An Introduction To Stochastic Differential Equations Version 1.2 ...
An Introduction to Stochastic
Differential Equations
Version 1.2
Lawrence C. Evans
Department of Mathematics
UC Berkeley
Chapter 1: Introduction
Chapter 2: A crash course in basic probability theory
Chapter 3: Brownian motion and “white noise”
Chapter 4: Stochastic integrals, Itˆo’s formula
Chapter 5: Stochastic differential equations
Chapter 6: Applications
Appendices
Exercises
References
1
PREFACE
These notes survey, without too many precise details, the basic theory of prob-
ability, random differential equations and some applications.
Stochastic differential equations is usually, and justly, regarded as a graduate
level subject. A really careful treatment assumes the students’ familiarity with
probability theory, measure theory, ordinary differential equations, and partial dif-
ferential equations as well.
But as an experiment I tried to design these lectures so that starting graduate
students (and maybe really strong undergraduates) can follow most of the theory,
at the cost of some omission of detail and precision. I for instance downplayed
most measure theoretic issues, but did emphasize the intuitive idea of σ–algebras as
“containing information”. Similarly, I “prove” many formulas by confirming them
in easy cases (for simple random variables or for step functions), and then just
stating that by approximation these rules hold in general. I also did not reproduce
in class some of the more complicated proofs provided in these notes, although I
did try to explain the guiding ideas.
My thanks especially to Lisa Goldberg, who several years ago presented my
class with several lectures on financial applications, and to Fraydoun Rezakhanlou,
who has taught from these notes and added several improvements.
I am also grateful to Jonathan Weare for several computer simulations illus-
trating the text. Thanks also to many readers who have found errors, especially
Robert Piche, who provided me with an extensive list of typos and suggestions that
I have incorporated into this latest version of the notes.
2
CHAPTER 1: INTRODUCTION
A. MOTIVATION
Fix a point x0 ∈ Rn and consider then the ordinary differential equation:
˙x(t) = b(x(t)) (t > 0)
(ODE)
x(0) = x0,
where b : Rn → Rn is a given, smooth vector field and the solution is the trajectory
x(·) : [0, ∞) → Rn.
x(t)
x0
Trajectory of the differential equation
Notation. x(t) is the state of the system at time t ≥ 0, ˙x(t) := d x(t).
dt
In many applications, however, the experimentally measured trajectories of
systems modeled by (ODE) do not in fact behave as predicted:
X(t)
x0
Sample path of the stochastic differential equation
Hence it seems reasonable to modify (ODE), somehow to include the possibility of
random effects disturbing the system. A formal way to do so is to write:
˙
X(t) = b(X(t)) + B(X(t))ξ(t) (t > 0)
(1)
X(0) = x0,
where B : Rn → Mn×m (= space of n × m matrices) and
ξ(·) := m-dimensional “white noise”.
This approach presents us with these mathematical problems:
• Define the “white noise” ξ(·) in a rigorous way.
• Define what it means for X(·) to solve (1).
• Show (1) has a solution, discuss uniqueness, asymptotic behavior, dependence
upon x0, b, B, etc.
3
B. SOME HEURISTICS
Let us first study (1) in the case m = n, x0 = 0, b ≡ 0, and B ≡ I. The
solution of (1) in this setting turns out to be the n-dimensional Wiener process, or
Brownian motion, denoted W(·). Thus we may symbolically write
˙
W(·) = ξ(·),
thereby asserting that “white noise” is the time derivative of the Wiener process.
Now return to the general case of the equation (1), write d instead of the dot:
dt
dX(t)
dW(t)
= b(X(t)) + B(X(t))
,
dt
dt
and finally multiply by “dt”:
dX(t) = b(X(t))dt + B(X(t))dW(t)
(SDE)
X(0) = x0.
This expression, properly interpreted, is a stochastic differential equation. We
say that X(·) solves (SDE) provided
t
t
(2)
X(t) = x0 +
b(X(s)) ds +
B(X(s)) dW
for all times t > 0 .
0
0
Now we must:
• Construct W(·): See Chapter 3.
• Define the stochastic integral t0 · · ·dW : See Chapter 4.
• Show (2) has a solution, etc.: See Chapter 5.
And once all this is accomplished, there will still remain these modeling problems:
• Does (SDE) truly model the physical situation?
• Is the term ξ(·) in (1) “really” white noise, or is it rather some ensemble of
smooth,
but highly oscillatory functions? See Chapter 6.
As we will see later these questions are subtle, and different answers can yield
completely different solutions of (SDE). Part of the trouble is the strange form of
the chain rule in the stochastic calculus:
C. IT ˆ
O’S FORMULA
Assume n = 1 and X(·) solves the SDE
(3)
dX = b(X)dt + dW.
Suppose next that u : R → R is a given smooth function. We ask: what stochastic
differential equation does
Y (t) := u(X(t)) (t ≥ 0)
solve? Offhand, we would guess from (3) that
dY = u′dX = u′bdt + u′dW,
4
according to the usual chain rule, where ′ = d . This is wrong, however ! In fact,
dx
as we will see,
(4)
dW ≈ (dt)1/2
in some sense. Consequently if we compute dY and keep all terms of order dt or
1
(dt) 2 , we obtain
1
dY = u′dX + u′ (dX)2 + . . .
2
1
= u′(bdt + dW ) + u′ (bdt + dW )2 + . . .
2
from (3)
1
=
u′b + u′
dt + u′dW +
2
{terms of order (dt)3/2 and higher}.
Here we used the “fact” that (dW )2 = dt, which follows from (4). Hence
1
dY =
u′b + u′
dt + u′dW,
2
with the extra term “ 1 u′ dt” not present in ordinary calculus.
2
A major goal of these notes is to provide a rigorous interpretation for calcula-
tions like these, involving stochastic differentials.
Example 1. According to Itˆo’s formula, the solution of the stochastic differ-
ential equation
dY = Y dW,
Y (0) = 1
is
Y (t) := eW (t)− t2 ,
and not what might seem the obvious guess, namely ˆ
Y (t) := eW (t).
Example 2. Let P (t) denote the (random) price of a stock at time t ≥ 0. A
standard model assumes that dP , the relative change of price, evolves according to
P
the SDE
dP = µdt + σdW
P
for certain constants µ > 0 and σ, called respectively the drift and the volatility of
the stock. In other words,
dP = µP dt + σP dW
P (0) = p0,
where p0 is the starting price. Using once again Itˆo’s formula we can check that the
solution is
P (t) = p
”t
0eσW (t)+“µ− σ2
2
.
5
A sample path for stock prices
6
CHAPTER 2: A CRASH COURSE IN BASIC PROBABILITY THEORY
A. Basic definitions
B. Expected value, variance
C. Distribution functions
D. Independence
E. Borel–Cantelli Lemma
F. Characteristic functions
G. Strong Law of Large Numbers, Central Limit Theorem
H. Conditional expectation
I. Martingales
This chapter is a very rapid introduction to the measure theoretic foundations
of probability theory. More details can be found in any good introductory text, for
instance Bremaud [Br], Chung [C] or Lamperti [L1].
A. BASIC DEFINITIONS Let us begin with a puzzle:
Bertrand’s paradox. Take a circle of radius 2 inches in the plane and choose
a chord of this circle at random. What is the probability this chord intersects the
concentric circle of radius 1 inch?
Solution #1 Any such chord (provided it does not hit the center) is uniquely
determined by the location of its midpoint.
Thus
area of inner circle
1
probability of hitting inner circle =
=
.
area of larger circle
4
Solution #2 By symmetry under rotation we may assume the chord is vertical.
The diameter of the large circle is 4 inches and the chord will hit the small circle if
it falls within its 2-inch diameter.
7
Hence
2 inches
1
probability of hitting inner circle =
=
.
4 inches
2
Solution #3 By symmetry we may assume one end of the chord is at the far
left point of the larger circle. The angle θ the chord makes with the horizontal lies
between ±π and the chord hits the inner circle if θ lies between
.
2
±π6
θ
Therefore
2π
1
probability of hitting inner circle = 6 =
.
2π
3
2
PROBABILITY SPACES. This example shows that we must carefully de-
fine what we mean by the term “random”. The correct way to do so is by introducing
as follows the precise mathematical structure of a probability space.
We start with a nonempty set, denoted Ω, certain subsets of which we will in a
moment interpret as being “events”.
DEFINITION. A σ-algebra is a collection U of subsets of Ω with these prop-
erties:
(i) ∅, Ω ∈ U.
(ii) If A ∈ U, then Ac ∈ U.
(iii) If A1, A2, · · · ∈ U, then
∞
∞
Ak,
Ak ∈ U.
k=1
k=1
Here Ac := Ω − A is the complement of A.
DEFINITION. Let U be a σ-algebra of subsets of Ω. We call P : U → [0, 1] a
probability measure provided:
(i) P (∅) = 0, P(Ω) = 1.
(ii) If A1, A2, · · · ∈ U, then
∞
∞
P (
Ak) ≤
P (Ak).
k=1
k=1
8
(iii) If A1, A2, . . . are disjoint sets in U, then
∞
∞
P (
Ak) =
P (Ak).
k=1
k=1
It follows that if A, B ∈ U, then
A ⊆ B implies P(A) ≤ P(B).
DEFINITION. A triple (Ω, U, P) is called a probability space provided Ω is
any set, U is a σ-algebra of subsets of Ω, and P is a probability measure on U.
Terminology. (i) A set A ∈ U is called an event; points ω ∈ Ω are sample
points.
(ii) P (A) is the probability of the event A.
(iii) A property which is true except for an event of probability zero is said to
hold almost surely (usually abbreviated “a.s.”).
Example 1. Let Ω = {ω1, ω2, . . ., ωN} be a finite set, and suppose we are
given numbers 0 ≤ pj ≤ 1 for j = 1, . . ., N, satisfying pj = 1. We take U
to comprise all subsets of Ω. For each set A = {ωj , ω , . . ., ω
1
j2
jm } ∈ U, with
1 ≤ j1 < j2 < . . .jm ≤ N, we define P(A) := pj + p +
.
1
j2
· · · + pjm
Example 2. The smallest σ-algebra containing all the open subsets of Rn is
called the Borel σ-algebra, denoted B. Assume that f is a nonnegative, integrable
function, such that
f dx = 1. We define
Rn
P (B) :=
f (x) dx
B
for each B ∈ B. Then (Rn, B, P) is a probability space. We call f the density of
the probability measure P .
Example 3. Suppose instead we fix a point z ∈ Rn, and now define
1
if z
P (B) :=
∈ B
0
if z /
∈ B
for sets B ∈ B. Then (Rn, B, P) is a probability space. We call P the Dirac mass
concentrated at the point z, and write P = δz.
A probability space is the proper setting for mathematical probability theory.
This means that we must first of all carefully identify an appropriate (Ω, U, P)
when we try to solve problems. The reader should convince himself or herself
that the three “solutions” to Bertrand’s paradox discussed above represent three
distinct interpretations of the phrase “at random”, that is, to three distinct models
of (Ω, U, P).
Here is another example.
9
Example 4 (Buffon’s needle problem). The plane is ruled by parallel
lines 2 inches apart and a 1-inch long needle is dropped at random on the plane.
What is the probability that it hits one of the parallel lines?
The first issue is to find some appropriate probability space (Ω, U, P). For this,
let
h = distance from the center of needle to nearest line,
θ = angle (≤ π) that the needle makes with the horizontal.
2
θ
h
needle
These fully determine the position of the needle, up to translations and reflec-
tion. Let us next take
π
Ω= [0, )
× [0, 1], U = Borel subsets of Ω,
2
values of h
values of θ
P(B)= 2·areaofB for each B
π
∈ U.
We denote by A the event that the needle hits a horizontal line. We can now check
that this happens provided
h
. Consequently A =
sin θ ≤ 12
π
{(θ, h) ∈ Ω| h ≤ sinθ
2 },
and so P (A) = 2(area of A) = 2
2 1 sin θ dθ = 1 .
π
π
0
2
π
RANDOM VARIABLES. We can think of the probability space as being an
essential mathematical construct, which is nevertheless not “directly observable”.
We are therefore interested in introducing mappings X from Ω to Rn, the values of
which we can observe.
Remember from Example 2 above that
B denotes the collection of Borel subsets of Rn, which is the
smallest σ-algebra of subsets of Rn containing all open sets.
We may henceforth informally just think of B as containing all the “nice, well-
behaved” subsets of Rn.
DEFINITION. Let (Ω, U, P) be a probability space. A mapping
X : Ω → Rn
is called an n-dimensional random variable if for each B ∈ B, we have
X−1(B) ∈ U.
We equivalently say that X is U-measurable.
10
Notation, comments. We usually write “X” and not “X(ω)”. This follows
the custom within probability theory of mostly not displaying the dependence of
random variables on the sample point ω ∈ Ω. We also denote P(X−1(B)) as
“P (X ∈ B)”, the probability that X is in B.
In these notes we will usually use capital letters to denote random variables.
Boldface usually means a vector-valued mapping.
We will also use without further comment various standard facts from measure
theory, for instance that sums and products of random variables are random vari-
ables.
Example 1. Let A ∈ U. Then the indicator function of A,
1
if ω
χ (ω) :=
∈ A
A
0
if ω /
∈ A,
is a random variable.
Example 2. More generally, if A1, A2, . . . , Am ∈ U, with Ω = ∪mi=1Ai, and
a1, a2, . . . , am are real numbers, then
m
X =
aiχAi
i=1
is a random variable, called a simple function.
LEMMA. Let X : Ω → Rn be a random variable. Then
U(X) := {X−1(B)| B ∈ B}
is a σ-algebra, called the σ-algebra generated by X. This is the smallest sub-σ-
algebra of U with respect to which X is measurable.
Proof. Check that {X−1(B) | B ∈ B} is a σ-algebra; clearly it is the smallest
σ-algebra with respect to which X is measurable.
IMPORTANT REMARK. It is essential to understand that, in probabilistic
terms, the σ-algebra U(X) can be interpreted as “containing all relevant informa-
tion” about the random variable X.
In particular, if a random variable Y is a function of X, that is, if
Y = Φ(X)
for some reasonable function Φ, then Y is U(X)-measurable.
Conversely, suppose Y : Ω → R is U(X)-measurable. Then there exists a
function Φ such that
Y = Φ(X).
Hence if Y is U(X)-measurable, Y is in fact a function of X. Consequently if we
know the value X(ω), we in principle know also Y (ω) = Φ(X(ω)), although we may
have no practical way to construct Φ.
STOCHASTIC PROCESSES. We introduce next random variables depend-
ing upon time.
11
DEFINITIONS. (i) A collection {X(t) | t ≥ 0} of random variables is called
a stochastic process.
(ii) For each point ω ∈ Ω, the mapping t → X(t, ω) is the corresponding sample
path.
The idea is that if we run an experiment and observe the random values of
X(·) as time evolves, we are in fact looking at a sample path {X(t, ω) | t ≥ 0} for
some fixed ω ∈ Ω. If we rerun the experiment, we will in general observe a different
sample path.
X(t,ω1)
time
X(t,ω2)
Two sample paths of a stochastic process
B. EXPECTED VALUE, VARIANCE
Integration with respect to a measure. If (Ω, U, P) is a probability space and
X =
k
a
is a real-valued simple random variable, we define the integral of
i=1
iχAi
X by
k
X dP :=
aiP (Ai).
Ω
i=1
If next X is a nonnegative random variable, we define
X dP :=
sup
Y dP.
Ω
Y ≤X,Y simple Ω
Finally if X : Ω → R is a random variable, we write
X dP :=
X+ dP − X− dP,
Ω
Ω
Ω
provided at least one of the integrals on the right is finite. Here X+ = max(X, 0)
and X− = max(−X, 0); so that X = X+ − X−.
Next, suppose X : Ω → Rn is a vector-valued random variable, X = (X1, X2, . . ., Xn).
Then we write
X dP =
X1 dP,
X2 dP, · · · , Xn dP .
Ω
Ω
Ω
Ω
12
We will assume without further comment the usual rules for these integrals.
DEFINITION. We call
E(X) :=
X dP
Ω
the expected value (or mean value) of X.
DEFINITION. We call
V (X) :=
|X − E(X)|2 dP
Ω
the variance of X, where | · | denotes the Euclidean norm.
Observe that
V (X) = E(|X − E(X)|2) = E(|X|2) − |E(X)|2.
LEMMA (Chebyshev’s inequality). If X is a random variable and 1 ≤
p < ∞, then
1
P (|X| ≥ λ) ≤ E(
λp
|X|p) for all λ > 0.
Proof. We have
E(|X|p) = |X|p dP ≥
|X|p dP ≥ λpP(|X| ≥ λ).
Ω
{|X|≥λ}
C. DISTRIBUTION FUNCTIONS
Let (Ω, U, P) be a probability space and suppose X : Ω → Rn is a random
variable.
Notation. Let x = (x1, . . . , xn) ∈ Rn, y = (y1, . . ., yn) ∈ Rn. Then
x ≤ y
means xi ≤ yi for i = 1, . . ., n.
DEFINITIONS. (i) The distribution function of X is the function FX : Rn →
[0, 1] defined by
FX(x) := P (X ≤ x) for all x ∈ Rn
(ii) If X1, . . . , Xm : Ω → Rn are random variables, their joint distribution
function is FX
: (Rn)m
1 ,...,Xm
→ [0, 1],
FX
(x
1 ,...,Xm
1, . . . , xm) := P (X1 ≤ x1, . . . , Xm ≤ xm)
for all xi ∈ Rn, i = 1, . . ., m.
13
x
X
Rn
Ω
DEFINITION. Suppose X : Ω → Rn is a random variable and F = FX its
distribution function. If there exists a nonnegative, integrable function f : Rn → R
such that
x1
xn
F (x) = F (x1, . . . , xn) =
· · ·
f (y1, . . . , yn) dyn . . . dy1,
−∞
−∞
then f is called the density function for X.
It follows then that
(1)
P (X ∈ B) = f(x) dx for all B ∈ B
B
This formula is important as the expression on the right hand side is an ordinary
integral, and can often be explicitly calculated.
Example 1. If X : Ω → R has density
1
f (x) = √
e− |x−m|2
2σ2
(x
2πσ2
∈ R),
we say X has a Gaussian (or normal) distribution, with mean m and variance σ2.
In this case let us write
X is an N (m, σ2) random variable.
Example 2. If X : Ω → Rn has density
1
f (x) =
e− 1 (x
2
−m)·C−1(x−m)
(x
((2π)n det C)1/2
∈ Rn)
for some m ∈ Rn and some positive definite, symmetric matrix C, we say X has
a Gaussian (or normal) distribution, with mean m and covariance matrix C. We
then write
X is an N (m, C) random variable.
14
LEMMA. Let X : Ω → Rn be a random variable, and assume that its distri-
bution function F = FX has the density f . Suppose g : Rn → R, and
Y = g(X)
is integrable. Then
E(Y ) =
g(x)f (x) dx.
Rn
In particular,
E(X) =
xf (x) dx and
V (X) =
|x − E(X)|2f(x)dx.
Rn
Rn
IMPORTANT REMARK. Hence we can compute E(X), V (X), etc. in
terms of integrals over Rn. This is an important observation, since as mentioned
before the probability space (Ω, U, P) is “unobservable”: all that we “see” are the
values X takes on in Rn. Indeed, all quantities of interest in probability theory can
be computed in Rn in terms of the density f .
Proof. Suppose first g is a simple function on Rn:
m
g =
biχB
(B
i
i ∈ B).
i=1
Then
m
m
E(g(X)) =
bi
χB (X) dP =
b
i
iP (X ∈ Bi).
i=1
Ω
i=1
But also
m
g(x)f (x) dx =
bi
f (x) dx
Rn
i=1
Bi
m
=
biP (X ∈ Bi) by (1).
i=1
Consequently the formula holds for all simple functions g and, by approximation,
it holds therefore for general functions g.
Example. If X is N (m, σ2), then
1
∞
E(X) = √
xe− (x−m)2
2σ2
dx = m
2πσ2 −∞
and
1
∞
V (X) = √
(x
2πσ2
− m)2e−(x−m)2
2σ2
dx = σ2.
−∞
Therefore m is indeed the mean, and σ2 the variance.
D. INDEPENDENCE
15
MOTIVATION. Let (Ω, U, P) be a probability space, and let A, B ∈ U be
two events, with P (B) > 0. We want to find a reasonable definition of
P (A | B), the probability of A, given B.
Think this way. Suppose some point ω ∈ Ω is selected “at random” and we are told
ω ∈ B. What then is the probability that ω ∈ A also?
A
ω
B
Ω
Since we know ω ∈ B, we can regard B as being a new probability space.
Therefore we can define ˜
Ω := B, ˜
U := {C ∩ B | C ∈ U} and ˜P := P ; so that
P (B)
˜
P ( ˜
Ω) = 1. Then the probability that ω lies in A is ˜
P (A ∩ B) = P(A∩B).
P (B)
This observation motivates the following
DEFINITION. We write
P (A
P (A | B) :=
∩ B) if P(B) > 0.
P (B)
Now what should it mean to say “A and B are independent”? This should mean
P (A | B) = P(A), since presumably any information that the event B has occurred
is irrelevant in determining the probability that A has occurred. Thus
P (A
P (A) = P (A | B) =
∩ B)
P (B)
and so
P (A ∩ B) = P(A)P(B)
if P (B) > 0. We take this for the definition, even if P (B) = 0:
DEFINITION. Two events A and B are called independent if
P (A ∩ B) = P(A)P(B).
This concept and its ramifications are the hallmarks of probability theory.
To gain some insight, the reader may wish to check that if A and B are inde-
pendent events, then so are Ac and B. Likewise, Ac and Bc are independent.
16
DEFINITION. Let A1, . . . , An, . . . be events. These events are independent
if for all choices 1 ≤ k1 < k2 < · · · < km, we have
P (Ak
) = P (A )P (A )
).
1 ∩ Ak2 ∩ · · · ∩ Akm
k1
k1 · · · P (Akm
It is important to extend this definition to σ-algebras:
DEFINITION. Let Ui ⊆ U be σ-algebras, for i = 1, . . .. We say that {Ui}∞i=1
are independent if for all choices of 1 ≤ k1 < k2 < · · · < km and of events Ak
,
i ∈ Uki
we have
P (Ak
) = P (A )P (A ) . . . P (A
).
1 ∩ Ak2 ∩ · · · ∩ Akm
k1
k2
km
Lastly, we transfer our definitions to random variables:
DEFINITION. Let Xi : Ω → Rn be random variables (i = 1, . . .). We say
the random variables X1, . . . are independent if for all integers k ≥ 2 and all choices
of Borel sets B1, . . . Bk ⊆ Rn:
P (X1 ∈ B1, X2 ∈ B2, . . ., Xk ∈ Bk) = P(X1 ∈ B1)P(X2 ∈ B2) · · ·P(Xk ∈ Bk).
This is equivalent to saying that the σ-algebras {U(Xi)}∞i=1 are independent.
Example. Take Ω = [0, 1), U the Borel subsets of [0, 1), and P Lebesgue
measure.
Define for n = 1, 2, . . .
1
if k
, k even
X
2n ≤ ω < k+1
2n
n(ω) :=
(0
−1 if k
, k odd
≤ ω < 1).
2n ≤ ω < k+1
2n
These are the Rademacher functions, which we assert are in fact independent ran-
dom variables. To prove this, it suffices to verify
P (X1 = e1, X2 = e2, . . . , Xk = ek) = P (X1 = e1)P (X2 = e2) · · ·P(Xk = ek),
for all choices of e1, . . . , ek ∈ {−1, 1}. This can be checked by showing that both
sides are equal to 2−k.
LEMMA. Let X1, . . . , Xm+n be independent Rk-valued random variables. Sup-
pose f : (Rk)n → R and g : (Rk)m → R. Then
Y := f (X1, . . . , Xn) and Z := g(Xn+1, . . . , Xn+m)
are independent.
We omit the proof, which may be found in Breiman [B].
17
THEOREM. The random variables X1, · · · , Xm : Ω → Rn are independent if
and only if
(2)
FX
(x
(x
(x
1 ,··· ,Xm
1, . . . , xm) = FX1
1) · · · FXm m)
for all xi ∈ Rn, i = 1, . . ., m.
If the random variables have densities, (2) is equivalent to
(3)
fX
(x
(x
(x
1 ,··· ,Xm
1, . . . , xm) = fX1
1) · · · fXm m)
for all xi ∈ Rn, i = 1, . . ., m,
where the functions f are the appropriate densities.
Proof. 1. Assume first that {Xk}m are independent. Then
k=1
FX
(x
1 ···Xm
1, . . . , xm) = P (X1 ≤ x1, . . . , Xm ≤ xm)
= P (X1 ≤ x1) · · ·P(Xm ≤ xm)
= FX (x
(x
1
1) · · · FXm m).
2. We prove the converse statement for the case that all the random variables
have densities. Select Ai ∈ U(Xi), i = 1, . . ., m. Then Ai = X−1(B
i
i) for some
Bi ∈ B. Hence
P (A1 ∩ · · · ∩ Am) = P(X1 ∈ B1, . . ., Xm ∈ Bm)
=
fX
(x
1 ···Xm
1, . . . , xm) dx1 · · · dxm
B1×...×Bm
=
fX (x
f
(x
1
1) dx1
. . .
Xm
m) dxm
by (3)
B1
Bm
= P (X1 ∈ B1) · · ·P(Xm ∈ Bm)
= P (A1) · · ·P(Am).
Therefore U(X1), · · · , U(Xm) are independent σ-algebras.
One of the most important properties of independent random variables is this:
THEOREM. If X1, . . . , Xm are independent, real-valued random variables,
with
E(|Xi|) < ∞ (i = 1, . . ., m),
then E(|X1 · · ·Xm|) < ∞ and
E(X1 · · ·Xm) = E(X1) · · ·E(Xm).
Proof. Suppose that each Xi is bounded and has a density. Then
E(X1 · · ·Xm) =
x1 · · ·xm fX
(x
1 ···Xm
1, . . . , xm) dx1 . . . xm
Rm
=
x1 fX (x
x
(x
1
1) dx1
· · ·
m fXm
m) dxm
by (3)
R
R
= E(X1) · · ·E(Xm).
18
THEOREM. If X1, . . . , Xm are independent, real-valued random variables,
with
V (Xi) < ∞ (i = 1, . . ., m),
then
V (X1 + · · · + Xm) = V (X1) + · · · + V (Xm).
Proof. Use induction, the case m = 2 holding as follows. Let m1 := EX1,
m2 := E(X2). Then E(X1 + X2) = m1 + m2 and
V (X1 + X2) =
(X1 + X2 − (m1 + m2))2 dP
Ω
=
(X1 − m1)2 dP + (X2 − m2)2 dP
Ω
Ω
+ 2
(X1 − m1)(X2 − m2) dP
Ω
= V (X1) + V (X2) + 2E(X1 − m1)E(X2 − m2),
=0
=0
where we used independence in the next last step.
E. BOREL–CANTELLI LEMMA
We introduce next a simple and very useful way to check if some sequence
A1, . . . , An, . . . of events “occurs infinitely often”.
DEFINITION. Let A1, . . . , An, . . . be events in a probability space. Then the
event
∞ ∞ Am = {ω ∈ Ω|ω belongs to infinitely many of the An},
n=1 m=n
is called “An infinitely often”, abbreviated “An i.o.”.
