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					                                       Brownian Motion

                  6

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                  0

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                  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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