Summary of Arbitrage Pricing Theory

Summary of Arbitrage Pricing TheoryA simple European derivative security makes a random payment at a time fixed in advance.. 22.1 Binomial model, Hedging Portfolio Let be the set of all

Summary of Arbitrage Pricing Theory

A simple European derivative security makes a random payment at a time fixed in advance The value at timetof such a security is the amount of wealth needed at timetin order to replicate the

security by trading in the market The hedging portfolio is a specification of how to do this trading.

22.1 Binomial model, Hedging Portfolio

Let be the set of all possible sequences ofncoin-tosses We have no probabilities at this point.

Letr0 ; u > r + 1 ; d = 1 =ube given (See Fig 2.1)

Evolution of the value of a portfolio:

X k+1 =
k S k+1 + (1 + r )( X k,
k S k ) :

Given a simple European derivative securityV ( ! 1 ;! 2 ), we want to start with a nonrandomX 0and use a portfolio processes

0 ;
1 ( H ) ;
1 ( T )

so that

X 2 ( ! 1 ;! 2 ) = V ( ! 1 ;! 2 ) 8! 1 ;! 2 : (four equations) There are four unknowns:X 0 ;
0 ;
1 ( H ) ;
1 ( T ) Solving the equations, we obtain:

223

X 1 ( ! 1 ) = 1 1 + r

2 6 4

1 + r,d

u,d X| 2 ( !{z1 ;H})

V ( !1;H)

+ u,(1 + r )

u,d X| 2 ( !{z1 ;T )}

V ( !1;T)

3 7

5;

X 0 = 1 1 + r



1 + r,d

u,d X 1 ( H ) + u,(1 + r )

u,d X 1 ( T )



;

1 ( ! 1 ) = X 2 ( ! 1 ;H ),X 2 ( ! 1 ;T )

S 2 ( ! 1 ;H ),S 2 ( ! 1 ;T ) ;

0 = X 1 ( H ),X 1 ( T )

S 1 ( H ),S 1 ( T ) :

The probabilities of the stock price paths are irrelevant, because we have a hedge which works on

every path From a practical point of view, what matters is that the paths in the model include all

the possibilities We want to find a description of the paths in the model They all have the property

(log S k+1,log S k ) 2 =



log S k+1

S k

2

= (log u ) 2

= (log u ) 2 :

Let = log u > 0 Then

nX ,1 k=0 (log S k+1,log S k ) 2 =  2 n:

The paths oflog S kaccumulate quadratic variation at rate 2per unit time.

If we changeu, then we change, and the pricing and hedging formulas on the previous page will give different results

We reiterate that the probabilities are only introduced as an aid to understanding and computation Recall:

X k+1 =
k S k+1 + (1 + r )( X k,
k S k ) :

Define

k = (1 + r ) k :

Then

X k+1 k+1 =
k S k+1

k+1 + X k

k ,
k S k

k ;

i.e.,

X k+1 k+1 ,

X k

k =
k



S k+1 k+1 ,

S k k



:

In continuous time, we will have the analogous equation

d

X ( t )

( t )



=
( t ) d

S ( t )

( t )



:

If we introduce a probability measure IP under which S  k

k is a martingale, then X  k

k will also be a

martingale, regardless of the portfolio used Indeed,

f

IE

X k+1 k+1

Fk



=fIE

X k

k +
k



S k+1 k+1 ,

S k k



Fk



= X k

k +
k

 f

IE

S k+1 k+1

Fk

 ,

S k k



:

=0

Suppose we want to haveX 2 = V, whereV is someF2-measurable random variable Then we must have

1

1 + rX 1 = X 1 1 = IEf 

X 2 2

F1



= IEf 

V 2

F1



;

X 0 = X 0

0 = IEf 

X 1 1



=fIE

V 2



:

To find the risk-neutral probability measureIPf under which S  k

k is a martingale, we denote p ~ =

f

IPf! k = Hg,q ~ = IPf f! k = Tg, and compute

f

IE

S k+1 k+1

Fk



= ~ pu S  k+1 k + ~ qd S  k+1 k

= 1 1 + r [~ pu + ~ qd ] S k k :

We need to choosep ~andq ~so that

~

pu + ~ qd = 1 + r;

~

p + ~ q = 1 :

The solution of these equations is

~

p = 1 + r,d

u,d ; q ~ = u,(1 + r )

u,d :

22.2 Setting up the continuous model

Now the stock price S ( t ) ; 0  t  T, is a continuous function of t We would like to hedge along every possible path ofS ( t ), but that is impossible Using the binomial model as a guide, we choose > 0and try to hedge along every pathS ( t )for which the quadratic variation oflog S ( t )

accumulates at rate 2per unit time These are the paths with volatility 2.

