[Stochastic Process] Brownian Motion

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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); tge 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}<u>Answer: </u> 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}{4sqrt{5}}).<b><span style="font-size: x-large;">5. Brownian Bridge</span></b><b><u>Basic Idea:</u> </b>condition on the final value of a Brownian motion process and derive the stochastic properties in between.<u style="font-weight: bold;">Main Results </u>:X(s) = x | X(t) = Bis normally distributed with<ul><li><span style="color: red;">Mean:frac{s}{t} cdot B</span></li><li><span style="color: red;">Variance:frac{s}{t} cdot (t-s)</span></li><li>The value of s with the highest variance iss = t/2. That is, we know the endpoints of the process, but we don't know exactly what happens in between.</li></ul><b>Note</b>: these results are for standard Brownian motion (SBM)<b><u>Proof:</u></b><div style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-fYCRQBP77lY/VHVa9KUoD4I/AAAAAAAAAUs/CDAwrhLMdK8/s1600/BMBridgeProof.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-fYCRQBP77lY/VHVa9KUoD4I/AAAAAAAAAUs/CDAwrhLMdK8/s1600/BMBridgeProof.PNG" height="640" width="528" /></a></div><div style="clear: both; text-align: center;"></div><div style="clear: both; text-align: left;"></div><div style="clear: both; text-align: left;"><u style="font-weight: bold;">Example 1: </u>Suppose you have a stock whose value follows Brownian motion withX(t)~N(0, sigma^2 t). If the stock is up10 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{2pi 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{2pi}} int^infty_{a/sqrt{t}} e^{-y^2/2t} dy.
By symmetry, we get Pr{T_a leq t}.