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<t_1<\cdots<t_k}, 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<t},

    \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<m}

\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]}.

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