# Brownian Motion

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```					                                       Brownian Motion

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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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