Post on 23-Jun-2020
Stochastic Integration and Itôs formula
Péter Medvegyev
January 2009
Medvegyev (CEU) Stochastic integration January 2009 1 / 64
Functions with nite variation
DenitionA function F on an interval [0, t] has nite variation, if
sup∆
∑k
F t∆k+1
F
t∆k
< ∞,
where∆ $ t0 = 0 < t1 < . . . < tk1 < tk = t
can be any partition of the interval [0, t] . A function has nite variation ifit has nite variation on any nite closed interval.The set of functions withnite variation is denoted by V .
TheoremF 2 V if and only if F = F1 F2, where F1 and F2 are increasingfunctions.
Medvegyev (CEU) Stochastic integration January 2009 2 / 64
Measures generated by functions with nite variation
TheoremIf F is increasing and right-continuous then there is a measure µ with theproperty that
µ ((a, b]) = F (b) F (a) .
DenitionR t0 YdF denotes the integral
R t0 Ydµ. (If it exits.)
Medvegyev (CEU) Stochastic integration January 2009 3 / 64
Processes with nite variation
DenitionA process X has nte variation if every trajectory of X has nite variation.If X has nite variation and the trajectories of Y are Borel-measurablethen one can dene the stochastic integral
(Y X ) (t,ω) $Z t
0Y (s,ω) dX (s)
where the integral is taken for every ω separatelly.
TheoremIf X is adapted and has nite variation and Y is left-continuous andadapted then Y X is also adapted. The same is true if Y isright-continuous.
Medvegyev (CEU) Stochastic integration January 2009 4 / 64
Processes with nite variation
We prove this when Y is continuous. In this case
(Y X ) (t) = lim∆!0∑k
Y (tk1) (X (tk ) X (tk1)) .
(By the continuity the step function ∑k Y (tk1) χ ((tk1, tk ]) convergesto Y and one can apply the dominated convergence theorem.) If ∆ is apartition of [0, t] then the sum is Ft -measurable and therefore the limit isalso Ft -measurable.
Medvegyev (CEU) Stochastic integration January 2009 5 / 64
Local martingales
DenitionA process X is a local martingale if
1 X is adapted and right-continuous with limits from the left,2 X = X (0) + L, and3 there is a sequence of stopping times τn % ∞ such that Lτn is amartingale for any n. In the denition one can assume that τn isbounded for any n. (Instead of τn one can take τ0n $ τn ^ n.)
Every martingale is a local martingale. One can take τn $ n. or one usethat the stopped martingales are martingales. If we study a martingale atxed time intervals we always see that it is a martingale. The denition oflocal martingale is a generaliation when we stop the process and observe itnot at xed times but at stopping times.
DenitionL denotes the set of local martingales with L (0) = 0.
Medvegyev (CEU) Stochastic integration January 2009 6 / 64
Fisks Theorem
Theorem (Fisk)If M is a local martingale and M 2 V is continuous, then the trajectoriesof M are almost surely constant. Particularily if M 2 L \ V and M iscontinuous then M = 0.
Let V (t) $ Var (M) (t) that is let V be the variation of M on [0, t] . V isalso continuous. Let
τn $ inf ft : jM (t)j > ng ^ inf ft : jV (t)j > ng .
As M and V are continuous jMτn j n and jV τn j n. As τn % ∞ it issu¢ cient to show that the trajectories of Mτn are constant. (Ifτn (ω)% T < ∞ for some ω then M (τn (ω)) = n% ∞ which isimpossible as M is continuous at T .)
Medvegyev (CEU) Stochastic integration January 2009 7 / 64
Fisks Theorem
Hence it is su¢ cient to prove the theorem when the processes V and Mare bounded. We can assume that M (0) = 0. By the Energy Equality if
t > s then E(M (t)M (s))2
= E
M2 (t)M (s)2
hence
EM2 (t)
= E
∑k
M2 (tk+1)M2 (tk )
!=
= E
∑k(M (tk+1)M (tk ))2
!
E maxkjM (tk+1)M (tk )j∑
kjM (tk+1)M (tk )j
!
EV (t) max
kjM (tk+1)M (tk )j
K E
maxkjM (tk+1)M (tk )j
.
Medvegyev (CEU) Stochastic integration January 2009 8 / 64
Fisks Theorem
The trajectories are continuous, so maxk jM (tk+1)M (tk )j ! 0.
maxkjM (tk+1)M (tk )j V (t) $ sup
∆∑ jM (tk+1)M (tk )j K ,
By the Dominated Convergence Theorem
EM2 (t)
K E
maxkjM (tk+1)M (tk )j
! 0,
that is M (t)a.s .= 0.
