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MATH 56A SPRING 2008 STOCHASTIC PROCESSES 1578. Brownian MotionWe will spend 5 days on this chapter and cover the following topics:8.1. introduction8.2. strong Markov property and the reflection principle8.3. fractal dimension of the zero set8.4. the heat equation8.5. recurrence and transience8.1. introduction. Brownian motion (named after Robert Brown) isalso called a Wiener process (named after Norbert Wiener). It refersto the fact that particles bounce around under a microscope. Theyseem to be coming back to the same place but they actually come to apoint just above or just below where they were b efore. You don’t seethe third dimension. (Random walk is recurrent in Z2and transient inZ3.)I started with general 1-dimensional Brownian motion. This meanswe are looking just at the x-coordinate of a particle moving in 3-dimensional space. You need to imagine that the particle is movingat random inside of a medium which is drifting.Xt= position of a particle on a line at time t. This is the x-coordinate of a particle in R3. The concept is verbally described asfollows.(1) memoryless If s > t then Xsdepends only on Xtand not onhow the particle got to the point Xt. More succinctly, Xs−Xtis independent of Ft(all Xufor u ≤ t).(2) time & space homogeneous The increment in position dep endsonly on the increment in time. I.e., the probability distributionof Xt+∆t−Xtdepends only on ∆t. (It is independent of t, Xt.)(3) Xtis continuous.8.1.1. mathematical definition.Definition 8.1. A random variable X : [0, ∞) → R (t %→ Xt) isBrownian motion with drift µ and variance σ2(the book calls this thevariance parameter) if(1) For all s1< t1≤ s2< t2≤ ··· ≤ sn< tn[This means(s1, t1], (s2, t2], ··· , (sn, tn] are disjoint, i.e., nonoverlapping.]Xt1− Xs1, Xt2− Xs2, ··· , Xtn− Xsnare independent random variables.(2) Xtis continuous (it doesn’t jump)158 BROWNIAN MOTION(3) Xt−Xsis normally distributed with mean (t−s)µ and variance|t − s|σ2. This is (t − s)σ2assuming t > s. So:Xt− Xs∼ N(µ(t − s), σ2|t − s|)∼ means “distributed as”.We can convert this to the standard normal distribution by subtract-ing the mean and dividing by the standard deviation:Xt− Xs− µ(t − s)σ!|t − s|∼ N(0, 1)Definition 8.2. Standard Brownian motion, denoted Wt, is Brownianmotion with three additional conditions:a) W0= 0 (Wtis “centered”)b) µ = 0 (no drift)c) σ2= 1.For example, if Xtis Brownian motion with drift µ and variance σ2thenWT=Xt− X0− µtσis standard Brownian. Solving this for Xtwe get:Xt= X0+ µt + σWtThe book assumes that X0= 0 and µ = 0. This focuses on thestochastic part of Xtwhich is σWt. (X0is F0-measurable and µt is notrandom.)Theorem 8.3. Condition (3) in the definition of Brownian motion isequivalent to the condition that Xt1−Xs1, Xt2−Xs2, ··· are i.i.d. withfinite mean and variance for nonoverlapping intervals (s1, t1], (s2, t2], ···of the same length:t1− s1= t2− s2= ···I pointed out that this condition follows from the assumptions ofbeing “memoryless” and “time & space homogeneous.” In other words,this theorem says that the verbal and mathematical descriptions ofBrownian motion agree.Proof. Choose ∆t small: ∆t =t−sN. ThenXt− Xs= (Xs+∆t− Xs) + (Xs+2∆t− Xs+∆t) + ···+ (Xt− Xt−∆t)The RHS is a sum of N i.i.d. random variables and we can take N → ∞without changing the LHS. Therefore, by the central limit theorem,MATH 56A SPRING 2008 STOCHASTIC PROCESSES 159Xt− Xsis normally distributed. We just need to compute its meanand standard deviation. But:E(Xt− Xs) = N"#$%t−s∆tE(Xs+∆t− Xs)E(Xt− Xs) = (t − s)E(Xs+∆t− Xs)∆t= (t − s)µwhere µ is defined byµ :=E(Xs+∆t− Xs)∆tThis does not depend of s, t or ∆t sinceE(Xt− Xs)t − s=E(Xs+∆t− Xs)∆tthe LHS does not depend on ∆t and the RHS does not depend on s ort.A similar trick works for the variance:Var(Xt− Xs) = N Var(Xs+∆t− Xs)Var(Xt− Xs) = (t − s)Var(Xs+∆t− Xs)∆t= (t − s)σ2where σ2is defined byσ2:=Var(Xs+∆t− Xs)∆tThis works because, again, there is no t in this expression.Putting these together we getXt− Xs∼ N((t − s)µ, (t − s)σ2)assuming t > s. !8.1.2. as a limit of random walks. Standard Brownian motion is a limit(as ∆t → 0) of random walks where:Time moves in integer multiples of ∆t andSpace is an integer multiple of√∆tThe Markov chain is: In each increment ∆t of time, you move ±√∆twith probability 1/2.Note that ∆t <<√∆t. For example,∆t = 10−6=11, 000, 000⇒√∆t = 10−3=11, 000= 103∆tThe cover of our book shows a graph of a typical motion of this Markovchain. But, since ∆t is going to 0, it would be more accurate to use160 BROWNIAN MOTION∆t = 10−6. Then,√∆t = 103∆t. So, a more accurate picture wouldbe:• ≈√103√∆t ≈ 31.6∆t = .0316W.001∼ N(0, .001)W.001≈ ±.0316103∆t = .001 = .03162√∆tSince X∆t= ±√∆t has mean 0 and variance ∆t, the CLT says thatXN ∆t˙∼N(0, N∆t)As ∆t → 0 (with t = N∆t fixed), Xtconverges toWt= Xt∼ N(0, t)MATH 56A SPRING 2008 STOCHASTIC PROCESSES 1618.1.3. L´evy’s theorem. I stated the following theorem without proofbecause I ran out of time. But the proof is easy.Theorem 8.4. Suppose that Xtis Brownian motion with drift µ andvariance σ2. Thena) Xt− µt is a continuous martingale.b) (Xt− µt)2− tσ2is a continuous martingale.Proof. (a) Let Mt= Xt− µt. Then:E(Mt|Fs) = E(Xt− µt |Fs) = Xs− µt + E(Xt− Xs)= Xt− µt + (t − s)µ = Xs− µs = Ms(b) Mt= Xt− µt is Brownian motion with zero drift. Therefore,E(Mt− Ms) = 0 and E((Mt− Ms)2) = σ2(t − s) for t > s. So,E(M2t− σ2t |Fs) = M2s− σ2t + E(M2t− M2s|Fs)But,M2t− M2s= (Mt− Ms)2+ 2M2(Mt− Ms)with expected valueE(M2t− M2s|Fs) = E((Mt− Ms)2|Fs) + 2MsE(Mt− Ms|Fs)" #$ %0= σ2(t − s)So,E(M2t− σ2t |Fs) = M2s− σ2t + σ2(t − s) = M2s− σ2smaking M2t− σ2t into a martingale. !So, you can see that this is easy. L´evy proved the difficult converseof this theorem. One of the most elegant proofs of this theorem byKunita and Watanabe is outlined in section 9.5 of the 2006 notes.Theorem 8.5 (L´evy). A continuous L2martingale Mtis standardBrownian motion if and only if M0= 0 and M2t− t is a

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