BOREL–CANTELLI LEMMA. If
∞ P (A
n=1
n) < ∞, then P (An i.o.) = 0.
Proof. By definition A
∞
n i.o. =
∞
A
n=1
m=n
m, and so for each n
∞
∞
P (An i.o.) ≤ P
Am
≤
P (Am).
m=n
m=n
The limit of the left-hand side is zero as n → ∞ because P(Am) < ∞.
APPLICATION. We illustrate a typical use of the Borel–Cantelli Lemma.
A sequence of random variables {Xk}∞ defined on some probability space
k=1
converges in probability to a random variable X, provided
lim P (|Xk − X| > ǫ) = 0
k→∞
for each ǫ > 0.
19
THEOREM. If Xk → X in probability, then there exists a subsequence {Xkj}∞j=1 ⊂
{Xk}∞ such that
k=1
Xk (ω)
j
→ X(ω) for almost every ω.
Proof. For each positive integer j we select kj so large that
1
1
P (|Xk
)
,
j − X| > j ≤ j2
and also . . . kj−1 < kj < . . ., kj → ∞. Let Aj := {|Xk
<
j −X| > 1j }. Since
1
j2
∞,
the Borel–Cantelli Lemma implies P (Aj i.o.) = 0. Therefore for almost all sample
points ω, |Xk (ω)
provided j
j
− X(ω)| ≤ 1j
≥ J, for some index J depending on
ω.
F. CHARACTERISTIC FUNCTIONS
It is convenient to introduce next a clever integral transform, which will later
provide us with a useful means to identify normal random variables.
DEFINITION. Let X be an Rn-valued random variable. Then
φX(λ) := E(eiλ·X)
(λ ∈ Rn)
is the characteristic function of X.
Example. If the real-valued random variable X is N (m, σ2), then
φX(λ) = eimλ− λ2σ2
2
(λ ∈ R).
To see this, let us suppose that m = 0, σ = 1 and calculate
∞
e −λ2
2
∞
φX(λ) =
eiλx 1
√ e−x22 dx = √
e− (x−iλ)2
2
dx.
−∞
2π
2π −∞
We move the path of integration in the complex plane from the line {Im(z) = −λ}
to the real axis, and recall that
∞ e−x22 dx = √2π. (Here Im(z) means the
−∞
imaginary part of the complex number z.) Hence φX(λ) = e− λ22 .
LEMMA. (i) If X1, . . . , Xm are independent random variables, then for each
λ ∈ Rn
φX
(λ) = φ
(λ) . . . φ
(λ).
1 +···+Xm
X1
Xm
(ii) If X is a real-valued random variable,
φ(k)(0) = ikE(Xk)
(k = 0, 1, . . . ).
(iii) If X and Y are random variables and
φX(λ) = φY(λ)
for all λ,
then
FX(x) = FY (x)
for all x.
20
Assertion (iii) says the characteristic function of X determines the distribution
of X.
Proof. 1. Let us calculate
φX
(λ) = E(eiλ·(X1+···+Xm))
1 +···+Xm
= E(eiλ·X1eiλ·X2 · · ·eiλ·Xm)
= E(eiλ·X1) · · ·E(eiλ·Xm) by independence
= φX (λ) . . . φ
(λ).
1
Xm
2. We have φ′(λ) = iE(XeiλX), and so φ′(0) = iE(X). The formulas in (ii) for
k = 2, . . . follow similarly.
3. See Breiman [B] for the proof of (iii).
Example. If X and Y are independent, real-valued random variables, and if
X is N (m1, σ21), Y is N(m2, σ22), then
X + Y is N (m1 + m2, σ21 + σ22).
To see this, just calculate
λ2 σ2
λ2σ2
1
2
φX+Y (λ) = φX(λ)φY (λ) = eim1λ− 2 eim2λ− 2
= ei(m1+m2)λ− λ2 (σ2+σ2)
2
1
2 .
G. STRONG LAW OF LARGE NUMBERS, CENTRAL LIMIT THEOREM
This section discusses a mathematical model for “repeated, independent exper-
iments”.
The idea is this. Suppose we are given a probability space and on it a real–valued
random variable X, which records the outcome of some sort of random experiment.
We can model repetitions of this experiment by introducing a sequence of random
variables X1, . . . , Xn, . . . , each of which “has the same probabilistic information as
X”:
DEFINITION. A sequence X1, . . . , Xn, . . . of random variables is called iden-
tically distributed if
FX (x) = F
(x) =
(x) = . . .
for all x.
1
X2
· · · = FXn
If we additionally assume that the random variables X1, . . . , Xn, . . . are inde-
pendent, we can regard this sequence as a model for repeated and independent runs
of the experiment, the outcomes of which we can measure. More precisely, imagine
that a “random” sample point ω ∈ Ω is given and we can observe the sequence of
values X1(ω), X2(ω), . . . , Xn(ω), . . . . What can we infer from these observations?
STRONG LAW OF LARGE NUMBERS. First we show that with prob-
ability one, we can deduce the common expected values of the random variables.
21
THEOREM (Strong Law of Large Numbers). Let X1, . . . , Xn, . . . be a
sequence of independent, identically distributed, integrable random variables defined
on the same probability space.
Write m := E(Xi) for i = 1, . . . . Then
X
P
lim
1 + · · · + Xn = m = 1.
n→∞
n
Proof. 1. Supposing that the random variables are real–valued entails no loss
of generality. We will as well suppose for simplicity that
E(X4i) < ∞ (i = 1, . . .).
We may also assume m = 0, as we could otherwise consider Xi − m in place of Xi.
2. Then
n
4
n
E
If i = j, k, or l, indep
Xi
E(XiXjXkXl).
i=1
endence impl = i,j,k,l=1
ies
E(XiXjXkXl) = E(Xi)E(XjXkXl).
=0
Consequently, since the Xi are identically distributed, we have
n
4
n
n
E
Xi
E(X4i) + 3
E(X2iX2j)
i=1
= i=1
i,j=1
i=j
= nE(X41) + 3(n2 − n)(E(X21))2
≤ n2C
for some constant C.
Now fix ε > 0. Then
1 n
n
P
X
X
n
i ≥ ε = P
i ≥ εn
i=1
i=1
4
1
n
≤
E
X
(εn)4
C 1
i
i=1
≤
.
ε4 n2
We used here the Chebyshev inequality. By the Borel–Cantelli Lemma, therefore,
1 n
P
X
n
i ≥ ε i.o. = 0.
i=1
22
3. Take ε = 1 . The foregoing says that
k
1 n
1
lim sup
Xi(ω) ≤ ,
n→∞ n
k
i=1
except possibly for ω lying in an event Bk, with P (Bk) = 0. Write B := ∪∞ B
k=1
k.
Then P (B) = 0 and
1 n
lim
Xi(ω) = 0
n→∞ n i=1
for each sample point ω /
∈ B.
FLUCTUATIONS, LAPLACE–DE MOIVRE THEOREM. The Strong
Law of Large Numbers says that for almost every sample point ω ∈ Ω,
X1(ω) + · · · + Xn(ω)
n
→ m as n → ∞.
We turn next to the Laplace–De Moivre Theorem, and its generalization the Central
Limit Theorem, which estimate the “fluctuations” we can expect in this limit.
Let us start with a simple calculation.
LEMMA. Suppose the real–valued random variables X1, . . . , Xn, . . . are inde-
pendent and identically distributed, with
P (Xi = 1) = p
P (Xi = 0) = q
for p, q ≥ 0, p + q = 1. Then
E(X1 + · · · + Xn) = np
V (X1 + · · · + Xn) = npq.
Proof. E(X1) =
X
Ω
1 dP = p and therefore E(X1 + · · · + Xn) = np. Also,
V (X1) =
(X1 − p)2 dP = (1 − p)2P(X1 = 1) + p2P(X1 = 0)
Ω
= q2p + p2q = qp.
By independence, V (X1 + · · · + Xn) = V (X1) + · · · + V (Xn) = npq.
We can imagine these random variables as modeling for example repeated tosses
of a biased coin, which has probability p of coming up heads, and probability q =
1 − p of coming up tails.
23
THEOREM (Laplace–De Moivre). Let X1, . . . , Xn be the independent,
identically distributed, real–valued random variables in the preceding Lemma. Define
the sums
Sn := X1 + · · · + Xn.
Then for all −∞ < a < b < +∞,
S
1
b
lim P
a
n − np
√
e− x2
2 dx.
n→∞
≤ √npq ≤ b = 2π a
A proof is in Appendix A.
Interpretation of the Laplace–De Moivre Theorem. In view of the
Lemma,
Sn − np S
√
=
n − E(Sn).
npq
V (Sn)1/2
Hence the Laplace–De Moivre Theorem says that the sums Sn, properly renormal-
ized, have a distribution which tends to the Gaussian N (0, 1) as n → ∞.
Consider in particular the situation p = q = 1 . Suppose a > 0; then
2
a√n
n
a√n
1
a
lim P
= √
e− x2
2
dx.
n→∞
− 2 ≤ Sn − 2 ≤ 2
2π −a
If we fix b > 0 and write a = 2b
√ , then for large n
n
2b
n
1
√n
P −b ≤ Sn −
e− x2
2 dx
2 ≤ b ≈ √2π − 2b
√n
→0 as n→∞.
Thus for almost every ω, 1 S
, in accord with the Strong Law of Large
n
n(ω) → 12
Numbers; but Sn(ω) − n “fluctuates” with probability 1 to exceed any finite bound
2
b.
CENTRAL LIMIT THEOREM. We now generalize the Laplace–De Moivre
Theorem:
THEOREM (Central Limit Theorem). Let X1, . . . , Xn, . . . be indepen-
dent, identically distributed, real-valued random variables with
E(Xi) = m, V (Xi) = σ2 > 0.
for i = 1, . . . . Set
Sn := X1 + · · · + Xn.
Then for all −∞ < a < b < +∞
S
1
b
(1)
lim P
a
n − nm
√
√
e− x22 dx.
n→∞
≤
nσ
≤ b = 2π a
24
Thus the conclusion of the Laplace–De Moivre Theorem holds not only for
the 0– or 1–valued random variable considered before, but for any sequence of
independent, identically distributed random variables with finite variance. We will
later invoke this assertion to motivate our requirement that Brownian motion be
normally distributed for each time t ≥ 0.
Outline of Proof. For simplicity assume m = 0, σ = 1, since we can always
rescale to this case. Then
λ
n
φ Sn (λ) = φ X (λ) . . . φ X (λ) =
φ
√
1
n
X1
√
n
√n
√n
n
for λ ∈ R, because the random variables are independent and identically distributed.
Now φ = φX satisfies
1
1
φ(µ) = φ(0) + φ′(0)µ + φ′ (0)µ2 + o(µ2)
as µ
2
→ 0,
with φ(0) = 1, φ′(0) = iE(X1) = 0, φ′′(0) = −E(X21) = −1. Consequently our
setting µ = λ
√ gives
n
λ
λ2
λ2
φX
= 1
+ o
,
1
√n
− 2n
n
and so
λ2
λ2
n
φ Sn (λ) =
1
+ o
√
−
→ e−λ22
n
2n
n
for all λ, as n → ∞. But e−λ22 is the characteristic function of an N(0, 1) random
variable. It turns out that this convergence of the characteristic functions implies
the limit (1): see Breiman [B] for more.
H. CONDITIONAL EXPECTATION
MOTIVATION. We earlier decided to define P (A | B), the probability of A,
given B, to be P (A∩B) , provided P (B) > 0. How then should we define
P (B)
E(X | B),
the expected value of the random variable X, given the event B? Remember that
we can think of B as the new probability space, with ˜
P =
P
. Thus if P (B) > 0,
P (B)
we should set
E(X | B) = mean value of X over B
1
=
X dP.
P (B) B
Next we pose a more interesting question. What is a reasonable definition of
E(X | Y ),
25
the expected value of the random variable X, given another random variable Y ? In
other words if “chance” selects a sample point ω ∈ Ω and all we know about ω is
the value Y (ω), what is our best guess as to the value X(ω)?
This turns out to be a subtle, but extremely important issue, for which we
provide two introductory discussions.
FIRST APPROACH TO CONDITIONAL EXPECTATION. We start with
an example.
Example. Assume we are given a probability space (Ω, U, P), on which is
defined a simple random variable Y . That is, Y =
m
a
, and so
i=1
iχAi
a
1 onA1
a2
on A2
Y =
.
.
.
am onAm,
for distinct real numbers a1, a2, . . . , am and disjoint events A1, A2, . . . , Am, each of
positive probability, whose union is Ω.
Next, let X be any other real–valued random variable on Ω. What is our best
guess of X, given Y ? Think about the problem this way: if we know the value
of Y (ω), we can tell which event A1, A2, . . . , Am contains ω. This, and only this,
known, our best estimate for X should then be the average value of X over each
appropriate event. That is, we should take
1 XdP on A
P(A
1
1) A1
1
X dP
on A
P (A
2
2)
A2
E(X | Y ) :=
.
.
.
1
X dP
on A
P (A
m.
m)
Am
We note for this example that
• E(X | Y ) is a random variable, and not a constant.
• E(X | Y ) is U(Y )-measurable.
• XdP = E(X
A
A
| Y )dP for all A ∈ U(Y ).
Let us take these properties as the definition in the general case:
DEFINITION. Let Y be a random variable. Then E(X | Y ) is any U(Y )-
measurable random variable such that
X dP =
E(X | Y ) dP for all A ∈ U(Y ).
A
A
Finally, notice that it is not really the values of Y that are important, but
rather just the σ-algebra it generates. This motivates the next
26
DEFINITION. Let (Ω, U, P) be a probability space and suppose V is a σ-
algebra, V ⊆ U. If X : Ω → Rn is an integrable random variable, we define
E(X | V)
to be any random variable on Ω such that
(i) E(X | V) is V-measurable, and
(ii)
X dP =
E(X
A
A
| V)dP for all A ∈ V.
Interpretation. We can understand E(X | V) as follows. We are given the “in-
formation” available in a σ-algebra V, from which we intend to build an estimate
of the random variable X. Condition (i) in the definition requires that E(X | V)
be constructed from the information in V, and (ii) requires that our estimate be
consistent with X, at least as regards integration over events in V. We will later see
that the conditional expectation E(X | V), so defined, has various additional nice
properties.
Remark. We can check without difficulty that
(i) E(X | Y ) = E(X | U(Y )).
(ii) E(E(X | V)) = E(X).
(iii) E(X) = E(X | W), where W = {∅, Ω} is the trivial σ-algebra.
THEOREM. Let X be an integrable random variable. Then for each σ-algebra
V ⊂ U, the conditional expectation E(X | V) exists and is unique up to V-measurable
sets of probability zero.
We omit the proof, which uses a few advanced concepts from measure theory.
SECOND APPROACH TO CONDITIONAL EXPECTATION. An ele-
gant alternative approach to conditional expectations is based upon projections
onto closed subspaces, and is motivated by this example:
Least squares method. Consider for the moment Rn and suppose that V is
a proper subspace.
Suppose we are given a vector x ∈ Rn. The least squares problem asks us to
find a vector z ∈ V so that
|z − x| = min|y − x|.
y∈V
It is not particularly difficult to show that, given x, there exists a unique vector
z ∈ V solving this minimization problem. We call v the projection of x onto V ,
(7)
z = projV (x).
Now we want to find formula characterizing z. For this take any other vector
w ∈ V . Define then
i(τ ) := |z + τw − x|2.
Since z + τ w ∈ V for all τ, we see that the function i(·) has a minimum at τ = 0.
Hence 0 = i′(0) = 2(z − x) · w; that is,
(8)
x · w = z · w for all w ∈ V.
27
x
z=projV(x)
0
V
The geometric interpretation is that the “error” x − z is perpendicular to the sub-
space V .
Projection of random variables. Motivated by the example above, we re-
turn now to conditional expectation. Let us take the linear space L2(Ω) = L2(Ω, U),
which consists of all real-valued, U–measurable random variables Y , such that
1
2
| Y | :=
Y 2 dP
< ∞.
Ω
We call | Y | the norm of Y ; and if X, Y ∈ L2(Ω), we define their inner product to
be
(X, Y ) :=
XY dP = E(XY ).
Ω
Next, take as before V to be a σ-algebra contained in U. Consider then
V := L2(Ω, V),
the space of square–integrable random variables that are V–measurable. This is a
closed subspace of L2(Ω). Consequently if X ∈ L2(Ω), we can define its projection
(9)
Z = projV (X),
by analogy with (7) in the finite dimensional case. Almost exactly as we established
(8) above, we can likewise show
(X, W ) = (Z, W ) for all W ∈ V.
Take in particular W = χA for any set A ∈ V. In view of the definition of the inner
product, it follows that
X dP =
Z dP
for all A ∈ V.
A
A
28
Since Z ∈ V is V-measurable, we see that Z is in fact E(X | V), as defined in the
earlier discussion. That is,
E(X | V) = projV (X).
We could therefore alternatively take the last identity as a definition of condi-
tional expectation. This point of view also makes it clear that Z = E(X | V) solves
the least squares problem:
| Z − X| = min | Y − X| ;
Y ∈V
and so E(X | V) can be interpreted as that V-measurable random variable which is
the best least squares approximation of the random variable X.
The two introductory discussions now completed, we turn next to examining
conditional expectation more closely.
THEOREM (Properties of conditional expectation).
(i) If X is V-measurable, then E(X | V) = X a.s.
(ii) If a, b are constants, E(aX + bY | V) = aE(X | V) + bE(Y | V) a.s.
(iii) If X is V-measurable and XY is integrable, then E(XY | V) = XE(Y | V)
a.s.
(iv) If X is independent of V, then E(X | V) = E(X) a.s.
(v) If W ⊆ V, we have
E(X | W) = E(E(X | V) | W) = E(E(X | W) | V) a.s.
(vi) The inequality X ≤ Y a.s. implies E(X | V) ≤ E(Y | V) a.s.
Proof.
1. Statement (i) is obvious, and (ii) is easy to check
2. By uniqueness a.s. of E(XY | V), it is enough in proving (iii) to show
(10)
XE(Y | V) dP = XY dP for all A ∈ V.
A
A
First suppose X =
m
b
, where B
i=1 iχB
i
i
∈ V for i = 1, . . ., m. Then
m
XE(Y | V) dP =
bi
E(Y | V) dP
A
i=1
A∩Bi
∈V
m
=
bi
Y dP =
XY dP.
i=1
A∩Bi
A
This proves (10) if X is a simple function. The general case follows by approxima-
tion.
29
3. To show (iv), it suffices to prove
E(X) dP =
X dP for all A
A
A
∈ V. Let
us compute:
X dP =
χAX dP = E(χAX) = E(X)P (A) =
E(X) dP,
A
Ω
A
the third equality owing to independence.
4. Assume W ⊆ V and let A ∈ W. Then
E(E(X | V) | W) dP = E(X | V) dP = X dP,
A
A
A
since A ∈ W ⊆ V. Thus E(X | W) = E(E(X | V) | W) a.s.
Furthermore, assertion (i) implies that E(E(X | W) | V) = E(X | W), since
E(X | W) is W-measurable and so also V-measurable. This establishes assertion
(v).
5. Finally, suppose X ≤ Y , and note that
E(Y | V) − E(X | V) dP = E(Y − X | V) dP
A
A
=
Y − X dP ≥ 0
A
for all A ∈ V. Take A := {E(Y | V) − E(X | V) ≤ 0}. This event lies in V, and we
deduce from the previous inequality that P (A) = 0.
LEMMA (Conditional Jensen’s Inequality). Suppose Φ : R → R is con-
vex, with E(|Φ(X)|) < ∞. Then
Φ(E(X | V)) ≤ E(Φ(X) | V).
We leave the proof as an exercise.
I. MARTINGALES
MOTIVATION. Suppose Y1, Y2, . . . are independent real-valued random vari-
ables, with
E(Yi) = 0 (i = 1, 2, . . . ).
Define the sum Sn := Y1 + · · · + Yn.
What is our best guess of Sn+k, given the values of S1, . . . , Sn? The answer is
E(Sn+k | S1, . . ., Sn) = E(Y1 + · · · + Yn | S1, . . ., Sn)
+ E(Y
(11)
n+1 + · · · + Yn+k | S1, . . . , Sn)
= Y1 + · · · + Yn + E(Yn+1 + · · · + Yn+k) = Sn.
=0
Thus the best estimate of the “future value” of Sn+k, given the history up to time
n, is just Sn.
If we interpret Yi as the payoff of a “fair” gambling game at time i, and therefore
Sn as the total winnings at time n, the calculation above says that at any time one’s
30
future expected winnings, given the winnings to date, is just the current amount of
money. So the formula (11) characterizes a “fair” game.
We incorporate these ideas into a formal definition:
DEFINITION. Let X1, . . . , Xn, . . . be a sequence of real-valued random vari-
ables, with E(|Xi|) < ∞ (i = 1, 2, . . .). If
Xk = E(Xj | X1, . . ., Xk) a.s. for all j ≥ k,
we call {Xi}∞i=1 a (discrete) martingale.
DEFINITION. Let X(·) be a real–valued stochastic process. Then
U(t) := U(X(s)| 0 ≤ s ≤ t),
the σ-algebra generated by the random variables X(s) for 0 ≤ s ≤ t, is called the
history of the process until (and including) time t ≥ 0.
DEFINITIONS. Let X(·) be a stochastic process, such that E(|X(t)|) < ∞
for all t ≥ 0.
(i) If
X(s) = E(X(t) | U(s)) a.s. for all t ≥ s ≥ 0,
then X(·) is called a martingale.
(ii) If
X(s) ≤ E(X(t) | U(s)) a.s. for all t ≥ s ≥ 0,
X(·) is a submartingale.
Example. Let W (·) be a 1-dimensional Wiener process, as defined later in
Chapter 3. Then
W (·) is a martingale.
To see this, write W(t) := U(W(s)| 0 ≤ s ≤ t), and let t ≥ s. Then
E(W (t) | W(s)) = E(W(t) − W(s) | W(s)) + E(W(s) | W(s))
= E(W (t) − W(s)) + W(s) = W(s) a.s.
(The reader should refer back to this calculation after reading Chapter 3.)
LEMMA. Suppose X(·) is a real-valued martingale and Φ : R → R is convex.
Then if E(|Φ(X(t))|) < ∞ for all t ≥ 0,
Φ(X(·)) is a submartingale.
We omit the proof, which uses Jensen’s inequality.
Martingales are important in probability theory mainly because they admit the
following powerful estimates:
31
THEOREM (Discrete martingale inequalities).
(i) If {Xn}∞n=1 is a submartingale, then
1
P
max Xk ≥ λ ≤ E(X+n)
1≤k≤n
λ
for all n = 1, . . . and λ > 0.
(ii) If {Xn}∞n=1 is a martingale and 1 < p < ∞, then
p
p
E
max |Xk|p ≤
E(|Xn|p)
1≤k≤n
p − 1
for all n = 1, . . . .
A proof is provided in Appendix B. Notice that (i) is a generalization of the
Chebyshev inequality. We can also extend these estimates to continuous–time mar-
tingales.
THEOREM (Martingale inequalities). Let X(·) be a stochastic process
with continuous sample paths a.s.
(i) If X(·) is a submartingale, then
1
P
max X(s) ≥ λ ≤ E(X(t)+) for all λ > 0, t ≥ 0.
0≤s≤t
λ
(ii) If X(·) is a martingale and 1 < p < ∞, then
p
p
E
max |X(s)|p ≤
E(|X(t)|p).
0≤s≤t
p − 1
Outline of Proof. Choose λ > 0, t > 0 and select 0 = t0 < t1 < · · · <
tn = t. We check that {X(ti)}ni=1 is a martingale and apply the discrete martingale
inequality. Next choose a finer and finer partition of [0, t] and pass to limits.
The proof of assertion (ii) is similar.
32
CHAPTER 3: BROWNIAN MOTION AND “WHITE NOISE”
A. Motivation and definitions
B. Construction of Brownian motion
C. Sample paths
D. Markov property
A. MOTIVATION AND DEFINITIONS
SOME HISTORY. R. Brown in 1826–27 observed the irregular motion of pollen
particles suspended in water. He and others noted that
• the path of a given particle is very irregular, having a tangent at no point,
and
• the motions of two distinct particles appear to be independent.
In 1900 L. Bachelier attempted to describe fluctuations in stock prices mathe-
matically and essentially discovered first certain results later rederived and extended
by A. Einstein in 1905. Einstein studied the Brownian phenomena this way. Let us
consider a long, thin tube filled with clear water, into which we inject at time t = 0
a unit amount of ink, at the location x = 0. Now let f (x, t) denote the density of
ink particles at position x ∈ R and time t ≥ 0. Initially we have
f (x, 0) = δ0, the unit mass at 0.
Next, suppose that the probability density of the event that an ink particle moves
from x to x + y in (small) time τ is ρ(τ, y). Then
∞
f (x, t + τ ) =
f (x − y, t)ρ(τ, y) dy
(1)
−∞
∞
1
=
f − fxy + fxxy2 + . . . ρ(τ, y) dy.
−∞
2
But since ρ is a probability density,
∞ ρ dy = 1; whereas ρ(τ,
−∞
−y) = ρ(τ, y) by
symmetry. Consequently
∞ yρ dy = 0. We further assume that ∞ y2ρ dy, the
−∞
−∞
variance of ρ, is linear in τ :
∞ y2ρdy = Dτ, D > 0.
−∞
We insert these identities into (1), thereby to obtain
f (x, t + τ ) − f(x, t) Df
=
xx(x, t)
τ
2
{+ higher order terms}.
Sending now τ → 0, we discover
D
ft =
f
2 xx
This is the diffusion equation, also known as the heat equation. This partial
differential equation, with the initial condition f (x, 0) = δ0, has the solution
1
f (x, t) =
e− x2
2Dt .
(2πDt)1/2
33
This says the probability density at time t is N (0, Dt), for some constant D.
In fact, Einstein computed:
R=gasconstant
RT
T = absolute temperature
D =
,
where
N
Af
f =frictioncoefficient
NA =Avogadro’snumber.
This equation and the observed properties of Brownian motion allowed J. Perrin to
compute NA (≈ 6×1023 = the number of molecules in a mole) and help to confirm
the atomic theory of matter.
N. Wiener in the 1920’s (and later) put the theory on a firm mathematical
basis. His ideas are at the heart of the mathematics in §B–D below.
RANDOM WALKS. A variant of Einstein’s argument follows. We intro-
duce a 2-dimensional rectangular lattice, comprising the sites {(m∆x, n∆t) | m =
0, ±1, ±2, . . .; n = 0, 1, 2, . . .}. Consider a particle starting at x = 0 and time t = 0,
and at each time n∆t moves to the left an amount ∆x with probability 1/2, to the
right an amount ∆x with probability 1/2. Let p(m, n) denote the probability that
the particle is at position m∆x at time n∆t. Then
0
m = 0
p(m, 0) =
1
m = 0.
Also
1
1
p(m, n + 1) =
p(m
p(m + 1, n),
2
− 1, n) + 2
and hence
1
p(m, n + 1) − p(m, n) = (p(m + 1, n)
2
− 2p(m, n) + p(m − 1, n)).
Now assume
(∆x)2 = D for some positive constant D.
∆t
This implies
p(m, n + 1) − p(m, n) D p(m + 1, n)
=
− 2p(m, n) + p(m − 1, n) .
