Itˆo’s Formula

This formula becomes mathematically respectable only after we integrate it.. The total error will have limit zero in the last step of the following argument... This term has zero quadrat

Itˆo’s Formula

15.1 Itˆo’s formula for one Brownian motion

We want a rule to “differentiate” expressions of the formf ( B ( t )), wheref ( x ) is a differentiable function IfB ( t )were also differentiable, then the ordinary chain rule would give

d dtf ( B ( t )) = f0( B ( t )) B0( t ) ; which could be written in differential notation as

df ( B ( t )) = f0

( B ( t )) B0

( t ) dt

= f0

( B ( t )) dB ( t )

However,B ( t )is not differentiable, and in particular has nonzero quadratic variation, so the correct formula has an extra term, namely,

df ( B ( t )) = f0( B ( t )) dB ( t ) + 1 2 f0( B ( t )) |{z}dt

dB(t) dB(t) :

This is It ˆo’s formula in differential form Integrating this, we obtain It ˆo’s formula in integral form:

f ( B ( t )),f ( B (0))

| {z }

f(0) =Z t

0 f0

( B ( u )) dB ( u ) + 1 2Z t

0 f00

( B ( u )) du:

Remark 15.1 (Differential vs Integral Forms) The mathematically meaningful form of It ˆo’s

for-mula is It ˆo’s forfor-mula in integral form:

f ( B ( t )),f ( B (0)) =Z t

0 f0( B ( u )) dB ( u ) + 1 2Z t

0 f00( B ( u )) du:

167

This is because we have solid definitions for both integrals appearing on the right-hand side The first,

Z t

0 f0

( B ( u )) dB ( u )

is an It ˆo integral, defined in the previous chapter The second,

Z t

0 f0( B ( u )) du;

is a Riemann integral, the type used in freshman calculus.

For paper and pencil computations, the more convenient form of It ˆo’s rule is It ˆo’s formula in

differ-ential form:

df ( B ( t )) = f0

( B ( t )) dB ( t ) + 1 2 f00

( B ( t )) dt:

There is an intuitive meaning but no solid definition for the termsdf ( B ( t )) ;dB ( t )anddtappearing

in this formula This formula becomes mathematically respectable only after we integrate it

15.2 Derivation of Itˆo’s formula

Considerf ( x ) = 1 2 x 2, so that

f0( x ) = x; f00( x ) = 1 : Letx k ;x k+1be numbers Taylor’s formula implies

f ( x k+1 ),f ( x k ) = ( x k+1,x k ) f0

( x k ) + 1 2 ( x k+1,x k ) 2 f00

( x k ) :

In this case, Taylor’s formula to second order is exact becausef is a quadratic function.

In the general case, the above equation is only approximate, and the error is of the order of( x k+1,

x k ) 3 The total error will have limit zero in the last step of the following argument.

FixT > 0and let =ft 0 ;t 1 ;::: ;t ngbe a partition of[0 ;T ] Using Taylor’s formula, we write:

f ( B ( T )),f ( B (0))

= 1

2 B 2 ( T ),

1

2 B 2 (0)

= nX ,1

k=0 [ f ( B ( t k+1 )),f ( B ( t k ))]

= nX ,1

k=0 [ B ( t k+1 ),B ( t k )] f0

( B ( t k )) + 1 2 nX ,1

k=0 [ B ( t k+1 ),B ( t k )] 2 f00

( B ( t k ))

= nX ,1

k=0 B ( t k )[ B ( t k+1 ),B ( t k )] + 1

2

nX ,1 k=0 [ B ( t k+1 ),B ( t k )] 2 :

We letjjjj!0to obtain

f ( B ( T )),f ( B (0)) =Z T

0 B ( u ) dB ( u ) + 1 2hBi( T )

| {z }

T

=Z T

0 f0( B ( u )) dB ( u ) + 1 2Z T

0 f00( B ( u ))

| {z }

1 du:

This is It ˆo’s formula in integral form for the special case

f ( x ) = 1 2 x 2 :

15.3 Geometric Brownian motion

Definition 15.1 (Geometric Brownian Motion) Geometric Brownian motion is

S ( t ) = S (0)expn

B ( t ) +

,1 2  2

to

; whereand > 0are constant

Define

f ( t;x ) = S (0)expn

x +

,1 2  2

to

; so

S ( t ) = f ( t;B ( t )) : Then

f t =

,1 2  2

f; f x = f; f xx =  2 f:

According to It ˆo’s formula,

dS ( t ) = df ( t;B ( t ))