Medvegyev (CEU) Stochastic integration January 2009 9 / 64
Quadratic variation
DenitionFor any process X let
T∆X (t) $ ∑
k
Xt∆k+1
X
t∆k
2where ∆ is a partition of [0, t]. We say that the quadratic variation of X isnite on [0, t] if whenever ∆ & 0 then T∆
X (t) is stochastically convergent.The limit is called the quadratic variation of X on the interval [0, t]. Thequadratic variation of X is denoted by [X ,X ] (t) , or simply [X ] (t) . If Xand Y are stochastic processes and
T∆ (s) $ ∑k
Xt∆k+1
X
t∆k
Yt∆k+1
Y
t∆k
.
is convergent in probability as ∆ & 0, then we call the limit the quadraticco-variation of X and Y on [0, t] and denote it as [X ,Y ] (t) .
Medvegyev (CEU) Stochastic integration January 2009 10 / 64
Quadratic variation
DenitionThe process (t,ω) 7! [X ,Y ] (t,ω) is called the quadratic co-variationprocess. The process (t,ω) 7! [X ] (t,ω) is called the quadratic variationof X .
The problem with the denition is that for every t the limit exists only inprobability so the limits are well-dened just as an equivalence class. Oneshould carefully select a representant for every t from each class. We willassume that [X ,Y ] are continuous or it is right-continuous with left limits.
Medvegyev (CEU) Stochastic integration January 2009 11 / 64
How to calculate the quadratic variation?
ExampleIf the trajectories of X are almost surely continuous and the trajectorieshave nite variation, then [X ]
a.s .= 0.
Assume that ∆ ! 0. ThenT∆ (t) = ∑
k(X (tk+1) X (tk ))2
(Var (X )) (t)maxkjX (tk+1) X (tk )j
a.s .! 0,
since the variation (Var (X )) (t) is nite, and the trajectories by theassumed continuity are uniformly continuous, hence for any nite interval
maxkjX (tk+1) X (tk )j
a.s .! 0.
Medvegyev (CEU) Stochastic integration January 2009 12 / 64
The quadratic variation of Poisson processes
ExampleThe quadratic variation of a Poisson process is the sum of the jumps,hence the quadratic variation is the same as the process itself.
If π is a Poisson-process, then for all ω on an interval [0, t] there are onlynite number of jumps. If the partition of [0, t] is ne enough, then
T∆ (t,ω) = ∑k(∆π (tk ,ω))
2 = the number of the jumps.
and obviously T∆ (t,ω)! [π] (t,ω) .
Medvegyev (CEU) Stochastic integration January 2009 13 / 64
The quadratic variation of Wiener processes
ExampleThe quadratic variation of the Wiener-process is t.
Denote ∆w (tk ) $ w (tk+1) w (tk ). By the denition of the Wienerprocess for a partition of the interval [0, t]
E
∑k(∆w (tk ))
2
!= ∑
k(tk+1 tk ) = t.
As the increments of w are independent and the expected value of theincrements are zero so as maxk jtk+1 tk j is going to zero
D2
∑k(∆w (tk ))
2
!= ∑
k
D2(∆w (tk ))
2=
= ∑k
D2(N (0, 1))2
(tk+1 tk )2
∑k
3 (tk+1 tk )2 3 t maxk(tk+1 tk )! 0.
Medvegyev (CEU) Stochastic integration January 2009 14 / 64
The quadratic variation of Wiener processes
Observe that we used that EN (0, 1)4
= 3.
By Tsebisevs inequality
P
∑k (∆w (tk ))2 t > ε
!=
= P
∑k (∆w (tk ))2 E
∑k(∆w (tk ))
2
! > ε
!
D2
∑k (∆w (tk ))2
ε2! 0
so the convergence in probability holds.
Medvegyev (CEU) Stochastic integration January 2009 15 / 64
Quadratic variation of Wiener processes
If [0, t] = [0, 1] , and ∆n is the partition of [0, 1] to n equal parts, then
w (tk+1) w (tk ) = N0,1n
,
so
∑k[w (tk+1) w (tk )]2 = ∑
k
N0,1n
2,
that is the sum of the square of independent random variables withvariance 1/n. As N (0, 1/n) = 1/
pnN (0, 1) ,obviously
∑k(w (tk+1) w (tk ))2 =
∑k N (0, 1)2
n.
By the theorem of large numbers the limit is 1. Which is in out case t.
