MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Lecture 6 Fall 2013 9/23/2013 Brownian motion. Introduction Content. 1. A heuristic construction of a Brownian motion from a random walk. 2. Deﬁnition and basic properties of a Brownian motion. 1 Historical notes • 1765 Jan Ingenhousz observations of carbon dust in alcohol. • 1828 Robert Brown observed that ”pollen grains suspended in water per form a continual swarming motion”. • 1900 Bachelier’s work ”The theory of speculation” on using Brownian motion to model stock prices. • 1905 Einstein and Smoluchovski. Physical interpretation of Brownian motion. • 1920’s Wiener concise mathematical description. 2 Construction of a Brownian motion from a random walk The developments in this lecture follow closely the book by Resnick [3]. In this section we provide a heuristic ”construction” of a Brownian motion from a random walk. The derivation below is not a proof. We will provide a rig orous construction of a Brownian motion when we study the weak convergence theory. The large deviations theory predicts exponential decay of probabilities P( 1≤i≤n Xn > an) when a > µ = E[X1 ] and E[eθX1 ] are ﬁnite. Naturally, the decay will be 1 slower the closer a is to µ. We considered only the case when a was a constant. But what if a is a function of n: a = an ? The Central Limit Theorem tells us that the ”decay” disappears when an ≈ √1n . Recall Theorem 1 (CLT). Given an i.i.d. sequence (Xn , n ≥ 1) with E[X1 ] = µ, var[X1 ] = σ 2 . For every constant a P a t2 1 1≤i≤n Xi − µn √ e− 2 dt. √ lim P( ≤ a) = n→∞ σ n 2π −∞ P = 1≤i≤n (Xi − µ). For Now let us look at a sequence of partial sums Sn P simplicity assume µ = 0 so that we look at Sn = 1≤i≤n Xi . Can we say anything about Sn as a function of n? In fact, let us make it a function o of a real √ Xi √ J variable t ∈ R+ and rescale it by n as follows. Deﬁne Bn (t) = 1≤i≤�nt n for every t ≥ 0. Denote by N (µ, σ 2 ) the distribution function of a normal r.v. with mean µ and variance σ 2 . 1. For every ﬁxed 0 ≤ s < t, by CLT we have the distribution of P lnsJ 0, a probability measure P satisfying the follow ing properties is called the Wiener measure: 1. P(B(0) = 0) = 1. 2. P has the independent increments property. Namely for every 0 ≤ t1 < · · · < tk < ∞ and x1 , . . . , xk−1 ∈ R, P(ω ∈ C[0, ∞) : B(t2 , ω) − B(t1 , ω) ≤ x1 , . . . , B(tk , ω) − B(tk−1 , ω) ≤ xk−1 ) P(ω ∈ C[0, ∞) : B(ti , ω) − B(ti−1 ), ω) ≤ xi−1 ) = 2≤i≤k 3. For every 0 ≤ s < t the distribution of B(t)−B(s) is normal N (0, σ 2 (t− s)). In particular, the variance is a linear function of the length of the time increment t − s and the increments are stationary. The stochastic process B described by this probability space (C[0, ∞), B, P) is called Brownian motion. When σ = 1, it is called the standard Brownian motion. 3 Theorem 2 (Existence of Wiener measure). For every σ ≥ 0 there exists a unique Wiener measure. (what is the Wiener measure when σ = 0?). As we mentioned before, we delay the proof of this fundamental result. For now just assume that the theorem holds and study the properties. Remarks : • In future we will not be explicitly writing samples ω when discussing Brownian motion. Also when we say B(t) is a Brownian motion, we un derstand it both as a Wiener measure or simply a sample of it, depending on the context. There should be no confusion. • It turns out that for any given σ such a probability measure is unique. On the other hand note that if B(t) is a Brownian motion, then −B(t) is also a Brownian motion. Simply check that all of the conditions of the Wiener measure hold. Why is there no contradiction? • Sometimes we will consider a Brownian motion which does not start at zero: B̄(0) = x for some value x = 0. We may deﬁne this process as x + B(t), where B is Brownian motion. Problem 1. 