∆t
2
(∆x)2
Let ∆t → 0, ∆x → 0, m∆x → x, n∆t → t, with (∆x)2
∆t
≡ D. Then presumably
p(m, n) → f(x, t), which we now interpret as the probability density that particle
is at x at time t. The above difference equation becomes formally in the limit
D
ft =
f
2 xx,
and so we arrive at the diffusion equation again.
34
MATHEMATICAL JUSTIFICATION. A more careful study of this tech-
nique of passing to limits with random walks on a lattice depends upon the Laplace–
De Moivre Theorem.
As above we assume the particle moves to the left or right a distance ∆x with
probability 1/2. Let X(t) denote the position of particle at time t = n∆t (n =
0, . . . ). Define
n
Sn :=
Xi,
i=1
where the Xi are independent random variables such that
P (Xi = 0) = 1/2
P (Xi = 1) = 1/2
for i = 1, . . . . Then V (Xi) = 1 .
4
Now Sn is the number of moves to the right by time t = n∆t. Consequently
X(t) = Sn∆x + (n − Sn)(−∆x) = (2Sn − n)∆x.
Note also
V (X(t)) = (∆x)2V (2Sn − n)
= (∆x)24V (Sn) = (∆x)24nV (X1)
(∆x)2
= (∆x)2n =
t.
∆t
Again assume (∆x)2 = D. Then
∆t
Sn
Sn
√
X(t) = (2S
− n2 √
− n2
n − n)∆x =
n∆x =
tD.
n
n
4
4
The Laplace–De Moivre Theorem thus implies
a
Sn
b
lim
P (a
− n2
n→∞
≤ X(t) ≤ b) = lim √
√
n→∞
tD ≤
n
≤ tD
4
t=n∆t, (∆x)2 =D
∆t
b
1
√tD
= √
e− x2
2 dx
2π
a
√tD
1
b
= √
e− x2
2Dt dx.
2πDt a
Once again, and rigorously this time, we obtain the N (0, Dt) distribution.
Inspired by all these considerations, we now introduce Brownian motion, for
which we take D = 1:
35
DEFINITION. A real-valued stochastic process W (·) is called a Brownian
motion or Wiener process if
(i) W (0) = 0 a.s.,
(ii) W (t) − W(s) is N(0, t − s) for all t ≥ s ≥ 0,
(iii) for all times 0 < t1 < t2 < · · · < tn, the random variables W(t1), W(t2) −
W (t1), . . . , W (tn) −W(tn−1) are independent (“independent increments”).
Notice in particular that
E(W (t)) = 0, E(W 2(t)) = t
for each time t ≥ 0.
The Central Limit Theorem provides some further motivation for our definition
of Brownian motion, since we can expect that any suitably scaled sum of indepen-
dent, random disturbances affecting the position of a moving particle will result in
a Gaussian distribution.
B. CONSTRUCTION OF BROWNIAN MOTION
COMPUTATION OF JOINT PROBABILITIES. From the definition
we know that if W (·) is a Brownian motion, then for all t > 0 and a ≤ b,
1
b
P (a ≤ W(t) ≤ b) = √
e− x2
2t dx,
2πt a
since W (t) is N (0, t).
Suppose we now choose times 0 < t1 < · · · < tn and real numbers ai ≤ bi, for
i = 1, . . . , n. What is the joint probability
P (a1 ≤ W(t1) ≤ b1, · · · , an ≤ W(tn) ≤ bn)?
In other words, what is the probability that a sample path of Brownian motion
takes values between ai and bi at time ti for each i = 1, . . . n?
a3
a5
a
a
1
2
b3
a4
b2
b
b
4
5
b1
t
t
1
2
t3
t
t
4
5
We can guess the answer as follows. We know
x2
b
1
1
e− 2t1
P (a1 ≤ W(t1) ≤ b1) =
√
dx1;
a
2πt
1
1
36
and given that W (t1) = x1, a1 ≤ x1 ≤ b1, then presumably the process is N(x1, t2−
t1) on the interval [t1, t2]. Thus the probability that a2 ≤ W(t2) ≤ b1, given that
W (t1) = x1, should equal
b2
1
e− |x2−x1|2
2(t2−t1) dx2.
a
2π(t
2
2 − t1)
Hence it should be that
b1
b2
P (a1 ≤ W(t1) ≤ b1, a2 ≤ W(t2) ≤ b2) =
g(x1, t1 | 0)g(x2, t2−t1 | x1) dx2dx1
a1
a2
for
1
g(x, t | y) := √ e−(x−y)2
2t
.
2πt
In general, we would therefore guess that
(2)
P (a1 ≤ W(t1) ≤ b1, . . ., an ≤ W(tn) ≤ bn) =
b1
bn
· · ·
g(x1, t1 | 0)g(x2, t2 − t1 | x1) . . .g(xn, tn − tn−1 | xn−1) dxn . . .dx1.
a1
an
The next assertion confirms and extends this formula.
THEOREM. Let W (·) be a one-dimensional Wiener process. Then for all
positive integers n, all choices of times 0 = t0 < t1 < · · · < tn and each function
f : Rn → R, we have
∞
∞
Ef (W (t1), . . . , W (tn)) =
· · ·
f (x1, . . . , xn)g(x1, t1 | 0)g(x2, t2 − t1 | x1)
−∞
−∞
. . . g(xn, tn − tn−1 | xn−1) dxn . . .dx1.
Our taking
f (x1, . . . , xn) = χ
(x
(x
[a
1)
n)
1 ,b1]
· · ·χ[an,bn]
gives (2).
Proof. Let us write Xi := W (ti), Yi := Xi − Xi−1 for i = 1, . . ., n. We also
define
h(y1, y2, . . . , yn) := f (y1, y1 + y2, . . . , y1 + · · · + yn).
37
Then
Ef (W (t1), . . . , W (tn)) = Eh(Y1, . . . , Yn)
∞
∞
=
· · ·
h(y1, . . . , yn)g(y1, t1 | 0)g(y2, t2 − t1 | 0)
−∞
−∞
. . . g(yn, tn − tn−1 | 0)dyn . . .dy1
∞
∞
=
· · ·
f (x1, . . . , xn)g(x1, t1 | 0)g(x2, t2 − t1 | x1)
−∞
−∞
. . . g(xn, tn − tn−1 | xn−1) dxn . . .dx1.
For the second equality we recalled that the random variables Yi = W (ti)−W(ti−1)
are independent for i = 1, . . . , n, and that each Yi is N (0, ti−ti−1). We also changed
variables using the identities yi = xi − xi−1 for i = 1, . . ., n and x0 = 0. The
Jacobian for this change of variables equals 1.
BUILDING A ONE-DIMENSIONAL WIENER PROCESS. The main
issue now is to demonstrate that a Brownian motion actually exists.
Our method will be to develop a formal expansion of white noise ξ(·) in terms
of a cleverly selected orthonormal basis of L2(0, 1), the space of all real-valued,
square–integrable funtions defined on (0, 1) . We will then integrate the resulting
expression in time, show that this series converges, and prove then that we have
built a Wiener process. This procedure is a form of “wavelet analysis”: see Pinsky
[P].
We start with an easy lemma.
LEMMA. Suppose W (·) is a one-dimensional Brownian motion. Then
E(W (t)W (s)) = t ∧ s = min{s, t} for t ≥ 0, s ≥ 0.
Proof. Assume t ≥ s ≥ 0. Then
E(W (t)W (s)) = E((W (s) + W (t) − W(s))W(s))
= E(W 2(s)) + E((W (t) − W(s))W(s))
= s + E(W (t) − W(s))E(W(s))
=0
=0
= s = t ∧ s,
since W (s) is N (0, s) and W (t) − W(s) is independent of W(s).
HEURISTICS. Remember from Chapter 1 that the formal time-derivative
˙
dW (t)
W (t) =
= ξ(t)
dt
is “1-dimensional white noise”. As we will see later however, for a.e. ω the sample
path t → W(t, ω) is in fact differentiable for no time t ≥ 0. Thus ˙W(t) = ξ(t) does
not really exist.
38
However, we do have the heuristic formula
(3)
“E(ξ(t)ξ(s)) = δ0(s − t)”,
where δ0 is the unit mass at 0. A formal “proof” is this. Suppose h > 0, fix t > 0,
and set
W (t + h)
W (s + h)
φ
− W(t)
− W(s)
h(s) := E
h
h
1
=
[E(W (t + h)W (s + h))
h2
− E(W(t + h)W(s)) − E(W(t)W(s + h)) + E(W(t)W(s))]
1
=
[((t + h)
h2
∧ (s + h)) − ((t + h) ∧ s) − (t ∧ (s + h)) + (t ∧ s)].
graph of φh
height = 1/h
t-h
t
t+h
Then φh(s) → 0 as h → 0, t = s. But φh(s) ds = 1, and so presumably
φh(s) → δ0(s − t) in some sense, as h → 0. In addition, we expect that φh(s) →
E(ξ(t)ξ(s)). This gives the formula (3) above.
Remark: Why ˙
W( · ) = ξ( · ) is called white noise. If X(·) is any real-valued
stochastic process with E(X2(t)) < ∞ for all t ≥ 0, we define
r(t, s) := E(X(t)X(s)) (t, s ≥ 0),
the autocorrelation function of X(·). If r(t, s) = c(t−s) for some function c : R → R
and if E(X(t)) = E(X(s)) for all t, s ≥ 0, X(·) is called stationary in the wide
sense. A white noise process ξ(·) is by definition Gaussian, wide sense stationary,
with c(·) = δ0.
In general we define
1
∞
f (λ) :=
e−iλtc(t) dt (λ
2π
∈ R)
−∞
to be the spectral density of a process X(·). For white noise, we have
1
∞
1
f (λ) =
e−iλtδ
for all λ.
2π
0 dt =
−∞
2π
39
Thus the spectral density of ξ(·) is flat; that is, all “frequencies” contribute equally
in the correlation function, just as—by analogy—all colors contribute equally to
make white light.
RANDOM FOURIER SERIES. Suppose now {ψn}∞n=0 is a complete, or-
thonormal basis of L2(0, 1), where ψn = ψn(t) are functions of 0 ≤ t ≤ 1 only and
so are not random variables. The orthonormality means that
1
ψn(s)ψm(s) ds = δmn
for all m, n.
0
We write formally
∞
(4)
ξ(t) =
Anψn(t) (0 ≤ t ≤ 1).
n=0
It is easy to see that then
1
An =
ξ(t)ψn(t) dt.
0
We expect that the An are independent and Gaussian, with E(An) = 0. Therefore
to be consistent we must have for m = n
1
1
0 = E(An)E(Am) = E(AnAm) =
E(ξ(t)ξ(s))ψn(t)ψm(s) dtds
0
0
1
1
=
δ0(s − t)ψn(t)ψm(s) dtds by (3)
0
0
1
=
ψn(s)ψm(s) ds.
0
But this is already automatically true as the ψn are orthogonal. Similarly,
1
E(A2n) =
ψ2n(s) ds = 1.
0
Consequently if the An are independent and N (0, 1), it is reasonable to believe that
formula (4) makes sense. But then the Brownian motion W (·) should be given by
t
∞
t
(5)
W (t) :=
ξ(s) ds =
An
ψn(s) ds.
0
n=0
0
This seems to be true for any orthonormal basis, and we will next make this rigorous
by choosing a particularly nice basis.
L´
EVY–CIESIELSKI CONSTRUCTION OF BROWNIAN MOTION
40
DEFINITION. The family {hk(·)}∞ of Haar functions are defined for 0
k=0
≤
t ≤ 1 as follows:
h0(t) := 1 for 0 ≤ t ≤ 1.
1 for 0
h
≤ t ≤ 12
1(t) :=
−1 for 1 < t
2
≤ 1.
If 2n ≤ k < 2n+1, n = 1, 2, . . ., we set
2n/2 for k−2n2n ≤t≤ k−2n+1/2
2n
hk(t) := −2n/2 for k−2n+1/2 < t ≤ k−2n+1
2n
2n
0 otherwise.
graph of hk
height = 2n/2
width = 2-(n+1)
Graph of a Haar function
LEMMA 1. The functions {hk(·)}∞ form a complete, orthonormal basis of
k=0
L2(0, 1).
Proof. 1. We have
1 h2 dt = 2n
1
+
1
= 1.
0
k
2n+1
2n+1
Note also that for all l > k, either hkhl = 0 for all t or else hk is constant on
the support of hl. In this second case
1
1
hlhk dt = ±2n/2
hl dt = 0.
0
0
2. Suppose f ∈ L2(0, 1), 1 fh
0
k dt = 0 for all k = 0, 1, . . . . We will prove f = 0
almost everywhere.
If n = 0, we have
1 f dt = 0. Let n = 1. Then 1/2 f dt = 1 f dt; and both
0
0
1/2
are equal to zero, since 0 = 1/2 f dt + 1 f dt = 1 f dt. Continuing in this way,
0
1/2
0
k+1
we deduce
2n+1 f dt = 0 for all 0
f dt = 0 for all dyadic
k
≤ k < 2n+1. Thus rs
2n+1
41
rationals 0 ≤ s ≤ r ≤ 1, and so for all 0 ≤ s ≤ r ≤ 1. But
d
r
f (r) =
f (t) dt = 0
a.e. r.
dr 0
DEFINITION. For k = 0, 1, 2, . . . ,
t
sk(t) :=
hk(s) ds (0 ≤ t ≤ 1)
0
is the kth–Schauder function.
height = 2-(n+2)/2
graph of sk
width = 2-n
Graph of a Schauder function
The graph of sk is a “tent” of height 2−n/2−1, lying above the interval [ k−2n , k−2n+1 ].
2n
2n
Consequently if 2n ≤ k < 2n+1, then
max |sk(t)| = 2−n/2−1.
0≤t≤1
Our goal is to define
∞
W (t) :=
Aksk(t)
k=0
for times 0 ≤ t ≤ 1, where the coefficients {Ak}∞ are independent, N(0, 1) random
k=0
variables defined on some probability space.
We must first of all check whether this series converges.
LEMMA 2. Let {ak}∞ be a sequence of real numbers such that
k=0
|ak| = O(kδ) as k → ∞
for some 0 ≤ δ < 1/2. Then the series∞ aksk(t)
k=0
converges uniformly for 0 ≤ t ≤ 1.
42
Proof. Fix ε > 0. Notice that for 2n ≤ k < 2n+1, the functions sk(·) have
disjoint supports. Set
bn :=
max
|ak| ≤ C(2n+1)δ.
2n≤k<2n+1
Then for 0 ≤ t ≤ 1,
∞
∞
|ak| sk(t)| ≤
bn
max
|sk(t)|
2n
k=2m
n=m
≤k<2n+1
0≤t≤1
∞
≤ C
(2n+1)δ2−n/2−1 < ε
n=m
for m large enough, since 0 ≤ δ < 1/2.
LEMMA 3. Suppose {Ak}∞ are independent, N(0, 1) random variables. Then
k=1
for almost every ω,
|Ak(ω)| = O( logk) as k → ∞.
In particular, the numbers {Ak(ω)}∞ almost surely satisfy the hypothesis of
k=1
Lemma 2 above.
Proof. For all x > 0, k = 2, . . . , we have
2
∞
P (|Ak| > x) = √
e− s22 ds
2π x
2
∞
≤ √ e−x24
e− s24 ds
2π
x
≤ Ce−x24 ,
for some constant C. Set x := 4√log k; then
1
P (|Ak| ≥ 4 log k) ≤ Ce−4logk = C .
k4
Since
1 <
k4
∞, the Borel–Cantelli Lemma implies
P (|Ak| ≥ 4 log k i.o.) = 0.
Therefore for almost every sample point ω, we have
|Ak(ω)| ≤ 4 log k provided k ≥ K,
where K depends on ω.
LEMMA 4.
∞ s
k=0 k(s)sk(t) = t ∧ s for each 0 ≤ s, t ≤ 1.
43
Proof. Define for 0 ≤ s ≤ 1,
1 0
φ
≤ τ ≤ s
s(τ ) :=
0 s < τ ≤ 1.
Then if s ≤ t, Lemma 1 implies
1
∞
s =
φtφs dτ =
akbk,
0
k=0
where
1
t
1
ak =
φthk dτ =
hk dτ = sk(t), bk =
φshk dτ = sk(s).
0
0
0
THEOREM. Let {Ak}∞ be a sequence of independent, N(0, 1) random vari-
k=0
ables defined on the same probability space. Then the sum
∞
W (t, ω) :=
Ak(ω)sk(t) ( 0 ≤ t ≤ 1)
k=0
converges uniformly in t, for a.e. ω. Furthermore
(i) W (·) is a Brownian motion for 0 ≤ t ≤ 1, and
(ii) for a.e. ω, the sample path t → W(t, ω) is continuous.
Proof. 1. The uniform convergence is a consequence of Lemmas 2 and 3; this
implies (ii).
2. To prove W (·) is a Brownian motion, we first note that clearly W(0) = 0
a.s.
We assert as well that W (t) − W(s) is N(0, t − s) for all 0 ≤ s ≤ t ≤ 1. To
prove this, let us compute
E(eiλ(W (t)−W (s))) = E(eiλ P∞ A
k=0
k (sk (t)−sk(s)))
∞
=
E(eiλAk(sk(t)−sk(s)))
by independence
k=0
∞
=
e− λ2 (s
2
k (t)−sk(s))2
since Ak is N (0, 1)
k=0
= e− λ2
(s
2 P∞
k=0
k (t)−sk(s))2
= e− λ2
s2 (t)
(s)
2 P∞
k=0
k
−2sk(t)sk(s)+s2k
= e− λ2 (t
2
−2s+s)
by Lemma 4
= e− λ2 (t
2
−s).
44
By uniqueness of characteristic functions, the increment W (t) −W(s) is N(0, t−s),
as asserted.
3. Next we claim for all m = 1, 2, . . . and for all 0 = t0 < t1 < · · · < tm ≤ 1,
that
m
λ2
j
(6)
E(ei Pm λ
(t
j=1
j (W (tj )−W (tj−1))) =
e− 2 j−tj−1).
j=1
Once this is proved, we will know from uniqueness of characteristic functions that
FW (t1),...,W(tm)−W(tm−1)(x1, . . ., xm) = FW(t1)(x1) · · ·FW(tm)−W(tm−1)(xm)
for all x1, . . . xm ∈ R. This proves that
W (t1), . . . , W (tm) − W(tm−1) are independent.
Thus (6) will establish the Theorem.
Now in the case m = 2, we have
E(ei[λ1W (t1)+λ2(W (t2)−W (t1))]) = E(ei[(λ1−λ2)W (t1)+λ2W (t2)])
= E(ei(λ1−λ2) P∞ A
A
k=0
k sk (t1 )+iλ2 P∞
k=0
k sk (t2 ))
∞
=
E(eiAk[(λ1−λ2)sk(t1)+λ2sk(t2)])
k=0
∞
=
e− 1 ((λ
2
1 −λ2)sk(t1)+λ2sk(t2))2
k=0
= e− 1
(λ
(t
s2 (t
2 P∞
k=0
1 −λ2)2s2k 1)+2(λ1−λ2)λ2sk(t1)sk(t2)+λ22 k 2)
= e− 1 [(λ
t
2
1 −λ2)2t1+2(λ1−λ2)λ2t1+λ22 2]
by Lemma 4
= e− 1 [λ2t
(t
2
1 1 +λ2
2
2 −t1)].
This is (6) for m = 2, and the general case follows similarly.
THEOREM (Existence of one-dimensional Brownian motion). Let
(Ω, U, P) be a probability space on which countably many N(0, 1), independent ran-
dom variables {An}∞n=1 are defined. Then there exists a 1-dimensional Brownian
motion W (·) defined for ω ∈ Ω, t ≥ 0.
Outline of proof. The theorem above demonstrated how to build a Brown-
ian motion on 0 ≤ t ≤ 1. As we can reindex the N(0, 1) random variables to obtain
countably many families of countably many random variables, we can therefore
build countably many independent Brownian motions W n(t) for 0 ≤ t ≤ 1.
We assemble these inductively by setting
W (t) := W (n − 1) + Wn(t − (n − 1)) for n − 1 ≤ t ≤ n.
Then W (·) is a one-dimensional Brownian motion, defined for all times t ≥ 0.
45
This theorem shows we can construct a Brownian motion defined on any prob-
ability space on which there exist countably many independent N (0, 1) random
variables.
We mostly followed Lamperti [L1] for the foregoing theory.
3. BROWNIAN MOTION IN Rn
It is straightforward to extend our definitions to Brownian motions taking values
in Rn.
DEFINITION. An Rn-valued stochastic process W(·) = (W1(·), . . ., Wn(·))
is an n-dimensional Wiener process (or Brownian motion) provided
(i) for each k = 1, . . . , n, W k(·) is a 1-dimensional Wiener process,
and
(ii) the σ-algebras Wk := U(Wk(t) | t ≥ 0) are independent, k = 1, . . ., n.
By the arguments above we can build a probability space and on it n inde-
pendent 1-dimensional Wiener processes W k(·) (k = 1, . . ., n). Then W(·) :=
(W 1(·), . . ., Wn(·)) is an n-dimensional Brownian motion.
LEMMA. If W(·) is an n-dimensional Wiener process, then
(i)
E(W k(t)W l(s)) = (t ∧ s)δkl (k, l = 1, . . ., n),
(ii) E((W k(t) −Wk(s))(Wl(t)−Wl(s))) = (t−s)δkl (k, l = 1, . . ., n; t ≥ s ≥ 0.)
Proof. If k = l, E(W k(t)W l(s)) = E(W k(t))E(W l(s)) = 0, by independence.
The proof of (ii) is similar.
THEOREM. (i) If W(·) is an n-dimensional Brownian motion, then W(t) is
N (0, tI) for each time t > 0. Therefore
1
P (W(t) ∈ A) =
e− |x|2
2t dx
(2πt)n/2 A
for each Borel subset A ⊆ Rn.
(ii) More generally, for each m = 1, 2, . . . and each function f : Rn × Rn ×
· · ·Rn → R, we have
(7)
Ef (W(t1), . . . , W(tm)) =
· · ·
f (x1, . . . , xm)g(x1, t1 | 0)g(x2, t2 − t1 | x1)
Rn
Rn
. . . g(xm, tm − tm−1 | xm−1) dxm . . .dx1.
where
1
g(x, t | y) :=
e− |x−y|2
2t
.
(2πt)n/2
46
Proof. For each time t > 0, the random variables W 1(t), . . . , W n(t) are inde-
pendent. Consequently for each point x = (x1, . . . , xn) ∈ Rn, we have
fW(t)(x1, . . . , xn) = fW 1(t)(x1) · · ·fWn(t)(xn)
1
x2
1
1
x2
n
=
e− 2t
e− 2t
(2πt)1/2
· · · (2πt)1/2
1
=
e− |x|2
2t
= g(x, t
(2πt)n/2
| 0).
We prove formula (7) as in the one-dimensional case.
C. SAMPLE PATH PROPERTIES
In this section we will demonstrate that for almost every ω, the sample path
t → W(t, ω) is uniformly H¨older continuous for each exponent γ < 1, but is nowhere
2
H¨older continuous with any exponent γ > 1 . In particular t
2
→ W(t, ω) almost
surely is nowhere differentiable and is of infinite variation for each time interval.
DEFINITIONS. (i) Let 0 < γ ≤ 1. A function f : [0, T] → R is called
uniformly H¨
older continuous with exponent γ > 0 if there exists a constant K such
that
|f(t) − f(s)| ≤ K|t − s|γ for all s, t ∈ [0, T].
(ii) We say f is H¨older continuous with exponent γ > 0 at the point s if there
exists a constant K such that
|f(t) − f(s)| ≤ K|t − s|γ for all t ∈ [0, T].
1. CONTINUITY OF SAMPLE PATHS.
A good general theorem to prove H¨
older continuity is this important theorem
of Kolmogorov:
THEOREM. Let X(·) be a stochastic process with continuous sample paths
a.s., such that
E(|X(t) − X(s)|β) ≤ C|t − s|1+α
for constants β, α > 0, C ≥ 0 and for all 0 ≤ t, s.
Then for each 0 < γ < α , T > 0, and almost every ω, there exists a constant
β
K = K(ω, γ, T ) such that
|X(t, ω) − X(s, ω)| ≤ K|t − s|γ for all 0 ≤ s, t ≤ T.
Hence the sample path t → X(t, ω) is uniformly H¨older continuous with expo-
nent γ on [0, T ].
47
APPLICATION TO BROWNIAN MOTION. Consider W(·), an n-
dimensional Brownian motion. We have for all integers m = 1, 2, . . .
1
E(|W(t) − W(s)|2m) = (2πr)n/2
|x|2me−|x|2
2r dx
for r = t − s > 0
Rn
1
x
=
rm
(2π)n/2
|y|2me−|y|22 dy
y = √
Rn
r
= Crm = C|t − s|m.
Thus the hypotheses of Kolmogorov’s theorem hold for β = 2m, α = m − 1. The
process W(·) is thus H¨older continuous a.s. for exponents
α
1
1
0 < γ <
=
for all m.
β
2 − 2m
Thus for almost all ω and any T > 0, the sample path t → W(t, ω) is uniformly
H¨
older continuous on [0, T ] for each exponent 0 < γ < 1/2.
Proof of Theorem. 1. For simplicity, take T = 1. Pick any
α
(8)
0 < γ <
.
β
Now define for n = 1, . . . ,
i + 1
i
1
An :=
X(
)
) >
for some integer 0
2n
− X(2n
2nγ
≤ i < 2n .
Then
2n−1
i + 1
i
1
P (An) ≤
P
X(
)
) >
2n
− X(2n
2nγ
i=0
2n−1
i + 1
i
β
1
−β
≤
E
X(
)
)
by Chebyshev’s inequality
2n
− X(2n
2nγ
i=0
2n−1
1
1+α
1
−β
≤ C
2n
2nγ
i=0
= C2n(−α+γβ).
Since (8) forces −α + γβ < 0, we deduce ∞ P(A
n=1
n) < ∞; whence the Borel–
Cantelli Lemma implies
P (An i.o.) = 0.
So for a.e. ω there exists m = m(ω) such that
i + 1
i
1
X(
, ω)
, ω)
for 0
2n
− X(2n
≤ 2nγ
≤ i ≤ 2n − 1
48
provided n ≥ m. But then we have
X( i+1 , ω)
, ω)
for 0
(9)
2n
− X( i2n
≤ K 12nγ
≤ i ≤ 2n − 1
for all n ≥ 0,
if we select K = K(ω) large enough.
2.* We now claim (9) implies the stated H¨
older continuity. To see this, fix
ω ∈ Ω for which (9) holds. Let t1, t2 ∈ [0, 1] be dyadic rationals, 0 < t2 − t1 < 1.
Select n ≥ 1 so that
(10)
2−n ≤ t < 2−(n−1) for t := t2 − t1.
We can write
t1 = i
(n < p
2n − 1
2p1 − · · · − 1
2pk
1 < · · · < pk)
t2 = j + 1 +
(n < q
2n
2q1
· · · + 12ql
1 < · · · < ql)
for
i
j
t1 ≤ 2n ≤ 2n ≤ t2.
Then
j − i
1
2n ≤ t < 2n−1
and so j = i or i + 1. In view of (9),
i
γ
|X(i/2n, ω) − X(j/2n, ω)| ≤ K − j
2n
≤ Ktγ.
