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Brownian Motion 6 4 2 0 2 4 6 812 10 8 6 4 2 0 2 4 6 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 1 / 37 Figure: Robert Brown 1773 – 1858 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 2 / 37 Figure: Louis Bachelier 1870 – 1946 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 3 / 37 Figure: Albert Einstein 1879 – 1955 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 4 / 37 Random Walk 1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 5 / 37 Random Walk 1 Consider a discrete–time random walk t Rt := Rt−1 + Xt i.e. Rt = Xs s=1 where Xt , t = 1, 2, . . . are independent identically distributed random “shocks”. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 5 / 37 Random Walk 1 Consider a discrete–time random walk t Rt := Rt−1 + Xt i.e. Rt = Xs s=1 where Xt , t = 1, 2, . . . are independent identically distributed random “shocks”. We seek a continuous–time version of this. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 5 / 37 Random Walk 1 Consider a discrete–time random walk t Rt := Rt−1 + Xt i.e. Rt = Xs s=1 where Xt , t = 1, 2, . . . are independent identically distributed random “shocks”. We seek a continuous–time version of this. Fix a time interval [0, T ], divided into N time steps, each of length ∆t := T /N. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 5 / 37 Random Walk 1 Consider a discrete–time random walk t Rt := Rt−1 + Xt i.e. Rt = Xs s=1 where Xt , t = 1, 2, . . . are independent identically distributed random “shocks”. We seek a continuous–time version of this. Fix a time interval [0, T ], divided into N time steps, each of length ∆t := T /N. Let P(Xt = ±∆x) = 1 . 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 5 / 37 Random Walk 1 Consider a discrete–time random walk t Rt := Rt−1 + Xt i.e. Rt = Xs s=1 where Xt , t = 1, 2, . . . are independent identically distributed random “shocks”. We seek a continuous–time version of this. Fix a time interval [0, T ], divided into N time steps, each of length ∆t := T /N. Let P(Xt = ±∆x) = 1 . 2 For t = 0, ∆t, 2∆t, . . . , N∆t = T , deﬁne n Rt = Xk where t = n∆t k=1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 5 / 37 Random Walk 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 6 / 37 Random Walk 2 Thus Rt+∆t = Rt ± ∆x P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 6 / 37 Random Walk 2 Thus Rt+∆t = Rt ± ∆x Then (∆x)2 E[Rt ] = 0 Var(Rt ) = n(∆x)2 = t ∆t P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 6 / 37 Random Walk 2 Thus Rt+∆t = Rt ± ∆x Then (∆x)2 E[Rt ] = 0 Var(Rt ) = n(∆x)2 = t ∆t We want that Var(Rt ) becomes neither 0 nor inﬁnite as ∆t → 0. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 6 / 37 Random Walk 2 Thus Rt+∆t = Rt ± ∆x Then (∆x)2 E[Rt ] = 0 Var(Rt ) = n(∆x)2 = t ∆t We want that Var(Rt ) becomes neither 0 nor inﬁnite as ∆t → 0. Normalizing, we require that (∆x)2 lim =1 so that Var(Rt ) = t ∆t→0 ∆t P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 6 / 37 Random Walk 2 Thus Rt+∆t = Rt ± ∆x Then (∆x)2 E[Rt ] = 0 Var(Rt ) = n(∆x)2 = t ∆t We want that Var(Rt ) becomes neither 0 nor inﬁnite as ∆t → 0. Normalizing, we require that (∆x)2 lim =1 so that Var(Rt ) = t ∆t→0 ∆t √ We therefore set ∆x := ∆t and look. . . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 6 / 37 Random Walk 3 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 7 / 37 Random Walk 3 Let u(t, x) be the density function of Rt : P(x ≤ Rt < x + dx) ≈ u(t, x)dx P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 7 / 37 Random Walk 3 Let u(t, x) be the density function of Rt : P(x ≤ Rt < x + dx) ≈ u(t, x)dx Then u(t + ∆t, x) = 1 u(t, x − ∆x) + 2 u(t, x + ∆x) 2 1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 7 / 37 Random Walk 3 Let u(t, x) be the density function of Rt : P(x ≤ Rt < x + dx) ≈ u(t, x)dx Then u(t + ∆t, x) = 1 u(t, x − ∆x) + 2 u(t, x + ∆x) 2 1 Taylor expanding up to o(∆t) = o((∆x)2 ), we obtain u(t, x) + ut (t, x)∆t = 1 [u(t, x) − ux (t, x)∆x + 1 uxx (t, x)(∆x)2 ] 2 2 + 2 [u(t, x) + ux (t, x)∆x + 1 uxx (t, x)(∆x)2 ] 1 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 7 / 37 Random Walk 3 Let u(t, x) be the density function of Rt : P(x ≤ Rt < x + dx) ≈ u(t, x)dx Then u(t + ∆t, x) = 1 u(t, x − ∆x) + 2 u(t, x + ∆x) 2 1 Taylor expanding up to o(∆t) = o((∆x)2 ), we obtain u(t, x) + ut (t, x)∆t = 1 [u(t, x) − ux (t, x)∆x + 1 uxx (t, x)(∆x)2 ] 2 2 + 2 [u(t, x) + ux (t, x)∆x + 1 uxx (t, x)(∆x)2 ] 1 2 Thus in the limit ∆t → 0, we see that u(t, x) satisﬁes the heat equation! 1 ut (t, x) = 2 uxx (t, x) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 7 / 37 Random Walk 4 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 8 / 37 Random Walk 4 Because u(t, x) is a probability density function, we must have ∞ u(t, x) dx = 1 all t ≥ 0 −∞ P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 8 / 37 Random Walk 4 Because u(t, x) is a probability density function, we must have ∞ u(t, x) dx = 1 all t ≥ 0 −∞ Moreover, for any “reasonable” function f , we have ∞ f (0) = E[f (R0 )] = f (x)u(0, x) dx −∞ so that u(0, x) = δ0 , the Dirac delta. