📂Stochastic Differential Equations

The Hyperfunctional Derivative of Brownian Motion is White Noise

Summary

The distributional derivative of Brownian motion is white noise.

Description

Brownian motion $B_{t}$ does not have a derivative in the traditional sense. Therefore, it can be defined as a stochastic process that satisfies the following condition $\xi$, which is defined as white noise:

$$\begin{align} E[\xi_{t}] &= 0, & \forall t \\ \Cov(\xi_{t}, \xi_{s}) &= \delta_{0} \end{align}$$

Here, $\Cov$ is the covariance, and $\delta$ is the Dirac delta function. If we extend $B_{t}$ as a distribution, it can be understood that its distributional derivative satisfies the definition of white noise. In other words, white noise is the weak derivative of Brownian motion.

Proof¹

Let $B(t,w) : [0, \infty) \times \Omega \to \mathbb{R}^{n}$ be a stochastic process of Brownian motion. For convenience, let’s denote it as $B_{t}(\omega) = B(t,\omega)$. Let’s define the following distribution $B_{t}$ for $B_{t}$:

$$B_{t} [\phi] := \int B_{t} \phi (t) dt, \quad \forall \phi \in \mathcal{D}$$

Where $\phi$ is a test function. The derivative of the distribution $B_{t}$ is defined as the following distribution according to the definition:

$$B_{t}^{\prime} [\phi] := - \int B_{t} \phi^{\prime}(t) dt, \quad \forall \phi \in \mathcal{D}$$

Let’s briefly denote this as $\xi (\phi) = B_{t}^{\prime} [\phi]$. Now, we need to show that $\xi (\phi)$ satisfies $(1)$ and $(2)$.

Basic Properties of Brownian Motion

[2] $E ( B_{t} ) = 0$

[4] $\Cov ( B_{t} , B_{s} ) = E (B_{t}B_{s}) = \min (t, s)$

• $E\left[ \xi (\phi) \right] = 0$

  $$\begin{align*} E[\xi (\phi)] &= E\left[ - \int B_{t} \phi^{\prime}(t) dt \right] \\ &= - \int E[B_{t}] \phi^{\prime}(t) dt \\ &= - \int 0 \cdot \phi^{\prime}(t) dt = 0 \end{align*}$$

  The second equals comes from $E$ being independent of $t$, and the third equality is derived from the property [2] of Brownian motion.

• $\Cov \left[ \xi, \xi \right] = \delta_{0}$

  According to property [4] of Brownian motion,

  $$\begin{align*} \Cov \left[ \xi (\phi), \xi (\psi) \right] &= E\left[ \xi (\phi) \xi (\psi) \right] \\ &= E\left[ \int B_{t}\phi^{\prime}(t)dt \int B_{s}\psi^{\prime}(s)ds \right] \\ &= E\left[ \int\int B_{t}B_{s}\phi^{\prime}(t)\psi^{\prime}(s) dtds \right] \\ &= \int\int E\left[ B_{t}B_{s} \right] \phi^{\prime}(t)\psi^{\prime}(s) dtds \\ &= \int\int \min(t,s) \phi^{\prime}(t)\psi^{\prime}(s) dtds \\ \end{align*}$$

  The fourth equals comes from $E$ being independent of $t$ and $s$, and the fifth equality is based on the property [4] of Brownian motion.

  When $s$ is fixed, $\min (t,s)$ is a function of $t$.

  $$\min (t,s) = \begin{cases} t & 0 \le t \le s \\ s & s \le t \end{cases}$$

  Therefore, computing the integral yields,

  $$\begin{align*} &\quad \int\int \min(t,s) \phi^{\prime}(t)\psi^{\prime}(s) dtds \\ &= \int \int_{0}^{s} t \phi^{\prime}(t) \psi^{\prime}(s) dtds + \int \int_{s}^{\infty} s \phi^{\prime}(t) \psi^{\prime}(s) dtds \\ &= \int \left( \int_{0}^{s} t \phi^{\prime}(t) dt \right) \psi^{\prime}(s)ds + \int s \int_{s}^{\infty} \phi^{\prime}(t) dt \psi^{\prime}(s) ds \\ &= \int \left( \left[ t\phi (t) \right]_{0}^{s} - \int_{0}^{s}\phi (t) dt \right) \psi^{\prime}(s)ds + \int s [\phi (t)]_{s}^{\infty} \psi^{\prime}(s) ds \\ &= \int s\phi (s) \psi^{\prime}(s)ds - \int \int_{0}^{s}\phi (t) dt \psi^{\prime}(s) ds - \int s \phi (s) \psi^{\prime}(s) dtds \\ &= - \int_{0}^{\infty} \int_{0}^{s}\phi (t) dt \psi^{\prime}(s) ds = - \int_{0}^{\infty} \left( \int_{0}^{s}\phi (t) dt \right) \psi^{\prime}(s) ds \\ &= - \left[ \left( \int_{0}^{s}\phi (t) dt \right) \psi (s) \right]_{0}^{\infty} + \int_{0}^{\infty} \dfrac{d}{ds}\left( \int_{0}^{s}\phi (t) dt \right) \psi (s) ds \\ &= \int_{0}^{\infty} \phi (s) \psi (s) ds \\ &= \phi (0) \psi (0) \\ &= \delta_{0}[\phi\psi] \\ \end{align*}$$

  $$\implies \Cov\left[ \xi, \xi \right] = \delta_{0}$$

  The third and sixth equalities use integration by parts. The seventh equality holds because of $\psi (\infty) = 0$ and $\displaystyle \int_{0}^{0}\phi (t)dt=0$.

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