1. Let Ω be the space of all (not necessarily continuous) functions ω : R+ → R. i Construct an example of a stochastic process in Ω which satisﬁes conditions (a)-(c) of the Brownian motion, but such that every path is almost surely discontinuous. i Construct an example of a stochastic process in Ω which satisﬁes conditions (a)-(c) of the Brownian motion, but such that every path is almost surely discontinuous in every point t ∈ [0, 1]. HINT: work with the Brownian motion. 2. Suppose B(t) is a stochastic process deﬁned on the set of all (not neces sarily continuous) functions x : R+ → R satisfying properties (a)-(c) of Deﬁnition 1. Prove that for every t ≥ 0, limn→∞ B(t+ n1 ) = B(t) almost surely. 4 Problem 2. Let AR be the set of all homogeneous linear functions x(t) = at where a varies over all values a ∈ R. B(t) denotes the standard Brownian motion. Prove that P(B ∈ AR ) = 0. 4 Properties We now derive several properties of a Brownian motion. We assume that B(t) is a standard Brownian motion. Joint distribution. Fix 0 < t1 < t2 < · · · < tk . Let us ﬁnd the joint distribution of the random vector (B(t1 ), . . . , B(tk )). Given x1 , . . . , xk ∈ R let us ﬁnd the joint density of (B(t1 ), B(t2 ), . . . , B(tk )) in (x1 , . . . , xk ). It is equal to the joint density of (B(t1 ), B(t2 ) − B(t1 ), . . . , B(tk ) − B(tk−1 )) in (x1 , x2 − x1 , . . . , xk − xk−1 ), which by independent Gaussian increments property of the Brownian motion is equal to kY −1 i=1 1 − 2π(ti+1 − ti ) e (xi+1 −xi )2 2(ti+1 −ti ) . Differential Property. For any s > 0, Bs (t) = B(t + s) − B(s), t ≥ 0 is a Brownian motion. Indeed Bs (0) = 0 and the process has independent increments Bs (t2 ) − Bs (t1 ) = B(t2 + s) − B(t1 + s) which have a Guassian distribution with variance t2 − t1 . Scaling. For every c, cB(t) is a Brownian motion with variance σ 2 = c2 . Indeed, continuity and the stationary independent increment properties as well as the Gaussian distribution of the increments, follow immediately. The variance of the increments cB(t2 ) − cB(t1 ) is c2 (t2 − t1 ). For every positive c > 0, B( ct ) is a Brownian motion with variance 1c . Indeed, the process is continuous. The increments are stationary, independent with Gaussian distribution. For every t1 < t2 , by deﬁnition of the standard Brownian motion, the variance of B( tc2 ) − B( tc1 ) is (t2 − t1 )/c = 1c (t2 − t1 ). Combining these two properties we obtain that nian motion. 5 √ cB( ct ) is also a standard Brow- Covariance. Fix 0 ≤ s ≤ t. Let us compute the covariance Cov(B(t), B(s)) = E[B(t)B(s)] − E[B(t)]E[B(s)] a = E[B(t)B(s)] = E[(B(s) + B(t) − B(s))B(s)] b = E[B 2 (s)] + E[B(t) − B(s)]E[B(s)] = s + 0 = s. Here (a) follows since E[B(t)] = 0 for all t and (b) follows since by deﬁnition of the standard Brownian motion we have E[B 2 (s)] = s and by independent increments property we have E[(B(t)−B(s))B(s)] = E[(B(t)−B(s))(B(s)− B(0))] = E[(B(t) − B(s)]E[(B(s) − B(0)] = 0, since increments have a zero mean Gaussian distribution. Time reversal. Given a standard Brownian motion B(t) consider the process B (1) (t) deﬁned by B (1) (t) = tB( 1t ) for all t > 0 and B (1) (0) = 0. In other words we reverse time by the transformation t → 1t . We claim that B (1) (t) is also a standard Brownian motion. Proof. We need to verify properties (a)-(c) of Deﬁnition 1 plus continuity. The continuity at any point t > 0 follows immediately since 1/t is continuous func tion. B is continuous by assumption, therefore tB( 1t ) is continuous for all t > 0. The continuity at t = 0 is the most difﬁcult part of the proof and we delay it till the end. For now let us check (a)-(c). (a) follows since we deﬁned B (1) (0) to be zero. We delay (b) till we establish normality in (c) (c) Take any s < t. Write tB( 1t ) − sB( 1s ) as 1 1 1 1 1 tB( ) − sB( ) = (t − s)B( ) + sB( ) − sB( ) t s t t s The distribution of B( 1t )−B( 1s ) is Gaussian with zero mean and variance 1s − 1t , since B is standard Brownian motion. By scaling property, the distribution of sB( 1t )−sB( 1s ) is zero mean Gaussian with variance s2 ( 1s − 1t ). The distribution of (t − s)B( 1t ) is zero mean Gaussian with variance (t − s)2 ( 1t ) and also it is independent from