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I understand that $(1 - B)x_{t}=x_{t}-x_{t-1}$. I understand that I can do algebra with $B$, treating it like any other variable. But can I do arithmetic operations with $B$? Like, can I subtract $B$ from 1? If so, what do I get? Does $B$ have a numerical meaning? Is it a scalar? A matrix? How do I compute $1 - B$?

The only numeric representation of $B$ I've found is in a tutorial where $B$ is a matrix. But that tutorial uses a notation I haven't seen anywhere else - it uses $(I - B)$ instead of $(1 - B)$, where $I$ is an identity matrix. When people write $(1 - B)$ do they normally mean $B$ to be a matrix or a scalar? Or is it just an operator and we can't really do math with it? In sum: what the heck is $B$?

Also, sometimes I see $B$ after a variable. Like in the series $1-dB+ \dfrac{d(d-1)}{2!}B^{2}...$ What does that mean? What does B do when it comes after something?

Parzival
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  • $B$ is a *linear operator* (on a linear space of sequences), in the meaning used in functional analysis. Matrices can be seen as a way of *representing* a linear operator between two finite-fimensional linear spaces, so in that setting $B$ could be represented by a matrix (but you would have to decide what to do at the ends ...). Because of this analogy, you can use some intuition from what you know about matrices (like products of linear operators do not always commute) – kjetil b halvorsen Apr 16 '21 at 17:28

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$B$ is not a number or a matrix. It's an operator. You can think of it as a function, or a mapping: it takes a time series and backshifts them, $Bx_t=x_{t-1}$. We could have used functional notation, $f(x_t) = x_{t-1}$, it's just that $B$ (for "backshift") became more common, and that by convention we do not use brackets. (Incidentally, sometimes you also see a nabla $\nabla$ instead of $B$.)

So just like for any function $g$ you can define the function $3g$ by $$(3g)(x):=3\times g(x),$$ i.e., multiply functions by a scalar, you can also multiply $B$ by a scalar: $$3Bx_t = 3\times (Bx_t) = 3x_{t-1}.$$

And just as we can concatenate functions, $$f^2(x) = f\circ f(x) = f\big(f(x)\big), $$ we can concatenate the backshift operator: $$B^2x_t=B(Bx_t)=Bx_{t-1}=x_{t-2}. $$

So in your example, $d$ is presumably some scalar variable, like the $3$ above. So $$\bigg(1-dB+ \dfrac{d(d-1)}{2!}B^{2}\bigg)x_t = x_t-dBx_t+\dfrac{d(d-1)}{2!}B^{2}x_t= x_t-dx_{t-1} +\dfrac{d(d-1)}{2!}x_{t-2}.$$

Stephan Kolassa
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