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 8 / 37 Random Walk 4 Because u(t, x) is a probability density function, we must have ∞ u(t, x) dx = 1 all t ≥ 0 −∞ Moreover, for any “reasonable” function f , we have ∞ f (0) = E[f (R0 )] = f (x)u(0, x) dx −∞ so that u(0, x) = δ0 , the Dirac delta. Thus u(t, x) is the fundamental solution (Green’s function) of the heat equation, well–known by physicists: 1 2 u(t, x) = √ e −x /2t 2πt P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 8 / 37 Random Walk 4 Because u(t, x) is a probability density function, we must have ∞ u(t, x) dx = 1 all t ≥ 0 −∞ Moreover, for any “reasonable” function f , we have ∞ f (0) = E[f (R0 )] = f (x)u(0, x) dx −∞ so that u(0, x) = δ0 , the Dirac delta. Thus u(t, x) is the fundamental solution (Green’s function) of the heat equation, well–known by physicists: 1 2 u(t, x) = √ e −x /2t 2πt And hence Rt is normally distributed with mean 0 and variance t. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 8 / 37 Figure: Andrei Kolmogorov 1903 – 1987 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 9 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] Events are modelled by subsets of Ω. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] Events are modelled by subsets of Ω. E = {2, 4, 6} ⊆ Ωd is the event that the outcome of the die is an even number. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] Events are modelled by subsets of Ω. E = {2, 4, 6} ⊆ Ωd is the event that the outcome of the die is an even number. [0.25, 0.5] ⊆ Ωr is the outcome that the random number lies between 0.25 and 0.5. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] Events are modelled by subsets of Ω. E = {2, 4, 6} ⊆ Ωd is the event that the outcome of the die is an even number. [0.25, 0.5] ⊆ Ωr is the outcome that the random number lies between 0.25 and 0.5. Events are organized into a σ–algebra F — the collection of events for which it can be decided whether or not they occurred. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] Events are modelled by subsets of Ω. E = {2, 4, 6} ⊆ Ωd is the event that the outcome of the die is an even number. [0.25, 0.5] ⊆ Ωr is the outcome that the random number lies between 0.25 and 0.5. Events are organized into a σ–algebra F — the collection of events for which it can be decided whether or not they occurred. Ω, ∅ ∈ F P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] Events are modelled by subsets of Ω. E = {2, 4, 6} ⊆ Ωd is the event that the outcome of the die is an even number. [0.25, 0.5] ⊆ Ωr is the outcome that the random number lies between 0.25 and 0.5. Events are organized into a σ–algebra F — the collection of events for which it can be decided whether or not they occurred. Ω, ∅ ∈ F If F ∈ F, then F c ∈ F P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 1 A probability space is a measure space, i.e. a triple (Ω, F, P). Ω is a set which models the possible outcomes of a random experiment Roll a fair die: Ωd := {1, 2, . . . , 6} Pick a random number from unit interval: Ωr := [0, 1] Events are modelled by subsets of Ω. E = {2, 4, 6} ⊆ Ωd is the event that the outcome of the die is an even number. [0.25, 0.5] ⊆ Ωr is the outcome that the random number lies between 0.25 and 0.5. Events are organized into a σ–algebra F — the collection of events for which it can be decided whether or not they occurred. Ω, ∅ ∈ F If F ∈ F, then F c ∈ F If Fn ∈ F for n ∈ N, then n Fn , n Fn ∈ F. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 10 / 37 Probability Theory 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 11 / 37 Probability Theory 2 The measure P : F → [0, 1] assigns to each event F ∈ F a probability P(F ) ∈ [0, 1], such that P(Ω) = 1 . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 11 / 37 Probability Theory 2 The measure P : F → [0, 1] assigns to each event F ∈ F a probability P(F ) ∈ [0, 1], such that P(Ω) = 1 . A random variable is a measurable function X : (Ω, F, P) → (R, B(R)). P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 11 / 37 Probability Theory 2 The measure P : F → [0, 1] assigns to each event F ∈ F a probability P(F ) ∈ [0, 1], such that P(Ω) = 1 . A random variable is a measurable function X : (Ω, F, P) → (R, B(R)). X is F–measurable if F contains enough information to compute X . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 11 / 37 Probability Theory 2 The measure P : F → [0, 1] assigns to each event F ∈ F a probability P(F ) ∈ [0, 1], such that P(Ω) = 1 . A random variable is a measurable function X : (Ω, F, P) → (R, B(R)). X is F–measurable if F contains enough information to compute X . We must be able to decide whether or not a < X < b for each a < b ∈ R. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 11 / 37 Probability Theory 2 The measure P : F → [0, 1] assigns to each event F ∈ F a probability P(F ) ∈ [0, 1], such that P(Ω) = 1 . A random variable is a measurable function X : (Ω, F, P) → (R, B(R)). X is F–measurable if F contains enough information to compute X . We must be able to decide whether or not a < X < b for each a < b ∈ R. Thus X −1 (a, b) = {ω ∈ Ω : a < X (ω) < b} must belong to F. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 11 / 37 Probability Theory 2 The measure P : F → [0, 1] assigns to each event F ∈ F a probability P(F ) ∈ [0, 1], such that P(Ω) = 1 . A random variable is a measurable function X : (Ω, F, P) → (R, B(R)). X is F–measurable if F contains enough information to compute X . We must be able to decide whether or not a < X < b for each a < b ∈ R. Thus X −1 (a, b) = {ω ∈ Ω : a < X (ω) < b} must belong to F. The expectation of a random variable is deﬁned to be its Lebesgue integral w.r.t measure P: EX := X dP Ω We also write E[X ; F ] := F X dP. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 11 / 37 Probability Theory 3 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 12 / 37 Probability Theory 3 The conditional probability of F given G (F , G ∈ F) is deﬁned by P(F ∩ G ) P(F |G ) := P(G ) This deﬁnes a new probability measure P(·|G ) whose mass is concentrated on the set G . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 12 / 37 Probability Theory 3 The conditional probability of F given G (F , G ∈ F) is deﬁned by P(F ∩ G ) P(F |G ) := P(G ) This deﬁnes a new probability measure P(·|G ) whose mass is concentrated on the set G . Two events F , G are independent (write F ⊥ G ) when P(F |G ) = P(F ), i.e. when P(F ∩ G ) = P(F ) · P(G ). P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 12 / 37 Probability Theory 3 The conditional probability of F given G (F , G ∈ F) is deﬁned by P(F ∩ G ) P(F |G ) := P(G ) This deﬁnes a new probability measure P(·|G ) whose mass is concentrated on the set G . Two events F , G are independent (write F ⊥ G ) when P(F |G ) = P(F ), i.e. when P(F ∩ G ) = P(F ) · P(G ). Similarly, if G, H are sub-σ-algebras of F, then G⊥H iﬀ ∀G ∈ G ∀H ∈ H (G ⊥ H) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 12 / 37 Probability Theory 3 The conditional probability of F given G (F , G ∈ F) is deﬁned by P(F ∩ G ) P(F |G ) := P(G ) This deﬁnes a new probability measure P(·|G ) whose mass is concentrated on the set G . Two events F , G are independent (write F ⊥ G ) when P(F |G ) = P(F ), i.e. when P(F ∩ G ) = P(F ) · P(G ). Similarly, if G, H are sub-σ-algebras of F, then G⊥H iﬀ ∀G ∈ G ∀H ∈ H (G ⊥ H) If X is a random variable and G an event, then E[X ; G ] E[X |G ] := X dP(·|G ) = P(G ) is the expectation of X given that G has occurred. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 12 / 37 Probability Theory 3 The conditional probability of F given G (F , G ∈ F) is deﬁned by P(F ∩ G ) P(F |G ) := P(G ) This deﬁnes a new probability measure P(·|G ) whose mass is concentrated on the set G . Two events F , G are independent (write F ⊥ G ) when P(F |G ) = P(F ), i.e. when P(F ∩ G ) = P(F ) · P(G ). Similarly, if G, H are sub-σ-algebras of F, then G⊥H iﬀ ∀G ∈ G ∀H ∈ H (G ⊥ H) If X is a random variable and G an event, then E[X ; G ] E[X |G ] := X dP(·|G ) = P(G ) is the expectation of X given that G has occurred. Roll a die, and let the outcome be X . Then E[X ] = 1 (1 + 2 + · · · + 6) = 3.5, but E[X |Even] = 4. 6 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 12 / 37 Probability Theory 4 Let X : (Ω, F, P) → R be a random variable. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 13 / 37 Probability Theory 4 Let X : (Ω, F, P) → R be a random variable. If we have no information, then the best estimate of X is its expectation E[X ]. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 13 / 37 Probability Theory 4 Let X : (Ω, F, P) → R be a random variable. If we have no information, then the best estimate of X is its expectation E[X ]. If we have all information in F, then the best estimate of X is X . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 13 / 37 Probability Theory 4 Let X : (Ω, F, P) → R be a random variable. If we have no information, then the best estimate of X is its expectation E[X ]. If we have all information in F, then the best estimate of X is X . If G is a sub-σ-algebra of F, then Z := E[X |G] is the “best” estimate of X given the the information in G. This means: P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 13 / 37 Probability Theory 4 Let X : (Ω, F, P) → R be a random variable. If we have no information, then the best estimate of X is its expectation E[X ]. If we have all information in F, then the best estimate of X is X . If G is a sub-σ-algebra of F, then Z := E[X |G] is the “best” estimate of X given the the information in G. This means: Z is G–measurable (so that we can compute it with information G). P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 13 / 37 Probability Theory 4 Let X : (Ω, F, P) → R be a random variable. If we have no information, then the best estimate of X is its expectation E[X ]. If we have all information in F, then the best estimate of X is X . If G is a sub-σ-algebra of F, then Z := E[X |G] is the “best” estimate of X given the the information in G. This means: Z is G–measurable (so that we can compute it with information G). For all G ∈ G we have E[Z |G ] = E[X |G ]. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 13 / 37 Figure: Joseph Doob 1910 –2004 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 14 / 37 Martingales 1 (Ω, F, P) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 15 / 37 Martingales 1 (Ω, F, P) Let F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Ft ⊆ . . . be an increasing sequence of sub-σ-algebras of F — a ﬁltration. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 15 / 37 Martingales 1 (Ω, F, P) Let F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Ft ⊆ . . . be an increasing sequence of sub-σ-algebras of F — a ﬁltration. Interpret Ft as the information available at time t. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 15 / 37 Martingales 1 (Ω, F, P) Let F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Ft ⊆ . . . be an increasing sequence of sub-σ-algebras of F — a ﬁltration. Interpret Ft as the information available at time t. An adapted stochastic process is a sequence (Xt )t of random variables, such that each Xt is Ft –measurable. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 15 / 37 Martingales 1 (Ω, F, P) Let F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Ft ⊆ . . . be an increasing sequence of sub-σ-algebras of F — a ﬁltration. Interpret Ft as the information available at time t. An adapted stochastic process is a sequence (Xt )t of random variables, such that each Xt is Ft –measurable. (Xt )t is called a martingale if the best estimate of any future value of X is its current value: E[Xt |Fs ] = Xs whenever s ≤ t P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 15 / 37 Martingales 1 (Ω, F, P) Let F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Ft ⊆ . . . be an increasing sequence of sub-σ-algebras of F — a ﬁltration. Interpret Ft as the information available at time t. An adapted stochastic process is a sequence (Xt )t of random variables, such that each Xt is Ft –measurable. (Xt )t is called a martingale if the best estimate of any future value of X is its current value: E[Xt |Fs ] = Xs whenever s ≤ t Equivalently, E[Xt+1 − Xt |Ft ] = 0 for all t. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 15 / 37 Martingales 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 16 / 37 Martingales 2 Example: If Rt is a symmetric random walk starting at zero with jumps Xt = ±1, then E[Rt+1 |Ft ] = E[Rt + Xt+1 |Ft ] = Rt + E[Xt+1 ] = Rt P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 16 / 37 Martingales 2 Example: If Rt is a symmetric random walk starting at zero with jumps Xt = ±1, then E[Rt+1 |Ft ] = E[Rt + Xt+1 |Ft ] = Rt + E[Xt+1 ] = Rt If we interpret Xt as a player’s fortune in a game of chance, then martingales model fair games. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 16 / 37 Martingales 2 Example: If Rt is a symmetric random walk starting at zero with jumps Xt = ±1, then E[Rt+1 |Ft ] = E[Rt + Xt+1 |Ft ] = Rt + E[Xt+1 ] = Rt If we interpret Xt as a player’s fortune in a game of chance, then martingales model fair games. The player expects neither to gain nor to lose on average. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 16 / 37 Martingales 3 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 17 / 37 Martingales 3 Let (Xt ) be a martingale. Interpret Xt as a player’s fortune at time n in a game of chance, betting unit stakes. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 17 / 37 Martingales 3 Let (Xt ) be a martingale. Interpret Xt as a player’s fortune at time n in a game of chance, betting unit stakes. If the player bets an amount Ct on the t th game, her gain/loss for that game will be Ct (Xt − Xt−1 ). P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 17 / 37 Martingales 3 Let (Xt ) be a martingale. Interpret Xt as a player’s fortune at time n in a game of chance, betting unit stakes. If the player bets an amount Ct on the t th game, her gain/loss for that game will be Ct (Xt − Xt−1 ). If (Ct )t is the player’s betting strategy, then her total gain/loss after the t th game will be an “integral”: t t Gt = Cs (Xs − Xs−1 ) = Cs ∆Xs “= Cs dXs ” s=1 s=1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 17 / 37 Martingales 3 Let (Xt ) be a martingale. Interpret Xt as a player’s fortune at time n in a game of chance, betting unit stakes. If the player bets an amount Ct on the t th game, her gain/loss for that game will be Ct (Xt − Xt−1 ). If (Ct )t is the player’s betting strategy, then her total gain/loss after the t th game will be an “integral”: t t Gt = Cs (Xs − Xs−1 ) = Cs ∆Xs “= Cs dXs ” s=1 s=1 To qualify as a betting strategy, bets must be placed before playing game, i.e. Ct must be Ft−1 –measurable — no peeking into the future! Then E[Gt+1 −Gt |Ft ] = E[Ct+1 (Xt+1 −Xt )|Ft ] = Ct+1 E[Xt+1 −Xt |Ft ] = 0 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 17 / 37 Martingales 4 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 18 / 37 Martingales 4 Doob Upcrossing Lemma: UT [a, b] = number of upcrossings of martingale X from below a to above b by time T . N 0 = E[ Ct ∆Xt ] ≥ (b − a)EUT [a, b] − E|XT − a| t=1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 18 / 37 Martingales 4 Doob Upcrossing Lemma: UT [a, b] = number of upcrossings of martingale X from below a to above b by time T . N 0 = E[ Ct ∆Xt ] ≥ (b − a)EUT [a, b] − E|XT − a| t=1 E|XT |+|a| Hence EUT [a, b] ≤ b−a . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 18 / 37 Martingales 5 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 19 / 37 Martingales 5 If supt E|Xt | < ∞ (i.e. X is L1 –bounded), then EU∞ [a, b] =↑ lim EUT [a, b] < ∞ T →∞ Hence P(U∞ [a, b] = ∞) = 0. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 19 / 37 Martingales 5 If supt E|Xt | < ∞ (i.e. X is L1 –bounded), then EU∞ [a, b] =↑ lim EUT [a, b] < ∞ T →∞ Hence P(U∞ [a, b] = ∞) = 0. Deﬁne A := {ω ∈ Ω : limt Xt (ω) does not exist} Then A= {ω ∈ Ω : lim inf Xt (ω) < a < b < lim sup Xt (ω)} t t a<b∈Q = {ω ∈ Ω : U∞ [a, b] = ∞} a<b∈Q P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 19 / 37 Martingales 5 If supt E|Xt | < ∞ (i.e. X is L1 –bounded), then EU∞ [a, b] =↑ lim EUT [a, b] < ∞ T →∞ Hence P(U∞ [a, b] = ∞) = 0. Deﬁne A := {ω ∈ Ω : limt Xt (ω) does not exist} Then A= {ω ∈ Ω : lim inf Xt (ω) < a < b < lim sup Xt (ω)} t t a<b∈Q = {ω ∈ Ω : U∞ [a, b] = ∞} a<b∈Q Hence P(A) = 0, i.e. lim Xt exists with probability 1. t→∞ P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 19 / 37 Martingales 5 If supt E|Xt | < ∞ (i.e. X is L1 –bounded), then EU∞ [a, b] =↑ lim EUT [a, b] < ∞ T →∞ Hence P(U∞ [a, b] = ∞) = 0. Deﬁne A := {ω ∈ Ω : limt Xt (ω) does not exist} Then A= {ω ∈ Ω : lim inf Xt (ω) < a < b < lim sup Xt (ω)} t t a<b∈Q = {ω ∈ Ω : U∞ [a, b] = ∞} a<b∈Q Hence P(A) = 0, i.e. lim Xt exists with probability 1. t→∞ Martingale Convergence Theorem: L1 –bounded martingales converge a.s. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 19 / 37 Martingales 6 (Ω, F, (F)t , P) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 20 / 37 Martingales 6 (Ω, F, (F)t , P) A stopping time is a non–negative random variable T such that {ω ∈ Ω : T (ω) ≤ t} ∈ Ft for all t ≥ 0 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 20 / 37 Martingales 6 (Ω, F, (F)t , P) A stopping time is a non–negative random variable T such that {ω ∈ Ω : T (ω) ≤ t} ∈ Ft for all t ≥ 0 For the symmetric random walk Rt , consider τ := inf{t : Rt = 1} σ := inf{t : Rt = max Rs } s≤10 τ is a stopping time; σ is not. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 20 / 37 Martingales 6 (Ω, F, (F)t , P) A stopping time is a non–negative random variable T such that {ω ∈ Ω : T (ω) ≤ t} ∈ Ft for all t ≥ 0 For the symmetric random walk Rt , consider τ := inf{t : Rt = 1} σ := inf{t : Rt = max Rs } s≤10 τ is a stopping time; σ is not. Note that Rτ = 1. Hence E[Rτ ] = 1 = 0 = R0 — the martingale property may break for stopping times. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 20 / 37 Martingales 7 (Ω, F, (F)t , P) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 21 / 37 Martingales 7 (Ω, F, (F)t , P) If T is a stopping time and Xt a martingale, then the stopped martingale is Xt if t < T XtT := Xt∧T = XT if T ≤ t P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 21 / 37 Martingales 7 (Ω, F, (F)t , P) If T is a stopping time and Xt a martingale, then the stopped martingale is Xt if t < T XtT := Xt∧T = XT if T ≤ t Optional Stopping Theorem: If Xt is a “nice” martingale, then the martingale property holds for stopping times T : T T E[XT ] = E[X∞ ] = E[ lim XtT ] = lim E[XtT ] = X0 = X0 t→∞ t→∞ Bounded or Lebesgue dominated suﬃce for “nice”. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 21 / 37 Figure: Norbert Wiener 1894 – 1964 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 22 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) A standard Ft –Brownian motion is an adapted process (Bt )t≥0 such that: P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) A standard Ft –Brownian motion is an adapted process (Bt )t≥0 such that: Increments independent of the past: Bt − Bs is independent of Fs P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) A standard Ft –Brownian motion is an adapted process (Bt )t≥0 such that: Increments independent of the past: Bt − Bs is independent of Fs Increments stationary and normal: Bt − Bt ∼ N(0, t − s) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) A standard Ft –Brownian motion is an adapted process (Bt )t≥0 such that: Increments independent of the past: Bt − Bs is independent of Fs Increments stationary and normal: Bt − Bt ∼ N(0, t − s) Continuous paths: t → Bt (ω) is continuous a.s. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) A standard Ft –Brownian motion is an adapted process (Bt )t≥0 such that: Increments independent of the past: Bt − Bs is independent of Fs Increments stationary and normal: Bt − Bt ∼ N(0, t − s) Continuous paths: t → Bt (ω) is continuous a.s. B0 (ω) = 0 a.s. — but may also start a BM at an arbitrary point x. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) A standard Ft –Brownian motion is an adapted process (Bt )t≥0 such that: Increments independent of the past: Bt − Bs is independent of Fs Increments stationary and normal: Bt − Bt ∼ N(0, t − s) Continuous paths: t → Bt (ω) is continuous a.s. B0 (ω) = 0 a.s. — but may also start a BM at an arbitrary point x. Brownian motion is a martingale: E[Bt |Fs ] − Bs = E[Bt − Bs |Fs ] = E[Bt − Bs ] = 0 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 1 (Ω, F, (Ft )t P) A standard Ft –Brownian motion is an adapted process (Bt )t≥0 such that: Increments independent of the past: Bt − Bs is independent of Fs Increments stationary and normal: Bt − Bt ∼ N(0, t − s) Continuous paths: t → Bt (ω) is continuous a.s. B0 (ω) = 0 a.s. — but may also start a BM at an arbitrary point x. Brownian motion is a martingale: E[Bt |Fs ] − Bs = E[Bt − Bs |Fs ] = E[Bt − Bs ] = 0 Bt2 − t is a martingale: 2 E[Bt2 − Bs |Fs ] = E[(Bt − Bs )2 |Fs ] + 2Bs E[Bt − Bs |Fs ] = E[(Bt − Bs )2 ] = t − s P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 23 / 37 Brownian Motion 2 (Ω, F, P) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) ˜ Bt = 1 Ba2 t (Scaling) a P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) ˜ Bt = 1 Ba2 t (Scaling) a tB 1 t>0 ˜ Bt = t (Time inversion) 0 t=0 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) ˜ Bt = 1 Ba2 t (Scaling) a tB 1 t>0 ˜ Bt = t (Time inversion) 0 t=0 lim sup Bt = +∞, lim inf Bt = −∞ a.s. t→∞ t→∞ P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) ˜ Bt = 1 Ba2 t (Scaling) a tB 1 t>0 ˜ Bt = t (Time inversion) 0 t=0 lim sup Bt = +∞, lim inf Bt = −∞ a.s. t→∞ t→∞ ˜ For each n, Bt := tB 1 has a zero in [n, ∞) a.s., so Bt has a zero in t 1 (0, n ]. Thus the set of zeroes of a Brownian path is perfect —closed with no isolated points — hence has cardinality |R|. