Admistrivia
Models of Brownian motion
Boundaries
So what about those boundaries? (Recall boundaries are what make
PDE/SDE/first-step-analysis difficult)
- Exact Brownian motion doesn't exist (velocities would be
infinite for example.)
- It is a very good approximation though
- What might boundaries look like for a random walk?
Discrete time revisited:
- No boundaries (state space -infinity to +infinity)
- absorbing boundary (i.e. gamblers ruin)
- Reflecting boundary (i.e. birth-death with imigration)
Reflecting boundary
Easy model: R(t) = |B(t)|
- E(R(t)|R(0)=0) = sqrt(2t/pi)
- p(y,t|x) = phit(y-x) + phit(-y-x)
Absorbing boundary
Since we will absorb at "zero" it is best not to start there.
- Start at B(0) = x.
- Recall tau, our first hitting time of B(t) = 0
- Let A(t) = B(t) up to time tau
- Let A(t) = 0 after tau.
Easy random variable model. How about probability density function?
- NOTE: I'll use infinitesimal approach. Book uses CDF approach.
- Use reflection principle to determine P(min X < 0 and y < X(t)
< y+h|X(0) = x)
- Draw picture
- Same probability as -y > X(t) > -y-h.
- so answer is phit(-x-y)h
- Use law of total probability to compute: P(min X > 0 and y < X(t)
< y+h|X(0) = x)
- P(y < X(t) < y+h|X(0) = x) = phit(y-x)h
- answer = (phit(y-x) - phit(-y-x))h
- Density = limit
Brownian Bridge
- aka tied down Brownian motion
- Condition on X(0) = 0, and X(1) = 0.
- E(X(t)) = 0.
- for 0 < s < t < 1: Cov(X(s),X(t)) = s(1-t).
- s = t: Var(X(s)) = s(1-s).
- E(X(t)|X(s)) = E(X(t)- beta X(s)+ beta X(s)hi
Hitting time distribution
- P(zero between t and t+s) = 2/pi arctan sqrt(s/t)
- Proof via graphs!
Number of zeros
- P(zero in interval epsilon3 to epsilon) = 1 - O(epsilon)
- Expected number "near" zero is infinite
- Worse than that: Call a zero at t, "delta-natural" if there is
another zero in (t,t+delta).
- Probability of being delta-natural is 1.
- Call a point a left-limit if it is natural for all delta.
- Probability of being left-limit is 1.
- If number of zeros is conuntable, probabilty of all being left-limits is 1.
- But, this means there can be no "last" zero.
- Hence there are an uncountable number of zeros.
Dean P. Foster
Last modified: Thu Apr 17 15:03:24 EDT 2008