Doob s inequality Let X(t) be a right continuous submartingale with respect to F(t), t 1 P(sup s t X(s) λ) 1 λ {sup s t X(s) λ} X + (t)dp 2 For 1 < p <, E[sup s t X(s) p ] ( p p 1) p E[ X(t) p ] 1 2: P(λ) = P( X λ), X = sup s t X(s), Y = X(t), p E[ X p ] = = p p 1 λ p dp(λ) = p λ p 1 P(λ)dλ λ p 1 ( 1 X YdP)dλ = p Y λ p 2 dλdp λ X>λ Y X p 1 dp p p 1 ( Y p dp) 1/p ( X p dp) 1 1 p () Stochastic Calculus March 4, 29 1 / 22
Existence and Uniqueness Theorem σ : R d [, T ] R d d, b : R d [, T ] R d be Borel measurable, A <, σ(x, t) + b(x, t) A(1 + x ) x R d, t T and Lipschitz; σ(x, t) σ(y, t) + b(x, t) b(y, t) A x y. x R d indep of B t, E[ x 2 ] <. Then there exists a unique solution X t on [, T ] to and E[ T X t 2 dt] <. dx t = b(x t, t)dt + σ(x t, t)db t, X = x Uniqueness means that if Xt 1 and Xt 2 are two solutions then P(Xt 1 = Xt 2, t T ) = 1 () Stochastic Calculus March 4, 29 2 / 22
dx t = σ(t, X t )db t + b(t, X t )dt t f (t, X t ) = f (, X ) + Lf (t, x) = 1 2 t + i,j=1 i,j=1 { } s f (s, X s ) + Lf (s, X s ) ds σ ij (s, X s ) x i f (s, X s )db j s a ij (t, x) 2 f x i x j (t, x) + i=1 b i (t, x) f x i (t, x) a ij = σ ik σ jk k=1 a = σσ T () Stochastic Calculus March 4, 29 3 / 22
Markov property X t can be obtained by solving the stochastic differential equation up to time s < t and then solving in [s, t] with initial condition X s By uniqueness this gives the same answer Define the transition probability p(s, x, t, A) = P(X s,x t A) where X s,x t is the solution starting at x at time s From the construction we have P(X,x t A F s ) = p(s, X,x s, t, A) which is the Markov property () Stochastic Calculus March 4, 29 4 / 22
Diffusions A diffusion is a Markov process with transition probabilities p(s, x, t, dy) satisfying, for each δ > as h, 1 i. p(t, x, t + h, dy) continuous paths h y x δ 1 ii. (y x)p(t, x, t + h, dy) b(t, x) h y x <δ 1 iii. (y i x i )(y j x j )p(t, x,, t + h, dy) a ij (t, x) h y x <δ () Stochastic Calculus March 4, 29 5 / 22
t { } t f (t, X t ) f (, X ) s + L f (s, X s )ds = f (s, X s ) σdb s L = 1 2 2 a ij (t, x) + x i x j i,j=1 M t = f (t, X t ) t { } s + L i=1 b i (t, x) x i = generator f (s, X s )ds is a martingale t { } = E[f (t, X t ) f (s, X s ) u + L f (u, X u )du F s ] s = f (t, y)p(s, x, t, y)dy f (s, x) t { u + L} f (u, y)p(s, x, u, y)dydu, s X s = x () Stochastic Calculus March 4, 29 6 / 22
For any f, = f (t, y)p(s, x, t, y)dy f (s, x) t { u + L} f (u, y)p(s, x, u, y)dydu s Fokker-Planck (Forward) Equation t p(s, x, t, y) = 1 2 i,j=1 i=1 = L yp(s, x, t, y) 2 y i y j (a i,j (t, y)p(s, x, t, y)) y i (b i (t, y)p(s, x, t, y)) lim p(s, x, t, y) = δ(y x). t s () Stochastic Calculus March 4, 29 7 / 22
Kolmogorov (Backward) Equation s p(s, x, t, y) = 1 2 + i,j=1 i=1 = L x p(s, x, t, y) a ij (s, x) 2 p(s, x, t, y) x i x j p(s, x, t, y) b i (s, x) x i lim p(s, x, t, y) = δ(y x). s t () Stochastic Calculus March 4, 29 8 / 22