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) ˜ Bt = 1 Ba2 t (Scaling) a tB 1 t>0 ˜ Bt = t (Time inversion) 0 t=0 lim sup Bt = +∞, lim inf Bt = −∞ a.s. t→∞ t→∞ ˜ For each n, Bt := tB 1 has a zero in [n, ∞) a.s., so Bt has a zero in t 1 (0, n ]. Thus the set of zeroes of a Brownian path is perfect —closed with no isolated points — hence has cardinality |R|. ˜ tB 1 dBt Bt −B0 ˜ dt |t=0 = lim = lim = lim Bs does not exist a.s. . t t→0 t t→0 t s→∞ P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) ˜ Bt = 1 Ba2 t (Scaling) a tB 1 t>0 ˜ Bt = t (Time inversion) 0 t=0 lim sup Bt = +∞, lim inf Bt = −∞ a.s. t→∞ t→∞ ˜ For each n, Bt := tB 1 has a zero in [n, ∞) a.s., so Bt has a zero in t 1 (0, n ]. Thus the set of zeroes of a Brownian path is perfect —closed with no isolated points — hence has cardinality |R|. ˜ tB 1 dBt Bt −B0 ˜ dt |t=0 = lim = lim = lim Bs does not exist a.s. . t t→0 t t→0 t s→∞ By time translation, for any point a ≥ 0, Bt is a.s. not diﬀerentiable at t = a. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 Brownian Motion 2 (Ω, F, P) Invariance Properties: If (Bt )t is a Brownian motion, then so are ˜ Bt := Ba+t − Ba (Time translation) ˜ Bt = 1 Ba2 t (Scaling) a tB 1 t>0 ˜ Bt = t (Time inversion) 0 t=0 lim sup Bt = +∞, lim inf Bt = −∞ a.s. t→∞ t→∞ ˜ For each n, Bt := tB 1 has a zero in [n, ∞) a.s., so Bt has a zero in t 1 (0, n ]. Thus the set of zeroes of a Brownian path is perfect —closed with no isolated points — hence has cardinality |R|. ˜ tB 1 dBt Bt −B0 ˜ dt |t=0 = lim = lim = lim Bs does not exist a.s. . t t→0 t t→0 t s→∞ By time translation, for any point a ≥ 0, Bt is a.s. not diﬀerentiable at t = a. In fact, almost surely, Brownian motion is nowhere diﬀerentiable. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 24 / 37 o Figure: Kiyoshi Itˆ 1915 – 2008 e Paul-Andr´ Meyer 1934 – 2003 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 25 / 37 Stochastic Calculus 1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 26 / 37 Stochastic Calculus 1 Along a partition PN : 0 = {0 = t0 ≤ t1 ≤ · · · ≤ tN = T } of [0, T ], consider N−1 f (t, Bt ) − f (0, B0 ) = f (tn+1 , Btn+1 ) − f (tn , Btn ) = ∆n f n=0 n P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 26 / 37 Stochastic Calculus 1 Along a partition PN : 0 = {0 = t0 ≤ t1 ≤ · · · ≤ tN = T } of [0, T ], consider N−1 f (t, Bt ) − f (0, B0 ) = f (tn+1 , Btn+1 ) − f (tn , Btn ) = ∆n f n=0 n Here ∆n f ≈ ft (tn , Btn )∆n t + fx (tn , Btn ) ∆n B + 1 fxx (tn , Btn ) (∆n B)2 2 and so f (t, Bt ) − f (0, B0 ) ≈ ft ∆ n t + fx ∆ n B + 1 2 fxx (∆n B)2 n n n P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 26 / 37 Stochastic Calculus 1 Along a partition PN : 0 = {0 = t0 ≤ t1 ≤ · · · ≤ tN = T } of [0, T ], consider N−1 f (t, Bt ) − f (0, B0 ) = f (tn+1 , Btn+1 ) − f (tn , Btn ) = ∆n f n=0 n Here ∆n f ≈ ft (tn , Btn )∆n t + fx (tn , Btn ) ∆n B + 1 fxx (tn , Btn ) (∆n B)2 2 and so f (t, Bt ) − f (0, B0 ) ≈ ft ∆ n t + fx ∆ n B + 1 2 fxx (∆n B)2 n n n The term N fx ∆n B = fx (tn , Btn )(Btn+1 − Btn ) n n=1 looks like the gain obtained by betting fx (tn , Btn ) on the martingale B. This is a martingale!! P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 26 / 37 Stochastic Calculus 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 27 / 37 Stochastic Calculus 2 For the last term, note that E(∆n B)2 = tn+1 − tn = ∆n t Var(∆n B)2 = 2∆n t 2 Thus as ∆t → 0, we see that (∆B)2 → dt. It follows that t 1 2 fxx (∆n B)2 → 1 2 fxx (s, Bs ) ds as ∆t → 0 n 0 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 27 / 37 Stochastic Calculus 2 For the last term, note that E(∆n B)2 = tn+1 − tn = ∆n t Var(∆n B)2 = 2∆n t 2 Thus as ∆t → 0, we see that (∆B)2 → dt. It follows that t 1 2 fxx (∆n B)2 → 1 2 fxx (s, Bs ) ds as ∆t → 0 n 0 o Hence we have the Itˆ formula: t t f (t, Bt ) = f (0, B0 ) + ft (s, Bs ) + 1 fxx (s, Bs ) ds + 2 fx (s, Bs ) dBs 0 0 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 27 / 37 Stochastic Calculus 2 For the last term, note that E(∆n B)2 = tn+1 − tn = ∆n t Var(∆n B)2 = 2∆n t 2 Thus as ∆t → 0, we see that (∆B)2 → dt. It follows that t 1 2 fxx (∆n B)2 → 1 2 fxx (s, Bs ) ds as ∆t → 0 n 0 o Hence we have the Itˆ formula: t t f (t, Bt ) = f (0, B0 ) + ft (s, Bs ) + 1 fxx (s, Bs ) ds + 2 fx (s, Bs ) dBs 0 0 t 1 t t Example: Bt2 = 0 2 · 2 ds + 0 2Bs dBs i.e. 0 Bs dBs = 1 [Bt2 − t]. 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 27 / 37 Stochastic Calculus 2 For the last term, note that E(∆n B)2 = tn+1 − tn = ∆n t Var(∆n B)2 = 2∆n t 2 Thus as ∆t → 0, we see that (∆B)2 → dt. It follows that t 1 2 fxx (∆n B)2 → 1 2 fxx (s, Bs ) ds as ∆t → 0 n 0 o Hence we have the Itˆ formula: t t f (t, Bt ) = f (0, B0 ) + ft (s, Bs ) + 1 fxx (s, Bs ) ds + 2 fx (s, Bs ) dBs 0 0 t t t 1 Example: Bt2 = 0 2 · 2 ds + 0 2Bs dBs i.e. 0 Bs dBs = 1 [Bt2 − t]. 