Brownian Motion

While stochastic differential equations with jumps are continuous-time and continuous-state, they are not quite what you’d imagine as a continuous process

Can we construct a process with independent and stationary increments that does not have state-dependent coefficients?

Definition

We start with assumptions about how this model should work,

• $X_0=0$
• $X_{t_i}-X_{s_i}$ and $X_{t_j}-X_{s_j}$ for $s_i\le t_i\le s_j\le t_j$ are independent
• The distribution of $X_t-X_s$ depends only on $t-s$
• $\mathbb{E}(X_t)=0$
• $X_t$ is a continuous function of $t$

These assumptions happen to uniquely describe the process, up to a scaling constant

What’s the distribution of $X_t$? Let’s start with $t=1$,
$X_1=[X_{1/n}-X_0]+[X_{2/n}-X_{1/n}]+\cdots +[X_{n/n}-X_{(n-1)/n}]$ for any $n$, i.e. $X_1$ is the sum of $n$ i.i.d. random variables

These increments are also small, meaning $M_n=\max \{|X_{1/n}-X_0|,\dots ,|X_{n/n}-X_{(n-1)/n}|\}\to 0$ as $n\to \infty$

Therefore, $X_t$ is normal

A Brownian motion or a Weiner process with variance parameter ${\sigma }^2$ is a stochastic process $X_t$ taking values in the real numbers satisfying,

1. $X_0=0$;
2. For any $s_1\le t_1\le s_2\le t_2\le \cdots \le s_n\le t_n$, the variables $X_{t_1}-X_{s_1},\dots ,X_{t_n}-X_{s_n}$ are independent;
3. For any $s<t$, the random variable $X_t-X_s$ has a normal distribution with mean 0 and variance $(t-s){\sigma }^2$;
4. The paths are continuous, i.e., the function $t\mapsto X_t$ is a continuous function of $t$

We include the normal distribution requirement, even though it follows naturally from the definition

Standard Brownian motion has ${\sigma }^2=1$
We can also speak of a Brownian motion with $X_0=x$

We can look at this motion as a limit of random walks

Suppose $S_n$ is an unbiased random walk on the integers
$S_n=Y_1+\cdots +Y_n$ where $\mathbb{P}\{Y_i=1\}=\mathbb{P}\{Y_i=-1\}=\frac{1}{2}$

We look at time increments of $\Delta t=1/N$ and set $W_{k\Delta t}^{(N)}=N^{-1/2}{S}_k$ ($N^{-1/2}$ normalizes $\text{Var}(S_N)=N$)), so the size of the jump in $\Delta t=1/N$ is $N^{-1/2}=(\Delta t{)}^{1/2}$

We consider the discrete approximation as a process for all values of $t$ (not just intervals of $\Delta t$) by linear interpolation, and as $N\to \infty$, the discrete approximation approaches a continuous process which is Brownian motion (the details of this limit are not important here)

$W_1^{(N)}=\frac{S_N}{\sqrt{N}}$ approaches a standard normal by CLT and $W_t^{(N)}$ approaches a normal distribution with mean 0 and variance $t$

$\mathbb{E}(|X_{t+\Delta t}-X_t{|}^2)=\Delta t$, meaning the typical size of an increment $|X_{t+\Delta t}-X_t|$ is about $\sqrt{\Delta t}$

$\frac{d{X}_t}{dt}={\lim }_{\Delta t\to 0}\frac{X_{t+\Delta t}-X_t}{\Delta t}$ does not exist, since $\sqrt{\Delta t}$ is much larger than $\Delta t$ for small values

The path of a Brownian motion $X_t$ is nowhere differentiable

This is strangely non-trivial to prove
Consider the following two statements,

• $\mathbb{P}\{X_t\ne 1\}=1$ for all $t$
• $\mathbb{P}\{X_t\ne 1\text{ for all }t\}=1$

The first statement is true, but the second statement is false! $X_0=0$ and likely some $X_t>1$, in which case there is a point that crosses over $X_t=1$