= f t dt + f x dB + 1 2 f xx dBdB| {z }

dt

= ( ,1 2  2 ) f dt + f dB + 1 2  2 f dt

= S ( t ) dt + S ( t ) dB ( t )

Thus, Geometric Brownian motion in differential form is

dS ( t ) = S ( t ) dt + S ( t ) dB ( t ) ;

and Geometric Brownian motion in integral form is

S ( t ) = S (0) +Z t

0 S ( u ) du +Z t

0 S ( u ) dB ( u ) :

15.4 Quadratic variation of geometric Brownian motion

In the integral form of Geometric Brownian motion,

S ( t ) = S (0)+Z t

0 S ( u ) du +Z t

0 S ( u ) dB ( u ) ; the Riemann integral

F ( t ) =Z t

0 S ( u ) du

is differentiable withF0( t ) = S ( t ) This term has zero quadratic variation The It ˆo integral

G ( t ) =Z t

0 S ( u ) dB ( u )

is not differentiable It has quadratic variation

hGi( t ) =Z t

0  2 S 2 ( u ) du:

Thus the quadratic variation ofS is given by the quadratic variation ofG In differential notation,

we write

dS ( t ) dS ( t ) = ( S ( t ) dt + S ( t ) dB ( t )) 2 =  2 S 2 ( t ) dt

15.5 Volatility of Geometric Brownian motion

Fix0  T 1  T 2 Let = ft 0 ;::: ;t ngbe a partition of[ T 1 ;T 2 ] The squared absolute sample

volatility ofSon[ T 1 ;T 2 ]is

1

T 2,T 1

nX ,1 k=0 [ S ( t k+1 ),S ( t k )] 2 '

1

T 2,T 1

T2 Z

T1

 2 S 2 ( u ) du ' 2 S 2 ( T 1 )

As T 2 # T 1, the above approximation becomes exact In other words, the instantaneous relative

volatility ofSis 2 This is usually called simply the volatility ofS

15.6 First derivation of the Black-Scholes formula

Wealth of an investor An investor begins with nonrandom initial wealth X 0 and at each timet, holds
( t )shares of stock Stock is modelled by a geometric Brownian motion:

dS ( t ) = S ( t ) dt + S ( t ) dB ( t ) :

( t ) can be random, but must be adapted The investor finances his investing by borrowing or lending at interest rater

LetX ( t )denote the wealth of the investor at timet Then

dX ( t ) =
( t ) dS ( t ) + r [ X ( t ),
( t ) S ( t )] dt

=
( t )[ S ( t ) dt + S ( t ) dB ( t )] + r [ X ( t ),
( t ) S ( t )] dt

= rX ( t ) dt +
( t ) S ( t ) ( ,r )

| {z }

Risk premium

dt +
( t ) S ( t ) dB ( t ) :

Value of an option Consider an European option which paysg ( S ( T ))at timeT Letv ( t;x )denote the value of this option at timetif the stock price isS ( t ) = x In other words, the value of the option at each timet2[0 ;T ]is

v ( t;S ( t )) : The differential of this value is

dv ( t;S ( t )) = v t dt + v x dS + 1 2 v xx dS dS

= v t dt + v x [ S dt + S dB ] + 1 2 v xx  2 S 2 dt

=h

v t + Sv x + 1 2  2 S 2 v xx

i

dt + Sv x dB

A hedging portfolio starts with some initial wealthX 0and invests so that the wealthX ( t )at each time tracksv ( t;S ( t )) We saw above that

dX ( t ) = [ rX +
( ,r ) S ] dt + S
dB:

To ensure thatX ( t ) = v ( t;S ( t ))for allt, we equate coefficients in their differentials Equating the

dBcoefficients, we obtain the
-hedging rule:

( t ) = v x ( t;S ( t )) : Equating thedtcoefficients, we obtain:

v t + Sv x + 1 2  2 S 2 v xx = rX +
( ,r ) S:

But we have set
= v x, and we are seeking to causeXto agree withv Making these substitutions,

we obtain

v t + Sv x + 1 2  2 S 2 v xx = rv + v x ( ,r ) S;

(wherev = v ( t;S ( t ))andS = S ( t )) which simplifies to

v t + rSv x + 1 2  2 S 2 v xx = rv:

In conclusion, we should letvbe the solution to the Black-Scholes partial differential equation

v t ( t;x ) + rxv x ( t;x ) + 1 2  2 x 2 v xx ( t;x ) = rv ( t;x )

satisfying the terminal condition

v ( T;x ) = g ( x ) :

If an investor starts withX 0 = v (0 ;S (0))and uses the hedge
( t ) = v x ( t;S ( t )), then he will have

