In my introductory book on time series analysis the backshift operator $\mathbf{B}$ is introduced using the following definition: $$ \mathbf{B}x_t=x_{t-1} $$ Then the author sets off to derive some properties of random walks, and there is this step, that is a complete mystery to me: $$ x_t=(1-\mathbf{B})^{-1}w_t\Rightarrow x_t=(1+\mathbf{B}+\mathbf{B}^2+\mathbf{B}^3+\dots)w_t $$

I do not understand how $(1-\mathbf{B})^{-1}$ is equivalent to $(1+\mathbf{B}+\mathbf{B}^2+\mathbf{B}^3+\dots)$. Maybe in the book the author did not provide some important properties of the backshift operator that were used deriving the above step.

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    $\begingroup$ What do you obtain when $1+b+b^2+\cdots$ is multiplied by $1-b$? $\endgroup$
    – whuber
    Commented Jul 15, 2014 at 16:06
  • $\begingroup$ @whuber now I get it. But is there a more formal way to prove it than to write $1+b+b^2+b^3+b^4+\dots-b-b^2-b^3-b^4-\dots=1$? $\endgroup$
    – Fazzolini
    Commented Jul 15, 2014 at 16:20
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    $\begingroup$ For complete rigor you need to understand $B$ as a linear operator on the Hilbert space $\mathcal{l}^1(\mathbb{Z,R})$. For partial rigor, understand $1+b+b^2+\cdots$ as the limiting value of partial sums $a_n=1+b+\cdots+b_n$ and compute $(1-b)a_n=1-b^{n+1}$. Its limiting value is $1$ provided the limiting value of $b^{n+1}$ is $0$. $\endgroup$
    – whuber
    Commented Jul 15, 2014 at 16:24
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    $\begingroup$ @whuber I have basic calculus, probability theory and linear algebra under my belt. What books would you suggest to read to be able to understand $B$ as a linear operator on the Hilbert space? $\endgroup$
    – Fazzolini
    Commented Jul 15, 2014 at 16:27
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    $\begingroup$ Walter Rudin, Real and Complex Analysis, has a very clear account of this. You can skip to the chapter on elementary Hilbert space theory without having to study the earlier chapters much (except to learn the notation). $\endgroup$
    – whuber
    Commented Jul 15, 2014 at 16:35

1 Answer 1


That step comes from the Taylor expansion of $\frac{1}{1-x}$, which is $1 + x + x^2 + ...$. Just substitute $x$ for the backward shift operator $B$ in the author's derivation and you'll arrive at the same result.

Have you taken a class on integral calculus? Usually you'll go through that derivation when you cover series. Here's mine:

Let $f(x) = \frac{1}{1-x}$. Then:

$f'(x) = -\frac{1}{(1-x)^2}$

$f''(x) = \frac{1}{(1-x)^3}$


$f^{(n)}(x) = (-1)^{2n+1}\frac{1}{(1-x)^{n+1}}$

The Maclaurin series (Taylor series centered at $x=0$) is therefore

$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + ... = \frac{1}{1-0} - \frac{1}{(1-0)^2}x^2 + \frac{1}{(1-0)^3}x^3 + ... = 1 + x + x^2 + ...$

Now replace $x$ in all of that with the backwards shift operator $B$ and you'll get the author's expression. Does that help?

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    $\begingroup$ Although this motivates the equation, it doesn't go very far towards demonstrating it--and perhaps even obscures the issue. The question concerns the meaning of convergence of the infinite series of shift operators. That meaning is quite different from convergence of the series of numbers (or functions) which appear in Taylor's Formula. Here, higher powers of $B$ just shift the time series $(x_t)$ by greater amounts. Thus, convergence must occur relative to some metric in which greater and greater shifts become ever closer to the zero operator. $\endgroup$
    – whuber
    Commented Jul 15, 2014 at 18:42
  • $\begingroup$ This answers my question for now, until I get my hands on that book on analysis @whuber recommended to me. $\endgroup$
    – Fazzolini
    Commented Jul 16, 2014 at 6:40
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    $\begingroup$ Sorry if I gave a bit of an oversimplified explanation. I didn't think OP had a really strong mathematical background and wanted to keep things as straightforward as possible. $\endgroup$ Commented Jul 19, 2014 at 19:54

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