We cannot simply take the union of an uncountable set

As an illustrative example, consider $\{-\infty <Y<\infty \}=\bigcup _{-\infty <y<\infty }\{Y=y\}$

Markov Property

Let ${\mathcal{F}}_t$ represent the information contained in $X_s,s\le t$
$E(X_t\mid {\mathcal{F}}_s)=E(X_s\mid {\mathcal{F}}_s)+E(X_t-X_s\mid {\mathcal{F}}_s)=X_s$

Rewritten, $E(X_t\mid {\mathcal{F}}_s)=X_s=E(X_t\mid X_s)$
This is the Markov property of Brownian motion, that we can predict $X_t$ given all the information up through time $s$ with just the value of the Brownian motion at time $s$

The general form of the Markov property is $E[f(X_t)\mid {\mathcal{F}}_s]=E[f(X_t)\mid X_s]$ for $s\le t$

For Brownian motion, this follows from a stronger property, that $X_{s+t}-X_s$ is a Brownian motion independent of ${\mathcal{F}}_s$, i.e. $X_{s+t}$ is a Brownian motion starting at $X_s$

$p_t(x,y)=\frac{1}{\sqrt{2\pi t}}{e}^{-(y-x{)}^2/2t}$, since $X_t-X_0$ is $\mathcal{N}(0,t)$
This satisfies Chapman-Kolmogorov, written in the continuous form as $p_{s+t}(x,y)={\int }_{-\infty }^{\infty }{p}_s(x,z)p_t(z,y)dz$

We want a stronger guarantee than the Markov property

A random variable $T\in [0,\infty ]$ is a stopping time for Brownian motion if $\{T\le t\}$ is measurable with respect to ${\mathcal{F}}_t$
We mainly consider $T_x=\inf \{t:X_t=x\}$

${\mathcal{F}}_T$ represents the information contained up through a stopping time $T$
$Y_t=X_{t+T}-X_T$ is the process beyond time $T$

Strong Markov Property: $Y_t$ is a Brownian motion independent of ${\mathcal{F}}_t$

This is more powerful because it states independence from a non-fixed time (a random variable)

Example: Calculate the probability that there is some $0\le t\le 1$ where $X_t\ge 1$

Let $T=T_1$ be the first time the Brownian motion equals 1
$\{X_t\ge 1\text{ for some }0\le t\le 1\}=\{T\le 1\}$
$\mathbb{P}\{T=1\}\le \mathbb{P}\{X_1=1\}=0$ so $\mathbb{P}\{T\le 1\}=\mathbb{P}\{T<1\}$
$\mathbb{P}\{X_1\ge 1\}={\int }_1^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}dx$
$=\mathbb{P}\{T\le 1\}\mathbb{P}\{X_1\ge 1\mid T\le 1\}$

This step uses the Strong Markov property,
$X_1-X_T=X_1-1$, so $\mathbb{P}\{X_1-1\ge 0\mid T\le 1\}=1/2$

$\mathbb{P}\{T\le 1\}=2\mathbb{P}\{X_1\ge 1\}=2{\int }_1^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}dx$

Reflection Principle: Suppose $X_t$ is a Brownian motion with ${\sigma }^2$ starting at $a$ and $a<b$. Then for any $t>0$, $\mathbb{P}\{X_s\ge b\text{ for some }0\le s\le t\}=2\mathbb{P}\{X_t\ge b\mid X_0=a\}=2{\int }_b^{\infty }\frac{1}{\sqrt{2\pi t{\sigma }^2}}{e}^{-(x-a{)}^2/(2{\sigma }^{2}t)}dx$

Example:
$\mathbb{P}\{X_s=0\text{ for some }1\le s\le t\}$

Suppose $X_1=b>0$
The probability $X_s=0$ for some $1\le s\le t$ is the same as the probability $X_s\le -b$ for some $0\le s\le t-1$, which by symmetry is the same as the probability $X_s\ge b$ for some $0\le s\le t-1$

