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Paul Lévy’s construction of Brownian motion

BROWNIAN MOTION
A real-valued stochastic process {B(t):t≥0}\{B(t): t \geq 0\} is called a standard Brownian Motion on [0,1][0, 1] with B0=0B_0 = 0 if the following properties holds
  • the process has independent increments, i.e., for all times 0≤t1<t2<⋯<tn0 \leq t_1 < t_2 < \cdots < t_n, the increments B(tn)−B(tn−1),B(tn−1)−B(tn−2),…,B(t2)−B(t1)B(t_n) - B(t_{n-1}), B(t_{n-1}) - B(t_{n-2}), \ldots, B(t_2) - B(t_1) are independent random variables,
  • for all t≥0t \geq 0 and h>0h > 0, the increments B(t+h)−B(t)B(t + h) - B(t) are normally distributed with expectation zero and variance hh,
  • almost surely, the function t↦B(t)t \mapsto B(t) is continuous.
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.
WIENER 1923
Standard Brownian motion exists.

The foolowing proof follows [MörtersPeres] pages 9-12

\quad Proof. We construct Brownian motion as a uniform limit of continuous functions, to ensure that it automatically has continuous paths.
The sequence of dyadic sets
The idea is to construct the Brownian Motion on the set of dyadic numbers
D=⋃n∈NDnandDn={k2n:0≤k≤2n}\begin{equation*}\mathcal{D} = \bigcup_{n \in \mathbb{N}} \mathcal{D}_n \quad \textrm{and} \quad \mathcal{D}_n = \left\{ \frac{k}{2^n} : 0 \leq k \leq 2^n \right\}\end{equation*}
and to interpolate the values on DD linearly and check that the uniform limit of these continuous functions exists is a Brownian motion.
To do this let (Ω,A,P) (\Omega, \mathcal{A}, P) be a probability space on which a collection {Zt:t∈D} \{Z_t : t \in \mathcal{D}\} of independent, standard normally distributed random variables can be defined.
Let B(0):=0 B(0) := 0 and B(1):=Z1 B(1) := Z_1 . For each n∈N n \in \mathbb{N} we define the random variables B(d), d∈Dn B(d), \, d \in \mathcal{D}_n , such that
  • for all r<s<t r < s < t in Dn \mathcal{D}_n , the random variable B(t)−B(s) B(t) - B(s) is normally distributed with mean zero and variance t−s t - s , and is independent of B(s)−B(r) B(s) - B(r) ,
  • the vectors (B(d):d∈Dn) (B(d) : d \in \mathcal{D}_n) and (Zt:t∈D∖Dn) (Z_t : t \in \mathcal{D} \setminus \mathcal{D}_n) are independent.
The one step of induction to construct B(d)B(d)
Note that we have already done this for n=0, D0={0,1} n=0, \, \mathcal{D}_0 = \{0,1\} . Proceeding inductively we may assume that we have succeeded in doing it for some n−1n - 1. We then define B(d)B(d) for d∈Dn∖Dn−1 d \in \mathcal{D}_n \setminus \mathcal{D}_{n-1} by
B(d)=B(d−2−n)+B(d+2−n)2+Zd2(n+1)/2.\begin{equation*}B(d) = \frac{B(d - 2^{-n}) + B(d + 2^{-n})}{2} + \frac{Z_d}{2^{(n+1)/2}}.\end{equation*}
Since d ± 2−n ∈ Dn−1d \, \pm \, 2^{-n} \, \in \, \mathcal{D}_{n-1}, then the first summand is the linear interpolation of the values of BB at the neighbouring points of dd in Dn−1 \mathcal{D}_{n-1} . Therefore (B(d):d∈Dn) (B(d) : d \in \mathcal{D}_n) is independent of {Zt:t∈D∖Dn} \{Z_t : t \in \mathcal{D} \setminus \mathcal{D}_n\} and the second property is fulfilled.
Now we need to verify that successive increments B(d)−B(d−2−n)B(d) - B(d - 2^{-n}) and B(d+2−n)−B(d)B(d + 2^{-n}) - B(d) for d∈Dn∖Dn−1d \in \mathcal{D}_n \setminus \mathcal{D}_{n-1} are independent.