Medvegyev (CEU) Stochastic integration January 2009 16 / 64
The variation of Wiener processes
Now consider the simply variation
∑kjw (tk+1) w (tk )j =
∑k jN (0, 1)jpn
By the central limit theorem if ∞ > M > 0 denotes the expected value ofjN (0, 1)j
∑k jN (0, 1)j nMpn
t N (0, 1) ,
hence
∑kjw (tk+1) w (tk )j t N (0, 1) +
nMpn! ∞.
On the other hand again by the theorem of large numbers
∑k jN (0, 1)jpn
=pn
∑k jN (0, 1)jn
!pnM ! ∞.
Medvegyev (CEU) Stochastic integration January 2009 17 / 64
Generalization
If for some continuous function S the quadratic variation is nite andpositive then as
∑k(∆S (tk+1))
2 maxkj∆S (tk+1)j∑
kj∆S (tk+1)j
and as maxk j∆S (tk+1)j ! 0, necessarily
∑kj∆S (tk+1)j ! ∞
so the simply variation is innite. The same argument is valid for the αand β variations. If S is continuous and for some α > 0.
lim∆&0∑k
(S (tk+1) S (tk ))α < ∞
then if β < α then
lim∆&0∑k
(S (tk+1) S (tk ))β = ∞
Medvegyev (CEU) Stochastic integration January 2009 18 / 64
Stochastic Integration
Assume that θ (t) is Ft -measurable. If S is a martingale then
E (θ (t) (S (t + h) S (t))) = E (θ (t)E (S (t + h) S (t) j Ft )) = 0.
so
E
∑k
θ (tk ) (S (tk+1) S (tk ))!= ∑
k
E (θ (tk ) (S (tk+1) S (tk ))) = 0,
so if we take the approximation point as the beginning of the interval onecan guess, that the RiemannStieltjes type approximating sum remains amartingale.
Medvegyev (CEU) Stochastic integration January 2009 19 / 64
Stochastic Integration
E
∑tkt
θ (tk ) (S (tk+1) S (tk )) j Fs
!=
= ∑tks
θ (tk ) (S (tk+1) S (tk )) + E( ∑s<tkt
θ(tk )(S(tk+1) S(tk )) j Fs ).
But by the martingale property of S
E (θ (tk ) (S (tk+1) S (tk )) j Fs ) == E (E (θ (tk ) (S (tk+1) S (tk )) j Ftk j Fs )) == E (θ (tk )E (S (tk+1) S (tk ) j Ftk j Fs )) =
= E (θ (tk ) 0) = 0.
Medvegyev (CEU) Stochastic integration January 2009 20 / 64
Stochastic Integration
DenitionIf S and θ are stochastic process, and if the limit
lim∆&0∑k
θt∆k
St∆k+1
S
t∆k
exists in probability, then the limit is called the stochastic integral of θwith respect of S and the integral is denoted byZ t
0θdS .
Medvegyev (CEU) Stochastic integration January 2009 21 / 64
Some remarks
It is very di¢ cult to describe the class of processes S and θ forwhich the integral exists. It is also very di¢ cult to analyze theproperties of the integral. We will assume that S and θ arecontinuous and S is a martingale and that the stochastic integralremains a martingale. This is not always true, but we will not dealwith this problem but simply skip and ignore it. If θ is adapted andcontinuous and S is a local martingale then the integral is a localmartingale. Of course there is a huge di¤erence between localmartingales and martingales, but it will lead us too far. A goodstudent should never trust in his or her professor, and always checkwhat the professor is saying. Be careful!!! If one is interested instochastic integration theory then one MUST consult the textbooksabout the subject. Our goal is just to give a heuristic introductionand show the importance of the basic ideas and not to give aprecise introduction.
Medvegyev (CEU) Stochastic integration January 2009 22 / 64
RiemannStieltjes
TheoremIf g is a continuous function on [a, b] and F has nite variation on [a, b] ,then the RiemannStieltjes integralZ b
agdF
exists.
Recall that RiemannStieltjes integralR ba gdF is dened as the limit of the
approximating sums
∑k
g (tk ) (F (xk+1) F (xk )) ,
where (xk )k is a partition of the interval [a, b] and the limit is taken asmaxk jxk+1 xk j ! 0. We emphasize that in the denition tk can be anarbitrary point in the subinterval [xk , xk+1].