$\mathbb{P}\{X_s=0\text{ for some }1\le s\le t\mid X_1=b\}=2\mathbb{P}\{X_t\ge b\mid X_0=0\}=2{\int }_b^{\infty }\frac{1}{\sqrt{2\pi (t-1)}}{e}^{-x^2/2(t-1)}dx$

We average over all values of $b$,
$\mathbb{P}\{X_s=0\text{ for some }1\le s\le t\}={\int }_{-\infty }^{\infty }{p}_1(0,b)\mathbb{P}\{X_s=0\text{ for some }1\le s\le t\mid X_1=b\}db=2{\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-b^2/2}\left[2{\int }_b^{\infty }\frac{1}{\sqrt{2\pi (t-1)}}{e}^{-x^2/2(t-1)}dx\right]db$

We can ultimately solve this with some substitutions, leading to $1-\frac{2}{\pi }{\text{tan}}^{-1}\frac{1}{\sqrt{t-1}}$

Scaling Properties

Consider $Z=\{t:X_t=0\}$
This is an interesting “fractal” of the real line

Scaling Properties: Suppose $X_t$ is a standard Brownian motion,

1. $Y_t=a^{-1/2}{X}_{at}$ is a standard Brownian motion, for $a>0$
2. $Y_t=t{X}_{1/t}$ is a standard Brownian motion

We showed before $\mathbb{P}\{Z\cap [1,t]\ne \varnothing \}=1-\frac{2}{\pi }{\tan }^{-1}\frac{1}{\sqrt{t-1}}$ which approaches 1 as $t\to \infty$
Along with the strong Markov property, this show that the Brownian motion returns to the origin infinite times

What happens near $t=0$? Our second scaling property implies that $Y_t=t{X}_{1/t}$ also returns to the origin infinite times near the origin

$Z$ has many interesting properties,

• It is closed, i.e. $t_i\in Z$ and $t_i\to t$ implies $t\in Z$ (from continuity)
• 0 is not an isolated point, i.e. there are positive numbers $t_i$ such that $t_i\to 0$ (in fact no points are isolated)
• $Z$ looks like the Cantor set

What is the dimension of $Z$?
Both Hausdorff dimensions and box dimensions can give rise to fractional dimensions, which are often referred to as fractal dimensions (we talk about box dimension)

Suppose we have a bounded set $A$ in ${\mathbb{R}}^d$ and we need to cover $A$ with $d$-dimensional balls with diameter $ϵ$… How many balls will it take? A line segment will take ${ϵ}^{-1}$ balls, a square ${ϵ}^{-2}$, and a standard $k$-dimensional set will take ${ϵ}^{-k}$ balls

Thus, the box dimension of a set $A$ is the number $D$ such that for small $ϵ$ the number of balls of diameter $ϵ$ needed to cover $A$ is on the order of ${ϵ}^{-D}$

Example:
Consider the Cantor set $A$, obtained by recursively removing the middle third of each interval

We can cover $A$ with $2^n$ intervals of length $3^{-n}$
$2^n=(3^{-n}{)}^{-D}$
$D=\frac{\ln 2}{\ln 3}$

How do we apply this to $Z$? Let’s cover $Z_1=Z\cap [0,1]$ with intervals of diameter $ϵ=1/n$

We will consider the $n$ intervals $\left[\frac{k-1}{n},\frac{k}{n}\right]$, $k=1,2,\dots ,n$
A particular interval is needed if $Z_1\cap [(k-1)/n,k/n]\ne \varnothing$
$P(k,n)=\mathbb{P}\left\{Z_1\cap \left[\frac{k-1}{n},\frac{k}{n}\right]\ne \varnothing \right\}$

Assume $k\ge 1$, $Y_t=((k-1)/n{)}^{-1/2}{X}_{nt(k-1)}$ is a standard Brownian motion

So $P(k,n)=\mathbb{P}\left\{Y_t=0\text{ for some }1\le t\le \frac{k}{k-1}\right\}$
This example solved the required probability, so we see $P(k,n)=1-\frac{2}{\pi }{\tan }^{-1}\sqrt{k-1}$