Example. Brownian motion d = 1 2 L = 1 2 x 2 Forward p(s, x, t, y t = 1 2 p(s, x, t, y) 2 y 2, t > s p(s, x, s, y) = δ(y x) Backward p(s, x, t, y s = 1 2 p(s, x, t, y) 2 x 2, s < t, p(t, x, t, y) = δ(y x) () Stochastic Calculus March 4, 29 9 / 22
Example. Ornstein-Uhlenbeck Process L = σ2 2 2 x 2 αx x Forward p(s, x, t, y t = 1 2 p(s, x, t, y) 2 y 2 + (αyp(s, x, t, y), t > s, y p(s, x, s, y) = δ(y x) Backward p(s, x, t, y s = 1 2 p(s, x, t, y) 2 x 2 αx p(s, x, t, y), s < t, x p(t, x, t, y) = δ(y x) () Stochastic Calculus March 4, 29 1 / 22
Formal derivation of the backward equation p(s, x, t, A) = p(s, x, s + h, dy)p(s + h, y, t, A) = { } p(s, x, s + h, dy) p(s + h, y, t, A) p(s, x, t, A) = { p(s, x, t, A) p(s, x, s + h, dy) h + s + 1 2 p(s, x, t, A) (y i x i ) x i i=1 (y i x i )(y j x j ) 2 p(s, x, t, A) } + x i x j i,j=1 p(s, x, t, A) = s i=1 p(s, x, t, A) b i (t, x) + 1 x i 2 i,j=1 a ij (t, x) 2 p(s, x, t, A) x i x j () Stochastic Calculus March 4, 29 11 / 22
Real derivation f (x) smooth s u = L su s < t u(t, x) = f (x) Ito s formula: u(s, X(s)) martingale up to time t u(s, x) = E s,x [u(s, X(s))] = E s,x [u(t, X(t))] = f (z)p(s, x, t, dz) Let f n (z) smooth functions tending to δ(y z) u(s, x) = p(s, x, t, y) if s u = L su s < t u(t, x) = δ(x y) () Stochastic Calculus March 4, 29 12 / 22
Existence result from PDE Suppose that a(t, x) and b(t, x) are bounded and that there are α >, γ (, 1], C < such that for all s, t, x, y R d, i. ξ T a(t, x)ξ α ξ 2, ξ R d, ii. a(s, x) a(t, y) + b(s, x) b(t, y) C( x y γ + t s γ ). Then the backward equation has a solution and furthermore p(s, x, t, A) = p(s, x, t, y)dy with p(s, x, t, y) jointly continuous in s, x, t, y. Furthermore, p(s, x, t, y) is the unique weak solution of the forward equation,i.e. f (t, y)p(s, x, t, y)dy f (s, x) = A t s { u + L } f (u, y)p(s, x, u, y)dydu () Stochastic Calculus March 4, 29 13 / 22
The solution X t, t of dx t = b(x t )dt + σ(x t )db t with X = x is a Markov process with infinitesimal generator Itô s formula L = 1 2 2 a ij (x) + x i x j i,j=1 f (t, X t ) = f (, X ) + t + i,j=1 t i=1 b i (x) x i, a = σσ. { } s f (s, X s ) + Lf (s, X s ) ds σ ij (s, X s ) x i f (s, X s )db j s () Stochastic Calculus March 4, 29 14 / 22
Under the previous conditions, the following are equivalent 1 B t, t, dx t = b(t, X t )dt + σ(t, X t )db t 2 For each λ R d, Z λ (t) = e λ{x t R t b(s,xs)ds} 1 R t 2 λt a(s,x s)λds is a martingale with respect to F t 3 For all smooth f (t, x), f (t, X t ) t is a martingale with respect to F t 4 For all smooth f (x), { s + L}f (s, X s )ds t f (X t ) Lf (X s )ds is a martingale with respect to F t () Stochastic Calculus March 4, 29 15 / 22
