# Derivative of the Backshift Operator?

Is there any meaningful sense in which the backshift operator can be said to have a derivative?

Here's my attempt at constructing one. Based on some helpful threads on this site, I represent the backshift operator $$B$$ as a matrix (showed here for an example time series of length $$3$$): $$\begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix} = \begin{pmatrix} 0\\ y_1\\ y_2 \end{pmatrix}\ .$$ I can take the derivative of the above equation with respect to $$y_1$$ and $$y_2$$ to obtain the vectors $$(0, 1, 0)$$ and $$(0, 0, 1)$$ respectively (the derivative w.r.t $$y_3$$ is undefined.) Extrapolating, and wishing to obtain an operator (instead of a vector) from the differentiation process, I conclude that the derivative of the backshift operator is a type of "shifted" indicator function $$I_{t}$$ which has a straightforward matrix representation. For example, I would write $$\frac{\partial B}{\partial y_1} = I_1 = \begin{pmatrix} 0 & 0 & 0\\ 1/y_1 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\ .$$ Does this calculation have any meaning, or are there some operator-space/common-sense rules which forbid some steps? References would be appreciated.

• The derivative with respect to $y_3$ is certainly defined: it's the zero vector. Commented Oct 20, 2022 at 6:30
• @GregMartin Thanks for the correction. Commented Oct 20, 2022 at 8:53

The backshift operator is a mapping (an "operator") between vector spaces, namely spaces of time series or sequences,

$$B\colon \mathbb{R}^\mathbb{N} \to \mathbb{R}^\mathbb{N}, (y_i)\mapsto (y_{i-1}).$$

Here, $$\mathbb{R}^\mathbb{N}$$ is the space of mappings from $$\mathbb{N}$$ to $$\mathbb{R}$$, i.e., of real-valued sequences indexed by natural numbers. Time series that are not infinite can be accommodated by having only finitely many observations nonzero. The space $$\mathbb{R}^\mathbb{N}$$ is naturally a real vector space: we can add time series and multiply them by real scalars.

In functional analysis, there is a notion of differential of an operator between normed vector spaces. If an operator $$f$$ admits a "local linear approximation" $$\varphi$$ near a point $$x$$, then we say that $$f$$ is differentiable and that $$\varphi$$ is its differential at $$x$$ (see, e.g., Coleman, 2012, section 2.2). Note that this generalizes the familiar notion of differentiability of mappings between finite dimensional spaces: a function $$f\colon\mathbb{R}^n\to\mathbb{R}^m$$ is differentiable at a point $$x$$ if and only if it admits a well-defined tangential subspace (tangent line in the most familiar case of $$n=m=1$$) near $$x$$.

Now, we observe that $$B$$ is already linear:

$$B\big(\lambda(y_i)\big) = \lambda B\big((y_i)\big)\quad\text{and}\quad B\big((y_i)+(z_i)\big) = B\big((y_i)\big)+B\big((z_i)\big).$$

Thus, a (unique!) "best approximation" to $$B$$ is $$B$$ itself (see here at Math.SE). Therefore, the differential of $$B$$ at all "points" (i.e., sequences) is $$B$$ itself. Note that we don't even need to think about what norm we put on our space $$\mathbb{R}^\mathbb{N}$$.

• For those looking for further intuition about why the derivative of a linear operator is the operator itself: math.stackexchange.com/a/1696767/49082 Commented Oct 20, 2022 at 8:55