Chapter 3

Stochastic Calculus (Lecture #3)
Siegfried Hörmann
Université libre de Bruxelles (ULB)
Spring 2014
Outline of the course
1. Stochastic processes in continuous time.
2. Brownian motion.
3. Itô integral: preliminaries.
4. Brownian motion calculus
5. Itô’s formula.
6. Stochastic differential equations.
7. Strong and weak solutions of SDEs.
8. Change of measure.
Itô integral: preliminaries
3.1. Functions of bounded variation.
3.2. Lebesgue-Stieltjes/Riemann-Stieltjes integral.
Rt
3.3. The stochastic integral 0 B(s)dB(s).
3.4. Integration of simple previsible processes.
3.5. Some martingale results.
Functions of bounded variation
Definition
The variation of a real function g over the interval [a, b] is
defined as
Vg ([a, b]) := sup
n
X
|g(ti ) − g(ti−1 )|,
i=1
where the supremum is taken over partitions
a = t0 < t1 < · · · < tn = b. If Vg (t) := Vg ([0, t]) < ∞ for all
t ≥ 0, then g is said to be a function of finite variation.
Functions of bounded variation
Example
If g is increasing then Vg (t) = g(t) − g(0). If g is decreasing
then Vg (t) = g(0) − g(t). (Exercise.)
Example
If g is differentiable with continuous derivative then
Z
Vg (t) =
t
|g 0 (s)|ds.
0
(Exercise.)
Example
The function g(t) = t sin(1/t) for t > 0 and g(0) = 0 is
continuous and differentiable at all points except 0, but has
infinite variation. (Exercise.)
Functions of bounded variation
Lemma (Jordan Decomposition)
Any function g of finite variation can be written as the difference
of two monotone functions. If g is right-continuous, then also
the two monotone functions can be chosen to be
right-continuous.
Proof. Take, for example,
h1 (t) =
1
1
(Vg (t) + g(t)) and h2 (t) = (Vg (t) − g(t)).
2
2
Remark
(1) The representation is not unique.
(2) If g(0) = 0 and if hi as above we have 0 ≤ hi (t) ≤ Vg (t).
Lebesgue-Stieltjes/Riemann-Stieltjes integral
Suppose that g is right-continuous and of bounded variation.
Then with h1 , h2 as before we can define the two (σ-finite)
measures
µ1 ((a, b]) := h1 (b) − h1 (a) and µ2 ((a, b]) := h2 (b) − h2 (a).
Then for f : [0, ∞) → R the Lebesgue-Stieltjes integral is
defined as
Z
Z
Z
fdg :=
fdµ1 −
fdµ2 ,
(0,t]
(0,t]
(0,t]
R
provided (0,t] |f |dµi < ∞ for i = 1, 2. One can show that this
definition is independent of the choice of h1 and h2 .
Lebesgue-Stieltjes/Riemann-Stieltjes integral
If f is continuous, then the Lebesgue-Stieltjes integral coincides
with the Riemann-Stieltjes integral
Z
t
fdg = lim
0
n
X
f (tk0 )(g(tk ) − g(tk −1 )),
tk −1 ≤ tk0 ≤ tk ,
k =1
where 0 = t0 < t1 < · · · < tn = t, defines a partition, whose
refinement tends to zero.
Remark
(1) The limit above is independent of the choice of tk0 .
(2) One can show: if the Riemann-Stieltjes integral exists for all
continuous f , then g is of finite variation.
Lebesgue-Stieltjes/Riemann-Stieltjes integral
(3) One can show: the Riemann-Stieltjes integral exists, if one
of the two functions f or g is continuous and the other one is of
bounded variation
(4) Due to (3): if f is of bounded variation and Bs a BM, then
Z
t
f (s)dBs (ω)
0
is defined for all ω.
=⇒ we could define the integral path-wise
(5) Partial integration: suppose that g and f are continuous and
of bounded variation. Then
Z
Z
f (t)g(t) = f (0)g(0) +
fdg +
gdf .
(0,t]
(0,t]
The stochastic integral
Rt
0
Bs dBs
One of the main targets of this course is to define
Z
t
Xs dBs .
0
By the previous statement, if Xs (ω) were of bounded variation
for all ω, we could define the integral pathwise as a
Riemann-Stieltjes integral. However, this would already exclude
the case where Xs = Bs as is implied by the following result.
Lemma
A BM path is a.s. not of bounded variation.
Proof. Fix ω and write Bt for Bt (ω). Then if VB(·) (t) were < ∞
n
X
(Btk − Btk −1 )2 ≤ VB(·) (t) × sup |Btk − Btk −1 | → 0.
k =1
1≤k ≤n
The stochastic integral
Rt
0
Bs dBs
Let us check what happens.
n
X
Btk −1 (Btk − Btk −1 )
k =1
n
n
1X
1X
(Btk + Btk −1 )(Btk − Btk −1 ) −
(Btk − Btk −1 )(Btk − Btk −1 )
=
2
2
=
=
1
2
k =1
n
X
k =1
(Bt2k − Bt2k −1 ) −
k =1
n
1X
1 2
B −
2 t
2
1
2
n
X
(Btk − Btk −1 )2
k =1
(Btk − Btk −1 )2
k =1
1
P 1
→ (Bt2 − hBit ) = (Bt2 − t).
