t, \end{align} which is simply an AR($$p$$) model. Thus, polynomials of the backward shift operator allow you to easily write down time series models. For instance, $$\big(1-A(B)\big) X_t=c+\big(1+C(B)\big)\varepsilon_t$$ is an ARMA($$p$$,$$q$$) model (if $$C$$ is a polynomial of order $$q$$).
• As a hint, you can express an ARIMA or ARFIMA model as $\big(1-A(B)\big)\big(1-L\big)^d X_t = c+\big(1+C(B)\big)\varepsilon_t$ where $d\in\mathbb{N}$ for an ARIMA and $d>0$ for an ARFIMA model. Similarly, an ARCH($p$) model is given by $\sigma_t^2=\omega+A(B)\varepsilon_t^2$ where $A(x)=a_1x+...+a_px^p$. – KeSchn Jul 21 at 15:27
• Moreover, would A(B)*B = A($B^2$)? – cem Jul 21 at 15:30
• No, unfortunately not. Let $A(x)=1+x$, Then, $\big(A(B)B\big)X_t=\big(B+B^2)X_t=X_{t-1}+X_{t-2}$ but $A(x^2)=1+x^2$ and thus, $A(B^2)X_t=X_t+X_{t-2}$ which are clearly different. – KeSchn Jul 21 at 15:34
• Yeah, sorry. I was messing around with the equations I got the same. where A(B)(1 - B)$X_t$ = [A(B) - A(B)B]$X_t$ = $a_1$( $x_1$ - $x_2$) + $a_2$($x_2$ - $x_3$) + .... + $a_p$ ($x_{t-p-1}$ - $x_{t-p}$) – cem Jul 21 at 15:36