Brownian Motion

Stochastic Process

A stochastic process on the probability space $(\Omega ,\Sigma ,\mathbb{P})$ is a collection of random variables $\{X_t{\}}_{t\in T},$ where $T$ is some indexing set.

Brownian Motion

A stochastic process $(B_t{)}_{t\in [0,1]}$ is called a standard Brownian motion on $[0,1]$ if it has the following properties:

1. $B_0=0$.
2. For any set of finite times, $0\le t_1<t_2<\cdots <t_N\le 1$, the increments of the process $(B_t{)}_{t\in [0,1]}$, $B_{t_2}-B_{t_1},B_{t_3}-B_{t_2},\dots ,B_{t_N}-B_{t_{N-1}},$ are all independent.
3. $B_t-B_s\sim \mathcal{N}(0,t-s)$ for any $s,t\in [0,1]$ with $s\le t$.
4. The paths of $B_t$ are almost surely continuous. That is, $\mathbb{P}(t\mapsto B_t\text{ is continuous})=1.$

Lemma

Let $(Z_n{)}_{n\in \mathbb{N}}$ be a sequence of independent Gaussian variables with $Z_n\sim \mathcal{N}(0,1).$ For almost every $\omega \in \Omega$, there exists a constant $C(\omega )>0$ for which$|Z_n(\omega )|\le C(\omega )\sqrt{\ln (n)}$for all $n\ge 2$.

Proof. As the variables $Z_n$ are normally distributed, for any $x\ge 1$,

$$\begin{array}{rl}\mathbb{P}(|Z_n|\ge x)& =\sqrt{\frac{2}{\pi }}{\int }_x^{\infty }{e}^{-y^2/2}dy\\ & \le \sqrt{\frac{2}{\pi }}{\int }_x^{\infty }y{e}^{-y^2/2}dy\\ & =\sqrt{\frac{2}{\pi }}{e}^{-x^2/2}\end{array}$$

for any $n\in \mathbb{N}$. This implies that for any $\alpha >1$ and $n\ge 2$,

$$\begin{array}{rl}\mathbb{P}\left(|Z_n|\ge \sqrt{2\alpha \ln (n)}\right)& \le \sqrt{\frac{2}{\pi }}{e}^{-\alpha \ln (n)}\\ & =\sqrt{\frac{2}{\pi }}{n}^{-\alpha },\end{array}$$

and therefore

$$\sum _{n=2}^{\infty }\mathbb{P}\left(|Z_n|\ge \sqrt{2\alpha \ln (n)}\right)<\infty .$$

The Borel—Cantelli lemma can then be applied to find that

$$\mathbb{P}\left(|Z_n|\ge \sqrt{2\alpha \ln (n)}\text{ for infinitely many }n\in \mathbb{N}\right)=0.$$

Equivalently, for almost every $\omega \in \Omega$, the bound

$$\frac{|Z_n(\omega )|}{\sqrt{\ln (n)}}\le \sqrt{2\alpha }$$

holds for all but finitely many $n\in \mathbb{N}$. This implies that for almost every $\omega \in \Omega$, the quantity

$$C(\omega ):=\underset{n\ge 2}{\sup }\frac{|Z_n(\omega )|}{\sqrt{\ln (n)}}$$

is finite. This quantity will give the result.

The following theorem asserts the existence of the Brownian motion process.

Theorem

Let $(Z_n{)}_{n\in \mathbb{N}}$ be a sequence of independent Gaussian variables with $Z_n\sim \mathcal{N}(0,1)$ for each $n\in \mathbb{N}$. The series $X_t:={\sum }_{n=0}^{\infty }⟨H_n,1_{[0,t]}⟩Z_n$ converges uniformly on $[0,1]$ almost surely. Moreover, the process $(X_t{)}_{t\in [0,1]}$ is a standard Brownian motion on $[0,1]$.

