Wiener measure

Def The Wiener measure on $C([0,1])$ is the probability measure $\mu$ on the measurable space $(C[0,1],\mathcal{B}(C[0,1]))$ with the following properties:

1. $x(t)\sim N(0,t)$ for all $t\in [0,1]$, that is

$$\mathbb{P}(x(t)\in I)={\int }_I\frac{1}{\sqrt{2\pi t}}{e}^{-u^2/2t}du$$

2. indipendence of the increments: given $0\le t_1<t_2\dots t_n\le 1$, the random variables

$$x(t_2)-x(t_1),\dots ,x(t_n)-x(t_{n-1})$$

are indipendent.

Remark By $x(0)\sim N(0,0)$ we mean $x(0)=0$ almost surely.

Existence §

We will follow this strategy: construct an explicit sequence of probability measures $({\mu }_n{)}_{n\ge 1}$ on $C[0,1]$, which will be shown to converge to the Wiener measure.

Let $({\xi }_k{)}_{k\ge 1}$ be a sequence of i.i.d real-valued random variables with

$$\mathbb{E}[{\xi }_1]=0\mathbb{E}[{\xi }_1^2]={\sigma }^2\in (0,\infty )$$

the result doesn not matter on the exact law, you could use ${\xi }_1\sim rad(1/2)$ or ${\xi }_1\sim N(0,{\sigma }^2)$.

We use this sequence to define a random walk on $\mathbb{R}$:

$$S_n=\sum _{k=1}^n{\xi }_k{S}_0=0$$

we make this random walk into a continuous function by linear interpolation, for $t\in [0,1]$ define

$$X_t^n:=\frac{S_{\lfloor nt\rfloor }}{\sigma \sqrt{n}}+(nt-\lfloor nt\rfloor )\frac{{\xi }_{\lfloor nt\rfloor +1}}{\sigma \sqrt{n}}$$

this is a sequence in $C[0,1]$, we donte by $({\mu }_n{)}_{n\ge 1}$ their associated sequence of probability measures.

Convergence of the FDDS §

We start by showing that the FDDs of $X^n$ converge to the ones of the Wiener measure. First let’s prove that $X_t^n⟹N(0,t)$ for any given $t\in (0,1]$, the case $t=0$ is obvious.

Let’s show that the interpolating part goes to zero in probability using Chebyschev inequality

$$\mathbb{P}(|\psi (n,t)|>ϵ)\le \frac{\mathbb{E}[\psi (n,t)2]}{{ϵ}^2}\le \frac{{\xi }^2}{{\sigma }^{2}n{ϵ}^2}=\frac{1}{n{ϵ}^2}\to 0$$

as $n\to \infty$. Then

$$\frac{S_{\lfloor nt\rfloor }}{\sigma \sqrt{n}}⟹N(0,t)$$

from the Central Limit Theorem.

We now prove that the FDDs converge, for $k=2$ (the general case follows from induction). We need to show that

$$\left(\frac{S_{\lfloor ns\rfloor }}{\sigma \sqrt{n}},\frac{S_{\lfloor nt\rfloor }}{\sigma \sqrt{n}}\right)⟹\left(B_s,B_t\right)$$

since we already proved that $\psi (n,t)\to 0$ in probability.

Remark $X_n⟹X$ and $Y_n⟹Y$ doesn not imply $(X_n,Y_n)⟹(X,Y)$. A counter example is $X_n=N(0,1)$ and $Y_n=(-1{)}^{n}N(0,1)$. However this holds when $X_n$ and $Y_n$, $X$ and $Y$ are indipendent. We can exploit this result using increments

$$\left(\frac{S_{\lfloor ns\rfloor }}{\sigma \sqrt{n}},\frac{S_{\lfloor nt\rfloor }}{\sigma \sqrt{n}}-\frac{S_{\lfloor ns\rfloor }}{\sigma \sqrt{n}}\right)⟹\left(B_s,B_t-B_s\right)$$

the functions $h(x,y-x)=(x,y)$ is continuous, hence by the continuous mapping theorem our goal follows.

Which follows from the convergence in distribution of the marginals

$$\frac{S_{\lfloor ns\rfloor }}{\sigma \sqrt{n}}⟹B_s$$ $$\frac{S_{\lfloor nt\rfloor }}{\sigma \sqrt{n}}-\frac{S_{\lfloor ns\rfloor }}{\sigma \sqrt{n}}⟹B_t-B_s$$

Relative compactness §