What is the "distributional derivative" of a Brownian motion?

Question

Let $\emptyset\ne I\subseteq\mathbb R$ be an open interval and $A:C_0^\infty(I)\to\mathbb R$ be a distribution. Then, $$\langle{\rm D}A,\varphi\rangle:=-\langle A,\varphi'\rangle\;\;\;\text{for }\varphi\in C_0^\infty(I)$$ is called distributional derivative of $A$.

Let $B=(B_t)_{t\ge 0}$ be a Brownian motion. How is the distributional derivative of $B$ with respect to $t$ defined?

I've found this term in the following paper: http://ac.els-cdn.com/S0377042708003178/1-s2.0-S0377042708003178-main.pdf?_tid=a704a996-acb7-11e5-b001-00000aacb360&acdnat=1451234228_e36ed3e5f5b1e386f60a272321cc6bab — 0xbadf00d, Dec 27 '15 at 16:34

score 6 · Accepted Answer · answered Dec 27 '15 at 16:34

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Presumably distribution derivative of $B$ refers to $\varphi\mapsto\int_0^\infty \varphi(s)\,dB_s=-\int_0^\infty \varphi'(s)B_s\,ds$, on the domain comprising smooth functions supported in compact subsets of $(0,\infty)$.

answered Dec 27 '15 at 16:34

John Dawkins

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Out of curiosity, is this the definition of so-called "white noise"? Also, is it known whether or not the distributional derivative of B is a continuous distribution-valued process? – Christian Bueno Aug 12 '17 at 20:20
no it's a distributional real-valued process – Abdelmalek Abdesselam Nov 28 '18 at 18:56

score 1 · Answer 2 · answered Mar 09 '17 at 12:08

The distributional derivative by your definition of a Brownian motion $B$ is just the "white noise process".

Indeed, the Brownian motion is nowhere differentiable almost surely, so we have to make use of the distributional derivative (generalized functions).

What is the "distributional derivative" of a Brownian motion?

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