X ( t ) = v ( t;S ( t ))for allt, and in particular,X ( T ) = g ( S ( T ))

15.7 Mean and variance of the Cox-Ingersoll-Ross process

The Cox-Ingersoll-Ross model for interest rates is

dr ( t ) = a ( b,cr ( t )) dt + q

r ( t ) dB ( t ) ; wherea;b;c;andr (0)are positive constants In integral form, this equation is

r ( t ) = r (0) + aZ t

0 ( b,cr ( u )) du + Z t

0

q

r ( u ) dB ( u ) :

We apply It ˆo’s formula to computedr 2 ( t ) This isdf ( r ( t )), wheref ( x ) = x 2 We obtain

dr 2 ( t ) = df ( r ( t ))

= f0

( r ( t )) dr ( t ) + 1 2 f00

( r ( t )) dr ( t ) dr ( t )

= 2 r ( t )



a ( b,cr ( t )) dt + q

r ( t ) dB ( t )



+



a ( b,cr ( t )) dt + q

r ( t ) dB ( t )

2

= 2 abr ( t ) dt,2 acr 2 ( t ) dt + 2 r3

( t ) dB ( t ) +  2 r ( t ) dt

= (2 ab +  2 ) r ( t ) dt,2 acr 2 ( t ) dt + 2 r3

2( t ) dB ( t )

The mean ofr ( t ) The integral form of the CIR equation is

r ( t ) = r (0) + aZ t

0 ( b,cr ( u )) du + Z t

0

q

r ( u ) dB ( u ) : Taking expectations and remembering that the expectation of an It ˆo integral is zero, we obtain

IEr ( t ) = r (0) + aZ t

0 ( b,cIEr ( u )) du:

Differentiation yields

d dtIEr ( t ) = a ( b,cIEr ( t )) = ab,acIEr ( t ) ; which implies that

d dt

h

e act IEr ( t )i

= e act

acIEr ( t ) + d

dtIEr ( t )



= e act ab:

Integration yields

e act IEr ( t ),r (0) = abZ t

0 e acu du = b

c ( e act,1) :

We solve forIEr ( t ):

IEr ( t ) = b

c + e,act

r (0),

b c

 :

Ifr (0) = b c, thenIEr ( t ) = b c for everyt Ifr (0)6= b c, thenr ( t )exhibits mean reversion:

lim

t IEr ( t ) = b

c:

Variance ofr ( t ) The integral form of the equation derived earlier fordr 2 ( t )is

r 2 ( t ) = r 2 (0) + (2 ab +  2 )Z t

0 r ( u ) du,2 acZ t

0 r 2 ( u ) du + 2 Z t

0 r3

2( u ) dB ( u ) : Taking expectations, we obtain

IEr 2 ( t ) = r 2 (0) + (2 ab +  2 )Z t

0 IEr ( u ) du,2 acZ t

0 IEr 2 ( u ) du:

Differentiation yields

d dtIEr 2 ( t ) = (2 ab +  2 ) IEr ( t ),2 acIEr 2 ( t ) ; which implies that

d dte 2act IEr 2 ( t ) = e 2act



2 acIEr 2 ( t ) + d

dtIEr 2 ( t )



= e 2act (2 ab +  2 ) IEr ( t ) : Using the formula already derived forIEr ( t ) and integrating the last equation, after considerable algebra we obtain

IEr 2 ( t ) = b 2

2 ac 2 + b 2

c 2 +

r (0),

b c



 2

ac + 2 c b

!

e,act

+

r (0),

b c

2  2

ace,2act +  2

ac

 b

2 c,r (0)

e,2act :

var r ( t ) = IEr 2 ( t ),( IEr ( t )) 2

= b 2

2 ac 2 +

r (0),

b c



 2

ace,act +  2

ac

 b

2 c ,r (0)

e,2act :

15.8 Multidimensional Brownian Motion

Definition 15.2 (d-dimensional Brownian Motion) A d-dimensional Brownian Motion is a

pro-cess

B ( t ) = ( B 1 ( t ) ;::: ;B d ( t ))

with the following properties:

 EachB k ( t )is a one-dimensional Brownian motion;

 Ifi6= j, then the processesB i ( t )andB j ( t )are independent

Associated with ad-dimensional Brownian motion, we have a filtrationfF( t )gsuch that

 For eacht, the random vectorB ( t )isF( t )-measurable;

 For eachtt 1 :::t n, the vector increments

B ( t 1 ),B ( t ) ;::: ;B ( t n ),B ( t n,1 )

are independent of ( t )

15.9 Cross-variations of Brownian motions

Because each componentB iis a one-dimensional Brownian motion, we have the informal equation

dB i ( t ) dB i ( t ) = dt:

