In 1923, Norbert Wiener demonstrated the existence of Standard Brownian motion in his paper "Differential Space." Subsequently, Paul Lévy has given the beautiful prove of this fact using dyadic numbers and properties of Gaussian variables. This article elaborates on Lévy's proof, supporting it by visually explaining key concepts.
Definition (Brownian Motion) A real-valued stochastic process {B(t):t≥0} is called a standard Brownian Motion on [0,1] with B0=0 if the following properties holds
the process has independent increments, i.e., for all times 0≤t1<t2<⋯<tn, the increments B(tn)−B(tn−1),B(tn−1)−B(tn−2),…,B(t2)−B(t1) are independent random variables,
for all t≥0 and h>0, the increments B(t+h)−B(t) are normally distributed with expectation zero and variance h,
almost surely, the function t↦B(t) is continuous.
Theorem (Wiener 1923) Standard Brownian motion exists.
Proof of Wiener Theorem
The foolowing proof follows [MörtersPeres] pages 9-12
Proof of Wiener Theorem
The foolowing proof follows [MörtersPeres] pages 9-12
Proof. We construct Brownian motion as a uniform limit of continuous functions, to ensure that it automatically has continuous paths.
Building Brownian Motion on Dyadic Set
The idea is to construct the Brownian Motion on the set of dyadic numbersD=n∈N⋃DnandDn={2nk:0≤k≤2n}and to interpolate the values on D linearly and check that the uniform limit of these continuous functions exists is a Brownian motion.
To do this let (Ω,A,P) be a probability space on which a collection {Zt:t∈D} of independent, standard normally distributed random variables can be defined. Let B(0):=0 and B(1):=Z1. For each n∈N we define the random variables B(d),d∈Dn, such that
for all r<s<t in Dn, the random variable B(t)−B(s) is normally distributed with mean zero and variance t−s, and is independent of B(s)−B(r),
the vectors (B(d):d∈Dn) and (Zt:t∈D∖Dn) are independent.
Note that we have already done this for n=0,D0={0,1}. Proceeding inductively we may assume that we have succeeded in doing it for some n−1. We then define B(d) for d∈Dn∖Dn−1 byB(d)=2B(d−2−n)+B(d+2−n)+2(n+1)/2Zd.Since d±2−n∈Dn−1, then the first summand is the linear interpolation of the values of B at the neighbouring points of d in Dn−1. Therefore (B(d):d∈Dn) is independent of {Zt:t∈D∖Dn} and the second property is fulfilled.
Now we need to verify that successive increments B(d)−B(d−2−n) and B(d+2−n)−B(d) for d∈Dn∖Dn−1 are independent.B(d)−B(d−2−n)B(d+2−n)−B(d)=2B(d+2−n)−B(d−2−n)+2(n+1)/2Zd.=2B(d+2−n)−B(d−2−n)−2(n+1)/2Zd.By our induction assumptions 21[B(d+2−n)−B(d−2−n)]∼N(0,2−n+1) and depends only on (Zt:t∈Dn−1), it is independent of Zd/2(n+1)/2. Hence their sum B(d)−B(d−2−n) and their difference B(d+2−n)−B(d) are independent and normally distributed with mean zero and variance 2−n. The same are true for any d∈Dn. Two any increments over disjoint intervals are thus independent by summing independent increments over intervals of the form (k2−n,(k+1)2−n).
Formally, we can define the functionsF0(t)=⎩⎨⎧Z10linear in between,for t=1,for t=0,and, for each n≥1,Fn(t)=⎩⎨⎧2−(n+1)/2Zt0linear between consecutive points in Dn.for t∈Dn∖Dn−1for t∈Dn−1We can define the partial sum, which for given m incorporates all induction steps up to mBtm=i=0∑mFi(t)
Uniform convergence Btm
We need to prove that with probability one (Btm)m∈N converges uniformly on [0,1]. To do it, we can start with evaluation of the successive differencet∈[0,1]sup∣Btm−Btm−1∣=t∈[0,1]sup∣Fm(t)∣=∥Fm∥≤2−(m+1)/2max{∣Zd∣,d∈Dm}
On the other hand, we can evaluate P(∣Zd∣≥cm) using the following estimation of the standard normal distribution:P(∣Zd∣≥λ)≤π2λσexp(−2σ2λ2),so that the series
P(2−(m+1)/2max{∣Zd∣,d∈Dm}>m21)≤2mP(∣Z1∣>m22(m+1)/2)≤2π1m222m−1exp(−m42m)The right-hand side is summable, for example, by ratio test. Then using the Borel-Cantelli lemma we can conclude that ∃n0∈N,∀m≥n0:t∈[0,1]sup∣Btm−Btm−1∣≤m21for large enoughma.s.and the uniform convergence follows from:t∈[0,1]sup∣Btm+n−Btm−1∣≤i=0∑nt∈[0,1]sup∣Btm+i−Btm+i−1∣≤i=0∑n(m+i)21Finally, the seriesB(t)=n=0∑∞Fn(t)is uniformly convergent on [0,1]. We denote the continuous limit by {B(t):t∈[0,1]}.
Continuous extension Bt to the whole [0,1]
It remains to check that the increments of this process have the right finite-dimensional distributions. This follows directly from the properties of B on the dense set D∩[0,1] and the continuity of the paths. Indeed, suppose that t1<t2<…<tn are in [0,1]. We find tki∈D such that limk→∞tki=ti and infer from the continuity of B that, for 1≤i≤n−1,B(ti+1)−B(ti)=k→∞lim(B(tki+1)−B(tki)).As limk→∞E[B(tki+1)−B(tki)]=0 andk→∞limCov(B(tki+1)−B(tki),B(tkj)−B(tkj−1))=k→∞lim1{j=i+1}(tki+1−tki)=1{j=i+1}(ti+1−ti),the increments B(ti+1)−B(ti) are independent Gaussian random variables with mean 0 and variance ti+1−ti, as required.
Extension Bt to R
We have thus constructed a continuous process B:[0,1]→R with the same finite-dimensional distributions as Brownian motion. Take a sequence B0,B1,… of independent C[0,1]-valued random variables with the distribution of this process, and define {B(t):t≥0} by gluing together the parts, more precisely byB(t)=B⌊t⌋(t−⌊t⌋)+ℓ=0∑⌊t⌋−1Bℓ(1), for all t≥0.This defines a continuous random function B:[0,∞)→R and one can see easily from what we have shown so far that it is a standard Brownian motion. □