Admistrivia
- Optional: read 6.6
- Read section 8.1
Brownian motion: 8.1 continued
Notation: little phi, big phi is normal distrubtion
Axioms
B(t) is a Brownian motion if:
- increments are normal with mean =0, var = t sigma2.
- increments are independent.
- B(0) = 0.
- (Difficult one) B(t) is continuous.
Typical Brownian motion: Chance of George II being elected
Covariance
Suppose T < s < t then
- Cov(B(s),B(t)|B(T)) = Cov(B(s)-B(T),B(t)-B(T)|B(T))
- = Cov(B(s)-B(T),B(t)-B(T)+ B(s)-B(s)|B(T))
- = Cov(B(s)-B(T),B(s)-B(T)|B(T)) + Cov(B(s)-B(T),B(t)-B(s)|B(T))
- = Var(B(s)-B(T),B(s)-B(T)|B(T)) + 0
- = (s-T)sigma2
Invariance principle
An alternative construction for a BM is:
- Let B(t) = sum(nt) X(i)/sqrt(n)
- Check assumptions:
- increments are close to normal (for large n) via CLT
- increments are independent by construction
- B(0) = 0.
- Continuous? Let the mathematicians worry about this one.
(If we assume "2+epsilon" moments it should be continuous.)
- Hence all "sums" look the same!
Brownian motion: 8.2
Hitting time
Definition: taux = min{t &ge 0: B(t) = x}.
- Well defined (because of continuity)
- may be infinite?
Reflection principle:
- B*(t) = x + (B(t)-x) for t &le taux
- B*(t) = x - (B(t)-x) for t > taux
- Note: B*(t) satisfies axioms of brownian motion
- same variance as before
- Draw picture (eg p 492)
Maximum of a brownian motion
- P(max B(s) > x) = P(taux < t)
- = P(B*(t) > x and B(t) < x) + P(B(t) > x and B*(t) < x)
- = P(B*(t) > x) + P(B(t) > x)
- = 2 P(B(t) > x)
- = 2(1 - Phit(x))
- = 2 Phit(-x)
Hitting time distribution
- P(taux < t) = P(max B(s) > x) = 2 Phit(-x)
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: Wed Apr 14 11:45:35 EDT 2004