Medvegyev (CEU) Stochastic integration January 2009 23 / 64
Cauchy criteria and integration
The well-known proof depends on the completeness of the real line, that ison the completeness of the domain of denition of the range of the g . Weshould just proof, that the approximating sequence
Sn $ ∑k
gt(n)k Fx (n)k+1
F
x (n)k
,
is a Cauchy-sequence. The g is continuous, so for any ε > 0, there is a δ,that if jx y j < δ, then jg (x) g (y)j < δ. If the partition ∆ is
su¢ ciently ne and (xk ) is the union ofx (n)k
and
x (m)k
then
jSn Sm j =
∑kg
ξ(n)k
g
ξ(m)k
(F (xk+1) F (xk ))
max
k
g ξ(n)k
g
ξ(m)k
∑kjF (xk+1) F (xk )j
εVar (F ) ,
where is the nal step we used that F has nite variation.Medvegyev (CEU) Stochastic integration January 2009 24 / 64
Orthogonal increments, the Energy Equality again
We want to generalize the argument to stochastic integrals. Observe thatif s < t then and if θ (t) is Ft -measurable and θ (s) is Fs -measurable then
E (θ (t) (S (t + h) S (t)) θ (s) (S (s + h) S (s))) == E (E (θ (t) (S (t + h) S (t)) θ (s) (S (s + h) S (s)) j Ft )) == E (θ (t) θ (s) (S (s + h) S (s))E ((S (t + h) S (t)) j Ft )) =
= E (θ (t) θ (s) (S (s + h) S (s)) 0) = 0,
since the product θ (t) θ (s) (S (s + h) S (s)) is Ft measurable.
Medvegyev (CEU) Stochastic integration January 2009 25 / 64
Orthogonal increments, the Energy Equality again
Hence calculating the L2-norm of the approximating sums
E
0@ ∑k
θ (tk ) (S (tk+1) S (tk ))!21A =
= E
∑k
∑j
θ (tk ) θ (tj ) (S (tk+1) S (tk ))(S (tj+1) S (tj )!=
= ∑k
E(θ (tk ))
2 (S (tk+1) S (tk ))2.
Medvegyev (CEU) Stochastic integration January 2009 26 / 64
Cauchy property for stochastic integrals
As in the proof of the existence of the RiemannStieltjes integral if thepartition is su¢ ciently ne and θ (t 0k ) $ X
t(n)k X
t(m)k
then
kSn Smk22 =
∑k θt 0k(S (tk+1) S (tk ))
2
2
=
= ∑k
E
θt 0k2
(S (tk+1) S (tk ))2
ε ∑k
E(S (tk+1) S (tk ))2
=
= ε ∑k
ES (tk+1)
2 E
S (tk )
2=
= ε ES2 (b)
E
S2 (a)
.
As L2 (Ω) is complete by the Cauchy-criteria the integralR ba XdS exits.
Medvegyev (CEU) Stochastic integration January 2009 27 / 64
Locally square integrable martingales
DenitionLet S 2 L. We say that S 2 H2
loc if there is a sequence of stopping time(τn) that Sτn 2 H2.
TheoremIf S 2 L and if S is continuous then S 2 H2
loc.
Example
If w is a Wiener process then S 2 H2loc but not S 2 H2. (Let τn n.)
Medvegyev (CEU) Stochastic integration January 2009 28 / 64
Existence of stochastic integrals
More precisely
TheoremIf X is continuous and adapted and S 2 H2
loc then the stochastic integralR ba XdS exists and the convergence holds in probability. The integralprocess
(X S) (t) $Z t
0XdS
is a local martingale.
Medvegyev (CEU) Stochastic integration January 2009 29 / 64
Existence of stochastic integrals
All integrals classical, stochastic are "basically" limits of some weightedaverages. Of course it is not clear how we dene the convergence. Onecan argue that ξn ! ξ if
P (fω : ξn (ω)! ξ (ω)g) = 1.
In this case the integral exists in almost sure sense. If the integrand iscontinuous and the trajectories of the integrator have nite variation, thenthe stochastic integral is convergent in this almost sure sense. But ascontinuous martingales, like the Wiener processes do not have trajectorieswith nite variation, one can guarantee only stochastic convergence, thatif for all ε > 0 on has.
P (jξn ξj > ε)! 0.
Medvegyev (CEU) Stochastic integration January 2009 30 / 64
Semimartingales
DenitionIf the stochastic process X is of the form
X (t,ω) = X (0,ω) + V (t,ω) + S (t,ω)
where V is adapted and has trajectories with nite variation and S is alocal martingale, then we say that X is a semi-martingale. If X is asemi-martingale, then one can deneZ b
aθdX $
Z b
aθdV +
Z b
aθdS ,
where the rst integral is a Stieltjes-intergal, convergent in almost suresense the second is a stochastic integral convergent is stochastic sense.
Medvegyev (CEU) Stochastic integration January 2009 31 / 64
Quadratic variation of semimartingales
DenitionIf X is a semi-martingale then the stochastic limit of
∑k(X (tk+1) X (tk ))2
is the quadratic variation of X .