However, we have:

Theorem 9.49 Ifi6= j,

dB i ( t ) dB j ( t ) = 0

Proof: Let =ft 0 ;::: ;t ngbe a partition of[0 ;T ] Fori6= j, define the sample cross variation

ofB iandB j on[0 ;T ]to be

C  = nX ,1 k=0 [ B i ( t k+1 ),B i ( t k )][ B j ( t k+1 ),B j ( t k )] : The increments appearing on the right-hand side of the above equation are all independent of one another and all have mean zero Therefore,

IEC  = 0 :

We computevar( C  ) First note that

C 2 = nX ,1

k=0



B i ( t k+1 ),B i ( t k )2

B j ( t k+1 ),B j ( t k )2

+ 2 nX ,1

`<k [ B i ( t `+1 ),B i ( t ` )][ B j ( t `+1 ),B j ( t ` )] : [ B i ( t k+1 ),B i ( t k )][ B j ( t k+1 ),B j ( t k )]

All the increments appearing in the sum of cross terms are independent of one another and have mean zero Therefore,

var( C  ) = IEC 2

= IE nX ,1 k=0 [ B i ( t k+1 ),B i ( t k )] 2 [ B j ( t k+1 ),B j ( t k )] 2 : But[ B i ( t k+1 ),B i ( t k )] 2 and[ B j ( t k+1 ),B j ( t k )] 2 are independent of one another, and each has expectation( t k+1,t k ) It follows that

var( C  ) = nX ,1

k=0 ( t k+1,t k ) 2  jjjj

nX ,1 k=0 ( t k+1,t k ) =jjjj:T:

Asjjjj!0, we havevar( C  )!0, soC  converges to the constantIEC  = 0

15.10 Multi-dimensional Itˆo formula

To keep the notation as simple as possible, we write the It ˆo formula for two processes driven by a

two-dimensional Brownian motion The formula generalizes to any number of processes driven by

a Brownian motion of any number (not necessarily the same number) of dimensions.

LetXandY be processes of the form

X ( t ) = X (0)+Z t

0 ( u ) du +Z t

0  11 ( u ) dB 1 ( u ) +Z t

0  12 ( u ) dB 2 ( u ) ;

Y ( t ) = Y (0) +Z t

0 ( u ) du +Z t

0  21 ( u ) dB 1 ( u ) +Z t

0  22 ( u ) dB 2 ( u ) : Such processes, consisting of a nonrandom initial condition, plus a Riemann integral, plus one or

more It ˆo integrals, are called semimartingales The integrands ( u ) ; ( u ) ;and ij ( u )can be any adapted processes The adaptedness of the integrands guarantees thatXandY are also adapted In differential notation, we write

dX = dt +  11 dB 1 +  12 dB 2 ;

dY = dt +  21 dB 1 +  22 dB 2 : Given these two semimartingalesXandY, the quadratic and cross variations are:

dX dX = ( dt +  11 dB 1 +  12 dB 2 ) 2 ;

=  211 dB| 1{zdB 1}

dt +2  11  12 dB| 1{zdB 2}

0 +  212 dB| 2{zdB 2}

dt

= (  211 +  212 ) 2 dt;

dY dY = ( dt +  21 dB 1 +  22 dB 2 ) 2

= (  221 +  222 ) 2 dt;

dX dY = ( dt +  11 dB 1 +  12 dB 2 )( dt +  21 dB 1 +  22 dB 2 )

= (  11  21 +  12  22 ) dt Letf ( t;x;y )be a function of three variables, and letX ( t )andY ( t )be semimartingales Then we have the corresponding It ˆo formula:

df ( t;x;y ) = f t dt + f x dX + f y dY + 1 2 [ f xx dX dX + 2 f xy dX dY + f yy dY dY ] :

In integral form, withXandY as decribed earlier and with all the variables filled in, this equation is

f ( t;X ( t ) ;Y ( t )),f (0 ;X (0) ;Y (0))

=Z t

0 [ f t + f x + f y + 1 2 (  211 +  212 ) f xx + (  11  21 +  12  22 ) f xy + 1 2 (  221 +  222 ) f yy ] du

+Z t

0 [  11 f x +  21 f y ] dB 1 + Z t

0 [  12 f x +  22 f y ] dB 2 ; wheref = f ( u;X ( u ) ;Y ( u ), fori;j 1 ; 2 , ij =  ij ( u ), andB i = B i ( u )

df ( t;x;y ) = f t dt + f x dX + f y dY + 2 [ f xx dX dX + f xy dX