Furthermore
1 γ
|X(i/2n − 1/2p1 − · · · − 1/2pr, ω) − X(i/2n − 1/2p1 − · · · − 1/2pr−1, ω)| ≤ K 2pr
for r = 1, . . . , k; and consequently
k
1 γ
|X(t1, ω) − X(i/2n, ω)| ≤ K
2pr
r=1
K ∞
1
≤
since p
2nγ
2rγ
r > n
r=1
C
= 2nγ ≤ Ctγ by (10).
In the same way we deduce
|X(t2, ω) − X(j/2n, ω)| ≤ Ctγ.
Add up the estimates above, to discover
|X(t1, ω) − X(t2, ω)| ≤ C|t1 − t2|γ
*Omit the second step in this proof on first reading.
49
for all dyadic rationals t1, t2 ∈ [0, 1] and some constant C = C(ω). Since t → X(t, ω)
is continuous for a.e. ω, the estimate above holds for all t1, t2 ∈ [0, 1].
Remark. The proof above can in fact be modified to show that if X(·) is a
stochastic process such that
E(|X(t) − X(s)|β) ≤ C|t − s|1+α (α, β > 0, C ≥ 0),
then X(·) has a version ˜X(·) such that a.e. sample path is H¨older continuous for
each exponent 0 < γ < α/β. (We call ˜
X(·) a version of X(·) if P(X(t) = ˜X(t)) = 1
for all t ≥ 0.)
So any Wiener process has a version with continuous sample paths a.s.
2. NOWHERE DIFFERENTIABILITY
Next we prove that sample paths of Brownian motion are with probability one
nowhere H¨older continuous with exponent greater than 1 , and thus are nowhere
2
differentiable.
THEOREM. (i) For each 1 < γ
2
≤ 1 and almost every ω, t → W(t, ω) is
nowhere H¨
older continuous with exponent γ.
(ii) In particular, for almost every ω, the sample path t → W(t, ω) is nowhere
differentiable and is of infinite variation on each subinterval.
Proof. (Dvoretzky, Erd¨os, Kakutani) 1. It suffices to consider a one-dimensional
Brownian motion, and we may for simplicity consider only times 0 ≤ t ≤ 1.
Fix an integer N so large that
1
N
γ −
> 1.
2
Now if the function t → W(t, ω) is H¨older continuous with exponent γ at some
point 0 ≤ s < 1, then
|W(t, ω) − W(s, ω)| ≤ K|t − s|γ for all t ∈ [0, 1] and some constant K.
For n ≫ 1, set i = [ns] + 1 and note that for j = i, i + 1, . . ., i + N − 1
j
j + 1
j
W ( , ω)
, ω)
, ω)
n
− W( n
≤ W(s, ω) − W(n
j + 1
+ W (s, ω) − W(
, ω)
n
j γ
j + 1 γ
≤ K s −
+ s
n
− n
M
≤ nγ
50
for some constant M . Thus
j
j + 1
M
ω ∈ AiM,n := W( )
)
for j = i, . . . , i + N
n − W ( n
≤ nγ
− 1
for some 1 ≤ i ≤ n, some M ≥ 1, and all large n.
Therefore the set of ω ∈ Ω such that W(ω, ·) is H¨older continuous with exponent
γ at some time 0 ≤ s < 1 is contained in
∞ ∞ ∞ n AiM,n.
M =1 k=1 n=k i=1
We will show this event has probability 0.
2. For all k and M ,
∞ n
n
P
AiM,n ≤ liminf P
AiM,n
n→∞
n=k i=1
i=1
n
≤ liminf
P (AiM,n)
n→∞ i=1
1
M
N
≤ liminf n P
)
,
n→∞
|W(n | ≤ nγ
since the random variables W ( j+1 )
) are N 0, 1 and independent. Now
n
− W( jn
n
1
M
√n Mn−γ
P
|W( )
=
e− nx2
2
dx
n | ≤ nγ
√2π −Mn−γ
1
M n1/2−γ
= √
e− y22 dy
2π −Mn1/2−γ
≤ Cn1/2−γ.
We use this calculation to deduce:
∞ n
P
AiM,n ≤ liminf nC[n1/2−γ]N = 0,
n→∞
n=k i=1
since N (γ − 1/2) > 1. This holds for all k, M. Thus
∞ ∞ ∞ n
P
AiM,n = 0,
M =1 k=1 n=k i=1
and assertion (i) of the Theorem follows.
3. If W (t, ω) is differentiable at s, then W (t, ω) would be H¨
older continuous
(with exponent 1) at s. But this is almost surely not so. If W (t, ω) were of finite
variation on some subinterval, it would then be differentiable almost everywhere
there.
51
Interpretation. The idea underlying the proof is that if
|W(t, ω) − W(s, ω)| ≤ K|t − s|γ for all t,
then
j
j + 1
M
|W( , ω)
, ω)
n
− W( n
| ≤ nγ
for all n ≫ 1 and at least N values of j. But these are independent events of small
probability. The probability that the above inequality holds for all these j’s is a
small number to the large power N , and is therefore extremely small.
A sample path of Brownian motion
D. MARKOV PROPERTY
DEFINITION. If V is a σ-algebra, V ⊆ U, then
P (A | V) := E(χA | V) for A ∈ U.
Therefore P (A | V) is a random variable, the conditional probability of A, given V.
DEFINITION. If X(·) is a stochastic process, the σ-algebra
U(s) := U(X(r)| 0 ≤ r ≤ s)
is called the history of the process up to and including time s.
We can informally interpret U(s) as recording the information available from
our observing X(r) for all times 0 ≤ r ≤ s.
52
DEFINITION. An Rn-valued stochastic process X(·) is called a Markov pro-
cess if
P (X(t) ∈ B | U(s)) = P(X(t) ∈ B | X(s)) a.s.
for all 0 ≤ s ≤ t and all Borel subset B of Rn.
The idea of this definition is that, given the current value X(s), you can predict
the probabilities of future values of X(t) just as well as if you knew the entire history
of the process before time s. Loosely speaking, the process only “knows” its value
at time s and does not “remember” how it got there.
THEOREM. Let W(·) be an n-dimensional Wiener process. Then W(·) is a
Markov process, and
1
(13)
P (W(t) ∈ B | W(s)) =
e− |x−W(s)|2
2(t−s)
dx a.s.
(2π(t − s))n/2 B
for all 0 ≤ s < t, and Borel sets B .
Note carefully that each side of this identity is a random variable.
Proof. We will only prove (13). Let A be a Borel set and write
1
Φ(y) :=
e− |x−y|2
2(t−s) dx.
(2π(t − s))n/2 A
As Φ(W(s)) is U(W(s)) measurable, we must show
(14)
χ{W(t)∈A}dP =
Φ(W(s)) dP
for all C ∈ U(W(s)).
C
C
Now if C ∈ U(W(s)), then C = {W(s) ∈ B} for some Borel set B ⊆ Rn. Hence
χ{W(t)∈A}dP = P(W(s) ∈ B, W(t) ∈ A)
C
=
g(y, s | 0)g(x,t− s | y) dxdy
B
A
=
g(y, s | 0)Φ(y) dy.
B
On the other hand,
Φ(W(s))dP =
χB(W(s))Φ(W(s)) dP
C
Ω
e− |y|2
2s
=
χB(y)Φ(y)
dy
Rn
(2πs)n/2
=
g(y, s | 0)Φ(y) dy,
B
and this last expression agrees with that above. This verifies (14), and so establishes
(13).
53
Interpretation. The Markov property partially explains the nondifferentia-
bility of sample paths for Brownian motion, as discussed before in §C.
If W(s, ω) = b, say, then the future behavior of W(t, ω) depends only upon
this fact and not on how W(t, ω) approached the point b as t → s−. Thus the path
“cannot remember” how to leave b in such a way that W(·, ω) will have a tangent
there.
54
CHAPTER 4: STOCHASTIC INTEGRALS, IT ˆ
O’S FORMULA
A. Motivation
B. Definition and properties of Itˆ
o integral
C. Indefinite Itˆ
o integrals
D. Itˆ
o’s formula
E. Itˆ
o integral in higher dimensions
A. MOTIVATION
Remember from Chapter 1 that we want to develop a theory of sto-
chastic differential equations of the form
dX = b(X, t)dt + B(X, t)dW
(SDE)
X(0) = X0,
which we will in Chapter 5 interpret to mean
t
t
(1)
X(t) = X0 +
b(X, s) ds +
B(X, s) dW
0
0
for all times t ≥ 0. But before we can study and solve such an integral
equation, we must first define
T
G dW
0
for some wide class of stochastic processes G, so that the right-hand side
of (1) at least makes sense. Observe also that this is not at all obvious.
For instance, since t → W(t, ω) is of infinite variation for almost every ω,
then
T G dW simply cannot be understood as an ordinary integral.
0
A FIRST DEFINITION. Suppose now n = m = 1. One possible defini-
tion is due to Paley, Wiener and Zygmund [P-W-Z]. Suppose g : [0, 1] → R
is continuously differentiable, with g(0) = g(1) = 0. Note carefully: g is
an ordinary, deterministic function and not a stochastic process. Then
let us define
1
1
g dW := −
g′W dt.
0
0
Note that
1 g dW is therefore a random variable. Let us check out the
0
properties following from this definition:
LEMMA (Properties of the Paley–Wiener–Zygmund integral).
(i) E
1 g dW = 0.
0
2
(ii) E
1 g dW
= 1 g2 dt.
0
0
55
Proof. 1. E
1 g dW =
g′E(W (t)) dt.
0
− 10
=0
2. To confirm (ii), we calculate
1
2
1
1
E
g dW
= E
g′(t)W (t) dt
g′(s)W (s) ds
0
0
0
1
1
=
g′(t)g′(s)E(W (t)W (s))dsdt
0
0
=t∧s
1
t
1
=
g′(t)
sg′(s) ds +
tg′(s) ds dt
0
0
t
1
t
=
g′(t) tg(t) −
g ds − tg(t) dt
0
0
1
t
1
=
g′(t) − g ds dt =
g2 dt.
0
0
0
Discussion. Suppose now g ∈ L2(0, 1). We can take a sequence of C1 functions
gn, as above, such that 1(g
0
n − g)2 dt → 0. In view of property (ii),
1
1
2
1
E
gm dW −
gn dW
=
(gm − gn)2 dt,
0
0
0
and therefore { 1 g
0
n dW }∞
n=1 is a Cauchy sequence in L2(Ω). Consequently we can
define
1
1
g dW := lim
gn dW.
0
n→∞ 0
The extended definition still satisfies properties (i) and (ii).
This is a reasonable definition of
1 g dW , except that this only makes sense
0
for functions g ∈ L2(0, 1), and not for stochastic processes. If we wish to define the
integral in (1),
t
B(X, s) dW,
0
then the integrand B(X, t) is a stochastic process and the definition above will not
suffice.
We must devise a definition for a wider class of integrands (although the def-
inition we finally decide on will agree with that of Paley, Wiener, Zygmund if g
happens to be a deterministic C1 function, with g(0) = g(1) = 0).
RIEMANN SUMS. To continue our study of stochastic integrals with ran-
dom integrands, let us think about what might be an appropriate definition for
T
W dW = ?,
0
56
where W (·) is a 1-dimensional Brownian motion. A reasonable procedure is to
construct a Riemann sum approximation, and then–if possible–to pass to limits.
DEFINITIONS. (i) If [0, T ] is an interval, a partition P of [0, T ] is a finite
collection of points in [0, T ]:
P := {0 = t0 < t1 < · · · < tm = T}.
(ii) Let the mesh size of P be |P| := max0≤k≤m−1 |tk+1 − tk|.
(iii) For fixed 0 ≤ λ ≤ 1 and P a given partition of [0, T], set
τk := (1 − λ)tk + λtk+1 (k = 0, . . ., m − 1).
For such a partition P and for 0 ≤ λ ≤ 1, we define
m−1
R = R(P, λ) :=
W (τk)(W (tk+1) − W(tk)).
k=0
This is the corresponding Riemann sum approximation of
T W dW . The key
0
question is this: what happens if |P| → 0, with λ fixed?
LEMMA (Quadratic variation). Let [a, b] be an interval in [0, ∞), and
suppose
P n := {a = tn0 < tn1 < · · · < tnm = b
n
}
are partitions of [a, b], with |Pn| → 0 as n → ∞. Then
mn−1(W(tnk+1)−W(tnk))2 → b−a
k=0
in L2(Ω) as n → ∞.
This assertion partly justifies the heuristic idea, introduced in Chapter 1, that
dW ≈ (dt)1/2.
Proof. Set Qn :=
mn−1(W (tn )
))2. Then
k=0
k+1 − W (tnk
mn−1
Qn − (b − a) =
((W (tnk+1) − W(tnk))2 − (tnk+1 − tnk)).
k=0
Hence
mn−1 mn−1
E((Qn − (b − a))2) =
E([(W (tnk+1) − W(tnk))2 − (tnk+1 − tnk)]
k=0
j=0
[(W (tnj+1) − W(tnj))2 − (tnj+1 − tnj)]).
For k = j, the term in the double sum is
E((W (tnk+1) − W(tnk))2 − (tnk+1 − tnk))E(· · ·),
57
according to the independent increments, and thus equals 0, as W (t) − W(s) is
N (0, t − s) for all t ≥ s ≥ 0. Hence
mn−1
E((Qn − (b − a))2) =
E((Y 2
k − 1)2(tnk+1 − tnk)2),
k=0
where
W (tn
)
)
Y
k+1 − W (tnk
k = Y n
k :=
is N (0, 1).
tnk+1 − tnk
Therefore for some constant C we have
mn−1
E((Qn − (a − b))2) ≤ C
(tnk+1 − tnk)2
k=0
≤ C | Pn | (b − a) → 0, as n → ∞.
Remark. Passing if necessary to a subsequence,
mn−1(W(tnk+1)−W(tnk))2 → b−a a.s.
k=0
Pick an ω for which this holds and also for which the sample path is uniformly
H¨older continuous with some exponent 0 < γ < 1 . Then
2
mn−1
b − a ≤ K limsup|Pn|γ
|W(tnk+1) − W(tnk)|
n→∞
k=0
for a constant K. Since |Pn| → 0, we see again that sample paths have infinite
variation with probability one:
m−1
sup
|W(tk+1) − W(tk)| = ∞.
P
k=0
Let us now return to the question posed above, as to the limit of the Riemann
sum approximations.
LEMMA. If P n denotes a partition of [0, T ] and 0 ≤ λ ≤ 1 is fixed, define
mn−1
Rn :=
W (τ n
k )(W (tn
k+1) − W (tnk)).
k=0
Then
W (T )2
1
lim Rn =
+ λ
T,
n→∞
2
− 2
58
the limit taken in L2(Ω). That is,
W (T )2
1
2
E
Rn −
T
2
− λ − 2
→ 0.
In particular the limit of the Riemann sum approximations depends upon the
choice of intermediate points tn
, where τ n = (1
+ λtn
.
k ≤ τn
k ≤ tnk+1
k
− λ)tnk
k+1
Proof. We have
mn−1
Rn :=
W (τ n
k )(W (tn
k+1) − W (tnk))
k=0
m
W 2(T )
1
n−1
=
(W (tn
2
− 2
k+1) − W (tnk))2
k=0
=:A
mn−1
mn−1
+
(W (τ n
k ) − W (tnk))2 +
(W (tnk+1) − W(τnk))(W(τnk) − W(tnk)).
k=0
k=0
=:B
=:C
According to the foregoing Lemma, A → T in L2(Ω) as n
2
→ ∞. A similar argument
shows that B → λT as n → ∞. Next we study the term C:
mn−1
E([
(W (tnk+1) − W(τnk))(W(τnk) − W(tnk))]2)
k=0
mn−1
=
E([W (tnk+1) − W(τnk)]2)E([W(τnk) − W(tnk)]2)
k=0
(independent increments)
mn−1
=
(1 − λ)(tnk+1 − tnk)λ(tnk+1 − tnk)
k=0
≤ λ(1 − λ)T|Pn | → 0.
Hence C → 0 in L2(Ω) as n → ∞.
We combine the limiting expressions for the terms A, B, C, and thereby establish
the Lemma.
It turns out that Itˆ
o’s definition (later, in §B) of T W dW corresponds to the
0
choice λ = 0. That is,
T
W 2(T )
T
W dW =
0
2
− 2
59
and, more generally,
r
W 2(r)
(r
W dW =
− W2(s)
− s) for all r ≥ s ≥ 0.
s
2
− 2
This is not what one would guess offhand. An alternative definition, due to Stratonovich,
takes λ = 1 ; so that
2
T
W 2(T )
W ◦ dW =
(Stratonovich integral).
0
2
See Chapter 6 for more.
More discussion. What are the advantages of taking λ = 0 and getting
T
W 2(T )
T
W dW =
?
0
2
− 2
First and most importantly, building the Riemann sum approximation by evaluating
the integrand at the left-hand endpoint τ n = tn on each subinterval [tn, tn
] will
k
k
k
k=1
ultimately permit the definition of
T
G dW
0
for a wide class of so-called “nonanticipating” stochastic processes G(·). Exact
definitions are later, but the idea is that t represents time, and since we do not know
what W (·) will do on [tn, tn ], it is best to use the known value of G(tn) in the
k
k+1
k
approximation. Indeed, G(·) will in general depend on Brownian motion W(·), and
we do not know at time tn its future value at the future time τ n = (1
+λtn
,
k
k
−λ)tnk
k+1
if λ > 0.
B. DEFINITION AND PROPERTIES OF IT ˆ
O’S INTEGRAL
Let W (·) be a 1-dimensional Brownian motion defined on some prob-
ability space (Ω, U, P).
DEFINITIONS. (i) The σ-algebra W(t) := U(W(s) | 0 ≤ s ≤ t) is called the
history of the Brownian motion up to (and including) time t.
(ii) The σ-algebra W+(t) := U(W(s)−W(t) | s ≥ t) is the future of the Brownian
motion beyond time t.
DEFINITION. A family F(·) of σ-algebras ⊆ U is called nonanticipating (with
respect to W (·)) if
(a) F(t) ⊇ F(s) for all t ≥ s ≥ 0
(b) F(t) ⊇ W(t) for all t ≥ 0
(c) F(t) is independent of W+(t) for all t ≥ 0.
We also refer to F(·) as a filtration.
IMPORTANT REMARK. We should informally think of F(t) as “containing
all information available to us at time t”. Our primary example will be F(t) :=
U(W(s) (0 ≤ s ≤ t), X0), where X0 is a random variable independent of W+(0).
60
This will be employed in Chapter 5, where X0 will be the (possibly random) initial
condition for a stochastic differential equation.
DEFINITION. A real-valued stochastic process G(·) is called nonanticipating
(with respect to F(·)) if for each time t ≥ 0, G(t) is F(t)–measurable.
The idea is that for each time t ≥ 0, the random variable G(t) “depends upon
only the information available in the σ-algebra F(t)”.
Discussion. We will actually need a slightly stronger notion, namely that G(·)
be progressively measurable. This is however a bit subtle to define, and we will not
do so here. The idea is that G(·) is nonanticipating and, in addition, is appropriately
jointly measurable in the variables t and ω together.
These measure theoretic issues can be confusing to students, and so we pause
here to emphasize the basic point, to be developed below. For progressively mea-
surable integrands G(·), we will be able to define, and understand, the stochastic
integral
T G dW in terms of some simple, useful and elegant formulas. In other
0
words, we will see that since at each moment of time “G depends only upon the
past history of the Brownian motion”, some nice identities hold, which would be
false if G “depends upon the future behavior of the Brownian motion”.
DEFINITIONS. (i) We denote by L2(0, T ) the space of all real–valued, pro-
gressively measurable stochastic processes G(·) such that
T
E
G2 dt
< ∞.
0
(ii) Likewise, L1(0, T ) is the space of all real–valued, progressively measurable pro-
cesses F (·) such that
T
E
|F| dt < ∞.
0
DEFINITION. A process G ∈ L2(0, T) is called a step process if there exists
a partition P = {0 = t0 < t1 < · · · < tm = T} such that
G(t) ≡ Gk
f or tk ≤ t < tk+1 (k = 0, . . ., m − 1).
Then each Gk is an F(tk)-measurable random variable, since G is nonanticipating.
DEFINITION. Let G ∈ L2(0, T) be a step process, as above. Then
T
m−1
G dW :=
Gk(W (tk+1) − W(tk))
0
k=0
is the Itˆ
o stochastic integral of G on the interval (0, T ).
Note carefully that this is a random variable.
61
LEMMA (Properties of stochastic integral for step processes). We
have for all constants a, b ∈ R and for all step processes G, H ∈ L2(0, T):
T
T
T
(i)
aG + bH dW = a
G dW + b
H dW,
0
0
0
T
(ii)
E
G dW
= 0,
0
2
T
T
(iii)
E
GdW
G2 dt .
0
=E 0
Proof. 1. The first assertion is easy to check.
Suppose next G(t) ≡ Gk for tk ≤ t < tk+1. Then
T
m−1
E
G dW
=
E(Gk(W (tk+1) − W(tk))).
0
k=0
Now Gk is F(tk)-measurable and F(tk) is independent of W+(tk). On the other
hand, W (tk+1)−W(tk) is W+(tk)-measurable, and so Gk is independent of W(tk+1)−
W (tk). Hence
E(Gk(W (tk+1) − W(tk))) = E(Gk)E(W(tk+1) − W(tk)).
=0
2. Furthermore,
2
T
m−1
E
GdW
E (GkGj(W (tk+1) − W(tk))(W(tj+1) − W(tj))).
0
=k,j=1
Now if j < k, then W (tk+1) − W(tk) is independent of GkGj(W(tj+1) − W(tj)).
Thus
E(GkGj(W (tk+1) − W(tk))(W(tj+1) − W(tj)))
= E(GkGj(W (tj+1) − W(tj)))E(W(tk+1) − W(tk)).
<∞
=0
62
Consequently
2
T
m−1
E
GdW
E(G2k(W (tk+1) − W(tk))2)
0
= k=0m−1
=
E(G2k)E((W (tk+1) − W(tk))2)
k=0
=tk+1−tk
T
= E
G2 dt .
0
APPROXIMATION BY STEP FUNCTIONS. The plan now is to ap-
proximate an arbitrary process G ∈ L2(0, T) by step processes in L2(0, T), and then
pass to limits to define the Itˆo integral of G.
LEMMA (Approximation by step processes). If G ∈ L2(0, T), there ex-
ists a sequence of bounded step processes Gn ∈ L2(0, T) such that
T
E
|G − Gn|2 dt → 0.
0
Outline of proof. We omit the proof, but the idea is this: if t → G(t, ω) is
continuous for almost every ω, we can set
k
k
k + 1
Gn(t) := G( ) for
, k = 0, . . . , [nT ].
n
n ≤ t <
n
For a general G ∈ L2(0, T), define
t
Gm(t) :=
mem(s−t)G(s) ds.
0
Then Gm ∈ L2(0, T), t → Gm(t, ω) is continuous for a.e. ω, and
T
|Gm − G|2dt → 0 a.s.
0
Now approximate Gm by step processes, as above.
DEFINITION. If G ∈ L2(0, T), take step processes Gn as above. Then
2
T
T
E
Gn−GmdW
(Gn − Gm)2 dt → 0 as n, m → ∞
0
=E 0
and so the limit
T
T
G dW := lim
Gn dW
0
n→∞ 0
63
exists in L2(Ω).
It is not hard to check that this definition does not depend upon the particular
sequence of step process approximations in L2(0, T ).
THEOREM (Properties of Itˆ
o Integral).
For all constants a, b ∈ R
and for all G, H ∈ L2(0, T), we have
T
T
T
(i)
aG + bH dW = a
G dW + b
H dW,
0
0
0
T
(ii)
E
G dW
= 0,
0
2
T
T
(iii)
E
GdW
G2 dt ,
0
=E 0
T
T
T
(iv)
E
G dW
H dW
= E
GH dt .
0
0
0
Proof. 1. Assertion (i) follows at once from the corresponding linearity prop-
erty for step processes.
Statements (ii) and (iii) are also easy consequences of the similar rules for step
processes.
2. Finally, assertion (iv) results from (iii) and the identity 2ab = (a+b)2−a2−b2,
and is left as an exercise.
EXTENDING THE DEFINITION. For many applications, it is important
to consider a wider class of integrands, instead of just L2(0, T ). To this end we define
M2(0, T ) to be the space of all real–valued, progressively measurable processes G(·)
such that
T
G2 dt < ∞ a.s.
0
It is possible to extend the definition of the Itˆ
o integral to cover G ∈ M2(0, T),
although we will not do so in these notes. The idea is to find a sequence of step
processes Gn ∈ M2(0, T) such that
T
(G − Gn)2 dt → 0 a.s. as n → ∞.
0
It turns out that we can then define
T
T
G dW := lim
Gn dW,
0
n→∞ 0
64
the expressions on the right converging in probability. See for instance Friedman
[F] or Gihman–Skorohod [G-S] for details.
More on Riemann sums. In particular, if G ∈ M2(0, T) and t → G(t, ω) is
continuous for a.e. ω, then
mn−1
T
G(tnk)(W (tnk+1) − W(tnk)) →
G dW
k=0
0
in probability, where P n = {0 = tn < · · · < tnm = T
n
} is any sequence of partitions,
with |Pn| → 0. This confirms the consistency of Itˆo’s integral with the earlier
calculations involving Riemann sums, evaluated at τ n = tn.
k
k
C. INDEFINITE IT ˆ
O INTEGRALS
DEFINITION. For G ∈ L2(0, T), set
t
I(t) :=
G dW
(0 ≤ t ≤ T),
0
the indefinite integral of G(·). Note I(0) = 0.
In this section we note some properties of the process I(·), namely that it is a
martingale and has continuous sample paths a.s. These facts will be quite useful for
proving Itˆo’s formula later in §D and in solving the stochastic differential equations
in Chapter 5.
THEOREM. (i) If G ∈ L2(0, T), then the indefinite integral I(·) is a martin-
gale.
(ii) Furthermore, I(·) has a version with continuous sample paths a.s.
Henceforth when we refer to I(·), we will always mean this version. We will not
prove assertion (i); a proof of (ii) is in Appendix C.
D. IT ˆ
O’S FORMULA
DEFINITION. Suppose that X(·) is a real–valued stochastic process satisfy-
ing
r
r
X(r) = X(s) +
F dt +
G dW
s
s
for some F ∈ L1(0, T), G ∈ L2(0, T) and all times 0 ≤ s ≤ r ≤ T. We say that
X(·) has the stochastic differential
dX = F dt + GdW
for 0 ≤ t ≤ T.
Note carefully that the differential symbols are simply an abbreviation for the
integral expressions above: strictly speaking “dX”, “dt”, and “dW ” have no mean-
ing alone.
65
THEOREM (Itˆ
o’s Formula). Suppose that X(·) has a stochastic differential
dX = F dt + GdW,
for F ∈ L1(0, T), G ∈ L2(0, T). Assume u : R × [0, T] → R is continuous and that
∂u , ∂u , ∂2u exist and are continuous.
∂t
∂x
∂x2
Set
Y (t) := u(X(t), t).
Then Y has the stochastic differential
dY = ∂u dt + ∂u dX + 1 ∂2u G2dt
∂t
∂x
2 ∂x2
(2)
= ∂u + ∂u F + 1 ∂2u G2 dt + ∂u GdW.
∂t
∂x
2 ∂x2
∂x
We call (2) Itˆo’s formula or Itˆo’s chain rule.
Remarks. (i) The argument of u, ∂u , etc. above is (X(t), t).