Proof. Fix $\omega \in \Omega$ for which $C(\omega )$, as defined in the previous lemma, is finite. It will be shown that the series $X_t(\omega )$ converges uniformly in $t$. For $M\ge 2$, the previous lemma leads to

$$\begin{array}{rl}\left|\sum _{n=M}^{\infty }⟨H_n,1_{[0,t]}⟩Z_n(\omega )\right|& \le C(\omega )\sum _{n=M}^{\infty }\sqrt{\ln (n)}\left|{\int }_0^t{H}_n(s)ds\right|\\ & \le C(\omega )\sum _{k=K(M)}^{\infty }\sum _{j=0}^{2^k-1}\sqrt{\ln (2^{k+1})}\left|{\int }_0^t{H}_{2^k+j}(s)ds\right|,\end{array}$$

where $K(M)$ is the largest integer for which $2^{K(M)}\le M.$ On noting that ${\int }_0^t{H}_{2^k+j}(s)ds$ is non-zero for at most two values of $j$,

$$\left|\sum _{n=M}^{\infty }⟨H_n,1_{[0,t]}⟩Z_n(\omega )\right|\le 2\sqrt{\ln (2)}\cdot C(\omega )\sum _{k=K(M)}^{\infty }\sqrt{k+1}\cdot 2^{k/2}{2}^{-k}.$$

This is enough to show uniform convergence in $t$ since the series ${\sum }_{k=K(M)}^{\infty }\sqrt{k+1}\cdot 2^{-k/2}$ can be made as small as desired, independently of $t$, by increasing $M$. It will now be shown that the process $(X_t{)}_{t\in [0,1]}$ is a standard Brownian motion on $[0,1]$. Fix a finite set of times $0\le t_1<t_2<\cdots <t_N\le 1$ for some $N\in \mathbb{N}$ and consider the increments $X_{t_2}-X_{t_1},\dots ,X_{t_N}-X_{t_{N-1}}.$ The joint probability distribution of $(X_{t_2}-X_{t_1},\dots ,X_{t_N}-X_{t_{N-1}})$ will be computed. Let ${\phi }_{X_{t_2}-X_{t_1},\dots ,X_{t_N}-X_{t_{N-1}}}$ denote the joint characteristic function of the increments. Fix $y\in {\mathbb{R}}^{N-1}$. Then

$$\begin{array}{rl}{\phi }_{X_{t_2}-X_{t_1},\dots ,X_{t_N}-X_{t_{N-1}}}(y)& =\mathbb{E}\left(\exp \left(i\sum _{j=1}^{N-1}{y}_j(X_{t_{j+1}}-X_{t_j})\right)\right)\\ & =\mathbb{E}\left(\exp \left(i\sum _{j=1}^{N-1}{y}_j\sum _{n=0}^{\infty }\left(⟨H_n,1_{[0,t_{j+1}]}⟩-⟨H_n,1_{[0,t_j]}⟩\right)Z_n\right)\right)\\ & =\mathbb{E}\left(\prod _{n=0}^{\infty }\exp \left(i{Z}_n\sum _{j=1}^{N-1}{y}_j⟨H_n,1_{[t_j,t_{j+1}]}⟩\right)\right).\end{array}$$

From the independence of the variables $(Z_n{)}_{n\in \mathbb{N}}$,

$$\begin{array}{rl}{\phi }_{X_{t_2}-X_{t_1},\dots ,X_{t_N}-X_{t_{N-1}}}(y)& =\prod _{n=0}^{\infty }\mathbb{E}\left(\exp \left(i{Z}_n\left(\sum _{j=1}^{N-1}{y}_j⟨H_n,1_{[t_j,t_{j+1}]}⟩\right)\right)\right)\\ & =\prod _{n=0}^{\infty }\left(\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }\exp \left(ix\sum _{j=1}^{N-1}{y}_j⟨H_n,1_{[t_j,t_{j+1}]}⟩\right)\exp \left(-\frac{x^2}{2}\right)dx\right)\\ & =\exp \left(-\frac{1}{2}\sum _{n=0}^{\infty }{\left(\sum _{j=1}^N{y}_j⟨H_n,1_{[t_j,t_{j+1}]}⟩\right)}^2\right)\\ & =\exp \left(-\frac{1}{2}\sum _{n=0}^{\infty }{⟨H_n,\sum _{j=1}^{N-1}{y}_j{1}_{[t_j,t_{j+1}]}⟩}^2\right).\end{array}$$