Brownian motion in R d 1 B t = (Bt 1,..., Bd t ), Bi t independent Brownian motions 2 B t Markov with P(B t A B s = x) = A 1 e y x 2 (2π(t s)) d/2 2(t s) dy 3 B t has stationary independent mean zero increments with E[ B t B s 2 ] = d(t s) 4 e λ B t 1 2 λ 2t is a martingale for any λ Note that 1 does not depend on the basis: If Bt 1,..., B2 t independent and O is orthogonal, then the coordinates of OB t are independent Brownian motions in fact Theorem Suppose X 1, X 2 independent and θ Nπ/2 such that X 1 cos θ + X 2 sin θ, X 1 sin θ + X 2 cos θ independent Then X 1, X 2 are Gaussians (Maxwell) () Stochastic Calculus March 4, 29 16 / 22
Dirichlet problem Given a bounded open subset G R d and a continuous function f : G R find a continuous function u : Ḡ R such that { u = in G u G = f u def = i=1 2 u x 2 i = 2d lim r r 2 ( 1 S(r, x) S(r,x) uds u(x) ) Lemma u harmonic in G u satisfies the mean value property: for all sufficiently small r >, 1 uds = u(x) S(r, x) S(r,x) () Stochastic Calculus March 4, 29 17 / 22
Lemma u harmonic in G u satisfies the mean value property: for all sufficiently small r >, 1 uds = u(x) S(r, x) S(r,x) Proof. Green s identity G v udx = G u vdx + { G v u n u v n log r log x log r log δ d = 2 G = {δ < x < r}, v = x 2 d r 2 d d > 2 δ 2 d r 2 d let ρ ds () Stochastic Calculus March 4, 29 18 / 22
B t d-dimensional Brownian motion starting at x G τ G = inf{t : B(t) G} u(x) = E x [f (B(τ G ))] Theorem If G nice then u solves the Dirichlet problem E x [f (B(τ G ))] = G f (y)π G (x, dy), π G (x, Γ) = P x (B(τ G ) Γ), Γ G Example. G = B(x, r), π G (x, Γ) = Γ, Γ S(x, r) S(x,r) Brownian motion is invariant under rotations π G (x, ) is invariant under rotations () Stochastic Calculus March 4, 29 19 / 22
Proposition G bounded open R d, f bounded measurable on G. Then u(x) = E x [f (B(τ G ))] is harmonic in G. Proof. B = B(x, r) G τ B τ G Strong Markov property: u(b(τ S )) = E x [f (B(τ G )) F τs ] u(x) = E x [f (B(τ G ))] = E x [E x [f (B(τ G )) F τs ]] So u satisfies the mean value property in G. = E x [u(b(τ S ))] = u(y)π S (x, dy) = S 1 S u(y)ds () Stochastic Calculus March 4, 29 2 / 22 S
a G To complete the proof that u solves the Dirichlet problem we need Proposition lim E x[f (B(τ G ))] = f (a) x a, x G It is not always true! If lim x a P x [τ G > ɛ] =, ɛ > then for any bdd mble function x G f : G R which is continuous at a, lim x a E x [f (B(τ G ))] = f (a) x G Proof. Need: lim x a, x G P x ( B(τ G ) x < δ) = 1 P x ( B(τ G ) x < δ) P x ( sup B(t) x < δ, τ G ɛ) t ɛ P x ( sup B(t) x < δ) P x (τ G ɛ) t ɛ 1 as x a, x G then ɛ () Stochastic Calculus March 4, 29 21 / 22
Proposition a G is regular if P a (σ G = ) = 1 σ G = inf{t > : B(t) G} a regular lim x a, x G E x [f (B τg )] = f (a) f bdd mble, cont at a Proof of Enough to prove P x (σ G < ɛ) lower semi-continuous in x Then lim sup x a P x (σ G < ɛ) P a (σ G < ɛ) = 1 and σ G τ G x G But p(, x, δ, y)p y ( s (, ɛ δ), B(s) G) continuous and P x (σ G < ɛ) as δ Examples 1 If G is a smooth manifold near a then a is regular by LIL 2 If cone C of height h > and vertex at a such that C {a} ḠC then a is a regular (exterior cone condition) 3 d 2 always,d 3 counterexamples () Stochastic Calculus March 4, 29 22 / 22