∂t
(ii) In view of our definitions, the expression (2) means for all 0 ≤ s ≤ r ≤ T,
Y (r) − Y (s) = u(X(r), r) − u(X(s), s)
r ∂u
∂u
1 ∂2u
=
(X, t) +
(X, t)F +
(X, t)G2 dt
(3)
s
∂t
∂x
2 ∂x2
r ∂u
+
(X, t)G dW
almost surely.
s
∂x
(iii) Since X(t) = X(0) + t F ds + t G dW , X(
0
0
·) has continuous sample paths
almost surely. Thus for almost every ω, the functions t → ∂u(X(t), t), ∂u(X(t), t),
∂t
∂x
∂2u (X(t), t) are continuous and so the integrals in (3) are defined.
∂x2
ILLUSTRATIONS OF IT ˆ
O’S FORMULA. We will prove Itˆ
o’s formula
below, but first here are some applications:
Example 1. Let X(·) = W(·), u(x) = xm. Then dX = dW and thus F ≡ 0,
G ≡ 1. Hence Itˆo’s formula gives
1
d(W m) = mW m−1dW + m(m
2
− 1)Wm−2dt.
In particular the case m = 2 reads
d(W 2) = 2W dW + dt.
This integrated is the identity
r
W 2(r)
(r
W dW =
− W2(s)
− s),
s
2
− 2
a formula we have established from first principles before.
66
Example 2. Again take X(·) = W(·), u(x, t) = eλx−λ2t2, F ≡ 0, G ≡ 1. Then
λ2
λ2
d eλW (t)− λ2t
2
=
− eλW(t)−λ2t2 + eλW(t)−λ2t2 dt + λeλW(t)−λ2t2dW
2
2
by Itˆ
o’s formula. Thus
dY = λY dW
Y (0) = 1.
This is a stochastic differential equation, about which more in Chapters 5 and 6.
In the Itˆ
o stochastic calculus the expression eλW (t)− λ2t
2
plays the role that eλt
plays in ordinary calculus. We build upon this observation:
Example 3. For n = 0, 1, . . . , define
(
h
−t)n
n(x, t) :=
ex2/2t dn
e−x2/2t ,
n!
dxn
the n-th Hermite polynomial. Then
h0(x, t) = 1, h1(x, t) = x
x2
t
x3
tx
h2(x, t) =
, h
2 − 2
3(x, t) = 6 − 2
x4
tx2
t2
h4(x, t) =
+
, etc.
24 − 4
8
THEOREM (Stochastic calculus with Hermite polynomials). We
have
t
hn(W, s) dW = hn+1(W (t), t) for t ≥ 0and n = 0, 1, . . .;
0
that is,
dhn+1(W, t) = hn(W, t) dW.
Consequently in the Itˆ
o stochastic calculus the expression hn(W (t), t) plays the
role that tn plays in ordinary calculus.
n!
Proof. (from McKean [McK]) Since
dn (e−(x−λt)2
2t
)
(e−x2/2t),
dλn
|λ=0 = (−t)n dn
dxn
we have
dn (eλx−λ2t2 )
(e−x2/2t)
dλn
|λ=0 = (−t)nex2/2t dn
dxn
= n!hn(x, t).
Hence
∞
eλx− λ2t
2
=
λnhn(x, t),
n=0
67
and so
∞
Y (t) = eλW (t)− λ2t
2
=
λnhn(W (t), t).
n=0
But Y (·) solves
dY = λY dW
Y (0) = 1;
that is,
t
Y (t) = 1 + λ
Y dW
for all t ≥ 0.
0
Plug in the expansion above for Y (t):
∞
t ∞
λnhn(W (t), t) = 1 + λ
λnhn(W (s), s) dW
n=0
0 n=0
∞
t
= 1 +
λn
hn−1(W(s), s) dW.
n=1
0
This identity holds for all λ and so the coefficients of λn on both sides are equal.
PROOF OF IT ˆ
O’S FORMULA. We now begin the proof of Itˆo’s formula,
by verifying directly two important special cases:
LEMMA (Two simple stochastic differentials). We have
(i) d(W 2) = 2W dW + dt,
and
(ii ) d(tW ) = W dt + tdW .
Proof. We have already established formula (i). To verify (ii), note that
r
mn−1
t dW = lim
tnk(W (tnk+1) − W(tnk)),
0
n→∞ k=0
where P n = {0 = tn0 < tn1 < · · · < tnm = r
n
} is a sequence of partitions of [0, r], with
|Pn| → 0. The limit above is taken in L2(Ω).
Similarly, since t → W(t) is continuous a.s.,
r
mn−1
W dt = lim
W (tnk+1)(tnk+1 − tnk),
0
n→∞ k=0
since for amost every ω the sum is an ordinary Riemann sum approximation and for
this we can take the right-hand endpoint tn
at which to evaluate the continuous
k+1
integrand.
We add these formulas to obtain
r
r
t dW +
W dt = rW (r).
0
0
68
These integral identities for all r ≥ 0 are abbreviated d(tW) = tdW + Wdt.
These special cases in hand, we now prove:
THEOREM (Itˆ
o product rule). Suppose
dX1 = F1dt + G1dW
(0 ≤ t ≤ T),
dX2 = F2dt + G2dW
for Fi ∈ L1(0, T), Gi ∈ L2(0, T) (i = 1, 2). Then
(4)
d(X1X2) = X2dX1 + X1dX2 + G1G2dt.
Remarks. (i) The expression G1G2dt here is the Itˆo correction term. The
integrated version of the product rule is the Itˆ
o integration-by-parts formula:
r
r
r
(5)
X2 dX1 = X1(r)X2(r) − X1(s)X2(s) −
X1 dX2 −
G1G2 dt.
s
s
s
(ii) If either G1 or G2 is identically equal to 0, we get the ordinary calculus
integration-by-parts formula. This confirms that the Paley–Wiener–Zygmund defi-
nition
1
1
g dW = −
g′W dt,
0
0
for deterministic C1 functions g, with g(0) = g(1) = 0, agrees with the Itˆ
o definition.
Proof. 1. Choose 0 ≤ r ≤ T.
First of all, assume for simplicity that X1(0) = X2(0) = 0, Fi(t) ≡ Fi, Gi(t) ≡
Gi, where Fi, Gi are time–independent, F(0)-measurable random variables (i =
1, 2). Then
Xi(t) = Fit + GiW (t) (t ≥ 0, i = 1, 2).
69
Thus
r
X2 dX1 + X1 dX2 + G1G2 dt
0
r
r
=
X1F2 + X2F1 dt +
X1G2 + X2G1 dW
0
0
r
+
G1G2 dt
0
r
=
(F1t + G1W )F2 + (F2t + G2W )F1 dt
0
r
+
(F1t + G1W )G2 + (F2t + G2W )G1 dW + G1G2r
0
r
r
= F1F2r2 + (G1F2 + G2F1)
W dt +
t dW
0
0
r
+ 2G1G2
W dW + G1G2r.
0
We now use the Lemma above to compute 2 r W dW = W 2(r)
W dt +
0
r
− r and r0
t dW = rW (r). Employing these identities, we deduce:
0
r
X2 dX1 + X1 dX2 + G1G2 dt
0
= F1F2r2 + (G1F2 + G2F1)rW (r) + G1G2W 2(r)
= X1(r)X2(r).
This is formula (5) for the special circumstance that s = 0, Xi(0) = 0, and Fi, Gi
time–independent random variables.
The case that s ≥ 0, X1(s), X2(s) are arbitrary, and Fi, Gi are constant F(s)-
measurable random variables has a similar proof.
2. If Fi, Gi are step processes, we apply Step 1 on each subinterval [tk, tk+1)
on which Fi and Gi are constant random variables, and add the resulting integral
expressions.
3. In the general situation, we select step processes F n
i
∈ L1(0, T), Gni ∈
L2(0, T ), with
E( T
0 |F n
i − Fi| dt) → 0
as n → ∞, i = 1, 2.
E( T (Gn
0
i − Gi)2 dt) → 0
Define
t
t
Xn
i (t) := Xi(0) +
F n
i ds +
Gni dW (i = 1, 2).
0
0
We apply Step 2 to Xn
i (·) on (s, r) and pass to limits, to obtain the formula
r
X1(r)X2(r) = X1(s)X2(s) +
X1dX2 + X2dX1 + G1G2 dt.
s
70
Now we are ready for
CONCLUSION OF THE PROOF OF IT ˆ
O’S FORMULA. Suppose
dX = F dt + GdW , with F ∈ L1(0, T), G ∈ L2(0, T).
1. We start with the case u(x) = xm, m = 0, 1, . . . , and first of all claim that
1
(6)
d(Xm) = mXm−1dX + m(m
2
− 1)Xm−2G2dt.
This is clear for m = 0, 1, and the case m = 2 follows from the Itˆ
o product formula.
Now assume the stated formula for m − 1:
1
d(Xm−1) = (m − 1)Xm−2dX + (m
2
− 1)(m − 2)Xm−3G2dt
1
= (m − 1)Xm−2(Fdt + GdW) + (m
2
− 1)(m − 2)Xm−3G2dt,
and we prove it for m:
d(Xm) = d(XXm−1)
= Xd(Xm−1) + Xm−1dX + (m − 1)Xm−2G2dt
(by the product rule)
1
= X (m − 1)Xm−2dX + (m
2
− 1)(m − 2)Xm−3G2dt
+ (m − 1)Xm−2G2dt + Xm−1dX
1
= mXm−1dX + m(m
2
− 1)Xm−2G2dt,
because m − 1 + 1(m
m(m
2
− 1)(m − 2) = 12
− 1). This proves (6).
Since Itˆo’s formula thus holds for the functions u(x) = xm, m = 0, 1, . . . and
since the operator “d” is linear, Itˆ
o’s formula is valid for all polynomials u in the
variable x.
2. Suppose now u(x, t) = f (x)g(t), where f and g are polynomials. Then
d(u(X, t)) = d(f (X)g)
= f (X)dg + gdf (X)
1
= f (X)g′dt + g[f ′(X)dX + f ′ (X)G2dt]
2
∂u
∂u
1 ∂2u
=
dt +
dX +
G2dt.
∂t
∂x
2 ∂x2
This calculation confirms Itˆ
o’s formula for u(x, t) = f (x)g(t), where f and g are
polynomials. Thus it is true as well for any function u having the form
m
u(x, t) =
f i(x)gi(t),
i=1
71
where f i and gi polynomials. That is, Itˆo’s formula is valid for all polynomial
functions u of the variables x, t.
3. Given u as in Itˆo’s formula, there exists a sequence of polynomials un such
that
un → u, ∂un
∂t
→ ∂u∂t
∂un
, ∂2un
,
∂x → ∂u
∂x
∂x2
→ ∂2u
∂x2
uniformly on compact subsets of R × [0, T]. Invoking Step 2, we know that for all
0 ≤ r ≤ T,
r ∂un
∂un
1 ∂2un
un(X(r), r) − un(X(0), 0) =
+
F +
G2 dt
0
∂t
∂x
2 ∂x2
r ∂un
+
G dW almost surely;
0
∂x
the argument of the partial derivatives of un is (X(t), t).
We may pass to limits as n → ∞ in this expression, thereby proving Itˆo’s
formula in general.
A similar proof gives this:
GENERALIZED IT ˆ
O FORMULA. Suppose dXi = F idt + GidW , with
for F i ∈ L1(0, T), Gi ∈ L2(0, T), for i = 1, . . ., n.
If u : Rn × [0, T] → R is continuous, with continuous partial derivatives ∂u,
∂t
∂u , ∂2u , (i, j = 1, . . . , n), then
∂xi
∂xi∂xj
∂u
n
∂u
1 n
∂2u
d(u(X1, . . . , Xn, t)) =
dt +
dXi +
GiGjdt.
∂t
∂x
2
∂x
i=1
i
i,j=1
i∂xj
E. IT ˆ
O’S INTEGRAL IN HIGHER DIMENSIONS
Notation. (i) Let W(·) = (W1(·), . . ., Wm(·)) be an m-dimensional Brownian
motion.
(ii) We assume F(·) is a family of nonanticipating σ-algebras, meaning that
(a) F(t) ⊇ F(s) for all t ≥ s ≥ 0
(b) F(t) ⊇ W(t) = U(W(s) | 0 ≤ s ≤ t)
(c) F(t) is independent of W+(t) := U(W(s) − W(t) | t ≤ s < ∞).
DEFINITIONS. (i) An Mn×m-valued stochastic process G = ((Gij)) belongs
to L2n×m(0, T) if
Gij ∈ L2(0, T)
(i = 1, . . . n; j = 1, . . . m).
(ii) An Rn-valued stochastic process F = (F 1, F 2, . . . , F n) belongs to L1n(0, T ) if
F i ∈ L1(0, T)
(i = 1, . . . n).
72
DEFINITION. If G ∈ L2n×m(0, T), then
T
G dW
0
is an Rn–valued random variable, whose i-th component is
m
T
Gij dW j (i = 1, . . . , n).
j=1
0
Approximating by step processes as before, we can establish this
LEMMA. If G ∈ L2n×m(0, T), then
T
E
G dW
= 0,
0
and
2
T
T
E
where
GdW
|G|2 dt ,
0
=E 0
|G|2 := 1≤i≤n |Gij|2.
1≤j≤m
DEFINITION. If X(·) = (X1(·), . . ., Xn(·)) is an Rn-valued stochastic pro-
cess such that
r
r
X(r) = X(s) +
F dt +
G dW
s
s
for some F ∈ L1n(0, T), G ∈ L2n×m(0, T) and all 0 ≤ s ≤ r ≤ T, we say X(·) has the
stochastic differential
dX = Fdt + GdW.
This means that
m
dXi = F idt +
GijdW j
for i = 1, . . . , n.
j=1
THEOREM (Itˆ
o’s formula in n-dimensions). Suppose that dX = Fdt +
GdW, as above. Let u : Rn×[0, T] be continuous, with continuous partial derivatives
∂u , ∂u , ∂2u , (i, j = 1, . . . , n). Then
∂t
∂xi
∂xi∂xj
d(u(X(t), t))= ∂u dt +
n
∂u dXi
∂t
i=1 ∂xi
(5)
+ 1
n
∂2u
m
GilGjldt,
2
i,j=1 ∂xi∂xj
l=1
where the argument of the partial derivatives of u is (X(t), t).
An outline of the proof follows some preliminary results:
73
LEMMA (Another simple stochastic differential). Let W (·) and ¯W(·)
be independent 1-dimensional Brownian motions. Then
d(W ¯
W ) = W d ¯
W + ¯
W dW.
Compare this to the case W = ¯
W . There is no correction term “dt” here, since
W, ¯
W are independent.
Proof. 1. To begin, set X(t) := W (t)+ ¯
W (t)
√
.
2
We claim that X(·) is a 1-dimensional Brownian motion. To see this, note
firstly that X(0) = 0 a.s. and X(·) has independent increments. Next observe that
since X is the sum of two independent, N (0, t ) random variables, X(t) is N (0, t). A
2
similar observation shows that X(t) −X(s) is N(0, t−s) for t ≥ s. This establishes
the claim.
2. From the 1-dimensional Itˆo calculus, we know
d(X2) =2XdX+dt,
d(W2) =2WdW +dt,
d(¯W2) =2¯Wd¯W +dt.
Thus
1
1
d(W ¯
W ) = d X2 − W2
¯
W 2
2
− 2
1
= 2XdX + dt − (2WdW + dt)
2
1
− (2 ¯Wd ¯W + dt)
2
= (W + ¯
W )(dW + d ¯
W ) − WdW − ¯Wd ¯W
= W d ¯
W + ¯
W dW.
We will also need the following modification of the product rule:
LEMMA (Itˆ
o product rule with several Brownian motions). Suppose
m
dX1 = F1dt +
Gk1dW k
k=1
and
m
dX2 = F2dt +
Gl2dW l,
l=1
where Fi ∈ L1(0, T) and Gki ∈ L2(0, T) for i = 1, 2; k = 1, . . ., m. Then
m
d(X1X2) = X1dX2 + X2dX1 +
Gk1Gk2dt.
k=1
74
The proof is a modification of that for the one–dimensional Itˆo product rule, as
before, with the new feature that
d(W iW j) = W idW j + W jdW i + δijdt,
according to the Lemma above.
The Itˆ
o formula in n-dimensions can now be proved by a suitable modification
of the one-dimensional proof. We first establish the formula for a multinomials
u = u(x) = xk1
1 . . . xkm
m , proving this by an induction on k1, . . . , km, using the
Lemma above. This done, the formula follows easily for polynomials u = u(x, t)
in the variables x = (x1, . . . , xn) and t, and then, after an approximation, for all
functions u as stated.
CLOSING REMARKS.
1. ALTERNATIVE NOTATION. When
dX = Fdt + GdW,
we sometimes write
m
Hij :=
GikGjk.
k=1
Then Itˆo’s formula reads
∂u
1
du(X, t) =
+ F
H : D2u dt + Du
∂t
· Du + 2
· GdW,
where Du =
∂u , . . . , ∂u
is the gradient of u in the x-variables, D2u =
∂2u
∂x1
∂xn
∂xi∂xj
is the Hessian matrix, and
n
F · Du =
F i ∂u ,
∂x
i=1
i
n
H : D2u =
Hij ∂2u ,
∂x
i,j=1
i∂xj
n
m
∂u
Du · GdW =
GikdW k.
∂xi
i=1 k=1
2. HOW TO REMEMBER IT ˆ
O’S FORMULA.
We may symbolically compute
n
n
∂u
∂u
1
∂2u
d(u(X, t)) =
dt +
dXi +
dXidXj,
∂t
∂xi
2
∂xi∂xj
i=1
i,j=1
and then simplify the term “dXidXj” by expanding it out and using the formal
multiplication rules
(dt)2 = 0, dtdW k = 0, dW kdW l = δkldt (k, l = 1, . . . , m).
75
The foregoing theory provides a rigorous meaning for all this.
76
CHAPTER 5: STOCHASTIC DIFFERENTIAL EQUATIONS
A. Definitions and examples
B. Existence and uniqueness of solutions
C. Properties of solutions
D. Linear stochastic differential equations
A. DEFINITIONS AND EXAMPLES
We are finally ready to study stochastic differential equations:
Notation. (i) Let W(·) be an m-dimensional Brownian motion and X0 an
n-dimensional random variable which is independent of W(·). We will henceforth
take
F(t) := U(X0, W(s) (0 ≤ s ≤ t)) (t ≥ 0),
the σ-algebra generated by X0 and the history of the Wiener process up to (and
including) time t.
(ii) Assume T > 0 is given, and
b : Rn × [0, T] → Rn,
B : Rn × [0, T] → Mn×m
are given functions. (Note carefully: these are not random variables.) We display
the components of these functions by writing
b11
. . .
b1m
.
.
.
b = (b1, b2, . . . , bn), B =
.. .. ..
bn1
. . .
bnm .
DEFINITION. We say that an Rn-valued stochastic process X(·) is a solution
of the Itˆ
o stochastic differential equation
dX = b(X, t)dt + B(X, t)dW
(SDE)
X(0) = X0
for 0 ≤ t ≤ T, provided
(i) X(·) is progressively measurable with respect to F(·),
(ii) F := b(X, t) ∈ L1n(0, T),
(iii) G := B(X, t) ∈ L2n×m(0, T),
and
(iv) X(t) = X0 + t b(X(s), s) ds + t B(X(s), s) dW a.s. for all 0
0
0
≤ t ≤ T.
Remarks. (i) A higher order SDE of the form
Y (n) = f (t, Y, . . . , Y (n−1)) + g(t, Y, . . . , Y (n−1))ξ,
77
where as usual ξ denotes “white noise”, can be rewritten into the form above by
the device of setting
Y(t)
X1(t)
˙
Y (t)
X2(t)
X(t) = . =
.
.
.
.
. .
Y (n−1)(t) Xn(t)
Then
X2
0
.
.
dX =
..
.
.
f (· · ·) dt + g(···)dW.
(ii) In view of (iii), we can always assume X(·) has continuous sample paths
almost surely.
EXAMPLES OF LINEAR STOCHASTIC DIFFERENTIAL EQUATIONS.
Example 1. Let m = n = 1 and suppose g is a continuous function (not a
random variable). Then the unique solution of
dX = gXdW
(1)
X(0) = 1
is
X(t) = e− 1
g2 ds+
g dW
2 R t
0
R t0
for 0 ≤ t ≤ T. To verify this, note that
1
t
t
Y (t) := −
g2 ds +
g dW
2 0
0
satisfies
1
dY = − g2dt + gdW.
2
Thus Itˆ
o’s lemma for u(x) = ex gives
∂u
1 ∂2u
dX =
dY +
g2dt
∂x
2 ∂x2
1
1
= eY
− g2dt + gdW + g2dt
2
2
= gXdW, as claimed.
We will prove uniqueness later, in §B.
78
Example 2. Similarly, the unique solution of
dX = f Xdt + gXdW
(2)
X(0) = 1
is
X(t) = eR t f
g2 ds+
g dW
0
−12
R t0
for 0 ≤ t ≤ T.
Example 3 (Stock prices). Let P (t) denote the price of a stock at time
t. We can model the evolution of P (t) in time by supposing that dP , the relative
P
change of price, evolves according to the SDE
dP = µdt + σdW
P
for certain constants µ > 0 and σ, called the drift and the volatility of the stock.
Hence
(3)
dP = µP dt + σP dW ;
and so
dP
1 σ2P 2dt
d(log(P )) =
by Itˆo’s formula
P − 2
P 2
σ2
=
µ −
dt + σdW.
2
Consequently
P (t) = p
”t
0eσW (t)+“µ− σ2
2
,
similarly to Example 2. Observe that the price is always positive, assuming the
initial price p0 is positive.
Since (3) implies
t
t
P (t) = p0 +
µP ds +
σP dW
0
0
and E
t σP dW = 0, we see that
0
t
E(P (t)) = p0 +
µE(P (s)) ds.
0
Hence
E(P (t)) = p0eµt for t ≥ 0.
The expected value of the stock price consequently agrees with the deterministic
solution of (3) corresponding to σ = 0.
79
Example 4 (Brownian bridge). The solution of the SDE
dB =
dt + dW
(0
(4)
− B
1−t
≤ t < 1)
B(0) = 0
is
t
1
B(t) = (1 − t)
dW
(0 ≤ t < 1),
0 1 − s
as we confirm by a direct calculation. It turns out also that limt→1− B(t) = 0 almost
surely. We call B(·) a Brownian bridge, between the origin at time 0 and at time 1.
A sample path of the Brownian bridge
Example 5 (Langevin’s equation). A possible improvement of our math-
ematical model of the motion of a Brownian particle models frictional forces as
follows for the one-dimensional case:
˙
X = −bX + σξ,
where ξ(·) is “white noise”, b > 0 is a coefficient of friction, and σ is a diffusion
coefficient. In this interpretation X(·) is the velocity of the Brownian particle: see
Example 6 for the position process Y (·). We interpret this to mean
dX =
(5)
−bXdt + σdW
X(0) = X0,
80
for some initial distribution X0, independent of the Brownian motion. This is the
Langevin equation.
The solution is
t
X(t) = e−btX0 + σ
e−b(t−s)dW
(t ≥ 0),
0
as is straightforward to verify. Observe that
E(X(t)) = e−btE(X0)
and
t
E(X2(t)) = E e−2btX20 + 2σe−btX0
e−b(t−s)dW
0
t
2
+ σ2
e−b(t−s)dW
0
t
= e−2btE(X20) + 2σe−btE(X0)E
e−b(t−s)dW
0
t
+ σ2
e−2b(t−s)ds
0
σ2
= e−2btE(X20) +
(1
2b
− e−2bt).
Thus the variance
V (X(t)) = E(X2(t)) − E(X(t))2
is given by
σ2
V (X(t)) = e−2btV (X0) +
(1
2b
− e−2bt),
assuming, of course, V (X0) < ∞. For any such initial condition X0 we therefore
have
E(X(t)) → 0
as t
V (X(t)) → σ2
→ ∞.
2b
From the explicit form of the solution we see that the distribution of X(t)
approaches N 0, σ2
as t
2b
→ ∞. We interpret this to mean that irrespective of the
initial distribution, the solution of the SDE for large time “settles down” into a
Gaussian distribution whose variance σ2 represents a balance between the random
2b
disturbing force σξ(·) and the frictional damping force −bX(·).
Example 6 (Ornstein–Uhlenbeck process). A better model of Brownian
movement is provided by the Ornstein–Uhlenbeck equation
¨
Y = −b ˙Y + σξ
Y (0) = Y0, ˙
Y (0) = Y1,
81
A simulation of Langevin’s equation
where Y (t) is the position of Brownian particle at time t, Y0 and Y1 are given
Gaussian random variables. As before b > 0 is the friction coefficient, σ is the
diffusion coefficient, and ξ(·) as usual is “white noise”.
Then X := ˙
Y , the velocity process, satisfies the Langevin equation
dX =
(6)
−bXdt + σdW
X(0) = Y1,
studied in Example 5. We assume Y1 to be normal, whence explicit formula for the
solution,
t
X(t) = e−btY1 + σ
e−b(t−s)dW,
0
shows X(t) to be Gaussian for all times t ≥ 0. Now the position process is
t
Y (t) = Y0 +
X ds.
0
Therefore
t
E(Y (t)) = E(Y0) +
E(X(s)) ds
0
t
= E(Y0) +
e−bsE(Y1) ds
0
1
= E(Y
− e−bt
0) +
E(Y
b
1);
82
and a somewhat lengthly calculation shows
σ2
σ2
V (Y (t)) = V (Y0) +
t +
(
b2
2b3 −3 + 4e−bt − e−2bt).
Nelson [N, p. 57] discusses this model as compared with Einstein’s .
Example 7 (Random harmonic oscillator). This is the SDE
¨
X = −λ2X − b ˙X + σξ
X(0) = X0, ˙
X(0) = X1,
where −λ2X represents a linear, restoring force and −b ˙X is a frictional damping
term.
An explicit solution can be worked out using the general formulas presented
below in §D. For the special case X1 = 0, b = 0, σ = 1, we have
1
t
X(t) = X0 cos(λt) +
sin(λ(t
λ
− s))dW.
0
B. EXISTENCE AND UNIQUENESS OF SOLUTIONS
In this section we address the problem of building solutions to stochastic dif-
ferential equations. We start with a simple case:
1. AN EXAMPLE IN ONE DIMENSION. Let us first suppose b : R → R
is C1, with |b′| ≤ L for some constant L, and try to solve the one–dimensional
stochastic differential equation
dX = b(X)dt + dW
(7)
X(0) = x
where x ∈ R.
Now the SDE means
t
X(t) = x +
b(X) ds + W (t),
0
for all times t ≥ 0, and this formulation suggests that we try a successive approxi-
mation method to construct a solution. So define X0(t) ≡ x, and then
t
Xn+1(t) := x +
b(Xn) ds + W (t)
(t ≥ 0)
0
for n = 0, 1, . . . . Next write
Dn(t) := max | Xn+1(s) − Xn(s)|
(n = 0, . . . ),
0≤s≤t
and notice that for a given continuous sample path of the Brownian motion, we
have
s
D0(t) = max
b(x) dr + W (s) ≤ C
0≤s≤t 0
83
for all times 0 ≤ t ≤ T, where C depends on ω.
