Construction of Brownian Motion

A stochastic process ${X=\{X_t:t\in\mathbb R_+\}}$ is said to be standard Brownian motion (or a Wiener process) if

1. ${X_0=0}$ almost surely.
2. ${X}$ has independent increments, i.e. for ${0\leqslant t_0, the random variables

$\displaystyle \left\{X_{t_{i+1}}-X_{t_i} : 0\leqslant i\leqslant k-1 \right\}$

are independent.

3. For ${0\leqslant s,

$\displaystyle X_t-X_s\sim\mathcal N(0,t-s).$

4. The map ${\omega\rightarrow X_t(\omega)}$ is almost surely continuous.

In this post I follow the method in Sidney Resnick’s Adventures in Stochastic Processes to construct Brownian motion on ${[0,1]}$ using i.i.d. standard normal random variables. The key to this construction is the following lemma:

Lemma 1 Suppose ${X(s)}$ and ${X(t)}$ are random variables such that

$\displaystyle X(t)-X(s)\sim\mathcal N(0,t-s).$

Then there exists a random variable ${X\left(\frac{t+s}2\right)}$ such that

$\displaystyle X\left(\frac{t+s}2\right) - X(s),\ X(t) - X\left(\frac{t+s}2\right)\stackrel{\mathrm{i.i.d.}}\sim\mathcal N\left(0,\frac{t-s}2\right).$

Proof: Define ${U:=X(t)-X(s)}$ and suppose ${V\sim\mathcal N(0,t-s)}$ is independent of ${U}$. Define ${X\left(\frac{t+s}2\right)}$ by

\displaystyle \begin{aligned} X(t) - X\left(\frac{t+s}2\right) &= \frac{U+V}2\\ X\left(\frac{t+s}2\right) - X(s) &= \frac{U-V}2, \end{aligned}

so that

\displaystyle \begin{aligned} X(t) - X(s) &= U\\ X(t) + X(s) - 2X\left(\frac{t+s}2\right) &= V. \end{aligned}

Then as ${U,V\stackrel{\mathrm{i.i.d.}}\sim\mathcal(0,t-s)}$, we have

$\displaystyle \mathbb E[(U+V)(U-V)] = \mathbb E[U^2] - \mathbb E[V^2] = 0$

so that ${X(t)-X\left(\frac{t+s}2\right)}$ and ${X\left(\frac{t+s}2\right)-X(s)}$ are uncorrelated and hence independent (as they are normally distributed).

$\Box$

Now let

$\displaystyle \left\{V(k2^{-n}):1\leqslant k\leqslant 2^n, n\geqslant 1\right\}$

be independent random variables with

$\displaystyle V\left(k2^{-(n+1)}\right)\sim\mathcal N(0,2^{-n}).$

Let ${X(0):=0}$, ${X(1):=V(1)}$, and define ${X(2^{-1})}$ using ${V(2^{-1})}$ so that ${X(2^{-1})-X(0)}$ and ${X(1)-X(2^{-1})}$ are i.i.d. ${\mathcal N(0,2^{-1})}$ random variables. Suppose

$\displaystyle \{ X(k2^{-n}): 0\leqslant k\leqslant 2^n\}$

are defined such that

$\displaystyle X(k2^{-n})-X((k-1)2^{-n})\stackrel{\mathrm{i.i.d.}}\sim\mathcal N(0,2^{-n}),\ 1\leqslant k\leqslant 2^n.$

For each ${k\leqslant 2^n}$, define ${X((2k+1)2^{-(n+1)})}$ such that

\displaystyle \begin{aligned} X\left((2k+1)2^{-(n+1)}\right)-X(k2^{-n}) &\stackrel d= X((k+1)2^{-n}) - X\left((2k+1)2^{-(n+1)} \right)\\ &\sim\mathcal N\left(0, 2^{-(n+1)}\right) \end{aligned}

and the sequence ${\left\{X\left(k2^{-(n+1)}\right):0\leqslant k\leqslant 2^{-(n+1)}\right\}}$ has independent increments. For each ${n=1,2,\ldots}$ and ${0\leqslant k\leqslant 2^n-1}$ define the processes

\displaystyle \begin{aligned} B^{(n,k)}(t,\omega) &= 2^n(X((k+1)2^{-n},\omega)-X(k2^{-n},\omega))t\\ &\quad -k\cdot X((k+1)2^{-n},\omega) - (k-1)(X(k2^{-n},\omega)). \end{aligned}

and

$\displaystyle B^{(n)}(t) = \sum_{k=0}^{2^n-1} B^{(n,k)}(t)$

That is,

$\displaystyle B^{(n)}(t,\omega) = X(t,\omega),\quad t\in\left\{k2^{-n}, 0\leqslant k\leqslant 2^{-n}\right\}$

and ${B^{(n)}}$ is linear on each interval ${\left[k2^{-n},(k+1)2^{-n}\right]}$. Let ${\Delta^{(n)}}$ be the maximum deviation of ${B^{(n)}}$ and ${B^{(n+1)}}$ on ${[0,1]}$, partitioned by the intervals ${\left[k2^{-n},(k+1)2^{-n}\right]}$, that is,