2
2
Compared to partial integration, we get extra term −t/2.
The stochastic integral
Now consider
Rt
0
Pn
Bs dBs
k =1 Btk0 (Btk
− Btk −1 ), with
tk0 = αtk −1 + (1 − α)tk ,
0 ≤ α ≤ 1.
Then
n
X
Btk0 (Btk − Btk −1 )
k =1
=
n
X
k =1
Btk −1 (Btk − Btk −1 ) +
n
X
(Btk0 − Btk −1 )(Btk − Btk −1 ).
k =1
The first summand converges to 12 (Bt2 − t) as we have shown.
The second, converges to (1 − α)t (exercise).
The stochastic integral
Hence
Pn
k =1 Btk0 (Btk
Rt
0
Bs dBs
− Btk −1 ), with
tk0 = αtk −1 + (1 − α)tk ,
0 ≤ α ≤ 1.
converges (in probability) to
1 2
B +
2 t
1
− α t.
2
Unlike the usual Riemann-Stieltjes integral, this limit depends
on the choice of the nodes tk0 .
The stochastic integral
Rt
0
Bs dBs
A way out of this problem is to fix the nodes. Different choices
of α yield different integrals.
1. The choice α = 0 gives rise to the Itô integral.
2. Another popular choice is α = 1/2, which corresponds to
the Stratonovich integral.
As we will see, the Itô integral has the important advantage,
that the resulting integrated process becomes a (local)
martingale, which is a fundamental property.
Integration of simple previsible processes
Our previous investigations motivate to define the stochastic
integral as a Riemann-Stieltjes integral, where the process is
evaluated on the left limit of the intervals.
Definition (Simple previsible)
A process Y is called simple previsible with respect to a
filtration (Ft )t≥0 , if it can be represented as
Yt = ξ0 I{t = 0} +
n
X
ξk I{t ∈ (tk −1 , tk ]},
k =1
where 0 = t0 ≤ t1 ≤ t2 ≤ · · · ≤ tn < ∞, ξk are bounded r.v.s and
Ftk −1 -measurable, and ξ0 is a bounded F0 -measurable r.v.
In the sequel (Ft )t≥0 will be the filtration of a BM B(t),
satisfying the usual conditions.
Integration of simple previsible processes
Example
Define Yt = ξk := max{min{Btk −1 , c} − c} (c > 0), if
tk −1 < t ≤ tk , k ≤ n, and Y0 = 0. Then Y is simple previsible.
We denote E the set of simple previsible processes (SPPs).
Definition (Stochastic integral of SPPs)
Let Y ∈ E. Then the stochastic integral of Y with respect to B is
defined as
Z ∞
n
X
IB (Y ) =
Ys dBs :=
ξk (Btk − Btk −1 ).
0
k =1
Integration of simple previsible processes
2
It is trivial to see that E IB (Y ) < ∞. Hence,
IB : E → L2 = L2 (Ω, A, P).
Here are some other simple properties:
Well def.: The definition is independent of the (non-unique)
representation of Y .
Linear: If X and Y are simple previsible and α and β real
numbers, then
IB (αX + βY ) = αIB (X ) + βIB (Y ).
2
R∞
Moments: EIB (Y ) = 0 and E IB (Y ) = E 0 Ys2 ds.
Integration of simple previsible processes
The first two properties are simple and left as an exercise.
We prove the moments properties.
Z ∞
X
E
Ys dBs =
E[ξk E[Btk − Btk −1 |Ftk −1 ]] = 0.
0
k ≤n
Furthermore
Z ∞
2 X X
E[ξk ξ` (Btk − Btk −1 )(Bt` − Bt`−1 )].
E
Ys dBs =
0
k ≤n `≤n
Integration of simple previsible processes
When k < `
E[ξk ξ` (Btk − Btk −1 )(Bt` − Bt`−1 )]
= E[E[ξk ξ` (Btk − Btk −1 )(Bt` − Bt`−1 )|Ft`−1 ]]
= E[ξk ξ` (Btk − Btk −1 )E[Bt` − Bt`−1 |Ft`−1 ]] = 0.
Same argument applies when ` < k . Hence,
Z
2
∞
E
Ys dBs
0
=
=
n
X
k =1
n
X
k =1
=
n
X
E[ξk2 (Btk − Btk −1 )2 ]
k =1
E[E[ξk2 (Btk − Btk −1 )2 |Ftk −1 ]] =
n
X
E[ξk2 E[(Btk − Btk −1 )2 |Ftk −1 ]]
k =1
"
E[ξk2 ](tk
− tk −1 ) = E
n
X
k =1
#
ξk2 (tk
Z
− tk −1 ) = E
0
∞
Ys2 ds.
Integration of simple previsible processes
The mapping IB is transforming a function (Ys ) into a (random)
number.