We now claim that
Ln
Dn(t) ≤ C tn
n!
for n = 0, 1, . . . , 0 ≤ t ≤ T. To see this note that
s
Dn(t) = max
b(Xn(r)) − b(Xn−1(r)) dr
0≤s≤t 0
t
≤ L Dn−1(s)ds
0
t
Ln−1sn−1
≤ L C
ds
by the induction assumption
0
(n − 1)!
Lntn
= C
.
n!
In view of the claim, for m ≥ n we have
∞ LkT k
max |Xm(t) − Xn(t)| ≤ C
0≤t≤T
k!
→ 0 as n → ∞.
k=n
Thus for almost every ω, Xn(·) converges uniformly for 0 ≤ t ≤ T to a limit process
X(·) which, as is easy to check, solves (7).
2. SOLVING SDE BY CHANGING VARIABLES. Next is a procedure for
solving SDE by means of a clever change of variables (McKean [McK, p. 60]).
Given a general one–dimensional SDE of the form
dX = b(X)dt + σ(X)dW
(8)
X(0) = x,
let us first solve
dY = f (Y )dt + dW
(9)
Y (0) = y,
where f will be selected later, and try to find a function u such that
X := u(Y )
solves our SDE (8). Note that we can in principle at least solve (9), according to
the previous example. Assuming for the moment u and f are known, we compute
using Itˆ
o’s formula that
1
dX = u′(Y )dY + u′ (Y )dt
2
1
= u′f + u′ dt + u′dW.
2
84
Thus X(·) solves (8) provided
u′(Y ) = σ(X) = σ(u(Y )),
u′(Y )f (Y ) + 1 u′ (Y ) = b(X) = b(u(Y )),
2
and
u(y) = x.
So let us first solve the ODE
u′(z) = σ(u(z))
(z
u(y) = x
∈ R),
where ′ = d , and then, once u is known, solve for
dz
1
1
f (z) =
b(u(z))
u′ (z) .
σ(u(z))
− 2
We will not discuss here conditions under which all of this is possible: see Lamperti
[L2].
Notice that both of the methods described above avoid all use of martingale
estimates.
3. A GENERAL EXISTENCE AND UNIQUENESS THEOREM
We start with a useful calculus lemma:
GRONWALL’S LEMMA. Let φ and f be nonnegative, continuous functions
defined for 0 ≤ t ≤ T, and let C0 ≥ 0 denote a constant. If
t
φ(t) ≤ C0 +
f φ ds
for all 0 ≤ t ≤ T,
0
then
φ(t) ≤ C
f ds
0eR t
0
for all 0 ≤ t ≤ T.
Proof. Set Φ(t) := C0 + t f φ ds. Then Φ′ = f φ
0
≤ fΦ, and so
′
e− R t f ds
f ds
f ds
0
Φ
= (Φ′ − fΦ)e−Rt0
≤ (fφ − fφ)e−Rt0
= 0.
Therefore
Φ(t)e− R t f ds
f ds
0
≤ Φ(0)e−R00
= C0,
and thus
φ(t) ≤ Φ(t) ≤ C
f ds
0eR t
0
.
85
EXISTENCE AND UNIQUENESS THEOREM. Suppose that b : Rn ×
[0, T ] → Rn and B : Rn × [0, T] → Mm×n are continuous and satisfy the following
conditions:
(a)
|b(x, t) − b(ˆx, t)| ≤ L|x − ˆx|
for all 0
|B(x, t) − B(ˆx, t)| ≤ L|x − ˆx|
≤ t ≤ T, x, ˆx ∈ Rn
(b)
|b(x, t)| ≤ L(1 + |x|) for all 0
|B(x, t)| ≤ L(1 + |x|)
≤ t ≤ T, x ∈ Rn,
for some constant L.
Let X0 be any Rn-valued random variable such that
(c)
E(|X0|2) < ∞
and
(d)
X0 is independent of W+(0),
where W(·) is a given m-dimensional Brownian motion.
Then there exists a unique solution X ∈ L2n(0, T) of the stochastic differential
equation:
dX = b(X, t)dt + B(X, t)dW
(0
(SDE)
≤ t ≤ T)
X(0) = X0.
Remarks. (i) “Unique” means that if X, ˆ
X ∈ L2n(0, T), with continuous sam-
ple paths almost surely, and both solve (SDE), then
P (X(t) = ˆ
X(t) for all 0 ≤ t ≤ T) = 1.
(ii) Hypotheses (a) says that b and B are uniformly Lipschitz continuous in the
variable x. Notice also that hypothesis (b) actually follows from (a).
Proof. 1. Uniqueness. Suppose X and ˆ
X are solutions, as above. Then for
all 0 ≤ t ≤ T,
t
t
X(t) − ˆX(t) =
b(X, s) − b( ˆX, s) ds +
B(X, s) − B( ˆX, s) dW.
0
0
Since (a + b)2 ≤ 2a2 + 2b2, we can estimate
t
2
E(|X(t) − ˆX(t)|2) ≤ 2E
b(X, s) − b( ˆX, s) ds
0
t
2
+ 2E
B(X, s) − B( ˆX, s) dW
.
0
86
The Cauchy–Schwarz inequality implies that
t
2
t
f ds
≤ t |f|2 ds
0
0
for any t > 0 and f : [0, t] → Rn. We use this to estimate
t
2
t
2
E
b(X, s) − b( ˆX, s) ds
≤ TE
b(X, s) − b( ˆX, s) ds
0
0
t
≤ L2T
E(|X − ˆX|2) ds.
0
Furthermore
t
2
t
2
E
B(X, s) − B( ˆX, s) dW
= E
B(X, s) − B( ˆX, s) ds
0
0
t
≤ L2
E(|X − ˆX|2) ds.
0
Therefore for some appropriate constant C we have
t
E(|X(t) − ˆX(t)|2) ≤ C
E(|X − ˆX|2) ds,
0
provided 0 ≤ t ≤ T. If we now set φ(t) := E(|X(t) − ˆX(t)|2), then the foregoing
reads
t
φ(t) ≤ C
φ(s) ds
for all 0 ≤ t ≤ T.
0
Therefore Gronwall’s Lemma, with C0 = 0, implies φ ≡ 0. Thus X(t) = ˆX(t) a.s.
for all 0 ≤ t ≤ T, and so X(r) = ˆX(r) for all rational 0 ≤ r ≤ T, except for some
set of probability zero. As X and ˆ
X have continuous sample paths almost surely,
P
max |X(t) − ˆX(t)| > 0 = 0.
0≤t≤T
2. Existence. We will utilize the iterative scheme introduced earlier. Define
X0(t) := X0
Xn+1(t) := X0 + t b(Xn(s), s) ds + t B(Xn(s), s) dW,
0
0
for n = 0, 1, . . . and 0 ≤ t ≤ T. Define also
dn(t) := E(|Xn+1(t) − Xn(t)|2).
We claim that
(M t)n+1
dn(t) ≤
for all n = 0, . . . , 0
(n + 1)!
≤ t ≤ T
87
for some constant M , depending on L, T and X0. Indeed for n = 0, we have
d0(t) = E(|X1(t) − X0(t)|2)
t
t
2
= E
b(X0, s) ds +
B(X0, s) dW
0
0
t
2
t
≤ 2E
L(1 + |X0|) ds
+ 2E
L2(1 + |X0|2) ds
0
0
≤ tM
for some large enough constant M . This confirms the claim for n = 0.
Next assume the claim is valid for some n − 1. Then
dn(t) = E(|Xn+1(t) − Xn(t)|2)
t
= E
b(Xn, s) − b(Xn−1, s) ds
0
t
2
+
B(Xn, s) − B(Xn−1, s) dW
0
t
≤ 2TL2E
|Xn − Xn−1|2 ds
0
t
+ 2L2E
|Xn − Xn−1|2 ds
0
t Mnsn
≤ 2L2(1 + T)
ds
by the induction hypothesis
0
n!
M n+1tn+1
≤
,
(n + 1)!
provided we choose M ≥ 2L2(1 + T). This proves the claim.
3. Now note
T
max |Xn+1(t) − Xn(t)|2 ≤ 2TL2
|Xn − Xn−1|2 ds
0≤t≤T
0
t
2
+ 2 max
B(Xn, s) − B(Xn−1, s) dW .
0≤t≤T 0
Consequently the martingale inequality from Chapter 2 implies
T
E
max |Xn+1(t) − Xn(t)|2 ≤ 2TL2
E(|Xn − Xn−1|2) ds
0≤t≤T
0
T
+ 8L2
E(|Xn − Xn−1|2) ds
0
(M T )n
≤ C
by the claim above.
n!
88
4. The Borel–Cantelli Lemma thus applies, since
1
P
max |Xn+1(t) − Xn(t)| >
|Xn+1(t) − Xn(t)|2
0≤t≤T
2n
≤ 22nE max
0≤t≤T
≤ 22nC(MT)n
n!
and
∞ 22n(MT)n <
n!
∞.
n=1
Thus
1
P
max |Xn+1(t) − Xn(t)| >
i.o.
= 0.
0≤t≤T
2n
In light of this, for almost every ω
n−1
Xn = X0 +
(Xj+1 − Xj)
j=0
converges uniformly on [0, T ] to a process X(·). We pass to limits in the definition
of Xn+1(·), to prove
t
t
X(t) = X0 +
b(X, s) ds +
B(X, s) dW
for 0 ≤ t ≤ T.
0
0
That is,
dX = b(X, t)dt + B(X, t)dW
(SDE)
X(0) = X0,
for times 0 ≤ t ≤ T.
5. We must still show X(·) ∈ L2n(0, T). We have
t
2
E(|Xn+1(t)|2) ≤ CE(|X0|2) + CE
b(Xn, s) ds
0
t
2
+ CE
B(Xn, s) dW
0
t
≤ C(1 + E(|X0|2)) + C
E(|Xn|2) ds,
0
where, as usual, “C” denotes various constants. By induction, therefore,
E(|Xn+1(t)|2) ≤ C + C2 + · · · + Cn+2 tn+1 (1 + E(
(n + 1)!
|X0|2)).
Consequently
E(|Xn+1(t)|2) ≤ C(1 + E(|X0|2))eCt.
Let n → ∞:
E(|X(t)|2) ≤ C(1 + E(|X0|2))eCt for all 0 ≤ t ≤ T;
89
and so X ∈ L2n(0, T).
C. PROPERTIES OF SOLUTIONS
In this section we mention, without proofs, a few properties of the solution to
various SDE.
THEOREM (Estimate on higher moments of solutions). Suppose that
b, B and X0 satisfy the hypotheses of the Existence and Uniqueness Theorem. If,
in addition,
E(|X0|2p) < ∞ for some integer p > 1,
then the solution X(·) of
dX = b(X, t)dt + B(X, t)dW
(SDE)
X(0) = X0
satisfies the estimates
(i)
E(|X(t)|2p) ≤ C2(1 + E(|X0|2p))eC1t
and
(ii)
E(|X(t) − X0|2p) ≤ C2(1 + E(|X0|2p))tpeC2t
for certain constants C1 and C2, depending only on T, L, m, n.
The estimates above on the moments of X(·) are fairly crude, but are never-
theless sometimes useful:
APPLICATION: SAMPLE PATH PROPERTIES. The possibility that
B ≡ 0 is not excluded, and consequently it could happen that the solution of our
SDE is really a solution of the ODE
˙
X = b(X, t),
with possibly random initial data. In this case the mapping t → X(t) will be smooth
if b is. On the other hand, if for some 1 ≤ i ≤ n
|bil(x, t)|2 > 0 for all x ∈ Rn, 0 ≤ t ≤ T,
1≤l≤m
then almost every sample path t → Xi(t) is nowhere differentiable for a.e. ω. We can
however use estimates (i) and (ii) above to check the hypotheses of Kolmogorov’s
Theorem from §C in Chapter 3. It follows that for almost all sample paths,
the mapping t → X(t) is H¨older continuous with each exponent less than 1,2
provided E(|X0|2p) < ∞ for each 1 ≤ p < ∞.
90
THEOREM (Dependence on parameters). Suppose for k = 1, 2, . . . that
bk, Bk and Xk0 satisfy the hypotheses of the Existence and Uniqueness Theorem,
with the same constant L. Assume further that
(a)
lim E(|Xk0 − X0|2) = 0,
k→∞
and for each M > 0,
(b)
lim max (|bk(x, t) − b(x, t)| + |Bk(x, t) − B(x, t)|) = 0.
k→∞ 0≤t≤T
|x|≤M
Finally suppose that Xk(·) solves
dXk = bk(Xk, t)dt + Bk(Xk, t)dW
Xk(0) = Xk0.
Then
lim E
max |Xk(t) − X(t)|2 = 0,
k→∞
0≤t≤T
where X is the unique solution of
dX = b(X, t)dt + B(X, t)dW
X(0) = X0.
Example (Small noise limits). In particular, for almost every ω the random
trajectories of the SDE
dXε = b(Xε)dt + εdW
Xε(0) = x0
converge uniformly on [0, T ] as ε → 0 to the deterministic trajectory of
˙x = b(x),
x(0) = x0.
D. LINEAR STOCHASTIC DIFFERENTIAL EQUATIONS
This section presents some fairly explicit formulas for solutions of linear SDE.
DEFINITION. The stochastic differential equation
dX = b(X, t)dt + B(X, t)dW
is linear provided the coefficients b and B have this form:
b(x, t) := c(t) + D(t)x,
for c : [0, T ] → Rn, D : [0, T] → Mn×n, and
B(x, t) := E(t) + F(t)x
for E : [0, T ] → Mn×m, F : [0, T] → L(Rn, Mn×m), the space of bounded linear
mappings from Rn to Mn×m.
91
DEFINITION. A linear SDE is called homogeneous if c ≡ E ≡ 0 for 0 ≤ t ≤
T . It is called linear in the narrow sense if F ≡ 0.
Remark. If
sup [|c(t)| + |D(t)| + |E(t)| + |F(t)|] < ∞,
0≤t≤T
then b and B satisfy the hypotheses of the Existence and Uniqueness Theorem.
Thus the linear SDE
dX = (c(t) + D(t)X)dt + (E(t) + F(t)X)dW
X(0) = X0
has a unique solution, provided E(|X0|2) < ∞, and X0 is independent of W+(0).
FORMULAS FOR SOLUTIONS: linear equations in narrow sense
Suppose first D ≡ D is constant. Then the solution of
dX = (c(t) + DX)dt + E(t)dW
(10)
X(0) = X0
is
t
(11)
X(t) = eDtX0 +
eD(t−s)(c(s)ds + E(s) dW),
0
where
∞ Dktk
eDt :=
.
k!
k=0
More generally, the solution of
dX = (c(t) + D(t)X)dt + E(t)dW
(12)
X(0) = X0
is
t
(13)
X(t) = Φ(t) X0 +
Φ(s)−1(c(s)ds + E(s) dW) ,
0
where Φ(·) is the fundamental matrix of the nonautonomous system of ODE
dΦ = D(t)Φ, Φ(0) = I.
dt
These assertions follow formally from standard formulas in ODE theory if we
write EdW = Eξdt, ξ as usual denoting white noise, and regard Eξ as an inhomo-
geneous term driving the ODE
˙
X = c(t) + D(t)X + E(t)ξ.
This will not be so if F(·) ≡ 0, owing to the extra term in Itˆo’s formula.
92
Observe also that formula (13) shows X(t) to be Gaussian if X0 is.
FORMULAS FOR SOLUTIONS: general scalar linear equations
Suppose now n = 1, but m ≥ 1 is arbitrary. Then the solution of
dX = (c(t) + d(t)X)dt +
m (el(t) + fl(t)X)dW l
(14)
l=1
X(0) = X0
is
t
m
X(t) = Φ(t) X0 +
Φ(s)−1 c(s) −
el(s)f l(s)
ds
0
(15)
l=1
t m
+
Φ(s)−1el(s) dW l,
0 l=1
where
t
m (fl)2
t m
Φ(t) := exp
d −
ds +
f l dW l
.
0
2
l=1
0 l=1
See Arnold [A, Chapter 8] for more formulas for solutions of general linear
equations.
3. SOME METHODS FOR SOLVING LINEAR SDE
For practice with Itˆo’s formula, let us derive some of the formulas stated above.
Example 1. Consider first the linear stochastic differential equation
dX = d(t)Xdt + f (t)XdW
(16)
X(0) = X0
for m = n = 1. We will try to find a solution having the product form
X(t) = X1(t)X2(t),
where
dX1 = f (t)X1dW
(17)
X1(0) = X0
and
dX2 = A(t)dt + B(t)dW
(18)
X2(0) = 1,
where the functions A and B are to be selected. Then
dX = d(X1X2)
= X1dX2 + X2dX1 + f (t)X1B(t)dt
= f (t)XdW + (X1dX2 + f (t)X1B(t)dt),
93
according to (17). Now we try to choose A, B so that
dX2 + f (t)B(t)dt = d(t)X2dt.
For this, B ≡ 0 and A(t) = d(t)X2(t) will work. Thus (18) reads
dX2 = d(t)X2dt
X2(0) = 1.
This is non-random: X
d(s) ds
2(t) = eR t
0
. Since the solution of (17) is
X
f (s) dW
f 2(s) ds
1(t) = X0eR t
0
−12 R t0
,
we conclude that
X(t) = X
f (s) dW +R t d(s)
f 2(s) ds
1(t)X2(t) = X0eR t
0
0
−12
,
a formula noted earlier.
Example 2. Consider next the general equation
dX = (c(t) + d(t)X)dt + (e(t) + f (t)X)dW
(19)
X(0) = X0,
again for m = n = 1. As above, we try for a solution of the form
X(t) = X1(t)X2(t),
where now
dX1 = d(t)X1dt + f (t)X1dW
(20)
X1(0) = 1
and
dX2 = A(t)dt + B(t)dW
(21)
X2(0) = X0,
the functions A, B to be chosen. Then
dX = X2dX1 + X1dX2 + f (t)X1B(t)dt
= d(t)Xdt + f (t)XdW
+ X1(A(t)dt + B(t)dW ) + f (t)X1B(t)dt.
We now require
X1(A(t)dt + B(t)dW ) + f (t)X1B(t)dt = c(t)dt + e(t)dW ;
and this identity will hold if we take
A(t) := [c(t) − f(t)e(t)](X1(t))−1
B(t) := e(t)(X1(t))−1.
94
Observe that since X
f dW +
d
f 2 ds
1(t) = eR t
0
R t0 − 12
, we have X1(t) > 0 almost surely.
Consequently
t
X2(t) = X0 +
[c(s) − f(s)e(s)](X1(s))−1 ds
0
t
+
e(s)(X1(s))−1 dW.
0
Employing this and the expression above for X1, we arrive at the formula, a special
case of (15):
X(t) = X1(t)X2(t)
t
1
t
= exp
d(s) − f2(s) ds +
f (s) dW
0
2
0
t
r
1
s
× X0 + exp −
d(r) − f2(r) dr −
f (r) dW
(c(s) − e(s)f(s)) ds
0
0
2
0
t
s
1
s
+
exp −
d(r) − f2(r) dr −
f (r) dW
e(s) dW.
0
0
2
0
Remark. There is great theoretical and practical interest in numerical methods
for simulation of solutions to random differential equations. The paper of Higham
[H] is a good introduction.
95
CHAPTER 6: APPLICATIONS
A. Stopping times
B. Applications to PDE, Feynman-Kac formula
C. Optimal stopping
D. Options pricing
E. The Stratonovich integral
This chapter is devoted to some applications and extensions of the theory de-
veloped earlier.
A. STOPPING TIMES
DEFINITIONS, BASIC PROPERTIES. Let (Ω, U, P) be a probability
space and F(·) a filtration of σ–algebras, as in Chapters 4 and 5. We introduce now
some random times that are well–behaved with respect to F(·):
DEFINITION. A random variable τ : Ω → [0, ∞] is called a stopping time
with respect to F(·) provided
{τ ≤ t} ∈ F(t) for all t ≥ 0.
This says that the set of all ω ∈ Ω such that τ(ω) ≤ t is an F(t)-measurable
set. Note that τ is allowed to take on the value +∞, and also that any constant
τ ≡ t0 is a stopping time.
THEOREM (Properties of stopping times). Let τ1 and τ2 be stopping
times with respect to F(·). Then
(i) {τ < t} ∈ F(t), and so {τ = t} ∈ F(t), for all times t ≥ 0.
(ii) τ1 ∧ τ2 := min(τ1, τ2), τ1 ∨ τ2 := max(τ1, τ2) are stopping times.
Proof. Observe that
∞
{τ < t} =
{τ ≤ t − 1/k}.
k=1 ∈F(t−1/k)⊆F(t)
Also, we have {τ1 ∧ τ2 ≤ t} = {τ1 ≤ t} ∪ {τ2 ≤ t} ∈ F(t), and furthermore
{τ1 ∨ τ2 ≤ t} = {τ1 ≤ t} ∩ {τ2 ≤ t} ∈ F(t).
The notion of stopping times comes up naturally in the study of stochastic dif-
ferential equations, as it allows us to investigate phenomena occuring over “random
time intervals”. An example will make this clearer:
Example (Hitting a set). Consider the solution X(·) of the SDE
dX(t) = b(t, X)dt + B(t, X)dW
X(0) = X0,
where b, B and X0 satisfy the hypotheses of the Existence and Uniqueness Theorem.
96
THEOREM. Let E be either a nonempty closed subset or a nonempty open
subset of Rn. Then
τ := inf{t ≥ 0 | X(t) ∈ E}
is a stopping time. (We put τ = +∞ for those sample paths of X(·) that never hit
E.)
X(τ)
E
Proof. Fix t ≥ 0; we must show {τ ≤ t} ∈ F(t). Take {ti}∞i=1 to be a
countable dense subset of [0, ∞). First we assume that E = U is an open set. Then
the event
{τ ≤ t} =
{X(ti) ∈ U}
ti≤t ∈F(ti)⊆F(t)
belongs to F(t).
Next we assume that E = C is a closed set. Set d(x, C) := dist(x, C) and define
the open sets
1
Un = {x : d(x, C) < n}.
The event
∞
{τ ≤ t} =
{X(ti) ∈ Un)}
n=1 ti≤t ∈F(ti)⊆F(t)
also belongs to F(t).
Discussion. The random variable
σ := sup{t ≥ 0 | X(t) ∈ E},
the last time that X(t) hits E, is in general not a stopping time. The heuristic
reason is that the event {σ ≤ t} would depend upon the entire future history of
process and thus would not in general be F(t)-measurable. (In applications F(t)
“contains the history of X(·) up to and including time t, but does not contain
information about the future”.)
The name “stopping time” comes from the example, where we sometimes think
of halting the sample path X(·) at the first time τ that it hits a set E. But there are
97
many examples where we do not really stop the process at time τ . Thus “stopping
time” is not a particularly good name and “Markov time” would be better.
STOCHASTIC INTEGRALS AND STOPPING TIMES. Our next
task is to consider stochastic integrals with random limits of integration and to
work out an Itˆ
o formula for these.
DEFINITION. If G ∈ L2(0, T) and τ is a stopping time with 0 ≤ τ ≤ T, we
define
τ
T
G dW :=
χ{t≤τ}G dW.
0
0
LEMMA (Itˆ
o integrals with stopping times). If G ∈ L2(0, T)and 0 ≤
τ ≤ T is a stopping time, then
τ
(i)
E
G dW
= 0
0
τ
τ
(ii)
E (
G dW )2
= E
G2 dt .
0
0
Proof. We have
τ
T
E
G dW
= E
χ{t≤τ}G dW = 0,
0
0
∈L2(0,T)
and
τ
T
E((
G dW )2) = E((
χ{t≤τ}G dW)2)
0
0
T
= E(
(χ{t≤τ}G)2 dt)
0
τ
= E(
G2 dt).
0
Similar formulas hold for vector–valued processes.
IT ˆ
O’S FORMULA WITH STOPPING TIMES. As usual, let W(·) de-
note m-dimensional Brownian motion. Recall next from Chapter 4 that if dX =
b(X, t)dt + B(X, t)dW, then for each C2 function u,
∂u
n
∂u
1 n
∂2u
m
(1)
du(X, t) =
dt +
dXi +
bikbjkdt.
∂t
∂x
2
∂x
i=1
i
i,j=1
i∂xj k=1
98
Written in integral form, this means:
t
∂u
t
(2)
u(X(t), t) − u(X(0), 0) =
+ Lu
ds +
Du · BdW,
0
∂t
0
for the differential operator
1 n
n
m
Lu :=
aiju
+
biu , aij =
bikbjk,
2
xixj
xi
i,j=1
i=1
k=1
and
m
n
Du · BdW =
ux bikdW k.
i
k=1 i=1
The argument of u in these integrals is (X(s), s). We call L the generator.
For a fixed ω ∈ Ω, formula (2) holds for all 0 ≤ t ≤ T. Thus we may set t = τ,
where τ is a stopping time, 0 ≤ τ ≤ T:τ ∂u
τ
u(X(τ ), τ ) − u(X(0), 0) =
+ Lu
ds +
Du · BdW.
0
∂t
0
Take expected value:
τ
∂u
(3)
E(u(X(τ ), τ )) − E(u(X(0), 0)) = E
+ Lu
ds .
0
∂t
We will see in the next section that this formula provides a very important
link between stochastic differential equations and (nonrandom) partial differential
equations.
BROWNIAN MOTION AND THE LAPLACIAN. The most important
case is X(·) = W(·), n-dimensional Brownian motion, the generator of which is
n
1
1
Lu =
u
=:
∆u.
2
xixi
2
i=1
The expression ∆u is called the Laplacian of u and occurs throughout mathematics
and physics. We will demonstrate in the next section some important links with
Brownian motion.
B. APPLICATIONS TO PDE, FEYNMAN–KAC FORMULA
PROBABILISTIC FORMULAS FOR SOLUTIONS OF PDE.
Example 1 (Expected hitting time to a boundary). Let U ⊂ Rn be a
bounded open set, with smooth boundary ∂U . According to standard PDE theory,
there exists a smooth solution u of the equation
∆u = 1
in U
(4)
−12 u = 0 on ∂U.
99
Our goal is to find a probabilistic representation formula for u. For this, fix
any point x ∈ U and consider then an n-dimensional Brownian motion W(·). Then
X(·) := W(·) + x represents a “Brownian motion starting at x”. Define
τx := first time X(·) hits ∂U.
THEOREM. We have
(5)
u(x) = E(τx)
for all x ∈ U.
In particular, u > 0 in U .
Proof. We employ formula (3), with Lu = 1 ∆u. We have for each n = 1, 2, . . .
2
τx∧n 1
E(u(X(τx ∧ n))) − E(u(X(0))) = E
∆u(X) ds .
0
2
Since 1 ∆u =
2
−1 and u is bounded,
lim E(τx
n→∞
∧ n) < ∞.
Thus τx is integrable. Thus if we let n → ∞ above, we get
τx
u(x) − E(u(X(τx))) = E
1 ds
= E(τx).
0
But u = 0 on ∂U , and so u(X(τx)) ≡ 0. Formula (5) follows.
Again recall that u is bounded on U . Hence
E(τx) < ∞, and so τx < ∞ a.s., for all x ∈ U.
This says that Brownian sample paths starting at any point x ∈ U will with proba-
bility 1 eventually hit ∂U .
Example 2 (Probabilistic representation of harmonic functions). Let U ⊂
Rn be a smooth, bounded domain and g : ∂U → R a given continuous function. It
is known from classical PDE theory that there exists a function u ∈ C2(U) ∩ C( ¯U)
satisfying the boundary value problem:
∆u = 0 in U
(6)
u = g
on ∂U .