As $(H_n{)}_{n\in \mathbb{N}}$ is a basis,

$$\begin{array}{rl}{\phi }_{X_{t_2}-X_{t_1},\dots ,X_{t_N}-X_{t_{N-1}}}(y)& =\exp \left(-\frac{1}{2}⟨\sum _{j=1}^{N-1}{y}_j{1}_{[t_j,t_{j+1}]},\sum _{j=1}^{N-1}{y}_j{1}_{[t_j,t_{j+1}]}⟩\right)\\ & =\exp \left(-\frac{1}{2}\sum _{j=1}^{N-1}{y}_j^2(t_{j+1}-t_j)\right)\\ & =\prod _{j=1}^{N-1}\exp \left(-\frac{1}{2}{y}_j^2(t_{j+1}-t_j)\right).\end{array}$$

The probability distribution of the increments is completely determined by the above characteristic function by taking the Fourier transform. The joint probability distribution is then given by

$$f_{X_{t_2}-X_{t_1},\dots ,X_{t_N}-X_{t_{N-1}}}(x)=\prod _{j=1}^{N-1}\frac{1}{{\left(\sqrt{2\pi (t_{j+1}-t_j)}\right)}^{N-1}}\exp \left(-\frac{x_j^2}{2(t_{j+1}-t_j)}\right).$$

This proves that the increments are independent and that $X_s-X_t$ is Gaussian distributed with mean $0$ and variance $s-t.$

Standard Brownian Motion

Let $(B_t^{(j)}{)}_{t\in [0,1]}$ be independent standard Brownian motions on $[0,1]$ for $j\in \mathbb{N}$. Define for $n\in \mathbb{N}$ and $s\in [n,n+1)$, $W_s:={\sum }_{j=0}^{n-1}{B}_1^{(j)}+B_{s-n}^{(n)}.$ $(W_s{)}_{s\ge 0}$ is called a standard Brownian motion on $[0,\infty )$.

Continuous Time Filtration

A family of sigma-algebras $({\mathcal{F}}_t{)}_{t\ge 0}$ is called a filtration for the probability space $(\Omega ,\Sigma ,\mathbb{P})$ if for any $t,s\ge 0$ with $t\ge s$ we have ${\mathcal{F}}_s\subset {\mathcal{F}}_t\subset \Sigma .$ A process $(X_t{)}_{t\ge 0}$ is said to be adapted to the filtration $({\mathcal{F}}_t{)}_{t\ge 0}$ if $X_t$ is ${\mathcal{F}}_t$-measurable for each $t\ge 0$.

Standard (augmented) Brownian Filtration

Let $(W_s{)}_{s\ge 0}$ be a standard Brownian motion on $[0,\infty )$. Define, for each $t\ge 0$, ${\mathcal{F}}_t^W:=\sigma (W_s:s\in [0,t])$ ${\mathcal{F}}_t^{W,+}:={\bigcap }_{s>t}{\mathcal{F}}_s^W{\mathcal{N}}_t:=\{A\in \Sigma :A\subset B\text{ for some }B\in {\mathcal{F}}_t\text{ with }\mathbb{P}(B)=0\}.$ The standard (augmented) Brownian filtration, $({\mathcal{F}}_t{)}_{t\ge 0}$, is defined by ${\mathcal{F}}_t:=\sigma \left({\mathcal{F}}_t^{W,+}\cup {\mathcal{N}}_t\right)\forall t\ge 0.$