We now wish
R t to obtain a random process instead, which should
represent 0 Ys dBs .
To this end we define
∞
Z
Ys I{s ≤ t}dBs .
IB (Y )t :=
0
For a simple function this gives
IB (Y )t =
n
X
k =1
ξk (Btk ∧t − Btk −1 ∧t ).
Integration of simple previsible processes
We obtain the following further properties.
Continuity The paths of IB (Y )t =
Martingale IB (Y )t an
L2 -bounded
Rt
0
Ys dBs are continuous.
martingale, i.e.
sup E IB (Y )t
2
<∞
t≥0
and for s ≤ t
E[IB (Y )t |Fs ] = IB (Y )s .
Integration of simple previsible processes
The continuity follows easily from the continuity of the BM
paths.
The L2 -boundedness follows from
Z ∞
2
sup E IB (Y )t = sup E
Ys2 I{s ≤ t}ds
t≥0
t≥0
0
Z
= sup E
t≥0
0
t
Ys2 ds
Z
=E
0
∞
Ys2 ds < ∞.
(In the last step we used the monotone convergence theorem.)
Integration of simple previsible processes
Finally, we can add t = tj and s = tk , k ≤ j, to the partition and
get


j
X
ξ` (Bt` − Bt`−1 )|Fs 
E[IB (Y )t |Fs ] = E 
`=1
= IB (Y )s +
j
X
E[ξ` (Bt` − Bt`−1 )|Fs ]
`=k +1
= IB (Y )s +
j
X
`=k +1
E[ξ` E[Bt` − Bt`−1 |Ft`−1 ]|Fs ] = IB (Y )s .
Some martingale results
Having developed the stochastic integral for the case where the
integrand is a simple previsible function, is a first step towards a
general definition. To extend the integral to a bigger class of
integrals (next lecture), we need some background information
about martingales.
Recall again that X = (Xt : t ≥ 0) is a martingale w.r.t. to
F = (Ft : t ≥ 0) if (i) E|Xt | < ∞ for all t ≥ 0, and (ii)
E[Xt |Fs ] = Xs a.s. for all 0 ≤ s ≤ t.
Examples: Let (Bt ) a Brownian motion and Ft = σ(Bs : s ≤ t).
Then the following process are martingales w.r.t. F:
1. Bt .
2. Bt2 − t.
3. exp(θBt − tθ2 /2).
4. If E|Y | < ∞, then E[Y |Ft ] is a martingale.
Some martingale results
Many important properties of martingales can be shown under
the assumption that the paths are càdlàg.
Definition (càdlàg)
A function f on [0, ∞) is called càdlàg (continue à droite, limite
à gauche), if it is right-continuous and if left limits exist.
Example
I
Continuous functions are càdlàg.
I
Distribution functions are càdlàg.
I
f : [0, 2] → R with f (t) = sin(1/(1 − t))I{t < 1} isn’t càdlàg.
Since one can prove that if the filtration F fulfills the usual
conditions, then every martingale X w.r.t. F possesses a
càdlàg version, this is not really a restriction.
Some martingale results
Theorem (Optional stopping theorem)
If X is a martingale with right-continuous paths and if T is a
bounded stopping time, then XT is integrable and FT
measurable. In addition we have
E[XT ] = E[X0 ]
Theorem (Optional sampling theorem)
If X is a martingale with right-continuous paths and if S ≤ T are
bounded stopping times, then
E[XT |FS ] = XS
a.s.
Some martingale results
Theorem (Doob’s inequality)
If X possesses right-continuous paths, then
"
#
E
sup |Xs |2 ≤ 4E|Xt |2 .
0≤s≤t
Remark
Doob’s inequality holds in a more general setting for p-th order
moments, p > 1.
We now provide some results when X is an L2 -bounded
martingale, i.e. a martingale X for which
sup EXt2 < ∞.
t≥0
Some martingale results
Theorem (Martingale convergence theorem)
If X possesses right-continuous paths, then
L2
Xt → X∞ .
We have
E[X∞ |Ft ] = Xt
a.s.
In addition, Doob’s inequality holds:
"
#
E
sup |Xs |2 ≤ 4E|X∞ |2 .
0≤s<∞
Exercises
1. Let (Mt ) be a square integrable martingale. Show that for
Y ∈ Fs and E(Y 2 ) < ∞ we have
E(Y (Mt − Ms )) = 0
for t ≥ s.
This shows that martingale increments are orthogonal.
P
2. Consider nk =1 Btk0 (Btk − Btk −1 ), with
tk0 = (1 − α)tk −1 + αtk ,
0 ≤ α ≤ 1.
Show that if the mesh of the partition tends to zero, then
the above term converges in probability to
1
1 2
Bt + (α − )t.
2
2
Exercises
3. Show that the integral of a simple previsible process is
independent of the representation of the integrand.
4. Show that the integral of a simple previsible process is
linear. I.e.
IB (αX + βY ) = αIB (X ) + βIB (Y ).
5. Show that if E|Y | < ∞, then E[Y |Ft ] is a martingale.

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