${\sum }_{k=1}^{n}P(k,n)={\sum }_{k=1}^n\left[1-\frac{2}{\pi }{\tan }^{-1}\sqrt{k-1}\right]$
${\tan }^{-1}\frac{1}{t}=\frac{\pi }{2}-t+O(t^2)$
For large $x$, ${\tan }^{-1}x\approx \frac{\pi }{2}-\frac{1}{x}$
Hence, ${\sum }_{k=1}^{n}P(k,n)\approx 1+{\sum }_{k=2}^n\frac{2}{\pi \sqrt{k-1}}\approx \frac{2}{\pi }{\int }_1^n(x-1{)}^{-1/2}dx\approx \frac{4}{\pi }\sqrt{n}$

So it takes on the order of $\sqrt{n}$ intervals of length $\frac{1}{n}$ to cover $Z_1$
$\sqrt{n}={\left(\frac{1}{n}\right)}^{-D}$
$D=\frac{1}{2}$

Brownian Motion in Several Dimensions

Suppose $X_t^1,\dots ,X_t^d$ are independent one-dimensional standard Brownian motions, then we call $X_t=(X_t^1,\dots ,X_t^d)$ a standard $d$-dimensional Brownian motion

This satisfies the following,

• $X_0=0$
• For any $s_1\le t_1\le s_2\le t_2\le \cdots \le s_n\le t_n$, the vector-valued random variables $X_{t_1}-X_{s_1},\dots ,X_{t_n}-X_{t_{n-1}}$ are independent
• The random variable $X_t-X_s$ has a joint normal distribution with mean 0 and covariance matrix $(t-s)\mathbf{I}$, i.e. joint density $\left(\frac{1}{\sqrt{2\pi r}}{e}^{-x_1^2/2r}\right)\dots \left(\frac{1}{\sqrt{2\pi r}}{e}^{-x_d^2/2r}\right)=\frac{1}{(2\pi r{)}^{d/2}}{e}^{-|x{|}^2/2r}$
• $X_t$ is a continuous function of $t$

$p_t(x,y),x,y\in {\mathbb{R}}^d$ is the probability density of $X_t$ assuming $X_0=x$
$p_t(x,y)=\frac{1}{(2\pi t{)}^{d/2}}{e}^{-|y-x{|}^2/2t}$

This still satisfies Chapman-Kolmogorov, $p_{s+t}(x,y)={\int }_{R^d}^{}{p}_s(x,z)p_t(z,y)d{z}_1\dots d{z}_d$

Brownian motion is related to the theory of diffusion
Suppose that a large number of particles are distributed in ${\mathbb{R}}^d$ according to density $f(y)$ and let $f(t,y)$ denote the density at time $t$, such that $f(0,y)=f(y)$

If a particle starts at position $x$, the probability density for its position at time $t$ is $p_t(x,y)$
Integrating gives $f(t,y)={\int }_{{\mathbb{R}}^d}^{}f(x)p_t(x,y)d{x}_1\dots d{x}_d$

Symmetry tells us $p_t(x,y)=p_t(y,x)$
$f(t,y)={\int }_{{\mathbb{R}}^d}^{}f(x)p_t(y,x)d{x}_1\dots d{x}_d$
Assuming $X_0=y$, this is the expected value of $f(X_t)$, which we denote as $f(t,y)={\mathbb{E}}^y[f(X_t)]$

We’d like to derive a differential equation for $f(t,x)$
Consider $\frac{\partial f}{\partial t}$ with $t=0,d=1$
The Taylor series for a function $f$ is $f(y)=f(x)+f^{\prime }(x)(y-x)+\frac{1}{2}{f}^{\prime \prime }(x)(y-x{)}^2+o((y-x{)}^2)$, where $o(\cdot )$ is an error term such that $\frac{o((y-x{)}^2)}{(y-x{)}^2}\to 0$ as $y\to x$

