BROWNIAN MOTION II D.Majumdar
DEFINITION Let (Ω, F, P) be a probability space. For each ω Ω, suppose there is a continuous function W(t) of t 0 that satisfies W(0) = 0 and that depends on ω. Then W(t), t 0 is a Brownian Motion if for all 0 = t 0 < t 1 <.. < t m the increments W t 1 = W t 1 W t 0, W t 2 W t 1,.3.1 Are independent and each of these increments are normally distributed with E W t i+1 W t i = 0 and Var W t i+1 W t i = t i+1 t i.3.2 The difference between W(t) and let s say W 100 (t) is that W 100 (t) has a natural time step of 1/100 and is linear between these time steps, on the other hand Brownian Motion has no linear pieces. W 100 (t) is approximately normal for each t, while Brownian Motion is exactly normal. We might want to determine the probability of the set A containing all ω Ω that result in a Brownian Motion path satisfying 0 W t 0.2 We can 1 st try to think of this from the perspective of the scaled Random Walk W 100 0.25.So the scaled Random Walk has to take the value between [0,0.2] in the 1 st 25 coin tosses. But the scaled Random Walk can only take the value of 0.1 in the specified range, and that is possible if there are 13 H s and 12 T s in the 1 st 25 coin tosses.
So A is the set of all infinite sequence of coin tosses with the property that in the 1 st 25 tosses there are 13 H s and 12 T s. P(A) = 0.1555 as seen before. In case of Brownian Motion we can define P{0 W 0.25 0.2} = 2 2π 0.2 0 e 2x 2 dx
DISTRIBUTION OF BROWNIAN MOTION As we know, the increments W t 1 = W t 1 W t 0, W t 2 W t 1.. Are independent and normally distributed, the R.V. s W t 1, W t 2, W t m are jointly normally distributed. The joint distribution of jointly normal random variables is determined by their means and covariance's. Each of the R.V. s W(t i ) has mean 0. For any two times 0 s < t, the covariance of W(s) and W(t) is E[W(s)W(t)] = E[{W(t)-W(s)+W(s)}W(s)] = E[{W(t)-W(s)}W(s)] + E[ W s 2 ] = E[W(s)] * E[W(t)-W(s)] + Var[W(s)] = 0 + s = s. So with this logic we can create the Covariance Matrix for the m-dim Random Vector t 1 t 1 (W t 1, W t 2, W t m ) as t 1 t m From here we can create the Moment Generating Function of the Random Vector. https://en.wikipedia.org/wiki/moment-generating_function In conclusion, the moment generating function for Brownian Motion (i.e. for the m-dim Random Vector [W t 1, W t 2, W t m ] is
M(u 1, u 2,.. u m ) = E[exp{u m W t m + u m 1 W t m 1 + u 1 W t 1 }] =exp{ 1 2 u 1 + u m 2 t 1 + 1 2 u 2 +. u m 2 t 2 t 1 +. + 1 2 u m 1 u m 2 (t m 1
FILTRATION FOR BROWNIAN MOTION Let (Ω, F, P) be a probability space, on which is defined a Brownian Motion W(t), t 0. A filtration for the Brownian Motion is a collection of σ algebras F(t), t 0, satisfying: 1. For 0 s < t, every set in F(s) is also in F(t). There is at least as much information available at time F(t) as there is at earlier time F(s) 2. For each t 0, the Brownian Motion W(t), at time t is F(t) measurable. In other words, the information available at time t is sufficient to evaluate the Brownian Motion W(t) at that time. 3. For 0 t < u, the increment W(u) W(t) is independent of F(t). In other words any increment of the Brownian Motion after time t is independent of the information available at time t. Let t, t 0, be a stochastic process. We say that t is adapted to the filtration F(t), if for each t 0 the R.V. t is F(t) measurable. Martingale Property for Brownian Motion Brownian Motion is a Martingale. The proof is similar to what we have seen before and can be found in Shreve in details.