Medvegyev (CEU) Stochastic integration January 2009 32 / 64
Quadratic variation of semimartingales
By simply calculation
∑k(X (tk+1) X (tk ))2 = ∑
k
(∆V (tk ))
2 + 2∆S (tk )∆V (tk ) + (∆S (tk ))2.
The limit of the third term is the quadratic variation of S , that is [S ] . Forthe rst tag
∑k(∆Vk )
2 maxkj∆Vk j∑
kj∆Vk j max
kj∆Vk jVar (V ) .
Assume that V is continuous. In this case maxk jVk j ! 0. As V has nitevariation , Var (V ) < ∞, hence ∑k (∆Vk )
2 ! 0. Similarly if S iscontinuous then∑k ∆S (tk )∆V (tk )
maxk j∆Sk j∑ j∆Vk j ! 0.
Medvegyev (CEU) Stochastic integration January 2009 33 / 64
Quadratic variation of semimartingales
TheoremFor continuous semi martingales the quadratic variation is the quadraticvariation of the local martingale part. More generally if
[X1,X2] $ limn!∞ ∑
k
∆X1t(n)k
∆X2t(n)k
is the quadratic covariation of the continuous semi martingalesX1 = X1 (0) + L1 + V1 and X2 = X2 (0) + L2 + V2 then
[X1,X2] (t) = [L1, L2] (t) ,
where Li is the local martingale part of the semi martingale Xi . That is thequadratic co-variation of continuous semi martingales is the quadraticcovariation of the local martingal parts.
Medvegyev (CEU) Stochastic integration January 2009 34 / 64
Quadratic variation of the stochastic integrals
Every stochastic integral is a semimartingale so it has a quadratic variation.
Theorem
If the integral (X S) (t) $R t0 X (s) dS (s) exists then
[X S ] (t) =Z t
0X 2 (s) d [S ] (s) .
[X S ] (t) ∑k(∆ (X S) (tk ))2 ∑
k(X (tk ) (∆S) (tk ))
2 =
= ∑k
X 2 (tk ) (∆S (tk ))2
∑k
X 2 (tk ) ([S ] (tk+1 [S (tk )]) Z t
0X 2 (s) d [S ] (s)
Medvegyev (CEU) Stochastic integration January 2009 35 / 64
Associativity rule for stochastic integrals
TheoremIf Y = X S then Z Y exists if and only if XZ S exists and in this case
Z Y $ Z (X S) = XZ S .
(Z Y ) (t) ∑k
Z (tk ) (Y (tk+1) Y (tk ))
∑k
Z (tk ) (X (tk ) (S (tk+1) S (tk ))) =
= ∑k(Z (tk )X (tk )) (S (tk+1) S (tk ))
Z t
0(ZX ) (s) dS (s) $ ((ZX ) S) (t) .
Medvegyev (CEU) Stochastic integration January 2009 36 / 64
Itôs formula
Theorem (Itôs formula)Let F be a function with n variables. If F is a twice continuouslydi¤erentiable function and if X $ (Xk )nk=1 are continuous semimartingales then
F (X (t)) F (X (0)) =n
∑k=1
Z t
0
∂F∂xk
(X )dXk +12 ∑i ,j
Z t
0
∂2F∂xi∂xj
(X )d [Xi ,Xj ].
Medvegyev (CEU) Stochastic integration January 2009 37 / 64
The proof of Itôs formula and the fundamental theorem ofthe calculus
The proof is quite di¢ cult, but some hints seems to be necessary
F (X (t)) F (X (0)) = ∑k(F (X (tk+1)) F (X (tk ))) .
During the proof of the fundamental theorem of the calculus one uses theapproximation
F (X (tk+1)) F (X (tk )) t F 0 (X (αk )) (X (tk+1) X (tk )) =
= ∑i
∑k
∂F∂xi(X (αk )) (Xi (tk+1) Xi (tk )) .
If Xi has nite variation and if X is continuous then
∑k
∂F∂xi(X (αk )) (Xi (tk+1) Xi (tk ))!
Z t
0
∂F∂xi(X (s)) dXi (s) .
Medvegyev (CEU) Stochastic integration January 2009 38 / 64
The major problem
The major problem is that the continuous local martingales does not havetrajectories with nite variation so the limit of the sum, approximating astochastic integral exists only when the test points αk are the rst pointsof the interval, which is obviously not the case now.