$\frac{\partial f}{\partial t}{|}_{t=0}={\lim }_{t\to 0}\frac{1}{t}{\mathbb{E}}^x[f(X_t)-f(X_0)]$
$={\lim }_{t\to 0}\frac{1}{t}\left[f^{\prime }(x){\mathbb{E}}^x[X_t-x]+\frac{1}{2}{f}^{\prime \prime }(x){\mathbb{E}}^x(X_t-x{)}^2+o((X_t-x{)}^2)\right]$
${\mathbb{E}}^x[X_t-x]=0$ and ${\mathbb{E}}^x[(X_t-x{)}^2]=\text{Var}(X_t)=t$
So we get $\frac{\partial f}{\partial t}{|}_{t=0}=\frac{1}{2}{f}^{\prime \prime }(x)$
The same thing works for all $t$, so $\frac{\partial f}{\partial t}=\frac{1}{2}\frac{{\partial }^{2}f}{\partial x^2}$
Generalizing $d$, $\frac{\partial f}{\partial t}=\frac{1}{2}\Delta f$ where $\Delta f(t,x_1,\dots ,x_d)={\sum }_{i=1}^d\frac{{\partial }^{2}f}{\partial x_i^2}$ is the Laplacian

This is known as the heat equation

Brownian motion with variance parameter ${\sigma }^2=D$ yields $\frac{\partial f}{\partial t}=\frac{D}{2}\Delta f$ with the diffusion constant

Consider a bounded region $B$ with boundary $\partial B$ with initial heat distribution $f(x),x\in B$ and fixed temperature $g(y),y\in \partial B$
If $u(t,x)$ denotes the temperature at $x$ at time $t$, then $u(t,x)$ satisfies

1. $\frac{\partial u}{\partial t}=\frac{D}{2}\Delta u,x\in B$
2. $u(t,x)=g(x),x\in \partial B$
3. $u(0,x)=f(x),x\in B$

We can write this in terms of Brownian motion,
Let $X_t$ be a $d$-dimensional Brownian motion with ${\sigma }^2=D$, and let $\tau ={\tau }_{\partial B}=\tau =\text{inf}\{t:X_t\in \partial B\}$,
$u(t,x)={\mathbb{E}}^x[f(X_t)I\{\tau >t\}+g(X_{\tau })I\{\tau \le t\}]$

As $t\to \infty$, we reach a steady-state distribution $v(x)$,

1. $\Delta v(x)=0,x\in B$
2. $v(x)=g(x),x\in \partial B$

$v(x)={\lim }_{t\to \infty }u(t,x)={\mathbb{E}}^x[g(X_{\tau })]$

Example:
Let $d=1$, $B=(a,b)$ with $0\le a\le b<\infty$ such that $\partial B=\{a,b\}$, and take $a<x<b$

Consider $\tau =\text{inf}\{t:X_t=a\text{ or }b\}$ where $X_t$ is a standard Brownian motion

Let $g$ be the function on $\partial B$ where $g(a)=0$ and $g(b)=1$
$v(x)={\mathbb{E}}^x[g(X_{\tau })]={\mathbb{P}}^x\{X_{\tau }=b\}$

$\frac{d^{2}v}{d{x}^2}=0$ for $a<x<b$
$v(a)=0$
$v(b)=1$
$v(x)=\frac{x-a}{b-a}$

We can extrapolate recurrence in one dimension from this,
$a=0$
${\lim }_{b\to \infty }v(x)=0$, i.e. $x$ always returns to $0$

Recurrence and Transience

Suppose $X_t$ is a standard $d$-dimensional Brownian motion and let $B=B(R_1,R_2)$ be the annulus $B=\{x\in {\mathbb{R}}^d:R_1<|x|<R_2\}$ with $\partial B=\{x\in {\mathbb{R}}^d:|x|=R_1\text{ or }|x|=R_2\}$

Let $f(x)=f(x,R_1,R_2)$ be the probability that a standard Brownian motion starting at $x$ hits the sphere defined by $R_2$ before $R_1$

$\tau ={\tau }_{\partial B}=\text{inf}\{t:X_t\in \partial B\}$
$f(x)={\mathbb{E}}^x[g(X_{\tau })]$ where $g(y)=1(|y|=R_2)$