We call u a harmonic function.
THEOREM. We have for each point x ∈ U
(7)
u(x) = E(g(X(τx))),
for X(·) := W(·) + x, Brownian motion starting at x.
100
Proof. As shown above,
τx 1
E(u(X(τx))) = E(u(X(0))) + E
∆u(X) ds = E(u(X(0))) = u(x),
0
2
the second equality valid since ∆u = 0 in U . Since u = g on ∂U , formula (7)
follows.
APPLICATION: In particular, if ∆u = 0 in some open set containing the
ball B(x, r), then
u(x) = E(u(X(τx))),
where τx now denotes the hitting time of Brownian motion starting at x to ∂B(x, r).
Since Brownian motion is isotropic in space, we may reasonably guess that the term
on the right hand side is just the average of u over the sphere ∂B(x, r), with respect
to surface measure. That is, we have the identity
1
(8)
u(x) =
u dS.
area of ∂B(x, r) ∂B(x,r)
This is the mean value formula for harmonic functions.
Example 3 (Hitting one part of a boundary first). Assume next that we can
write ∂U as the union of two disjoint parts Γ1, Γ2. Let u solve the PDE
∆u=0 inU
u=1 on Γ1
u=0 on Γ2.
THEOREM. For each point x ∈ U, u(x) is the probability that a Brownian
motion starting at x hits Γ1 before hitting Γ2.
Γ1
x
Γ2
Proof. Apply (7) for
1
on Γ1
g =
0
on Γ2.
Then
u(x) = E(g(X(τx))) = probability of hitting Γ1 before Γ2.
101
FEYNMAN–KAC FORMULA. Now we extend Example #2 above to obtain
a probabilistic representation for the unique solution of the PDE
−1∆u + cu = f in U
2
u = 0 on ∂U.
We assume c, f are smooth functions, with c ≥ 0 in U.
THEOREM (Feynman–Kac formula). For each x ∈ U,
τx
u(x) = E
f (X(t))e− R t c(X) ds
0
dt
0
where, as before, X(·) := W(·) + x is a Brownian motion starting at x, and τx
denotes the first hitting time of ∂U .
Proof. We know E(τx) < ∞. Since c ≥ 0, the integral above converges.
First look at the process
Y (t) := eZ(t),
for Z(t) := − t c(X) ds. Then
0
dZ = −c(X)dt,
and so Itˆo’s formula yields
dY = −c(X)Y dt.
Hence the Itˆo product rule implies
d u(X)e− R t c(X) ds
c(X) ds
0
= (du(X))e− R t0
+ u(X)d e− R t c(X)ds
0
1
n
∂u(X)
=
∆u(X)dt +
dW i
e− R t c(X)ds
0
2
∂x
i=1
i
+ u(X)(−c(X)dt)e−Rt c(X)ds
0
.
We use formula (3) for τ = τx, and take the expected value, obtaining
E u(X(τ
c(X) ds
x))e− R τx
0
− E(u(X(0)))
τx
1
= E
∆u(X) − c(X)u(X) e−Rt c(X)ds
0
dt .
0
2
Since u solves (8), this simplifies to give
τx
u(x) = E(u(X(0))) = E
f (X)e− R t c(X) ds
0
dt ,
0
as claimed.
An interpretation. We can explain this formula as describing a Brownian motion
with “killing”, as follows.
102
Suppose that the Brownian particles may disappear at a random killing time σ,
for example by being absorbed into the medium within which it is moving. Assume
further that the probability of its being killed in a short time interval [t, t + h] is
c(X(t))h + o(h).
Then the probability of the particle surviving until time t is approximately equal to
(1 − c(X(t1))h)(1 − c(X(t2))h) . . .(1 − c(X(tn))h),
where 0 = t0 < t1 < · · · < tn = t, h = tk+1 − tk. As h → 0, this converges to
e− R t c(X) ds
0
.
Hence it should be that
u(x) = average of f (X(·)) over all sample paths which survive to hit ∂U
τx
= E
f (X)e− R t c(X) ds
0
dt .
0
Remark. If we consider in these examples the solution of the SDE
dX = b(X)dt + B(X)dW
X(0) = x,
we can obtain similar formulas, where now
τx = hitting time of ∂U for X(·)
and 1 ∆u is replaced by the operator
2
1 n
n
Lu :=
aiju
+
biu .
2
xixj
xi
i,j=1
i=1
Note, however, we need to know that the various PDE have smooth solutions. This
need not always be the case for degenerate elliptic operators L.
C. OPTIMAL STOPPING The general mathematical setting for many con-
trol theory problems is this. We are given some “system” whose state evolves in
time according to a differential equation (deterministic or stochastic). Given also
are certain controls which affect somehow the behavior of the system: these controls
typically either modify some parameters in the dynamics or else stop the process,
or both. Finally we are given a cost criterion, depending upon our choice of control
and the corresponding state of the system.
The goal is to discover an optimal choice of controls, to minimize the cost
criterion.
The easiest stochastic control problem of the general type outlined above occurs
when we cannot directly affect the SDE controlling the evolution of X(·) and can
only decide at each instance whether or not to stop. A typical such problem follows.
103
STOPPING A STOCHASTIC DIFFERENTIAL EQUATION. Let
U ⊂ Rm be a bounded, smooth domain. Suppose b : Rn → Rn, B : Rn → Mn×m
satisfy the usual assumptions.
Then for each x ∈ U the stochastic differential equation
dX = b(X)dt + B(X)dW
X0 = x
has a unique solution. Let τ = τx denote the hitting time of ∂U . Let θ be any
stopping time with respect to F(·), and for each such θ define the expected cost of
stopping X(·) at time θ ∧ τ to be
θ∧τ
(9)
Jx(θ) := E(
f (X(s)) ds + g(X(θ ∧ τ))).
0
The idea is that if we stop at the possibly random time θ < τ , then the cost is a
given function g of the current state of X(θ). If instead we do not stop the process
before it hits ∂U , that is, if θ ≥ τ, the cost is g(X(τ)). In addition there is a running
cost per unit time f of keeping the system in operation until time θ ∧ τ.
OPTIMAL STOPPING. The main question is this: does there exist an
optimal stopping time θ∗ = θ∗x, for which
Jx(θ∗) =
min
Jx(θ)?
θ stopping
time
And if so, how can we find θ∗? It turns out to be very difficult to try to design θ∗
directly. A much better idea is to turn attention to the value function
(10)
u(x) := inf Jx(θ),
θ
and to try to figure out what u is as a function of x ∈ U. Note that u(x) is the
minimum expected cost, given we start the process at x. It turns out that once we
know u, we will be then be able to construct an optimal θ∗. This approach is called
dynamic programming.
OPTIMALITY CONDITIONS. So assume u is defined above and suppose
u is smooth enough to justify the following calculations. We wish to determine the
properties of this function.
First of all, notice that we could just take θ ≡ 0 in the definition (10). That is,
we could just stop immediately and incur the cost g(X(0)) = g(x). Hence
(11)
u(x) ≤ g(x) for each point x ∈ U.
Furthermore, τ ≡ 0 if x ∈ ∂U, and so
(12)
u(x) = g(x) for each point x ∈ ∂U.
Next take any point x ∈ U and fix some small number δ > 0. Now if we do not
stop the system for time δ, then according to (SDE) the new state of the system
104
at time δ will be X(δ). Then, given that we are at the point X(δ), the best we can
achieve in minimizing the cost thereafter must be
u(X(δ)).
So if we choose not to stop the system for time δ, and assuming we do not hit ∂U ,
our cost is at least
δ
E(
f (X) ds + u(X(δ))).
0
Since u(x) is the infimum of costs over all stopping times, we therefore have
δ
u(x) ≤ E( f(X) ds + u(X(δ))).
0
Now by Itˆo’s formula
δ
E(u(X(δ))) = u(x) + E(
Lu(X) ds),
0
for
1 n
n
m
Lu =
aij ∂2u
+
bi ∂u ,
aij =
bikbjk.
2
∂xi∂xj
∂xi
i,j=1
i=1
k=1
Hence
δ
0 ≤ E( f(X) + Lu(X) ds).
0
Divide by δ > 0, and then let δ → 0:
0 ≤ f(x) + Lu(x).
Equivalently, we have
(13)
M u ≤ f in U,
where M u := −Lu.
Finally we observe that if in (11) a strict inequality held, that is, if
u(x) < g(x) at some point x ∈ U,
then it is optimal not to stop the process at once. Thus it is plausible to think
that we should leave the system going, for at least some very small time δ. In this
circumstance we then would have an equality in the formula above; and so
(14)
M u = f
at those points where u < g.
In summary, we combine (11)–(14) to find that if the formal reasoning above is
valid, then the value function u satisfies:
max(M u
(15)
− f, u − g) = 0 in U
u = g
on ∂U
These are the optimality conditions.
105
SOLVING FOR THE VALUE FUNCTION. Our rigorous study of the
stopping time problem now begins by showing first that there exists a unique solu-
tion u of (15) and second that this u is in fact minθ Jx(θ). Then we will use u to
design θ∗, an optimal stopping time.
THEOREM. Suppose f, g are given smooth functions. There exists a unique
funtion u, with bounded second derivatives, such that:
(i) u ≤ g in U,
(ii) M u ≤ f almost everywhere in U,
(iii) max(M u − f, u − g) = 0 almost everywhere in U,
(iv) u = g on ∂U .
In general u /
∈ C2(U).
The idea of the proof is to approximate (15) by a penalized problem of this form:
M uε + βε(uε − g) = f in U
uε = g
on ∂U ,
where βε : R → R is a smooth, convex function, β′ε ≥ 0, and βε ≡ 0 for x ≤ 0,
limǫ→0 βε(x) = ∞ for x > 0. Then uε → u. It will in practice be difficult to find
a precise formula for u, but computers can provide accurate numerical approxima-
tions.
DESIGNING AN OPTIMAL STOPPING POLICY. Now we show that
our solution of (15) is in fact the value function, and along the way we will learn
how to design an optimal stopping strategy θ∗.
First note that the stopping set
S := {x ∈ U | u(x) = g(x)}
is closed. Define for each x ∈ ¯U,
θ∗ = first hitting time of S.
THEOREM. Let u be the solution of (15). Then
u(x) = Jx(θ∗) = inf Jx(θ)
θ
for all x ∈ ¯U.
This says that we should first compute the solution to (15) to find S, define θ∗
as above, and then we should run X(·) until it hits S (or else exits from U).
Proof. 1. Define the continuation set
C := U − S = {x ∈ U | u(x) < g(x)}.
On this set Lu = f , and furthermore u = g on ∂C. Since τ ∧ θ∗ is the exit time
from C, we have for x ∈ Cτ∧θ∗
u(x) = E(
f (X(s)) ds + g(X(θ∗ ∧ τ))) = Jx(θ∗).
0
106
On the other hand, if x ∈ S, τ ∧ θ∗ = 0; and so
u(x) = g(x) = Jx(θ∗).
Thus for all x ∈ ¯U, we have u(x) = Jx(θ∗).
2. Now let θ be any other stopping time. We need to show
u(x) = Jx(θ∗) ≤ Jx(θ).
Now by Itˆ
o’s formula
τ ∧θ
u(x) = E(
M u(X) ds + u(X(τ ∧ θ)))
0
But M u ≤ f and u ≤ g in ¯U. Hence
τ ∧θ
u(x) ≤ E(
f (X) ds + g(X(τ ∧ θ))) = Jx(θ).
0
But since u(x) = Jx(θ∗), we consequently have
u(x) = Jx(θ∗) = min Jx(θ),
θ
as asserted.
D. OPTIONS PRICING
In this section we outline an application to mathematical finance, mostly fol-
lowing Baxter–Rennie [B-R] and the class lectures of L. Goldberg. Another basic
reference is Hull [Hu].
THE BASIC PROBLEM. Let us consider a given security, say a stock,
whose price at time t is S(t). We suppose that S evolves according to the SDE
introduced in Chapter 5:
dS = µSdt + σSdW
(16)
S(0) = s0,
where µ > 0 is the drift and σ = 0 the volatility. The initial price s0 is known.
A derivative is a financial instrument whose payoff depends upon (i.e., is derived
from) the behavior of S(·). We will investigate a European call option, which is the
right to buy one share of the stock S, at the price p at time T . The number p is
called the strike price and T > 0 the strike (or expiration) time. The basic question
is this:
What is the “proper price” at time t = 0 of this option?
In other words, if you run a financial firm and wish to sell your customers this call
option, how much should you charge? (We are looking for the “break–even” price,
for which the firm neither makes nor loses money.)
107
ARBITRAGE AND HEDGING. To simplify, we assume hereafter that the
prevailing, no-risk interest rate is the constant r > 0. This means that $1 put in a
bank at time t = 0 becomes $erT at time t = T . Equivalently, $1 at time t = T is
worth only $e−rT at time t = 0.
As for the problem of pricing our call option, a first guess might be that the
proper price should be
(17)
e−rT E((S(T ) − p)+),
for x+ := max(x, 0). The reasoning behind this guess is that if S(T ) < p, then the
option is worthless. If S(T ) > p, we can buy a share for the price p, immediately
sell at price S(T ), and thereby make a profit of (S(T ) − p)+. We average this over
all sample paths and multiply by the discount factor e−rT , to arrive at (17).
As reasonable as this may all seem, (17) is in fact not the proper price. Other
forces are at work in financial markets. Indeed the fundamental factor in options
pricings is arbitrage, meaning the possibility of risk-free profits.
We must price our option so as to create no arbitrage opportunities for others.
To convert this principle into mathematics, we introduce also the notion of
hedging. This means somehow eliminating our risk as the seller of the call option.
The exact details appear below, but the basic idea is that we can in effect “duplicate”
our option by a portfolio consisting of (continually changing) holdings of a risk–free
bond and of the stock on which the call is written.
A PARTIAL DIFFERENTIAL EQUATION. We demonstrate next how
use these principles to convert our pricing problem into a PDE. We introduce for
s ≥ 0 and 0 ≤ t ≤ T, the unknown price function
(18)
u(s, t), denoting the proper price of the option at time t, given that S(t) = s.
Then u(s0, 0) is the price we are seeking.
Boundary conditions. We need to calculate u. For this, notice first that at the
expiration time T , we have
(19)
u(s, T ) = (s − p)+
(s ≥ 0).
Furthermore, if s = 0, then S(t) = 0 for all 0 ≤ t ≤ T and so
(20)
u(0, t) = 0
(0 ≤ t ≤ T).
We seek how a PDE u solves for s > 0, 0 ≤ t ≤ T.
Duplicating an option, self-financing. To go further, define the process
(21)
C(t) := u(S(t), t)
(0 ≤ t ≤ T).
108
Thus C(t) is the current price of the option at time t, and is random since the stock
price S(t) is random. According to Itˆo’s formula and (16)
1
dC = utdt + usdS + u
2 ss(dS)2
(22)
σ2
= (ut + µSus +
S2u
2
ss)dt + σSusdW.
Now comes the key idea: we propose to “duplicate” C by a portfolio consisting of
shares of S and of a bond B. More precisely, assume that B is a risk-free investment,
which therefore grows at the prevailing interest rate r:
dB = rBdt
(23)
B(0) = 1.
This just means B(t) = ert, of course. We will try to find processes φ and ψ so that
(24)
C = φS + ψB
(0 ≤ t ≤ T).
Discussion. The point is that if we can construct φ, ψ so that (24) holds, we can
eliminate all risk. To see this more clearly, imagine that your financial firm sells a
call option, as above. The firm thereby incurs the risk that at time T , the stock
price S(T ) will exceed p, and so the buyer will exercise the option. But if in the
meantime the firm has constructed the portfolio (24), the profits from it will exactly
equal the funds needed to pay the customer. Conversely, if the option is worthless
at time T , the portfolio will have no profit.
But to make this work, the financial firm should not have to inject any new
money into the hedging scheme, beyond the initial investment to set it up. We
ensure this by requiring that the portfolio represented on the right-hand side of
(24) be self-financing. This means that the changes in the value of the portfolio
should depend only upon the changes in S, B. We therefore require that
(25)
dC = φdS + ψdB
(0 ≤ t ≤ T).
Remark (discrete version of self-financing). Roughly speaking, a portfolio is
self-financing if it is financially self contained. To understand this better, let us
consider a different model in which time is discrete, and the values of the stock and
bond at a time ti are given by Si and Bi respectively. Here {ti}Ni=0 is an increasing
sequence of times and we suppose that each time step ti+1 −ti is small. A portfolio
can now be thought of as a sequence {(φi, ψi)}Ni=0, corresponding to our changing
holdings of the stock S and the bond B over each time interval.
Now for a given time interval (ti, ti+1), Ci = φiSi + ψiBi is the opening value
of the portfolio and Ci+1 = φiSi+1 + ψiBi+1 represents the closing value. The
self-financing condition means that the financing gap Ci+1 −Ci of cash (that would
otherwise have to be injected to pay for our construction strategy) must be zero.This
is equivalent to saying that
Ci+1 − Ci = φi(Si+1 − Si) + ψi(Bi+1 − Bi),
109
the continuous version of which is condition (25).
Combining formulas (22), (23) and (25) provides the identity
σ2
(u
S2u
(26)
t + µSus + 2
ss)dt + σSusdW
= φ(µSdt + σSdW ) + ψrBdt.
So if (24) holds, (26) must be valid, and we are trying to select φ, ψ to make all this
so. We observe in particular that the terms multiplying dW on each side of (26)
will match provided we take
(27)
φ(t) := us(S(t), t)
(0 ≤ t ≤ T).
Then (26) simplifies, to read
σ2
(ut +
S2u
2
ss)dt = rψBdt.
But ψB = C − φS = u − usS, according to (24), (27). Consequently,
σ2
(28)
(ut + rSus +
S2u
2
ss − ru)dt = 0.
The argument of u and its partial derivatives is (S(t), t).
Consequently, to make sure that (21) is valid, we ask that the function u =
u(s, t) solve the Black–Scholes–Merton PDE
σ2
(29)
ut + rsus +
s2u
2
ss − ru = 0
(0 ≤ t ≤ T).
The main outcome of all our financial reasoning is the derivation of this partial
differential equation. Observe that the parameter µ does not appear.
More on self-financing. Before going on, we return to the self-financing condition
(25). The Itˆ
o product rule and (24) imply
dC = φdS + ψdB + Sdφ + Bdψ + dφ dS.
To ensure (25), we consequently must make sure that
(30)
Sdφ + Bdψ + dφ dS = 0,
where we recall φ = us(S(t), t). Now dφ dS = σ2S2ussdt. Thus (30) is valid
provided
(31)
dψ = −B−1(Sdφ + σ2S2ussdt).
We can confirm this by noting that (24), (27) imply
ψ = B−1(C − φS) = e−rt(u(S, t) − us(S, t)S).
A direct calculation using (28) verifies (31).
110
SUMMARY. To price our call option, we solve the boundary-value problem
ut + rsus + σ2 s2u
2
ss − ru = 0 (s > 0, 0 ≤ t ≤ T )
(32)
u=(s−p)+
(s > 0, t = T )
u=0
(s = 0, 0 ≤ t ≤ T).
Remember that u(s0, 0) is the price we are trying to find. It turns out that this
problem can be solved explicitly, although we omit the details here: see for instance
Baxter–Rennie [B-R].
E. THE STRATONOVICH INTEGRAL
We next discuss the Stratonovich stochastic calculus, which is an alternative to
Itˆ
o’s approach. Most of the following material is from Arnold [A, Chapter 10].
1. Motivation. Let us consider first of all the formal random differential
equation
˙
X = d(t)X + f (t)Xξ
(33)
X(0) = X0,
where m = n = 1 and ξ(·) is 1-dimensional “white noise”. If we interpret this
rigorously as the stochastic differential equation:
dX = d(t)Xdt + f (t)XdW
(34)
X(0) = X0,
we then recall from Chapter 5 that the unique solution is
(35)
X(t) = X
d(s)
f 2(s) ds+R t f(s) dW
0eR t
0
−12
0
.
On the other hand perhaps (33) is a proposed mathematical model of some
physical process and we are not really sure whether ξ(·) is “really” white noise.
It could perhaps be instead some process with smooth (but highly complicated)
sample paths. How would this possibility change the solution?
APPROXIMATING WHITE NOISE. More precisely, suppose that {ξk(·)}∞k=1
is a sequence of stochastic processes satisfying:
(a) E(ξk(t)) = 0,
(b) E(ξk(t)ξk(s)) := dk(t − s),
(c) ξk(t) is Gaussian for all t ≥ 0,
(d) t → ξk(t) is smooth for all ω,
where we suppose that the functions dk(·) converge as k → ∞ to δ0, the Dirac
measure at 0.
In light of the formal definition of the white noise ξ(·) as a Gaussian process
with Eξ(t) = 0, E(ξ(t)ξ(s)) = δ0(t − s), the ξk(·) are thus presumably smooth
approximations of ξ(·).
LIMITS OF SOLUTIONS. Now consider the problem
˙
Xk = d(t)Xk + f (t)Xkξk
(36)
Xk(0) = X0.
111
For each ω this is just a regular ODE, whose solution is
Xk(t) := X
d(s) ds+R t f(s)ξk(s) ds
0eR t
0
0
.
Next look at
t
Zk(t) :=
f (s)ξk(s) ds.
0
For each time t ≥ 0, this is a Gaussian random variable, with
E(Zk(t)) = 0.
Furthermore,
t
s
E(Zk(t)Zk(s)) =
f (τ )f (σ)dk(τ − σ) dσdτ
0
0
t
s
→
f (τ )f (σ)δ0(τ − σ) dσdτ
0
0
t∧s
=
f 2(τ ) dτ.
0
Hence as k → ∞, Zk(t) converges in L2 to a process whose distributions agree with
those
t f(s) dW . And therefore Xk(t) converges to a process whose distributions
0
agree with
(37)
ˆ
X(t) := X
d(s) ds+R t f(s) dW
0eR t
0
0
.
This does not agree with the solution (35)!
Discussion. Thus if we regard (33) as an Itˆo SDE with ξ(·) a “true” white
noise, (35) is our solution. But if we approximate ξ(·) by smooth processes ξk(·),
solve the approximate problems (36) and pass to limits with the approximate so-
lutions Xk(·), we get a different solution. This means that (33) is unstable with
respect to changes in the random term ξ(·). This conclusion has important conse-
quences in questions of modeling, since it may be unclear experimentally whether
we really have ξ(·) or instead ξk(·) in (33) and similar problems.
In view of all this, it is appropriate to ask if there is some way to redefine
the stochastic integral so these difficulties do not come up. One answer is the
Stratonovich integral.
2. Definition of Stratonovich integral.
A one-dimensional example. Recall that in Chapter 4 we defined for 1-
dimensional Brownian motion
T
mn−1
W 2(T )
W dW := lim
W (tn
− T
k )(W (tn
k+1) − W (tnk)) =
,
0
|Pn|→0
2
k=0
where P n := {0 = tn0 < tn1 < · · · < tnm = T
n
} is a partition of [0, T]. This
corresponds to a sequence of Riemann sum approximations, where the integrand is
evaluated at the left-hand endpoint of each subinterval [tn, tn
].
k
k+1
112
The corresponding Stratonovich integral is instead defined this way:
T
mn−1 W (tn ) + W (tn)
W 2(T )
W ◦ dW := lim
k+1
k
(W (tnk+1) − W(tnk)) =
.
0
|Pn|→0
2
2
k=0
(Observe the notational change: we hereafter write a small circle before the dW to
signify the Stratonovich integral.) According to calculations in Chapter 4, we also
have
T
mn−1
tn
+ tn
W ◦ dW = lim
W
k+1
k
(W (tnk+1) − W(tnk)).
0
|Pn|→0
2
k=0
Therefore for this case the Stratonovich integral corresponds to a Riemann sum
approximation where we evaluate the integrand at the midpoint of each subinterval
[tn, tn
].
k
k+1
We generalize this example and so introduce the
DEFINITION. Let W(·) be an n-dimensional Brownian motion and let B :
Rn × [0, T] → Mn×n be a C1 function such that
T
E
|B(W, t)|2 dt < ∞.
0
Then we define
T
mn−1
W(tn
) + W(tn)
B(W, t) ◦ dW := lim
B
k+1
k , tnk (W(tnk+1) − W(tnk)).
0
|Pn|→0
2
k=0
It can be shown that this limit exists in L2(Ω).
A CONVERSION FORMULA. Remember that Itˆ
o’s integral can be com-
puted this way:
T
mn−1
B(W, t) dW = lim
B(W(tnk), tnk)(W(tnk+1) − W(tnk)).
0
|Pn|→0 k=0
This is in general not equal to the Stratonovich integral, but there is a conversion
formula
i
i
T
T
1
T
n
∂bij
(38)
B(W, t) ◦ dW =
B(W, t) dW
+
(W, t) dt,
0
0
2 0
∂x
j=1
j
113
for i = 1, . . . , n. Here vi means the ith–component of the vector function v. This
formula is proved by noting
T
T
B(W, t) ◦ dW −
B(W, t) dW
0
0
mn−1
W(tn
) + W(tn)
= lim
B
k+1
k , tnk − B(W(tnk),tnk)
|Pn|→0
2
k=0
· (W(tnk+1) − W(tnk))
and using the Mean Value Theorem plus some usual methods for evaluating the
limit. We omit details.
Special case. If n = 1, then
T
T
1
T ∂b
b(W, t) ◦ dW =
b(W, t) dW +
(W, t) dt.
0
0
2 0 ∂x
Assume now B : Rn ×[0, T] → Mn×m and W(·) is an m-dimensional Brownian
motion. We make this informal
DEFINITION. If X(·) is a stochastic process with values in Rn, we define
T
mn−1
X(tn
) + X(tn)
B(X, t) ◦ dW := lim
B
k+1
k , tnk (W(tnk+1) − W(tnk)),
0
|Pn|→0
2
k=0
provided this limit exists in L2(Ω) for all sequences of partitions P n, with |Pn| → 0.
3. Stratonovich chain rule.
DEFINITION. Suppose that the process X(·) solves the Stratonovich integral
equation
t
t
X(t) = X(0) +
b(X, s) ds +
B(X, s) ◦ dW (0 ≤ t ≤ T)
0
0
for b : Rn × [0, T] → Rn and B : Rn × [0, T] → Mn×m. We then write
dX = b(X, t)dt + B(X, t) ◦ dW,
the second term on the right being the Stratonovich stochastic differential.
114
THEOREM (Stratonovich chain rule). Assume
dX = b(X, t)dt + B(X, t) ◦ dW
and suppose u : Rn × [0, T] → R is smooth. Define
Y (t) := u(X(t), t).
Then
n
∂u
∂u
dY =
dt +
∂t
∂x ◦ dXi
i=1
i
∂u
n
∂u
n
m
∂u
=
+
bi
dt +
bik
∂t
∂x
∂x
◦ dWk.
i=1
i
i=1
i
k=1
Thus the ordinary chain rule holds for Stratonovich stochastic differentials, and
there is no additional term involving
∂2u
as there is for Itˆo’s formula. We omit
∂xi∂xj
the proof, which is similar to that for the Itˆo rule. The main difference is that we
make use of the formula
T W
W 2(T ) in the approximations.