FIRST ORDER VARIATION
We would like to compute the amount of up and down oscillation undergone by this function between times 0 and T, with the down moves adding to rather than subtracting from the up moves. We call this the 1 st order variation FV T f.this can be written as:- FV T f = [f t 1 f(t 0 )] [f t 2 f(t 1 )] + [f T f(t 2 )] = 0 t 1 f t dt + t1 t 2 f t dt + t2 T f t dt = 0 T f t dt.4.2 In general, to compute the 1 st order variation of a function up to time T, we 1 st choose a partition Π = t 0, t 1,. t n of 0, T where 0 = t 0 < t 1,. t n = T. It is not required for the partition points to be equally spaced but they can be. Π = max j=0,.n 1 (t j+1 t j ). We define :- FV T f = lim σ n 1 j=0 f t j+1 Π 0 f t j 4.3 1 st we need to verify if 4.3 is consistent with 4.2 or not. Mean Value Theorem If f is a continuous function on the closed interval [a,b], and differentiable on the open interval (a,b), then there exists a point c in (a,b) such that: f c = f b f(a) b a https://en.wikipedia.org/wiki/mean_value_theorem
By using Mean Value Theorem, f t j+1 f t j = f (t j )(t j+1 t j ) FV T f = lim Π 0 σ n 1 j=0 f t j (t j+1 t j ), this is nothing but the Riemann Sum for the integral of the function f (t). Hence we have :- FV T f = lim σ n 1 j=0 f t T j (t j+1 t j ) = Π 0 0 f t dt which is 4.2
QUADRATIC VARIATION Let f(t) be a function defined for 0 t T. The quadratic variation of f up to time T is [f,f](t) = lim σ n 1 2 j=0 f t j+1 f t j.4.5 Π 0 Where Π = {t 0, t 1,. t n } and 0=t 0 <.. < t n = T Suppose the function f has continuous derivative. Then, σ n 1 2 j=0 f t j+1 f t j = n 1 σj=0 f t j 2 ൫tj+1
We usually work with functions that have continuous derivatives, and their Quadratic Variations are 0. For this reason we never consider Quadratic Variation in ordinary calculus. The paths of the Brownian Motion, on the other hand, cannot be differentiated w.r.t. the time variable. For functions that do not have derivatives, the Mean Value Theorem can fail and previous proof also fails. Let s think of the function f(t) = t. We can easily see that we wont be able to find a point c such that f f b f(a) c = such that this relationship is valid. b a f (t) = -1 for t<0 and = 1 for t > 0 and undefined at t=0, which is like a point. You can think of the paths of the Brownian Motion to be very pointy. For a path of the Brownian Motion W(t) there is no value of t for which d W(t) is defined. dt Thereon 4.3:- Let W be a Brownian Motion. Then [W,W](T) = T for all T 0 almost surely. Proof Let Π = {t 0, t 1,. t n } and 0=t 0 <.. < t n = T. We define the sample quadratic variation to be Q Π = σ n 1 j=0 (W(t j+1 ) W(t j ))^2 We need to show that this R.V. converges to T as Π 0. We will show that this has Expected Value = T and Variance Converges to 0. Hence it is converging to it s expected value T regardless of it s path. This type of convergence is also known as L 2 convergence.
We say that X n converges to X in L P or on the p-th moment, p > 0, (X n Lp X) if, lim E X n X p = 0 n E W t j+1 W t j 2 = Var[W tj+1 W t j ] = t j+1 - t j.4.6 E[Q Π ] = E[σ n 1 j=0 (W(t j+1 ) W(t j ))^2] = σ n 1 j=0 E[(W(t j+1 ) W(t j ))^2 ] = σ n 1 j=0 ൫t j+1
Var[Q Π ] = σ n 1 2 j=0 Var W t j+1 W t j = n 1 σj=0 2൫t j+1
CROSS VARIATION We can compute the Cross Variation of W(t) with t and the quadratic variation of t with itself, which are denoted as :- lim σ n 1 j=0 W t j+1 W t j t j+1 t j = 0 4.12 Π 0 lim σ n 1 2 j=0 t j+1 t j = 0..4.13 Π 0 Proof W t j+1 W t j t j+1 t j max 0 k n 1 W t k+1 W t k t j+1 t j σ n 1 n 1 j=0 W t j+1 W t j t j+1 t j σ j=0 max W t k+1 W t k ൫t j+1 0 k n 1
QUADRATIC VARIATION OF TIME n 1 t j+1 t j 2 σ j=0 Π 0 Hence 4.13 is proved max 0 k n 1 t k+1 t k σ n 1 j=0 t j+1 t j = Π *T -> 0 as Hence we can informally write dw(t)dt = 0 and dt dt = 0.4.14
QUIZ 1. Let W(t) be a Brownian Motion. Is W 2 t t a martingale? 2. Prove:- a) σ n 1 j=0 W(t j+1 ) W(t j ) = b) σ n 1 j=0 W t j+1 W(t j ) 3 = 0
QUIZ SOLUTION 1. Let 0 s t be given. Then E[{W 2 t t } F(s)] = E[W 2 t F(s)] E[t F(s)] = E[ W t W s + W s - t = E[ W t W s 2 + W 2 s + 2W s W t W s F s t = E[ W t W s 2 ] + W 2 s + 2W(s)E[W(t) F(s)] - 2W 2 s - t = t s - W 2 s + 2W(s)W(s) t = W 2 s - s Hence proved that W 2 t t 2 F s
QUIZ SOLUTION 2 a)σ n 1 j=0 W(t j+1 ) W(t j ) = Let s assume there exists A F, such that P A > 0 and for every ω A lim σ n j=0 n 1 j=0 n 1 W(t j+1 ) W(t j ) < n 1 W t j+1 W t j 2 max 0 k n 1 W t k+1 W t k j=0 W(t j+1 ) W(t j ) = 0 As W is continuous, when Π 0, then max 0 k n 1 W t k+1 W t k 0. This is only possible when the other term is finite. On the other hand 0* can give you anything from 0 to n 1 But this contradicts the fact that σ j=0 Hence lim σ n 1 n j=0 W(t j+1 ) W(t j ) = W t j+1 W t j 2 = T
QUIZ SOLUTION σ n 1 j=0 W t j+1 W(t j ) 3 = 0 n 1 σ j=0 W t j+1 W t j 3 max W t k+1 W t k σ n 1 j=0 ቀW t j+1 0 k n 1