[Stochastic Process] Brownian Motion

Course notes for "Stochastic Process"

2014 Fall 

1. Motivation

Brownian motion can be thought of a symmetric random walk where the jumps sizes are very small and where jumps occur very frequently.

• Each jump size are $\Delta x$
• The time before two jumps are $\Delta t$.

1.1 What's the mean and variance?

Let $X_i$ denote whether the i-th jump is to the right(+1) or to the left(-1), we have $X_i =1, -1$ with probability $\frac{1}{2}$ and $\frac{1}{2}$ respectively.
Thus, $E[X_i] = 0$ and $Var[X_i] = E[X^2_i] - E[X_i]^2 = 1$.

Let $X$ denote the state of  Markov Chain after n jumps, then
$X = \Delta x \cdot (X_1+X_2+\cdots+X_n)$
The $X(t)$ denote the continuous Markov Chain after n jumps, then
$X(t) = \Delta x \cdot (X_1 + X_2 + \cdots + X_{|t/\Delta t|})$

Then we have
$E[X(t)] = 0$ and $Var[X(t)] = (\Delta x)^2 \cdot \frac{t}{\Delta t}$.

Let $\Delta x = \sigma \sqrt{\Delta t} \to 0$,  then
$V[X(t)] = (\Delta x)^2 \cdot \frac{t}{\Delta} \to \sigma^2 t$.

2. Properties of Brownian Motion

• (1)  $X(0) = 0$
• (2)  $X(t) ~ N(0, \sigma^2 t)$
• (3)  X(t) has independent increments
         i.e., $X(t_2) - X(t_1)$ is independent of $X(t_4) - X(t_3)$ assuming the intervals of $[t_1,t_2]$ and $[t_3,t_4]$ are disjoint.
• (4)  X(t) has stationary increments
         i.e., $X(t_2) - X(t_1)$ has the same distribution as $X(t_4) - X(t_3)$ if $t_2 - t_1 = t_4 - t_3$.

Example: What's the distribution of $X(t_2) - X(t_1)$?

Answer: by the stationary property, we have $X(t_2) - X(t_1)$ ~ $N(0, \sigma^2(t_2 - t_1))$.


3. Standard Brownian Motion (SBM)

• SBM ~ $N(0,1)$
• Let $Y(t) = \frac{X(t)}{\sigma} = \sqrt{\Delta t}$, then $V[Y(t)] = 1$.

4. Brownian Motion with Drift

Definition 1: Let $B(t)$ be the standard Brownian motion. Let $X(t) = \sigma B(t) + \mu t$, then $X(t)$ is Brownian motion with drift $\mu$.

Definition 2: ${X(t); t\ge 0}$. $X(t)$ is Brownian Motion with drift $\mu$ and variance parameter $\sigma^2$ if

• $X(0) = 0$
• $\{X(t); t \geq 0\}$ has stationary and independent increments
• $X(t)$ ~ $N(\mu t , \sigma^2 t)$

Example: Let $X(t)$ be Brownian motion with $\sigma = 2$ and drift $\mu = 0.1$. What is $Pr\{X(30) >0 | X(10) = -3\}
Answer:    
                     $Pr\{X(30) >0 | X(10) = -3\}$
                =  $Pr\{X(30) - X(20) >3 | X(10) =3 \}$
                =  $Pr\{X(30) - X(10) >3  \}$  (independent property)
                =  $Pr\{X(20) - X(0) >3  \}$ (stationary property)
                =  $Pr\{X(20)>3  \}$  (X(0) = 0)
                =  $Pr\{N(2,80) > 3\} = Pr\{X(0,1) > \frac{3-2}{\sqrt{80}}$ = $1-\Phi(\frac{1}{4\sqrt{5}})$.


5. Brownian Bridge
Basic Idea: condition on the final value of a Brownian motion process and derive the stochastic properties in between.

Main Results : $X(s) = x | X(t) = B$ is normally distributed with

• Mean: $\frac{s}{t} \cdot B$
• Variance: $\frac{s}{t} \cdot (t-s)$
• The value of s with the highest variance is $s = t/2$. That is, we know the endpoints of the process, but we don't know exactly what happens in between.

Note: these results are for standard Brownian motion (SBM)

Proof:

Example 1:  Suppose you have a stock whose value follows Brownian motion with $X(t)$ ~ $N(0, \sigma^2 t)$. If the stock is up $10 after 6 hours, what is the probability that the stack was above its starting value after 3 hours?

Answer: $Pr ( X(3) > 0 | X(6) = 10 ) = P( \sigma \cdot Y(3) >0 | \sigma \cdot Y(6) = 10) = P( Y(3) > 0 | Y(6) = 2.5$

This is a Brownian bridge process where $(Y(3) | Y(6) = 2.5)$ ~ $N(3/6 \cdot 2.5, 3 \cdot 6 \cdot 3)$ = $N(5/4, 3/2)$.

Thus $P( Y(3) > 0 | Y(6) = 2.5) = P(N(1.25,1.5) > 0$ 

Example 2: If a bicycle race between two competitors, Let $X(t)$ denote the amount of time (in seconds) by which the racer that started in the insider position is ahead when 100t percent of the race has been completed, and suppose that $X(t), 0 \leq t \leq 1$, can be effectively modeled as a Brownian Motion process with variance parameter $\sigma^2$.

6. First Passage Time

Let $T_a$ denote the first time that standard Brownian motion hits level a (starting at X(0) = 0), assuming a >0, then we have
$ Pr\{X(t) \geq a\} = Pr\{X(t) \geq a | T_a \leq t\} \cdot Pr\{T_a \leq t\} + Pr\{X(t) \geq a | T_a > t\}Pr\{T_a \geq t\}$

• $Pr\{X(t) \geq a | T_a \leq t\} = \frac{1}{2}$: you know that at some time before t, the process hits a. From that point forward, you are just as likely to be above a as below a. 
• $ Pr\{X(t) \geq a | T_a > t\} = 0$: $X(t)$ cannot be above a, because the first passage time to a is after t.

Thus, $Pr\{X(t) \geq a\} = \frac{1}{2} Pr\{T_a \leq t\}$, 

$Pr\{T_a \leq t\} = 2 Pr\{ X(t) \geq a\} = \frac{2}{\sqrt{2\pi t}} \int^\infty_{a} e^{-x^2/2t} dx$.

Change variables: $y = \frac{x}{\sqrt{t}}$, we have $Pr\{T_a \leq t\} = \frac{2}{\sqrt{2\pi}} \int^\infty_{a/\sqrt{t}} e^{-y^2/2t} dy$.
By symmetry, we get $Pr\{T_a \leq t\}$.