Medvegyev (CEU) Stochastic integration January 2009 39 / 64
The proof of Itôs formula the Taylor approximation
We can also use the better approximation
F 0 (X (tk )) (X (tk+1) X (tk )) +12F 00 (X (αk )) ((X (tk+1) X (tk )))
as well. The second derivative is a quadratic form
F 00(X (αk )(X (tk+1)X (tk )) = ∑j
∑i
∂2F∂xi∂xj
(X (αk ))∆Xi (tk+1)∆Xj (tk+1).
As now the test point in the rst part is tk , the rst point of the interval,the limit of the sums of the rst order terms is now the stochastic integralZ b
aF 0 (X ) dX $ ∑
i
Z b
a
∂F∂xi(X ) dXi .
What is the limit of the second order terms?
Medvegyev (CEU) Stochastic integration January 2009 40 / 64
The limit of the second order terms
As∆Xi (tk+1)∆Xj (tk+1) t ∆ [Xi ,Xj ] (tk+1)
so
∂2F∂xi∂xj
(X (αk ))∆Xi (tk+1)∆Xj (tk+1) t∂2F
∂xi∂xj(X (αk ))∆ [Xi ,Xj ] (tk+1)
hence for the seond order terms
∑k
∂2F∂xi∂xj
(X (αk ))∆Xi (tk+1)∆Xj (tk+1) t
t ∑k
∂2F∂xi∂xj
(X (αk ))∆[Xi ,Xj ](tk+1)!
!Z b
a
∂2F∂xi∂xj
(X (t))d [Xi ,Xj ].
Medvegyev (CEU) Stochastic integration January 2009 41 / 64
Itôs formula
Itôs formula is just a generalization of the fundamental theorem of thecalculus. In the classical analysis the value of the second order terms iszero, but in the stochastic analysis the second order terms are notnegligible. The reason for this is that the trajectories of the localmartingale part of the semi martingales, at least in the continuous case, donot have nite variation.
Medvegyev (CEU) Stochastic integration January 2009 42 / 64
Time dependent Itôs formula
An important special case is f (s,X (s)) :
f (T ,X (T )) f (t,X (t)) =Z T
t
∂f∂s(s,X (s))ds +
Z T
t
∂f∂x(s,X (s))dX (s)+
+12
Z T
t
∂2f∂s2(s,X (s))d [s ] +
12
Z T
t
∂2f∂x2
(s,X (s))d [X ] (s)+
+2 12
Z T
t
∂2f∂s∂x
(s,X (s))d [s,X ] .
Trivially [s ] = 0. And as j∑k akbk j q
∑k a2k
q∑k b
2k
j[s,X ]j t∑k (sk+1 sk ) (X (sk+1) X (sk ))
r∑
k(sk+1 sk )2
q(X (sk+1) X (sk ))2
!! [s ] [X ] = 0.
Medvegyev (CEU) Stochastic integration January 2009 43 / 64
Time dependent Itôs formula
TheoremIf t < T and f (t, x) is a twice continuously di¤erentiable function thenfor any continuous semi martingale X
f (T ,X (T )) f (t,X (t)) =
=Z T
t
∂f∂s(s,X (s)) ds +
Z T
t
∂f∂x(s,X (s)) dX (s) +
+12
Z T
t
∂2f∂x2
(s,X (s)) d [X ] (s) .
Medvegyev (CEU) Stochastic integration January 2009 44 / 64
Time dependent Itôs formula for Wiener processes
TheoremIf t < T and f (t, x) is a twice continuously di¤erentiable function and wis a Wiener process then
f (T ,w (T )) f (t,w (t)) =
=Z T
t
∂f∂s(s,w (s)) +
12
∂2f∂x2
(s,w (s)) ds +Z T
t
∂f∂x(s,X (s)) dw (s) .
The formula is very ofter is written with the next symbolism:
df =
∂f∂s+12
∂2f∂x2
ds +
∂f∂xdw .
Medvegyev (CEU) Stochastic integration January 2009 45 / 64
Calculation of expected value
ExampleCalculate the expected value of the stochastic process sinw .
By Itôs formula
sinw (t) sinw (0) =Z t
0cosw (s) dw (s) +
12
Z t
0 sin (w (s)) d hw (s)i =
=Z t
0cosw (s) dw (s) 1
2
Z t
0sin (w (s)) ds.
The expected value of the stochastic integrals are zero, since thestochastic integrals are martingales which are zero at time zero.
Medvegyev (CEU) Stochastic integration January 2009 46 / 64
Calculation of expected value
Calculating the expected value
E (sinw (t)) = EZ t
0cosw (s) dw (s)
12EZ t
0sin (w (s)) ds
=
= 12
Z t
0E (sin (w (s))) ds.