0
◦ dW = 12
More discussion. Next let us return to the motivational example we began
with. We have seen that if the differential equation (33) is interpreted to mean
dX = d(t)Xdt + f (t)XdW
(Itˆo’s sense),
X(0) = X0,
then
X(t) = X
d(s)
f 2(s) ds+R t f(s) dW
0eR t
0
−12
0
.
However, if we interpret (33) to mean
dX = d(t)Xdt + f (t)X ◦ dW (Stratonovich’s sense)
X(0) = X0,
the solution is
˜
X(t) = X
d(s) ds+R t f(s) dW
0eR t
0
0
,
as is easily checked using the Stratonovich calculus described above.
This solution ˜
X(·) is also the solution obtained by approximating the “white
noise” ξ(·) by smooth processes ξk(·) and passing to limits. This suggests that in-
terpreting (16) and similar formal random differential equations in the Stratonovich
sense will provide solutions which are stable with respect to perturbations in the
random terms. This is indeed the case: See the articles [S1-2] by Sussmann.
Note also that these considerations clarify a bit the problems of interpreting
mathematically the formal random differential equation (33), but do not say which
interpretation is physically correct. This is a question of modeling and is not, strictly
speaking, a mathematical issue.
CONVERSION RULES FOR SDE.
115
Let W(·) be an m-dimensional Wiener process and suppose b : Rn × [0, T] →
Rn, B : Rn × [0, T] → Mn×m satisfy the hypotheses of the basic existence and
uniqueness theorem. Then X(·) solves the Itˆo stochastic differential equation
dX = b(X, t)dt + B(X, t)dW
X(0) = X0
if and only if X(·) solves the Stratonovich stochastic differential equation
dX = b(X, t) − 1c(X, t) dt + B(X, t)
2
◦ dW
X(0) = X0,
for
m
n
∂bik
ci(x, t) =
(x, t)bjk(x, t)
(1
∂x
≤ i ≤ n).
j
k=1 j=1
A special case. For m = n = 1, this says
dX = b(X)dt + σ(X)dW
if and only if
1
dX = (b(X) − σ′(X)σ(X))dt + σ(X)
2
◦ dW.
4. Summary We conclude these lectures by summarizing the advantages of
each definition of the stochastic integral:
Advantages of Itˆ
o integral
2
1. Simple formulas: E
t G dW = 0, E
t G dW
= E
t G2 dt .
0
0
0
2. I(t) = t G dW is a martingale.
0
Advantages of Stratonovich integral
1. Ordinary chain rule holds.
2. Solutions of stochastic differential equations interpreted in Stratonovich
sense are stable with respect to changes in random terms.
116
APPENDICES
Appendix A: Proof of Laplace–De Moivre Theorem (from §G in Chapter
2)
Proof. 1. Set S∗n := Sn−np
√
, this being a random variable taking on the value
npq
xk = k−np
√
(k = 0, . . . , n) with probability p
pkqn−k.
npq
n(k) =
n
k
Look at the interval
−np
√
, nq
. The points x
npq
√npq
k divide this interval into n
subintervals of length
1
h := √ .
npq
Now if n goes to ∞, and at the same time k changes so that |xk| is bounded, then
k = np + x √
k
npq → ∞
and
n − k = nq − x √
k
npq → ∞.
2. We recall next Stirling’s formula, which says says
n! = e−nnn√2πn (1 + o(1)) as n → ∞,
where “o(1)” denotes a term which goes to 0 as n → ∞. (See Mermin [M] for a
nice discussion.) Hence as n → ∞
n
n!
pn(k) =
pkqn−k =
pkqn−k
k
k!(n − k)!
e−nnn√2πnpkqn−k
(1)
=
(1 + o(1))
e−kkk√2πke−(n−k)(n − k)(n−k) 2π(n − k)
1
n
np k
nq
n−k
= √
(1 + o(1)).
2π
k(n − k) k
n − k
3. Observe next that if x = xk = k−np
√
, then
npq
q
q
k
k
1 +
x = 1 +
− np =
np
np
√npq
np
and
p
n
1 −
x =
− k.
nq
nq
117
Note also log(1 ± y) = ±y − y2 + O(y3) as y
2
→ 0. Hence
np k
k
log
=
k
−k log np
q
= −k log 1 +
x
np
q
q
= −(np + x√npq)
x
x2
+ O n− 12 .
np − 2np
Similarly,
nq
n−k
p
p
log
=
x
x2
+ O n− 12 .
n − k
−(nq − x√npq) − nq − 2nq
Add these expressions and simplify, to discover
np k
nq
n−k
x2
lim
log
=
.
n
k
n
− 2
k
→∞
−np
− k
√npq →x
Consequently
np k
nq
n−k
(2)
lim
= e− x2
2 .
n
k
n
k
→∞
−np
− k
√npq →x
4. Finally, observe
n
1
(3)
=
(1 + o(1)) = h(1 + o(1)),
k(n − k) √npq
since k = np + x√npq, n − k = nq − x√npq.
Now
P (a ≤ S∗n ≤ b) =
pn(k)
a≤xk≤b
xk= k−np
√npq
for a < b. In view of (1)−(3), the latter expression is a Riemann sum approximation
as n → ∞ of the integral
1
b
√
e− x2
2
dx.
2π a
Appendix B: Proof of discrete martingale inequalities (from §I in Chap-
ter 2)
118
Proof. 1. Define
k−1
Ak :=
{Xj ≤ λ} ∩ {Xk > λ} (k = 1, . . ., n).
j=1
Then
n
A :=
max Xk > λ =
Ak .
1≤k≤n
k=1
disjoint union
Since λP (Ak) ≤
X
A
k dP , we have
k
n
n
(4)
λP (A) = λ
P (Ak) ≤
E(χA X
k
k).
k=1
k=1
Therefore
n
E(X+
n ) ≥
E(X+
n χA )
k
k=1
n
=
E(E(X+
n χAk | X1, . . . , Xk))
k=1
n
=
E(χA E(X+
k
n | X1, . . . , Xk))
k=1
n
≥
E(χA E(X
k
n | X1, . . . , Xk))
k=1
n
≥
E(χA X
k
k)
by the submartingale property
k=1
≥ λP(A) by (4).
2. Notice next that the proof above in fact demonstrates
λP
max Xk > λ ≤
X+
n dP.
1≤k≤n
{max1≤k≤n Xk>λ}
Apply this to the submartingale |Xk|:
(5)
λP (X > λ) ≤
Y dP,
{X>λ}
119
for X := max1≤k≤n |Xk|, Y := |Xn|. Now take some 1 < p < ∞. Then
∞
E(|X|p) = −
λp dP (λ) for P (λ) := P (X > λ)
0
∞
= p
λp−1P (λ) dλ
0
∞
≤ p
λp−1
1
Y dP
dλ
by (5)
0
λ {X>λ}
X
= p
Y
λp−2dλ
dP
Ω
0
p
=
Y Xp−1 dP
p − 1 Ω
p
1/p
1−1/p
≤
Y p dP
Xp dP
.
p − 1 Ω
Ω
Appendix C: Proof of continuity of indefinite Itˆ
o integral (from §C in
Chapter 4)
Proof. We will assume assertion (i) of the Theorem in §C of Chapter 4, which
states that the indefinite integral I(·) is a martingale.
There exist step processes Gn ∈ L2(0, T), such that
T
E
(Gn − G)2 dt → 0.
0
Write In(t) := t Gn dW , for 0
for tn
, then
0
≤ t ≤ T. If Gn(s) ≡ Gnk
k ≤ s < tnk+1
k−1
In(t) =
Gni(W (tni+1) − W(tni)) + Gnk(W(t) − W(tnk))
i=0
for tn
. Therefore In(
k ≤ t < tnk+1
·) has continuous sample paths a.s., since Brownian
motion does. Since In(·) is a martingale, it follows that |In−Im|2 is a submartingale.
The martingale inequality now implies
P
sup |In(t) − Im(t)| > ε = P sup |In(t) − Im(t)|2 > ε2
0≤t≤T
0≤t≤T
1
≤ E(
ε2
|In(T) − Im(T)|2)
1
T
=
E
ε2
|Gn − Gm|2 dt .
0
120
Choose ε = 1 . Then there exists n
2k
k such that
1
T
P
sup |In(t) − Im(t)| >
|Gn(t) − Gm(t)|2dt
0≤t≤T
2k
≤ 22kE 0
1
≤
for m, n
k2
≥ nk.
We may assume nk+1 ≥ nk ≥ nk−1 ≥ . . ., and nk → ∞. Let
1
Ak :=
sup |Ink(t) − Ink+1(t)| >
.
0≤t≤T
2k
Then
1
P (Ak) ≤ .
k2
Thus by the Borel–Cantelli Lemma, P (Ak i.o.) = 0; which is to say, for almost all
ω
1
sup |Ink(t, ω) − Ink+1(t, ω)| ≤
provided k ≥ k0(ω).
0≤t≤T
2k
Hence Ink (·, ω) converges uniformly on [0, T] for almost every ω, and therefore
J(t, ω) := limk→∞ Ink(t, ω) is continuous for amost every ω. As In(t) → I(t) in
L2(Ω) for all 0 ≤ t ≤ T, we deduce as well that J(t) = I(t) amost every for all
0 ≤ t ≤ T. In other words, J(·) is a version of I(·). Since for almost every ω, J(·, ω)
is the uniform limit of continuous functions, J(·) has continuous sample paths a.s.
121
EXERCISES
(1) Show, using the formal manipulations for Itˆo’s formula discussed in Chapter
1, that
Y (t) := eW (t)− t2
solves the stochastic differential equation
dY = Y dW,
Y (0) = 1.
(Hint: If X(t) := W (t) − t, then dX =
+ dW .)
2
−dt2
(2) Show that
P (t) = p
”t
0eσW (t)+“µ− σ2
2
,
solves
dP = µP dt + σP dW,
P (0) = p0.
(3) Let Ω be any set and A any collection of subsets of Ω. Show that there
exists a unique smallest σ-algebra U of subsets of Ω containing A. We call
U the σ-algebra generated by A.
(Hint: Take the intersection of all the σ-algebras containing A.)
(4) Let X =
k
a
be a simple random variable, where the real numbers
i=1
iχAi
ai are distinct, the events Ai are pairwise disjoint, and Ω = ∪ki=1Ai. Let
U(X) be the σ-algebra generated by X.
(i) Describe precisely which sets are in U(X).
(ii) Suppose the random variable Y is U(X)-measurable. Show that Y
is constant on each set Ai.
(iii) Show that therefore Y can be written as a function of X.
(5) Verify:
∞
1
∞
e−x2 dx = √π,
√
xe− (x−m)2
2σ2
dx = m,
−∞
2πσ2 −∞
1
∞
√
(x − m)2e−(x−m)2
2σ2
dx = σ2.
2πσ2 −∞
(6) (i) Suppose A and B are independent events in some probability space.
Show that Ac and B are independent. Likewise, show that Ac and Bc are
independent.
(ii) Suppose that A1, A2, . . . , Am are disjoint events, each of positive
probability, such that Ω = ∪mj=1Aj. Prove Bayes’ formula:
P (B
P (A
m
| Ak)P(Ak)
k | B) =
(k = 1, . . . , m),
P (B
j=1
| Aj)P(Aj)
provided P (B) > 0.
122
(7) During the Fall, 1999 semester 105 women applied to UC Sunnydale, of
whom 76 were accepted, and 400 men applied, of whom 230 were accepted.
During the Spring, 2000 semester, 300 women applied, of whom 100
were accepted, and 112 men applied, of whom 21 were accepted.
Calculate numerically
a. the probability of a female applicant being accepted during the fall,
b. the probability of a male applicant being accepted during the fall,
c. the probability of a female applicant being accepted during the spring,
d. the probability of a male applicant being accepted during the spring.
Consider now the total applicant pool for both semesters together, and
calculate
e. the probability of a female applicant being accepted,
f. the probability of a male applicant being accepted.
Are the University’s admission policies biased towards females? or
males?
(8) Let X be a real–valued, N (0, 1) random variable, and set Y := X2. Calcu-
late the density g of the distribution function for Y .
(Hint: You must find g so that P (−∞ < Y ≤ a) = a g dy for all a.)
−∞
(9) Take Ω = [0, 1] ×[0, 1], with U the Borel sets and P Lebesgue measure. Let
g : [0, 1] → R be a continuous function.
Define the random variables
X1(ω) := g(x1), X2(ω) := g(x2) for ω = (x1, x2) ∈ Ω.
Show that X1 and X2 are independent and identically distributed.
(10) (i) Let (Ω, U, P) be a probability space and A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ . . . be
events. Show that
∞
P
An
= lim P (Am).
m
n=1
→∞
(Hint: Look at the disjoint events Bn := An+1 − An.)
(ii) Likewise, show that if A1 ⊇ A2 ⊇ · · · ⊇ An ⊇ . . ., then
∞
P
An
= lim P (Am).
m
n=1
→∞
(11) Let f : [0, 1] → R be continuous and define the Bernstein polynomial
n
k
n
bn(x) :=
f
xk(1
n
k
− x)n−k.
k=0
Prove that bn → f uniformly on [0, 1] as n → ∞, by providing the details
for the following steps.
123
(i) Since f is uniformly continuous, for each ǫ > 0 there exists δ(ǫ) > 0
such that |f(x) − f(y)| ≤ ǫ if |x − y| ≤ δ(ǫ).
(ii) Given x ∈ [0, 1], take a sequence of independent random variables
Xk such that P (Xk = 1) = x, P (Xk = 0) = 1−x. Write Sn = X1+· · ·+Xn.
Then bn(x) = E(f ( Sn )).
n
(iii) Therefore
S
|b
n
n(x) − f(x)| ≤ E(|f(
)
n
− f(x)|)
S
S
=
|f( n) − f(x)| dP +
|f( n) − f(x)| dP,
A
n
Ac
n
for A = {ω ∈ Ω | |Snn − x| ≤ δ(ǫ)}.
(iv) Then show
2M
S
2M
|b
n
n(x) − f(x)| ≤ ǫ +
V (
) = ǫ +
V (X
δ(ǫ)2
n
nδ(ǫ)2
1),
for M = max |f|. Conclude that bn → f uniformly.
(12) Let X and Y be independent random variables, and suppose that fX and
fY are the density functions for X, Y . Show that the density function for
X + Y is
∞
fX+Y (z) =
fX(z − y)fY (y) dy.
−∞
(Hint: If g : R → R, we have
∞
∞
E(g(X + Y )) =
fX,Y (x, y)g(x + y) dxdy,
−∞ −∞
where fX,Y is the joint density function of X, Y .)
(13) Let X and Y be two independent positive random variables, each with
density
e−x
if x
f (x) =
≥ 0
0
if x < 0.
Find the density of X + Y .
(14) Show that
1
1
1
x
1
lim
f ( 1 + . . . xn ) dx1dx2 . . . dxn = f ( )
n→∞
· · ·
0
0
0
n
2
for each continuous function f .
(Hint: P (|x1+...xn
V ( x1+...xn ) =
1
.)
n
− 12| > ǫ) ≤ 1ǫ2
n
12ǫ2n
(15) Prove that
(i) E(E(X | V)) = E(X).
(ii) E(X) = E(X | W), where W = {∅, Ω} is the trivial σ-algebra.
124
(16) Let X, Y be two real–valued random variables and suppose their joint dis-
tribution function has the density f (x, y) . Show that
E(X|Y ) = Φ(Y ) a.s.
for
∞ xf(x, y) dx
Φ(y) = −∞
∞
.
f (x, y) dx
−∞
(Hints: Φ(Y ) is a function of Y and so is U(Y )–measurable. Therefore we
must show that
(∗)
X dP =
Φ(Y ) dP
for all A ∈ U(Y ).
A
A
Now A = Y −1(B) for some Borel subset of R. So the left hand side of
(∗) is
∞
(∗∗)
X dP =
χB(Y )X dP =
xf (x, y) dydx.
A
Ω
−∞ B
The right hand side of (∗) is ∞
Φ(Y ) dP =
Φ(y)f (x, y) dydx,
A
−∞ B
which equals the right hand side of (∗∗). Fill in the details.)
(17) A smooth function Φ : R → R is called convex if Φ′′(x) ≥ 0 for all x ∈ R.
(i) Show that if Φ is convex, then
Φ(y) ≥ Φ(x) + Φ′(x)(y − x) for all x, y ∈ R.
(ii) Show that
x + y
1
1
Φ(
)
Φ(x) + Φ(y) for all x, y
2
≤ 2
2
∈ R.
(iii) A smooth function Φ : Rn → R is called convex if the matrix ((Φx )) is
ixj
nonnegative definite for all x ∈ Rn. (This means that n Φ ξ
i,j=1
xixj iξj ≥ 0
for all ξ ∈ Rn.) Prove
x + y
1
1
Φ(y) ≥ Φ(x) + DΦ(x) · (y − x) and Φ(
)
Φ(x) + Φ(y)
2
≤ 2
2
for all x, y ∈ Rn. (Here “D” denotes the gradient.)
(18) (i) Prove Jensen’s inequality:
Φ(E(X)) ≤ E(Φ(X))
for a random variable X : Ω → R, where Φ is convex. (Hint: Use assertion
(iii) from the previous problem.)
(ii) Prove the conditional Jensen’s inequality:
Φ(E(X|V)) ≤ E(Φ(X)|V).
125
(19) Let W (·) be a one-dimensional Brownian motion. Show
(2k)!tk
E(W 2k(t)) =
.
2kk!
(20) Show that if W(·) is an n-dimensional Brownian motion, then so are
(i) W(t + s) − W(s) for all s ≥ 0,
(ii) cW(t/c2)
for all c > 0
(“Brownian scaling”).
(21) Let W (·) be a one-dimensional Brownian motion, and define
¯
tW ( 1 ) for t > 0
W (t) :=
t
0
for t = 0.
Show that ¯
W (t) − ¯W(s) is N(0, t − s) for times 0 ≤ s ≤ t. ( ¯W(·) also
has independent increments and so is a one-dimensional Brownian motion.
You do not need to show this.)
(22) Define X(t) :=
t W (s) ds, where W (
0
·) is a one-dimensional Brownian
motion. Show that
t3
E(X2(t)) =
for each t > 0.
3
(23) Define X(t) as in the previous problem. Show that
λ2 t3
E(eλX(t)) = e 6
for each t > 0.
(Hint: X(t) is a Gaussian random variable, the variance of which we know
from the previous homework problem.)
(24) Define U (t) := e−tW (e2t), where W (·) is a one-dimensional Brownian mo-
tion. Show that
E(U (t)U (s)) = e−|t−s|
for all − ∞ < s, t < ∞.
(25) Let W (·) be a one-dimensional Brownian motion. Show that
W (m)
lim
= 0
almost surely.
m→∞
m
(Hint: Fix ǫ > 0 and define the event Am := {|W(m)
m
| ≥ ǫ}. Then Am =
{|X| ≥ √mǫ} for the N(0, 1) random variable X = W(m)
√
. Apply the
m
Borel–Cantelli Lemma.)
(26) (i) Let 0 < γ ≤ 1. Show that if f : [0, T] → Rn is uniformly H¨older
continuous with exponent γ, it is also is uniformly H¨
older continuous with
each exponent 0 < δ < γ.
(ii) Show that f (t) = tγ is uniformly H¨
older continuous with exponent γ
on the interval [0, 1].
126
(27) Let 0 < γ < 1 . These notes show that if W (
2
·) is a one–dimensional Brown-
ian motion, then for almost every ω there exists a constant K, depending
on ω, such that
(∗)
|W(t, ω) − W(s, ω)| ≤ K|t − s|γ for all 0 ≤ s, t ≤ 1.
Show that there does not exist a constant K such that (∗) holds for almost
all ω.
(28) Prove that if G, H ∈ L2(0, T), then
T
T
T
E
G dW
H dW
= E
GH dt .
0
0
0
(Hint: 2ab = (a + b)2 − a2 − b2.)
(29) Let (Ω, U, P) be a probability space, and take F(·) to be a filtration of σ–
algebras. Assume X be an integrable random variable, and define X(t) :=
E(X|F(t)) for times t ≥ 0.
Show that X(·) is a martingale.
(30) Show directly that I(t) := W 2(t) − t is a martingale.
(Hint: W 2(t) = (W (t)−W(s))2−W2(s)+2W(t)W(s). Take the conditional
expectation with respect to W(s), the history of W(·), and then condition
with respect to the history of I(·).)
(31) Suppose X(·) is a real-valued martingale and Φ : R → R is convex. Assume
also E(|Φ(X(t))|) < ∞ for all t ≥ 0. Show that
Φ(X(·)) is a submartingale.
(Hint: Use the conditional Jensen’s inequality.)
t
(32) Use the Itˆo chain rule to show that Y (t) := e 2 cos(W (t)) is a martingale.
(33) Let W(·) = (W1, . . ., Wn) be an n-dimensional Brownian motion, and
write Y (t) := |W(t)|2 −nt for times t ≥ 0. Show that Y (·) is a martingale.
(Hint: Compute dY .)
(34) Show that
T
1
T
W 2 dW =
W 3(T ) −
W dt
0
3
0
and
T
1
3
T
W 3 dW =
W 4(T ) −
W 2 dt.
0
4
2 0
(35) Recall from the notes that
Y := eR t g dW
g2 ds
0
−12 R t0
satisfies
dY = gY dW.
127
Use this to prove
1
E(eR T g dW
g2 ds
0
) = e 2 R T
0
.
(36) Let u = u(x, t) be a smooth solution of the backwards diffusion equation
∂u
1 ∂2u
+
= 0,
∂t
2 ∂x2
and suppose W (·) is a one-dimensional Brownian motion.
Show that for each time t > 0:
E(u(W (t), t)) = u(0, 0).
(37) Calculate E(B2(t)) for the Brownian bridge B(·), and show in particular
that E(B2(t)) → 0 as t → 1−.
(38) Let X solve the Langevin equation, and suppose that X0 is an N (0, σ2 )
2b
random variable. Show that
σ2
E(X(s)X(t)) =
e−b|t−s|.
2b
(39) (i) Consider the ODE
˙x = x2
(t > 0)
x(0) = x0.
Show that if x0 > 0, the solution “blows up to infinity” in finite time.
(ii) Next, look at the ODE
˙x = x 12
(t > 0)
x(0) = 0.
Show that this problem has infinitely many solutions.
(Hint: x ≡ 0 is a solution. Find also a solution which is positive for times
t > 0, and then combine these solutions to find ones which are zero for
some time and then become positive.)
(40) (i) Use the substituion X = u(W ) to solve the SDE
dX = −1e−2Xdt + e−XdW
2
X(0) = x0.
(ii) Show that the solution blows up at a finite, random time.
(41) Solve the SDE dX = −Xdt + e−tdW.
(42) Let W = (W 1, W 2, . . . , W n) be an n-dimensional Brownian motion and
write
1
n
2
R := |W| =
(W i)2
.
i=1
128
Show that R solves the stochastic Bessel equation
n
W i
n
dR =
dW i +
− 1dt.
R
2R
i=1
(43) (i) Show that X = (cos(W ), sin(W )) solves the SDE system
dX1 = −1X1dt
2
− X2dW
dX2 = −1X2dt + X1dW
2
(ii) Show also that if X = (X1, X2) is any other solution, then |X| is
constant in time.
(44) Solve the system
dX1 = dt + dW 1
dX2 = X1dW 2,
where W = (W 1, W 2) is a Brownian motion.
(45) Solve
dX1 = X2dt + dW 1
dX2 = X1dt + dW 2.
(46) Solve
dX = 1 σ′(X)σ(X)dt + σ(X)dW
2
X(0) = 0
where W is a one–dimensional Brownian motion and σ is a smooth, positive
function.
(Hint: Let f (x) :=
x dy
and set g := f −1, the inverse function of f .
0 σ(y)
Show X := g(W ).)
(47) Let τ be the first time a one–dimensional Brownian motion hits the half-
open interval (a, b]. Show τ is a stopping time.
(48) Let W denote an n–dimensional Brownian motion, for n ≥ 3. Write X =
W + x0, where the point x0 lies in the region U = {0 < R1 < |x| < R2}
Calculate explicitly the probability that X will hit the outer sphere {|x| =
R2} before hitting the inner sphere {|x| = R1}.
(Hint: Check that
1
Φ(x) = |x|n−2
satisfies ∆Φ = 0 for x = 0. Modify Φ to build a function u which equals 0
on the inner sphere and 1 on the outer sphere.)
129
References
[A]
L. Arnold, Stochastic Differential Equations: Theory and Applications, Wiley, 1974.
[B-R]
M. Baxter and A. Rennie, Financial Calculus: An Introduction to Derivative Pricing,
Cambridge U. Press, 1996.
[B-S]
A. N. Borodin and P. Salminen, Handbook of Brownian Motion: Facts and Formulae,
Birkhauser, 1996.
[B]
L. Breiman, Probability, Addison–Wesley, 1968.
[Br]
P. Bremaud, An Introduction to Probabilistic Modeling, Springer, 1988.
[C]
K. L. Chung, Elementary Probability Theory with Stochastic Processes, Springer, 1975.
[D]
M. H. A. Davis, Linear Estimation and Stochastic Control, Chapman and Hall.
[F]
A. Friedman, Stochastic Differential Equations and Applications, Vol. 1 and 2, Academic
Press.
[Fr]
M. Freidlin, Functional Integration and Partial Differential Equations, Princeton U.
Press, 1985.
[G]
C. W. Gardiner, Handbook of Stochastic Methods for Physics, Chemistry, and the Natural
Sciences, Springer, 1983.
[G-S]
I. I. Gihman and A. V. Skorohod, Stochastic Differential Equations, Springer, 1972.
[G]
D. Gillespie, The mathematics of Brownian motion and Johnson noise, American J.
Physics 64 (1996), 225–240.
[H]
D. J. Higham, An algorithmic introduction to numerical simulation of stochastic differ-
ential equations, SIAM Review 43 (2001), 525–546.
[Hu]
J. C. Hull, Options, Futures and Other Derivatives (4th ed), Prentice Hall, 1999.
[K]
N. V. Krylov, Introduction to the Theory of Diffusion Processes, American Math Society,
1995.
[L1]
J. Lamperti, Probability, Benjamin.
[L2]
J. Lamperti, A simple construction of certain diffusion processes, J. Math. Kyˆ
oto (1964),
161–170.
[Ml]
A. G. Malliaris, Itˆ
o’s calculus in financial decision making, SIAM Review 25 (1983),
481–496.
[M]
D. Mermin, Stirling’s formula!, American J. Physics 52 (1984), 362–365.
[McK]
H. McKean, Stochastic Integrals, Academic Press, 1969.
[N]
E. Nelson, Dynamical Theories of Brownian Motion, Princeton University Press, 1967.
[O]
B. K. Oksendal, Stochastic Differential Equations: An Introduction with Applications,
4th ed., Springer, 1995.
[P-W-Z] R. Paley, N. Wiener, and A. Zygmund, Notes on random functions, Math. Z. 37 (1959),
647–668.
[P]
M. Pinsky, Introduction to Fourier Analysis and Wavelets, Brooks/Cole, 2002.
[S]
D. Stroock, Probability Theory: An Analytic View, Cambridge U. Press, 1993.
[S1]
H. Sussmann, An interpretation of stochastic differential equations as ordinary differen-
tial equations which depend on the sample point., Bulletin AMS 83 (1977), 296–298.
[S2]
H. Sussmann, On the gap between deterministic and stochastic ordinary differential equa-
tions, Ann. Probability 6 (1978), 19–41.
130