To start, we have,
$\Delta f(x)=0$, $x\in B$ and $f(y)=0$ for $|y|=R_1$ and $1$ for $|y|=R_2$

Due to symmetry, we can also write $f(x)=\varphi (|x|)$ for some $\varphi$
Written in spherical coordinates,
$\Delta \varphi (r)=\frac{d^2\varphi }{d{r}^2}+\frac{d-1}{r}\frac{d\varphi }{dr}$
${\varphi }^{\prime \prime }(r)+\frac{d-1}{r}{\varphi }^{\prime }(r)=0$
$\varphi (r)=\{\begin{array}{ll}{c}_1\ln r+c_2,& d=2\\ c_1{r}^{2-d}+c_2,& d\ge 3\end{array}$

We use our boundary conditions to obtain,
$f(x)=\frac{\ln |x|-\ln R_1}{\ln R_2-\ln R_1}$, for $d=2$
$f(x)=\frac{R_1^{2-d}-|x{|}^{2-d}}{R_1^{2-d}-R_2^{2-d}}$, for $d\ge 3$

Given this equation, what’s the probability that $d=2$ Brownian motion never returns to the disc of radius $ϵ$?

${\lim }_{R_2\to \infty }{\mathbb{P}}^x\{|X_t|=R_2\text{ before }|X_t|=ϵ\}={\lim }_{R_2\to \infty }\frac{\ln |x|-\ln ϵ}{\ln R_2-\ln ϵ}=0$

So for discs with positive radius, Brownian motion is recurrent, but can $d=2$ Brownian motion return to exactly $X_t=0$?

${\lim }_{ϵ\to 0}\left[1-\frac{\ln |x|-\ln ϵ}{\ln R_2-\ln ϵ}\right]=0$

No, Brownian motion in two dimensions is neighborhood recurrent but not point recurrent

For $d\ge 3$,
${\lim }_{R_2\to \infty }\frac{{ϵ}^{2-d}-|x{|}^{2-d}}{{ϵ}^{2-d}-R_2^{2-d}}=1-{\left(\frac{ϵ}{|x|}\right)}^{d-2}>0$

So Brownian motion with $d\ge 3$ is transient

Fractal Nature

We already talked about the fractal nature of the zero-set of Brownian motion, but there is more to say about fractals

Let $A=\{x\in {\mathbb{R}}^d:X_t=x\text{ for some }t\}$ with $d\ge 2$
We can bound the set $A_1=A\cap \{x:|x|\le 1\}$

How many balls of diameter $ϵ$ do we need to cover $A_1$?

For $d=2$, consider the ball of radius $1$, which needs on order of ${ϵ}^{-d}$ to be covered. By our argument in the previous section, the Brownian motion will visit all open balls hence we need all ${ϵ}^{-2}$ and the box dimension of $A$ is $2$

For $d>2$, the probability a ball with diameter $ϵ/2$ around a point $x$ is visiting is $(ϵ/2|x|{)}^{d-2}$ which is about a constant times ${ϵ}^{d-2}$ (considering the scaling in terms of $ϵ$), thus the total number of balls needed is ${ϵ}^{d-2}{ϵ}^{-d}={ϵ}^{-2}$

We conclude the path of a $d$-dimensional Brownian motion $(d\ge 2)$ has fractal dimension two

Closely related to the fractal nature of Brownian motion is the scaling rule listed earlier, if $X_t$ is a standard one-dimensional Brownian motion and $b>0$, then $Y_t=b^{-1/2}{X}_{bt}$ is also a standard Brownian motion

We examine whether there is some process by which $Y_t=b^{-\lambda }{X}_{bt}$ has the same distribution as $X_t$ for $\lambda \ne 1/2$, i.e. whether there are possible alternative scaling rules

If $X_t$ has finite variance then $\text{Var}(X_1)=\text{Var}[X_{1/n}+(X_{2/n}-X_{1/n})+\cdots +(X_{n/n}-X_{(n-1)/n})]$
$=n\text{Var}(X_{1/n})=n\text{Var}(n^{-\lambda }{X}_1)=n^{1-2\lambda }\text{Var}(X_1)$, implying $\lambda =1/2$