B(d)−B(d−2−n)=B(d+2−n)−B(d−2−n)2+Zd2(n+1)/2.B(d+2−n)−B(d)=B(d+2−n)−B(d−2−n)2−Zd2(n+1)/2.\begin{align*}B(d) - B(d - 2^{-n}) &= \frac{B(d + 2^{-n}) - B(d - 2^{-n})}{2} + \frac{Z_d}{2^{(n+1)/2}}. \\B(d + 2^{-n}) - B(d) &= \frac{B(d + 2^{-n}) - B(d - 2^{-n})}{2} - \frac{Z_d}{2^{(n+1)/2}}.\end{align*}
By our induction assumptions 12[B(d+2−n)−B(d−2−n)]∼N(0,2−n+1)\frac{1}{2}[B(d+2^{-n}) - B(d-2^{-n})] \sim \mathcal{N}(0,2^{-n+1}) and depends only on (Zt:t∈Dn−1)(Z_t : t \in \mathcal{D}_{n-1}), it is independent of Zd/2(n+1)/2Z_d/2^{(n+1)/2}. Hence their sum B(d)−B(d−2−n)B(d) - B(d - 2^{-n}) and their difference B(d+2−n)−B(d)B(d + 2^{-n}) - B(d) are independent and normally distributed with mean zero and variance 2−n2^{-n}. The same are true for any d∈Dnd \in \mathcal{D}_{n}. Two any increments over disjoint intervals are thus independent by summing independent increments over intervals of the form (k2−n,(k+1)2−n)(k 2^{-n}, (k+1) 2^{-n}).
The illustration of Dynkin's Theorem
Formally, we can define the functions
F0(t)={Z1for t=1,0for t=0,linear in between,\begin{equation*}F_0(t) = \begin{cases}Z_1 & \text{for } t = 1, \\0 & \text{for } t = 0, \\\text{linear in between},\end{cases}\end{equation*}
and for any nn
Fn(t)={2−(n+1)/2Ztfor t∈Dn∖Dn−10for t∈Dn−1linear between consecutive points in Dn.\begin{equation*}F_n(t) = \begin{cases}2^{-(n+1)/2} Z_t & \text{for } t \in D_n \setminus D_{n-1} \\0 & \text{for } t \in D_{n-1} \\\text{linear between consecutive points in } D_n.\end{cases}\end{equation*}
Finally we can define the approximation sum for Brownian motion, which sums all steps up to nn
Btn=∑i=0nFi(t)\begin{equation*}B_t^n = \sum_{i=0}^{n} F_i(t)\end{equation*}
We need to prove that with probability one (Btn)n∈N(B_t^n)_{n \in \mathbb{N}} converges uniformly on [0,1][0, 1]. To do it, we can start with evaluation of the successive difference
sup⁡t∈[0,1]∣Btm−Btm−1∣=sup⁡t∈[0,1]∣Fm(t)∣=∥Fm∥≤2−(m+1)/2max⁡{∣Zd∣, d∈Dm}\begin{equation*}\sup_{t \in [0,1]} | B_t^{m} - B_t^{m-1} | = \sup_{t \in [0,1]} | F_m(t) | = \| F_m \| \leq 2^{-(m+1)/2} \max \{ |Z_d|, \, d \in \mathcal{D}_m \}\end{equation*}
On the other hand, if Zd∼N(0,σ2)Z_d \sim \mathcal N(0,\sigma^2), then
P(∣Zd∣≥λ)=2∫λ∞1σ2πexp⁡ ⁣(−x22σ2) dx.\begin{equation*}\mathbb P(|Z_d|\ge \lambda) =2\int_\lambda^\infty \frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{x^2}{2\sigma^2}\right)\,dx.\end{equation*}
using the standard estimate on the gaussian ditribution
∫λ∞exp⁡ ⁣(−x22σ2) dx≤σ2λexp⁡ ⁣(−λ22σ2)⟹P(∣Zd∣≥λ)≤21σ2π⋅σ2λexp⁡ ⁣(−λ22σ2)≤2πσλexp⁡ ⁣(−λ22σ2).\begin{equation*}\int_\lambda^\infty \exp\!\left(-\frac{x^2}{2\sigma^2}\right)\,dx \le \frac{\sigma^2}{\lambda}\exp\!\left(-\frac{\lambda^2}{2\sigma^2}\right) \Longrightarrow \mathbb P(|Z_d|\ge \lambda)\le2\frac{1}{\sigma\sqrt{2\pi}}\cdot\frac{\sigma^2}{\lambda}\exp\!\left(-\frac{\lambda^2}{2\sigma^2}\right) \le\sqrt{\frac{2}{\pi}}\frac{\sigma}{\lambda}\exp\!\left(-\frac{\lambda^2}{2\sigma^2}\right).\end{equation*}
Now applying this we can bound 2−(m+1)/2max⁡{∣Zd∣, d∈Dm}2^{-(m+1)/2} \max \{ |Z_d|, \, d \in \mathcal{D}_m \}