To generate these paths, we use Brownian motion, rather than coin-tossing To introduce Brownian motion, we need a probability measure However, the only thing about this probability measure which ultimately matters is the set of paths to which it assigns probability zero

LetB ( t ) ; 0  t  T, be a Brownian motion defined on a probability space ;F;P) For any

2IR, the paths of

t + B ( t )

accumulate quadratic variation at rate 2per unit time We want to define

S ( t ) = S (0)expft + B ( t )g;

so that the paths of

log S ( t ) = log S (0) + t + B ( t )

accumulate quadratic variation at rate 2per unit time Surprisingly, the choice ofin this definition

is irrelevant Roughly, the reason for this is the following: Choose! 1 2 Then, for 12IR,

 1 t + B ( t;! 1 ) ; 0tT;

is a continuous function oft If we replace 1by 2, then 2 t + B ( t;! 1 )is a different function However, there is an! 22 such that

 1 t + B ( t;! 1 ) =  2 t + B ( t;! 2 ) ; 0tT:

In other words, regardless of whether we use 1or 2in the definition ofS ( t ), we will see the same paths The mathematically precise statement is the following:

If a set of stock price paths has a positive probability whenS ( t )is defined by

S ( t ) = S (0)expf 1 t + B ( t )g;

then this set of paths has positive probability whenS ( t )is defined by

S ( t ) = S (0)expf 2 t + B ( t )g:

Since we are interested in hedging along every path, except possibly for a set of paths which has probability zero, the choice ofis irrelevant

The most convenient choice ofis

 = r,1 2  2 ;

so

S ( t ) = S (0)expfrt + B ( t ),1 2  2 tg;

and

e,rt S ( t ) = S (0)expfB ( t ),1 2  2 tg

is a martingale underIP With this choice of,

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

andIP is the risk-neutral measure If a different choice ofis made, we have

S ( t ) = S (0)expft + B ( t )g;

dS ( t ) = (  + 1 2  2 )

| {z }

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

= rS ( t ) dt + h

,r

 dt + dB ( t )i

:

d B(t)e e

B has the same paths as B We can change to the risk-neutral measure fIP, under whichBe is a Brownian motion, and then proceed as ifhad been chosen to be equal tor, 1 2  2.

22.3 Risk-neutral pricing and hedging

LetfIP denote the risk-neutral measure Then

dS ( t ) = rS ( t ) dt + S ( t ) d Be( t ) ;

whereBe is a Brownian motion underfIP Set

( t ) = e rt :

Then

d

S ( t )

( t )



= S  ( ( t t ) ) d Be( t ) ;

so S(t)

(t)is a martingale underfIP

Evolution of the value of a portfolio:

dX ( t ) =
( t ) dS ( t ) + r ( X ( t ),
( t ) S ( t )) dt; (3.1) which is equivalent to

d

X ( t )

( t )



=
( t ) d

S ( t )

( t )



(3.2)

=
( t ) S  ( ( t t ) ) d Be( t ) :

Regardless of the portfolio used, X(t)

(t) is a martingale underfIP Now supposeV is a givenF( T )-measurable random variable, the payoff of a simple European derivative security We want to find the portfolio process
( T ) ; 0  t  T, and initial portfolio valueX (0)so thatX ( T ) = V BecauseX(t)

(t) must be a martingale, we must have

X ( t )

( t ) =fIE

V ( T )

F( t )

This is the risk-neutral pricing formula We have the following sequence:

1 V is given,

2 DefineX ( t ) ; 0tT, by (3.3) (not by (3.1) or (3.2), because we do not yet have
( t ))

3 Construct
( t ) so that (3.2) (or equivalently, (3.1)) is satisfied by theX ( t ) ; 0  t  T, defined in step 2

To carry out step 3, we first use the tower property to show thatX(t)

(t) defined by (3.3) is a martingale underfIP We next use the corollary to the Martingale Representation Theorem (Homework Problem 4.5) to show that

d

X ( t )

( t )



= ( t ) d Be( t ) (3.4) for some proecss Comparing (3.4), which we know, and (3.2), which we want, we decide to define

( t ) = ( t ) ( t )

Then (3.4) implies (3.2), which implies (3.1), which implies thatX ( t ) ; 0t T, is the value of the portfolio process
( t ) ; 0tT

From (3.3), the definition ofX, we see that the hedging portfolio must begin with value

X (0) =fIE

V ( T )



;

and it will end with value

X ( T ) = ( T ) IEf 

V ( T )

F( T )

= ( T ) V

( T ) = V:

Remark 22.1 Although we have takenrand to be constant, the risk-neutral pricing formula is still “valid” whenrand are processes adapted to the filtration generated byB If they depend on eitherBeor onS, they are adapted to the filtration generated byB The “validity” of the risk-neutral pricing formula means:

1 If you start with

X (0) = IEf 

V ( T )