2 o For multidimensional Brownian motions, the Itˆ formula is: t t ∂ f (t, Bt ) = f (0, B0 ) + ∂t + 1 ∆ f (s, Bs ) ds + 2 f (s, Bs ) · dBs 0 0 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 27 / 37 e Figure: Paul L´vy 1886 – 1971 o Paul Erd¨s 1913 – 1996 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 28 / 37 Transience and Recurrence 1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 29 / 37 Transience and Recurrence 1 A harmonic function is a function f (x) that satisﬁes Laplace’s equartion n ∂2f ∆f (x) = =0 ∂xi2 i=1 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 29 / 37 Transience and Recurrence 1 A harmonic function is a function f (x) that satisﬁes Laplace’s equartion n ∂2f ∆f (x) = =0 ∂xi2 i=1 o By Itˆ, if f is harmonic (with no explicit time-dependence) then t f (Bt ) = f (B0 ) + f (Bs ) · dBs is a martingale ! 0 . . . well. . . a local martingale. . . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 29 / 37 Transience and Recurrence 1 A harmonic function is a function f (x) that satisﬁes Laplace’s equartion n ∂2f ∆f (x) = =0 ∂xi2 i=1 o By Itˆ, if f is harmonic (with no explicit time-dependence) then t f (Bt ) = f (B0 ) + f (Bs ) · dBs is a martingale ! 0 . . . well. . . a local martingale. . . In particular, if Bt is a BM starting at x, then f (x) = f (B0 ) = Ex [f (Bt )] P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 29 / 37 Transience and Recurrence 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 30 / 37 Transience and Recurrence 2 Consider now a Brownian motion starting at a point x in side an annulus bounded by circles of radii r , R respectively. Deﬁne Tr = inf{t : |Bt | = r } Tr = inf{t : |Bt | = R} τ = Tr ∧ TR P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 30 / 37 Transience and Recurrence 3 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 31 / 37 Transience and Recurrence 3 If f is harmonic, then f (x) = Ex [f (Bτ )] P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 31 / 37 Transience and Recurrence 3 If f is harmonic, then f (x) = Ex [f (Bτ )] Now note that |x| d =1 f (x) = ln |x| d =2 2−d |x| d ≥3 are spherically symmetric harmonic functions. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 31 / 37 Transience and Recurrence 3 If f is harmonic, then f (x) = Ex [f (Bτ )] Now note that |x| d =1 f (x) = ln |x| d =2 2−d |x| d ≥3 are spherically symmetric harmonic functions. For such f we have f (x) = Ex [f (Bτ )] = f (r )Px (Tr < TR ) + f (R)Px (TR < Tr ) Using P(TR < Tr ) = 1 − P(Tr < TR ) and rearranging, we see that f (R) − f (x) Px (Tr < TR ) = f (R) − f (r ) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 31 / 37 Transience and Recurrence 4 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 32 / 37 Transience and Recurrence 4 Letting R → ∞, we obtain ln |R| − ln |x| R→∞ ln |R| − ln |r | = 1 lim d =2 P(Tr < ∞) = d−2 − |x|d−2 lim R |x|d−2 = d−2 d ≥3 R→∞ R d−2 − r d−2 |r | P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 32 / 37 Transience and Recurrence 4 Letting R → ∞, we obtain ln |R| − ln |x| R→∞ ln |R| − ln |r | = 1 lim d =2 P(Tr < ∞) = d−2 − |x|d−2 lim R |x|d−2 = d−2 d ≥3 R→∞ R d−2 − r d−2 |r | Thus in dimension d = 2, Brownian motion is recurrent: Every open set will be hit with probability 1, again and again. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 32 / 37 Transience and Recurrence 4 Letting R → ∞, we obtain ln |R| − ln |x| R→∞ ln |R| − ln |r | = 1 lim d =2 P(Tr < ∞) = d−2 − |x|d−2 lim R |x|d−2 = d−2 d ≥3 R→∞ R d−2 − r d−2 |r | Thus in dimension d = 2, Brownian motion is recurrent: Every open set will be hit with probability 1, again and again. In dimensions d ≥ 3 Brownian motion is transient. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 32 / 37 Application: Fundamental Theorem of Algebra P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 33 / 37 Application: Fundamental Theorem of Algebra Suppose now that z = x + iy and that f (z) = Re f (z) + Im f (z) = u(x, y ) + iv (x, y ) is holomorphic. Then u, v are harmonic (Cauchy–Riemann eqn.) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 33 / 37 Application: Fundamental Theorem of Algebra Suppose now that z = x + iy and that f (z) = Re f (z) + Im f (z) = u(x, y ) + iv (x, y ) is holomorphic. Then u, v are harmonic (Cauchy–Riemann eqn.) 1 If p(z) is a polynomial without roots, then f (z) = p(z) is a bounded holomorphic function (entire). P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 33 / 37 Application: Fundamental Theorem of Algebra Suppose now that z = x + iy and that f (z) = Re f (z) + Im f (z) = u(x, y ) + iv (x, y ) is holomorphic. Then u, v are harmonic (Cauchy–Riemann eqn.) 1 If p(z) is a polynomial without roots, then f (z) = p(z) is a bounded holomorphic function (entire). Since f is non–constant, there exist α < β and open discs D1 , D2 ⊆ C such that u(x, y ) < α on D1 β < u(x, y ) on D2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 33 / 37 Application: Fundamental Theorem of Algebra Suppose now that z = x + iy and that f (z) = Re f (z) + Im f (z) = u(x, y ) + iv (x, y ) is holomorphic. Then u, v are harmonic (Cauchy–Riemann eqn.) 1 If p(z) is a polynomial without roots, then f (z) = p(z) is a bounded holomorphic function (entire). Since f is non–constant, there exist α < β and open discs D1 , D2 ⊆ C such that u(x, y ) < α on D1 β < u(x, y ) on D2 But u(Bt1 , Bt2 ) is a bounded martingale. By the MCT, lim u(Bt1 , Bt2 ) t→∞ exists a.s. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 33 / 37 Application: Fundamental Theorem of Algebra Suppose now that z = x + iy and that f (z) = Re f (z) + Im f (z) = u(x, y ) + iv (x, y ) is holomorphic. Then u, v are harmonic (Cauchy–Riemann eqn.) 1 If p(z) is a polynomial without roots, then f (z) = p(z) is a bounded holomorphic function (entire). Since f is non–constant, there exist α < β and open discs D1 , D2 ⊆ C such that u(x, y ) < α on D1 β < u(x, y ) on D2 But u(Bt1 , Bt2 ) is a bounded martingale. By the MCT, lim u(Bt1 , Bt2 ) t→∞ exists a.s. By recurrence, lim inf u(Bt1 , Bt2 ) ≤ α < β ≤ lim sup u(Bt1 , Bt2 ). t→∞ t→∞ P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 33 / 37 Application: Fundamental Theorem of Algebra Suppose now that z = x + iy and that f (z) = Re f (z) + Im f (z) = u(x, y ) + iv (x, y ) is holomorphic. Then u, v are harmonic (Cauchy–Riemann eqn.) 1 If p(z) is a polynomial without roots, then f (z) = p(z) is a bounded holomorphic function (entire). Since f is non–constant, there exist α < β and open discs D1 , D2 ⊆ C such that u(x, y ) < α on D1 β < u(x, y ) on D2 But u(Bt1 , Bt2 ) is a bounded martingale. By the MCT, lim u(Bt1 , Bt2 ) t→∞ exists a.s. By recurrence, lim inf u(Bt1 , Bt2 ) ≤ α < β ≤ lim sup u(Bt1 , Bt2 ). t→∞ t→∞ Hence it is impossible that f is entire, i.e. p(z) has a root! P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 33 / 37 Figure: Richard Feynman 1918 – 1988 c Mark Kaˇ 1914 – 1984 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 34 / 37 c Application: Feynman–Kaˇ Formula P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 35 / 37 c Application: Feynman–Kaˇ Formula Consider the initial value problem for u(t, x) — heat equation on the real line: ∂u 1 ∂2u = u(0, x) = Φ(x) ∂t 2 ∂x 2 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 35 / 37 c Application: Feynman–Kaˇ Formula Consider the initial value problem for u(t, x) — heat equation on the real line: ∂u 1 ∂2u = u(0, x) = Φ(x) ∂t 2 ∂x 2 ∂f 1 ∂2f Fix T > 0 and let f (t, x) := u(T − t, x). Then ∂t + 2 ∂x 2 = 0. P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 35 / 37 c Application: Feynman–Kaˇ Formula Consider the initial value problem for u(t, x) — heat equation on the real line: ∂u 1 ∂2u = u(0, x) = Φ(x) ∂t 2 ∂x 2 ∂f 1 ∂2f Fix T > 0 and let f (t, x) := u(T − t, x). Then ∂t + 2 ∂x 2 = 0. Now let Bt be a BM starting at x and deﬁne Mt := f (t, Bt ). Then M0 = f (0, B0 ) = u(T , x) and MT = u(0, BT ) = Φ(BT ). P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 35 / 37 c Application: Feynman–Kaˇ Formula Consider the initial value problem for u(t, x) — heat equation on the real line: ∂u 1 ∂2u = u(0, x) = Φ(x) ∂t 2 ∂x 2 ∂f 1 ∂2f Fix T > 0 and let f (t, x) := u(T − t, x). Then ∂t + 2 ∂x 2 = 0. Now let Bt be a BM starting at x and deﬁne Mt := f (t, Bt ). Then M0 = f (0, B0 ) = u(T , x) and MT = u(0, BT ) = Φ(BT ). o By Itˆ’s formula MT = f (T , BT ) T T ∂ 1 ∂2 ∂ = f (0, B0 ) + + f (t, Bt ) dt + f (t, Bt ) dBt 0 ∂t 2 ∂x 2 0 ∂x T ∂ = u(T , x) + 0 + f (t, Bt ) dBt 0 ∂x P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 35 / 37 c Application: Feynman–Kaˇ Formula P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 36 / 37 c Application: Feynman–Kaˇ Formula Take expectations, and use the martingale property of stochastic integrals: T ∂ E[MT ] = u(T , x) + E f (t, Bt ) dBt = u(T , x) 0 ∂x and since MT = Φ(BT ) we have u(T , x) = Ex [Φ(BT )] P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 36 / 37 c Application: Feynman–Kaˇ Formula Take expectations, and use the martingale property of stochastic integrals: T ∂ E[MT ] = u(T , x) + E f (t, Bt ) dBt = u(T , x) 0 ∂x and since MT = Φ(BT ) we have u(T , x) = Ex [Φ(BT )] To solve the heat equation u(T , x): P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 36 / 37 c Application: Feynman–Kaˇ Formula Take expectations, and use the martingale property of stochastic integrals: T ∂ E[MT ] = u(T , x) + E f (t, Bt ) dBt = u(T , x) 0 ∂x and since MT = Φ(BT ) we have u(T , x) = Ex [Φ(BT )] To solve the heat equation u(T , x): Run a Brownian motion starting x until time T . P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 36 / 37 c Application: Feynman–Kaˇ Formula Take expectations, and use the martingale property of stochastic integrals: T ∂ E[MT ] = u(T , x) + E f (t, Bt ) dBt = u(T , x) 0 ∂x and since MT = Φ(BT ) we have u(T , x) = Ex [Φ(BT )] To solve the heat equation u(T , x): Run a Brownian motion starting x until time T . Evaluate at the initial condition Φ(BT ) P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 36 / 37 c Application: Feynman–Kaˇ Formula Take expectations, and use the martingale property of stochastic integrals: T ∂ E[MT ] = u(T , x) + E f (t, Bt ) dBt = u(T , x) 0 ∂x and since MT = Φ(BT ) we have u(T , x) = Ex [Φ(BT )] To solve the heat equation u(T , x): Run a Brownian motion starting x until time T . Evaluate at the initial condition Φ(BT ) Average: u(T , x) = Ex [Φ(BT )] P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 36 / 37 P. Ouwehand (Stellenbosch Univ.) Brownian Motion 27 August 2010 37 / 37

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posted: | 3/29/2011 |

language: | Afrikaans |

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