If f (t) $ E (sin (w (t))) , then
f (t) = 12
Z t
0f (s) ds.
Di¤erentiating both sides
ddsf = 1
2f , f (0) = 0.
Solving it f (t) = 0.
Medvegyev (CEU) Stochastic integration January 2009 47 / 64
Explanation
This is not a big surprise since we are looking for the value of the integral
1p2πt
Z ∞
∞exp
x
2
2t
sin xdx ,
which by the evenness of the sine function is obviously zero.
Medvegyev (CEU) Stochastic integration January 2009 48 / 64
Calculation of expected value
ExampleCalculate the expected value of the stochastic process cosw .
Using Itôs formula
cosw (t) cosw (0) = Z t
0sinw (s) dw (s) +
12
Z t
0 cos (w (s)) ds.
Taking expected value
E( cosw(t) 1 = E(Z t
0sinw(s)dw(s)) 1
2E(Z t
0cosw(s)ds) =
= 12
Z t
0E (cosw (s)) ds.
If f (t) $ E (cosw (t)) thenddtf (t) = 1
2f (s) f (0) = 1,
and the solution is f (t) = exp 12 t.
Medvegyev (CEU) Stochastic integration January 2009 49 / 64
Calculation of expected value
Example
Calculate the expected value of exp (w).
Again by Itôs formula
exp (w (t)) 1 =Z t
0exp (w (s)) dw (s) +
12
Z t
0exp (w (s)) ds.
Taking expected value
E (exp (w (t))) 1 = EZ t
0exp (w (s)) dw (s)
+12EZ t
0exp (w (s)) ds
.
The di¤erential equation isddtf (t) =
12f (t) , f (0) = 1.
The solution is
f (t) = exp12t.
E.g.
1p2π
Z ∞
∞exp (s) exp
s
2
2
ds =
1p2π
Z ∞
∞exp
12
h(s 1)2 1
ids =
= e1/2.
Medvegyev (CEU) Stochastic integration January 2009 50 / 64
Calculation of expected value
ExampleCalculate the expected value of the lognormal distribution.
Using Itôs formula
exp (µ+ σw (t)) exp (µ) = σZ t
0exp (µ+ σw (s)) dw +
+12
σ2Z t
0exp (µ+ σw (s)) ds.
Taking expected value and using that the expected value of the stochasticintegral is zero
ddtf =
σ2
2f (t) , f (0) = exp (µ) .
The solution of the equation is
f (t) = exp
σ2
2t + µ
.
Medvegyev (CEU) Stochastic integration January 2009 51 / 64
Calculation of quadratic variation
Example
Calculate the quadratic variation [sinw ] (t).
By Itôs formula
sinw (t) sinw (0) =Z t
0cosw (s) dw (s) 1
2
Z t
0sinw (s) d hwi (s) =
=Z t
0cosw (s) dw (s) 1
2
Z t
0sinw (s) ds.
We should take only the quadratic variation of the stochastic integral,since the other terms are zero, so
[sinw ] (t) =Z t
0cos2 w (s) ds.
Medvegyev (CEU) Stochastic integration January 2009 52 / 64
Calculation of expected value
Example
Calculate E ([sinw ])
Using the results of the previous example
E ([sinw ] (t)) = EZ t
0cos2 w (s) ds
=Z t
0Ecos2 w (s)
ds =
=Z t
0E1+ cos 2w (s)
2
ds =
12
Z t
01ds +
Z t
0E (cos 2w (s)) ds.
The second term can be calculated by Itôs formula
cos 2w (s) 1 = 2Z s
0sin 2w (u) dw (u) 1
24Z s
0cos 2w (u) du.
Taking expected value
f (s) 1 = 2Z s
0f (u) du,
sof (s) = exp (2s) .
hence the expected value is
12t +
Z t
0exp (2s) ds = 1
2t + [exp (2s)]t0 =
12t + exp (2t) 1.
Medvegyev (CEU) Stochastic integration January 2009 53 / 64
The BlackScholes equation
The BlackScholes partial di¤erential equation is
∂f∂t+ rS
∂f∂S+12
σ2S2∂2f∂S2
= rf ,
with the conditionf (T ,S) = max fS X , 0g .
This is the special case of the Cauchy problem
∂f∂t+ µ
∂f∂x+12
σ2∂2f∂x2
+ rf + k = 0,
f (T , x) = Φ (x) .
Medvegyev (CEU) Stochastic integration January 2009 54 / 64
Solution of partial di¤erential equations
We will study the equations
∂f∂t+ µ
∂f∂x+12
σ2∂2f∂x2
= 0,
f (T , x) = Φ (x) .