;

then there is a hedging portfolio
( t ) ; 0tT, such thatX ( T ) = V;

2 At each timet, the valueX ( t )of the hedging portfolio in 1 satisfies

X ( t )

( t ) = IEf 

V ( T )

F( t )

:

Remark 22.2 In general, when there are multiple assets and/or multiple Brownian motions, the

risk-neutral pricing formula is valid provided there is a unique risk-neutral measure A probability

measure is said to be risk-neutral provided

 it has the same probability-zero sets as the original measure;

 it makes all the discounted asset prices be martingales

To see if the risk-neutral measure is unique, compute the differential of all discounted asset prices and check if there is more than one way to define Be so that all these differentials have onlyd Be

terms

22.4 Implementation of risk-neutral pricing and hedging

To get a computable result from the general risk-neutral pricing formula

X ( t )

( t ) = IEf 

V ( T )

F( t )

;

one uses the Markov property We need to identify some state variables, the stock price and possibly

other variables, so that

X ( t ) = ( t ) IEf 

V ( T )

F( t )

is a function of these variables

Example 22.1 Assumerandare constant, andV = h(S(T)) We can take the stock price to be the state variable Define

v(t;x) = IEe t;x

h

e,r (T,t)h(S(T))i

:

Then

X(t) = er t

e

IE



e,r Th(S(T))

F(t)



= v(t;S(t));

andX(t)

(t) = e,r tv(t;S(t))is a martingale underIPe

Example 22.2 Assumerandare constant

V = h Z

T

0

S(u) du

!

:

TakeS(t)andY (t) =R

t

0S(u) duto be the state variables Define

v(t;x;y) = IEe t;x;y

h

e,r (T ,t)h(Y (T))i

;

where

Y (T) = y +Z

T

S(u) du:

X(t) = er t

e

IE



e,r Th(S(T))

F(t)



= v(t;S(t);Y (t))

and

X(t) (t) = e,r tv(t;S(t);Y (t))

is a martingale underIPe

Example 22.3 (Homework problem 4.2)

dS(t) = r(t;Y (t)) S(t)dt + (t;Y (t))S(t) d B(t);e

e

B(t);

V = h(S(T)):

TakeS(t)andY (t)to be the state variables Define

v(t;x;y) = IEe t;x;y

2 6 6 6 4

exp

( , Z T

t

r(u;Y (u)) du

)

(t) (T)

h(S(T))

3 7 7 7 5

:

Then

X(t) = (t) IEe



h(S(T)) (T)

F(t)



= IEe

"

exp

( , Z T

t

r(u;Y (u)) du

)

h(S(T))

F(t)

#

= v(t;S(t);Y (t));

and

X(t) (t) = exp

 , Z t

0

r(u;Y (u)) du



v(t;S(t);Y (t))

is a martingale underIPe

In every case, we get an expression involvingv to be a martingale We take the differential and set thedtterm to zero This gives us a partial differential equation for v, and this equation must hold wherever the state processes can be Thed Be term in the differential of the equation is the differential of a martingale, and since the martingale is

X ( t )

( t ) = X (0) +Z t

0
( u ) S  ( ( u u ) ) d Be( u )

we can solve for
( t ) This is the argument which uses (3.4) to obtain (3.5)

Example 22.4 (Continuation of Example 22.3)

X(t) (t) = exp

 , Z t

0

r(u;Y (u)) du



1= (t)

v(t;S(t);Y (t))

is a martingale underIPe We have

d



X(t)

(t)



= 1 (t)

 ,r(t;Y (t))v(t;S(t);Y (t)) dt + vtdt + vxdS + vydY +1

2vxxdS dS + vxydS dY +1

2vy ydY dY



= 1 (t)



(,rv + vt+ rSvx+ vy+1

22

S2

vxx xy+1

2 2

vy y) dt + (Svx y) d Be



The partial differential equation satisfied byvis

,rv + vt+ rxvx+ vy+1

22x2vxx xy+1

2

2vy y= 0

where it should be noted thatv = v(t;x;y), and all other variables are functions of(t;y) We have

d



X(t) (t)



= 1 (t)[Svx y] d B(t);e

where = (t;Y (t)), ,v = v(t;S(t);Y (t)), andS = S(t) We want to choose
(t)so that (see (3.2))

d



X(t) (t)



=
(t)(t;Y (t))S(t) (t) d B(t):e

Therefore, we should take
(t)to be

(t) = vx

(t;Y (t)) S(t)vy(t;S(t);Y (t)):

If a set of stock... along every path, except possibly for a set of paths which has probability zero, the choice of< h3>is irrelevant

The most convenient choice of< /i>is

 = r,1... t  T, is a continuous function of t We would like to hedge along every possible path of< h3>S ( t ), but that is impossible Using the binomial model