Stochastic Integration and Itô™s formulamedvegyev.uni-corvinus.hu/St3.pdf · Medvegyev (CEU)...

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.


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.)


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.


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.


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.


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 .)


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

.


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.


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) .


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.


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.


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) .


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.


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.


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.


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 ! ∞.


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 = ∞


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.



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.



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 .


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.


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].


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.


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.


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.


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.)


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.


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.


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.


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 .



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.



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.


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)


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) .


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 ].


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) .


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.


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?


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 ].


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.


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.


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) .


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 .


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.



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.


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.



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.



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.


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.



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

.


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.



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.


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.


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) .


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) .


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.


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.



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 .



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.


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 ))) .


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.


Example I

∂h∂t+12

σ2∂2h∂x2

= σ2 + σ2 = 0

h (T , x) = σ2 (T T ) + x2 = x2.


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)

.


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.


Stochastic Integration and Itô™s formulamedvegyev.uni-corvinus.hu/St3.pdf · Medvegyev (CEU)...

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Transcript of Stochastic Integration and Itô™s formulamedvegyev.uni-corvinus.hu/St3.pdf · Medvegyev (CEU)...