Let $M_n=\max \{|X_{1/n}|,|X_{2/n}-X_{1/n}|,\dots ,|X_{n/n}-X_{(n-1)/n}|\}$
If the paths have jumps then $\mathbb{P}\{M_n\ge ϵ\}$ ought not to go to zero as $n\to \infty$, however they should still be limited at some point

$\mathbb{P}\{M_n\le r\}=\mathbb{P}\{|X_{j/n}-X_{(j-1)/n}|\le r\text{ for }j=1,\dots ,n\}$
$=\mathbb{P}\{|X_{1/n}|\le r{\}}^n$
$=\mathbb{P}\{n^{-\lambda }|X_1|\le r{\}}^n=\mathbb{P}\{|X_1|\le r{n}^{\lambda }{\}}^n$

The book implies we can extrapolate from this that $\mathbb{P}\{|X_1|\ge y\}\sim c{y}^{-1/\lambda }$ is a good candidate

If $\lambda <1/2$ then this $X_1$ has a finite variance and therefore $\lambda =1/2$
For $\lambda >1/2$, there are examples called symmetric stable distributions, with corresponding processes called symmetric stable processes

These densities can only be given explicitly for $\lambda =1$, which is the Cauchy distribution with $f(x)=\frac{1}{\pi (1+x^2)}$

Brownian Motion with Drift

Consider a $d$-dimensional Brownian motion $X_t$ with variance parameter ${\sigma }^2$ starting at $x\in {\mathbb{R}}^d$, let $\mu \in {\mathbb{R}}^d$ and $Y_t=X_t+t\mu$

We call $Y_t$ a $d$-dimensional Brownian motion with drift $\mu$ and variance parameter ${\sigma }^2$ starting at $x$

$Y_t$ satisfies,

1. $Y_0=x$
2. Disjoint increments are independent
3. $Y_t-Y_s$ has a normal distribution with mean $\mu (t-s)$ and covariance matrix ${\sigma }^2(t-s)\mathbf{I}$
4. $Y_t$ is a continuous function of $t$

The motion of $Y_t$ is essentially a “straight line” with fluctuations, i.e. $\mathbb{E}(Y_t)=t\mu$

$p_t(x,y)=\frac{1}{(2\pi {\sigma }^{2}t{)}^{d/2}}{e}^{-|y-x-t\mu {|}^2/2t{\sigma }^2}$ which satisfies the Chapman-Kolmogorov equation

Considering $d=1$, $t=0$, we write $f$ in a Taylor series about $x$,
$f(y)=f(x)+f^{\prime }(x)(y-x)+\frac{1}{2}{f}^{\prime \prime }(x)(y-x{)}^2+o((y-x{)}^2)$

${\mathbb{E}}^x[f(Y_t)]=f(x)+f^{\prime }(x){\mathbb{E}}^x[Y_t-x]+\frac{1}{2}{f}^{\prime \prime }(x){\mathbb{E}}^x[(Y_t-x{)}^2]+o(\mathbb{E}(Y_t-x{)}^2)$
${\mathbb{E}}^x[Y_t-x]=\mathbb{E}[X_t+t\mu ]=t\mu$
${\mathbb{E}}^x[(Y_t-x{)}^2]=\mathbb{E}[(X_t+t\mu {)}^2]=(t\mu {)}^2+{\sigma }^{2}t$
$\frac{\partial f}{\partial t}{|}_{t=0}={\lim }_{t\to 0}\frac{{\mathbb{E}}^x[f(Y_t)]-{\mathbb{E}}^x[f(Y_0)]}{t}=\mu f^{\prime }(x)+\frac{{\sigma }^2}{2}{f}^{\prime \prime }(x)$

Compared to the earlier calculations without drift, the addition of drift added a first derivative

In $d$ dimensions, $\frac{\partial f}{\partial t}={\sum }_{i=1}^d{\mu }_i\frac{\partial f}{\partial x_i}+\frac{{\sigma }^2}{2}\Delta f$