sB( 1t ) − sB( 1s ) by independent increments properties of 6 the Brownian motion. Therefore tB( 1t ) − sB( 1s ) is zero mean Gaussian with variance 1 1 1 s2 ( − ) + (t − s)2 ( ) = t − s. s t t This proves (c). We now return to (b). Take any t1 < t2 < t3 . We established in (c) that all the differences B (1) (t2 ) − B (1) (t1 ), B (1) (t3 ) − B (1) (t2 ), B (1) (t3 ) − B (1) (t1 ) = B (1) (t3 ) − B (1) (t2 ) + B (1) (t2 ) − B (1) (t1 ) are zero mean Gaussian with vari ances t2 − t1 , t3 − t2 and t3 − t1 respectively. In particular the variance of B (1) (t3 ) − B (1) (t1 ) is the sum of the variances of B (1) (t3 ) − B (1) (t2 ) and B (1) (t2 ) − B (1) (t1 ). This implies that the covariance of the summands is zero. Moreover, from part (b) it is not difﬁcult to establish that B (1) (t3 ) − B (1) (t2 ) and B (1) (t2 ) − B (1) (t1 ) are jointly Gaussian. Recall, that two jointly Gaussian random variables are independent if and only if their covariance is zero. It remains to prove the continuity at zero of B (1) (t). We need to show the continuity almost surely, so that the zero measure set corresponding to the sam ples ω ∈ C[0, ∞) where the continuity does not hold, can be thrown away. Thus, we need to show that the probability measure of the set 1 A = {ω ∈ C[0, ∞) : lim tB( , ω) = 0} t→0 t is equal to unity. We will use Strong Law of Large Numbers (SLLN). First set t = 1/n and consider tB( 1t ) = B (n)/n. Because of the independent Gaussian incre ments property B(n) = 1≤i≤n (B(i) − B(i − 1)) is the sum of independent i.i.d. standard normal random variables. By SLLN we have then B(n)/n → E[B(1) − B(0)] = 0 a.s. We showed convergence to zero along the sequence t = 1/n almost surely. Now we need to take care of the other values of t, or equivalently, values s ∈ [n, n + 1). For any such s we have | B(s) B(n) B(s) B(n) B(n) B(n) − |≤| − |+| − | s n s s s n 1 1 1 ≤ |B(n)|| − | + sup |B(s) − B(n)| s n n n≤s≤n+1 ≤ |B(n)| 1 + sup |B(s) − B(n)|. n2 n n≤s≤n+1 We know from SLLN that B(n)/n → 0 a.s. Moreover then B(n)/n2 → 0. 7 (1) a.s. Now consider the second term and set Zn = supn≤s≤n+1 |B(s) − B(n)|. We claim that for every E > 0, P(Zn /n > E i.o.) = P(ω ∈ C[0, ∞) : Zn (ω)/n > E i.o.) = 0 (2) where i.o. stands for inﬁnitely often. Suppose (2) was indeed the case. The equality means that for almost all samples ω the inequality Zn (ω)/n > E hap pens for at most ﬁnitely many n. This means exactly that for almost all ω (that is a.s.) Zn (ω)/n → 0 as n → ∞. Combining with (1) we would conclude that a.s. B(s) B(n) − | → 0, sup | s n n≤s≤n+1 as n → ∞. Since we already know that B(n)/n → 0 we would conclude that a.s. lims→∞ B(s)/s = 0 and this means almost sure continuity of B (1) (t) at zero. It remains to show (2). We observe that due to the independent stationary increments property, the distribution of Zn is the same as that of Z1 . This is the distribution of the maximum of the absolute value of a standard Brownian motion during the interval [0, 1]. In the following lecture we will show that this maximum has ﬁnite expectation: E[|Z1 |] < ∞. On the other hand Z ∞ ∞ Z (n+1)� 0 E[|Z1 |] = P(|Z1 | > x)dx = P(|Z1 | > x)dx 0 n=0 n� ∞ 0 ≥E P(|Z1 | > (n + 1)E). n=0 We conclude that the sum in the right-hand side is ﬁnite. By i.i.d. of Zn ∞ 0 n=1 ∞ 0 |Z1 | |Zn | P( > E) = P( > E) n n n=1 Thus the sum on the left-hand side is ﬁnite. Now we use the Borel-Cantelli Lemma to conclude that (2) indeed holds. 5 Additional reading materials • Sections 6.1 and 6.4 from Chapter 6 of Resnick’s book ”Adventures in Stochastic Processes” in the course packet. • Durrett [2], Section 7.1 • Billingsley [1], Chapter 8. 8 References [1] P. Billingsley, Convergence of probability measures, Wiley-Interscience publication, 1999. [2] R. Durrett, Probability: theory and examples, Duxbury Press, second edi tion, 1996. [3] S. Resnick, Adventures in stochastic processes, Birkhuser Boston, Inc., 1992. 9

Lecture 6: Intro Brownian Motion

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