8.2. strong Markov property and reflection principle. These are concepts that you can use to compute probabilities for Brownian motion.
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1 62 BROWNIAN MOTION 8.2. strong Markov property and reflection principle. These are concepts that you can use to compute proailities for Brownian motion strong Markov property. a) Brownian motion satisfies the Markov property: For t > s, X t depends only on X s the ook tries to write this mathematically. In class we decided that the statement is F s is given y X s In fact (a fortiori), X t X s is independent of F s. ) The strong Markov property says the same thing with s replaced y a stopping time T : If T is a stopping time and t > T then X t depends only on F T In fact, we have the following theorem: Theorem 8.6. Suppose that X t is Brownian motion. If t > T, T a stopping time, then X t X T is independent of F T. An example of a stopping time is the first time that X t reaches reflection principle. Suppose X t is Brownian motion with zero drift (µ = 0). Then we want to calculate the proaility that, starting at X 0 = 0, it will reach X s = at some time 0 < s < t. P(X s = for some 0 < s < t X 0 = 0) =? Let T = first time that X T =. Then X s reaches for s < t if T < t. So, this is the same as P(T < t) The strong Markov property implies that X t X T F T. We also know that X t X T is normal: is independent of X t X T N(0, σ 2 (t T )) (assuming that t > T ). Since the mean is zero, it is positive half the time and negative half the time (and the proaility of eing exactly zero is 0): P(X t X T > 0) = 2 P(X t X T 0) = 2
2 MATH 56A SPRING 2008 STOCHASTIC PROCESSES 63 reflection Half the time X will reach and go up, half the time it will reach and go down. So, P(T < t) = 2P(T < t and X t > X T = ) But X t is continuous. So, the intermediate value theorem (IMT) tells us that the second condition implies the first: If X t > and X 0 = 0 then 0 < s < so that X s =. So, This is given y an integral P(T < t X 0 = 0) = 2P(X t > X 0 = 0) = 2 f t (x) dx where f t is the density function for X t X density function for normal distriution. Since the density function is In other words, X t X 0 N(0, σ 2 t) f t (x) = 2πσ2 t e x2 /2σ 2 t f t (x)ds = P(x < X t x + dx X 0 = 0) Comine this with the refection principle gives: P(T < t) = 2 2πσ2 t e x2 /2σ 2t dt
3 64 BROWNIAN MOTION To compute this (or look it up in a tale) we should convert to standard normal. But first, I redid the reflection principle with 0, replace with a, : reflection P(X s = for some 0 < s < t X 0 = a) = 2P(X t > X 0 = a) y the reflection principle = 2 f t (x a) dx since X t X 0 = X t a N(0, σ 2 t) this integral is = 2 2πσ2 t e (x a)2 /2σ 2t dx conversion to standard normal. To convert to standard normal (N(0, )) we should sutract the mean and divide y the standard deviation: y = x a σ t, dx dy = σ t This converts the integral into: P(X s = for some 0 < s < t X 0 = a) = 2 I will use the areviation: φ t (x) = 2πt e x2 /2t a σ t e y2 /2 dy 2π }{{} φ (y) This is the density function for X t X 0 if X t is standard Brownian motion.
4 MATH 56A SPRING 2008 STOCHASTIC PROCESSES Chapmann-Kolmogorov equation. This is an ovious equation which I proved using the theorem that the density function of a sum of two random variales is the convolution of the density functions. First some notation: p t (x, y) = proaility density of going from x to y in time t Multiply dy to get an actual proaility: p t (x, y)dy = P(y < X s+t y + dy X s = x) So, p t (x, y) = f t (y x). Theorem 8.7 (Chapmann-Kolmogorov). p s+t (x, y) = p s (x, z)p t (z, y) dz Proof. Since p s+t (x, y) = f s+t (y x) we can rewrite this as: f s+t (y x) = f s (z x)f t (y z)dz But (z x) + (y z) = y x. So, the RHS is the convolution of f s and f t. But, X s+t X 0 = (X s X 0 ) + (X s+t X s ) So, I.e., ( ) ( ) ( ) density of density of density of = X s+t X 0 X s X 0 X s+t X s f s+t = f s f t So, LHS=RHS. The reason that this is supposed to e ovious is that, in order to go from x to y in time s + t you have to first go to some z at time s and then get from z to y in the remaining time t. Since z could e anything you integrate over all z. This integral is the continuous version of matrix multiplication.
5 66 BROWNIAN MOTION return proaility. We used Chapmann-Kolmogorov to compute return proaility. For this I assumed that X t = W t is standard Brownian. So, µ = 0, σ =, X 0 = 0. We want the return proaility: P(X s = 0 for some < s < t X 0 = 0) Later we will replace with an aritrary numer. For this prolem we use the reflection principle twice: Half the time X will e positive: P(X > 0) = 2 And, given that X = > 0 and X s = 0 then half the time X t < 0. So, y the reflection principle: P(X s = 0 for some < s < t X 0 = 0) = 4P(X > 0 and X t < 0) By Chapmann-Kolmogorov (with x, y, z = X 0, X t, X resp.) this is = 4 0 P( < X + d }{{} φ ()d and X t X < ) }{{} ( ) But these two conditions are independent. So, we multiply the proailities: ( ) = P(X t X < ) = Converting to normal with y = P( < X + d) = φ ()d ( ) = x φ t (x)dx = t, this is φ (y)dy t φ t (x)dx
6 MATH 56A SPRING 2008 STOCHASTIC PROCESSES 67 So, the answer is ans = 4 φ ()φ (y) dyd =0 y= t φ ()φ (y) = 2π e 2 /2 2π e y2 /2 = 2 +y 2 2π e 2 Convert to polar coordinates: 2 + y 2 = r 2, ddy = rdrdθ. Then ans = 4 π/2 tan t 0 2π e r2 /2 rdrdθ The limits of integration are given y the picture The integral is easy to calculate 4 /2 0 2π e r2 r dr = 2 π So, ans = π/2 2 tan π dθ = 2 π t [ ] π 2 tan = 2 t π tan t The expression is - (something) where the something is the proaility that the event does not occur. So, 2 π tan = P(X s 0 for all < s < t) t
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