P(2−(m+1)/2max⁡{∣Zd∣, d∈Dm}⏟2m elements>1m2)≤2mP(∣Z1∣>2(m+1)/2m2)≤12πm22m−12exp⁡(−2mm4)\begin{equation*}\mathbb{P} \left( 2^{-(m+1)/2} \max \underbrace{\{ |Z_d|, \, d \in \mathcal{D}_m \}}_{2^m \text{ elements}} > \frac{1}{m^2}\right) \leq 2^m \mathbb{P} \left( |Z_1| > \frac{2^{(m+1)/2}}{m^2} \right) \leq \frac{1}{\sqrt{2 \pi}} m^2 2^{\frac{m-1}{2}} \exp\left(-\frac{2^{m}}{m^4}\right)\end{equation*}
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\exists n_0 \in \mathbb{N}, \forall m \geq n_0:
sup⁡t∈[0,1]∣Btm−Btm−1∣≤1m2for large enough m a.s.\begin{equation*}\sup_{t \in [0,1]} | B_t^{m} - B_t^{m-1} | \leq \frac{1}{m^2} \quad \textrm{for large enough} \, m \, \textrm{a.s.}\end{equation*}
and we get the Cauchy sequence, which imply the limit B(t)B(t) on [0,1][0,1]
sup⁡t∈[0,1]∣Btm+n−Btm−1∣≤∑i=0nsup⁡t∈[0,1]∣Btm+i−Btm+i−1∣≤∑i=0n1(m+i)2⟹B(t)=∑n=0∞Fn(t)\begin{equation*}\sup_{t \in [0,1]} | B_t^{m+n} - B_t^{m-1} | \leq \sum^n_{i=0} \sup_{t \in [0,1]} | B_t^{m+i} - B_t^{m+i-1} | \leq \sum^n_{i=0} \frac{1}{(m+i)^2} \quad \Longrightarrow \quad B(t) = \sum_{n=0}^{\infty} F_n(t)\end{equation*}
It remains to check that the increments of this process have the right finite-dimensional distributions. This follows directly from the properties of BB on the dense set D∩[0,1]\mathcal{D} \cap [0,1] and the continuity of the paths. Indeed, suppose that t1<t2<…<tnt_1 < t_2 < \ldots < t_n are in [0,1][0,1]. We find tki∈Dt_{k_i} \in \mathcal{D} such that lim⁡k→∞tki=ti\lim_{k \to \infty} t_{k_i} = t_i and infer from the continuity of BB that, for 1≤i≤n−11 \leq i \leq n-1,
B(ti+1)−B(ti)=lim⁡k→∞(B(tki+1)−B(tki)).\begin{equation*}B(t_{i+1}) - B(t_i) = \lim_{k \to \infty} (B(t_{k_{i+1}}) - B(t_{k_i})).\end{equation*}
As lim⁡k→∞E[B(tki+1)−B(tki)]=0\lim_{k \to \infty} \mathbb{E}[B(t_{k_{i+1}}) - B(t_{k_i})] = 0 and
lim⁡k→∞Cov(B(tki+1)−B(tki),B(tkj)−B(tkj−1))=lim⁡k→∞1{j=i+1}(tki+1−tki)=1{j=i+1}(ti+1−ti),\begin{equation*}\lim_{k \to \infty} \text{Cov}(B(t_{k_{i+1}}) - B(t_{k_i}), B(t_{k_j}) - B(t_{k_{j-1}})) = \lim_{k \to \infty} 1_{\{j=i+1\}}(t_{k_{i+1}} - t_{k_i}) = 1_{\{j=i+1\}}(t_{i+1} - t_i),\end{equation*}
the increments B(ti+1)−B(ti)B(t_{i+1}) - B(t_i) are independent Gaussian random variables with mean 0 and variance ti+1−tit_{i+1} - t_i, as required.
We have thus constructed a continuous process B:[0,1]→RB: [0,1] \rightarrow \mathbb{R} with the same finite-dimensional distributions as Brownian motion. Take a sequence B0,B1,…B_0, B_1, \ldots of independent C[0,1]C[0,1]-valued random variables with the distribution of this process, and define {B(t):t≥0}\{B(t) : t \geq 0\} by gluing together the parts, more precisely by
B(t)=B⌊t⌋(t−⌊t⌋)+∑ℓ=0⌊t⌋−1Bℓ(1), for all t≥0.\begin{equation*}B(t) = B_{\lfloor t \rfloor}(t - \lfloor t \rfloor) + \sum_{\ell=0}^{\lfloor t \rfloor - 1} B_{\ell}(1), \text{ for all } t \geq 0.\end{equation*}
This defines a continuous random function B:[0,∞)→RB: [0, \infty) \rightarrow \mathbb{R} and one can see easily from what we have shown so far that it is a standard Brownian motion. □\Box

References