$\displaystyle \Delta^{(n)}(\omega) = \max_{0\leqslant k\leqslant 2^n-1}\;\max_{k2^{-n}\leqslant t\leqslant (k+1)2^{-n}}\left|B^{(n+1)}(t,\omega)-B^{(n)}(t,\omega)\right|.$

For each ${n}$ we have

\displaystyle \begin{aligned} &\quad\max_{k2^{-n}\leqslant t\leqslant (k+1)2^{-n}}\left|B^{(n+1)}(t,\omega)-B^{(n)}(t,\omega)\right|\\ &= \left|\frac{B^{(n)}((k+1)2^{-n}) + B^{(n)}(k2^{-n}) }2 - B^{(n+1)}\left((2k+1)2^{-(n+1)}\right) \right|\\ &=\left|\frac{X((k+1)2^{-n}) + X(k2^{-n}) }2 - X\left((2k+1)2^{-(n+1)}\right) \right|\\ &=\frac12|V\left((2k+1)2^{-(n+1)}\right)|, \end{aligned}

where ${V\left((2k+1)2^{-(n+1)}\right)\sim\mathcal N(0,2^{-n})}$, and so

$\displaystyle \Delta^{(n)}(\omega) = \frac12\max_{0\leqslant k\leqslant 2^n-1}\left|V\left((2k+1)2^{-(n+1)}\right)\right|.$

For ${x\geqslant1}$ we compute

\displaystyle \begin{aligned} \mathbb P\left(\Delta^{(n)}> \frac{x/2}{2^{n/2}}\right) &= \mathbb P\left(\frac12\max_{0\leqslant k\leqslant 2^n-1}\left|V\left((2k+1)2^{-(n+1)}\right)\right| > \frac12\frac x{2^{n/2}}\right)\\ &= \mathbb P\left(\bigcup_{k=0}^{2^n-1}\left\{\left|V\left((2k+1)2^{-(n+1)}\right)\right|>2^{-\frac n2}x\right\}\right) \\ &\leqslant 2^n \mathbb P(|V(2^{-n})|>2^{-\frac n2}x)\\ &= 2^{n+1}\mathbb P(V(2^{-n})/2^{-\frac n2}>x)\\ &= 2^{n+1}\mathbb P(Z > x)\\ &\leqslant 2^{n+1}\left(\frac{\phi(x)}x\right)\\ &=\frac{2^{n+1}e^{-\frac12 x^2}}{\sqrt{2\pi}x}, \end{aligned}

where ${Z\sim\mathcal N(0,1)}$ and ${\phi}$ is the probability density of ${Z}$. Let ${x=2n^{\frac12}}$, then we have

\displaystyle \begin{aligned} \mathbb P\left(\Delta^{(n)}> 2^{-\frac12} n^{\frac12}\right) &\leqslant \frac{2^{n+1}e^{-\frac12\left(2n^{\frac12}\right)^2}}{\sqrt{2\pi}\left(2n^{\frac12}\right)}\\ &= \left(\frac1{\sqrt\pi e^n} \right)\left(\frac 2e \right)^n\\ &\leqslant \left(\frac 2e \right)^n. \end{aligned}

Since

$\displaystyle \sum_{n=0}^\infty \left(\frac 2e \right)^n,$

from the first Borel-Cantelli lemma we have

$\displaystyle \mathbb P\left(\limsup_{n\rightarrow\infty} \left\{\Delta^{(n)}> 2^{-\frac12} n^{\frac12}\right\}\right) = 0,$

from which

$\displaystyle \mathbb P\left(\sum_{n=1}^\infty \Delta^{(n)}<\infty \right)=1.$

This implies that the sequence of functions ${\{B^{(n)}\}}$ on ${C[0,1]}$ is Cauchy, as for ${n

\displaystyle \begin{aligned} \|B^{(n)}-B^{(m)}\|_\infty &= \sup_{t\in[0,1]} \left|B^{(n)}(t)-B^{(m)}(t)| \right|\\ &\leqslant \sum_{j=n}^{m-1}\Delta^{(j)}\\&\stackrel{n,m\rightarrow\infty}\longrightarrow0. \end{aligned}

Since ${C[0,1]}$ is complete, we conclude that ${B:=\lim_{n\rightarrow\infty}B^{(n)}}$ exists, with probability ${1}$, and by construction is Brownian motion on ${[0,1]}$.