To nd f (t, x) for every (t, x) let us try to solve the stochastic di¤erentialequation
dX (s) = µ (s,X (s)) ds + σ (s,X (s)) dw (s) , X (t) = x .
This means that for every t < T
X (T ) x = X (T )X (t) =Z T
tµ (s,X (s)) ds+
Z T
tσ (s,X (s)) dw (s) .
Medvegyev (CEU) Stochastic integration January 2009 55 / 64
Solution of partial di¤erential equations
If we introduce the operator
Af $ µ∂f∂x+12
σ2∂2f∂x2
then the PDE is∂f∂t+ Af = 0.
Medvegyev (CEU) Stochastic integration January 2009 56 / 64
Application of Itôs formula
Let f (t, x) 2 C 2 be a solution of the equation. By the time-dependentItôs formula
df =∂f∂sds +
∂f∂xdX +
12
∂2f∂x2
d [X ] .
Using that X satises the stochastic di¤erential equation and using theassociativity of the stochastic integration
∂f∂xdX =
∂f∂x(µds + σdw) = µ
∂f∂xds + σ
∂f∂xdw .
Using that the quadratic variation of a semi martingale is the quadraticvariation of the local martingale part
[X ] = [µds + σdw ] =
= [µds ] + 2 [µds, σdw ] + [σdw ] =
= [σdw ] = σ2d [w ] = σ2ds.
Medvegyev (CEU) Stochastic integration January 2009 57 / 64
Application of Itôs formula
Using the f satises the PDE
df =∂f∂sds +
∂f∂xdX +
12
∂2f∂x2
d [X ] =
=∂f∂sds + µ
∂f∂xds + σ
∂f∂xdw +
12
σ2∂2f∂x2
ds =
=
∂f∂s+ µ
∂f∂x+12
σ2∂2f∂x2
ds + σ
∂f∂xdw =
=
∂f∂s+ Af
ds + σ
∂f∂xdw = σ
∂f∂xdw .
Medvegyev (CEU) Stochastic integration January 2009 58 / 64
Application of Itôs formula
Using the usual notation this means that
f (T ,X (T )) f (t,X (t)) =Z T
t
∂f∂s+ Af
ds +
Z T
t
∂f∂x
σdw =
=Z T
t0ds +
Z T
t
∂f∂x
σdw
=Z T
t
∂f∂x
σdw .
Taking the expected value on both sides
E (f (T ,X (T )) f (t,X (t))) = EZ T
t
∂f∂x
σdw= 0.
Medvegyev (CEU) Stochastic integration January 2009 59 / 64
The formula
But X (t) = x , so f (t,X (t)) = f (t, x) is a constant. Asf (T , x) = Φ (x) we get the following formuala
Theorem
f (t, x) = E (f (T ,X (T ))) = E (Φ (X (T ))) .
Medvegyev (CEU) Stochastic integration January 2009 60 / 64
Example I
ExampleLet us solve the equation
∂h∂t+12
σ2∂2h∂x2
= 0, h (T , x) = x2
The SDE isdX (s) = 0 ds + σdw (s) , X (t) = x .
The solution isX (s) = x + σ (w (s) w (t)) .
Hence X (s) = Nx , σ
ps t
. So the solution is
h (t, x) = EX 2 (T )
= D2 (X (T )) + E2 (X (T )) =
= σ2 (T t) + x2.
Medvegyev (CEU) Stochastic integration January 2009 61 / 64
Example I
∂h∂t+12
σ2∂2h∂x2
= σ2 + σ2 = 0
h (T , x) = σ2 (T T ) + x2 = x2.
Medvegyev (CEU) Stochastic integration January 2009 62 / 64
Example II
ExampleLet us solve the equation
∂h∂t+12x2
∂2h∂x2
= 0, h (T , x) = ln x2
The SDE is
dX (s) = 0 ds + X (s) dw (s) = X (s) dw (s) , X (t) = x ,
The solution is
X (s) = x exp12(s t) + w (s t)
.
Medvegyev (CEU) Stochastic integration January 2009 63 / 64
Example II
lnX 2 (T )
= (T t) + 2N
0,pT t
+ ln x2,
that is
h (t, x) = E (T t) + 2N
0,pT t
+ ln x2
=
= (t T ) + ln x2.
Which is really the solution
∂h∂t+12x2
∂2h∂x2
= 1+12x21x22x0=
= 1+ x21x
0= 1+ x2
1x2
= 0.
h (T , x) = (T T ) + ln x2 = ln x2.
Medvegyev (CEU) Stochastic integration January 2009 64 / 64