Exercise 13.1. The price evolution of two assets are modelled with Geometrical Brownian motions :

Hedging Exercises

Exercise 13.1. The price evolution of two assets are modelled with Geometrical Brownian motions :

dS

(t) =

S

(t) dt +

S

(t) dW

(t) ;

dS

(t) =

S

(t) dt +

S

(t) dW

(t) +

p

1

S

(t) dW

(t)

where W

and W

are two independent ( ; F ; fF

: t 0 g ; P ) Brownian motions. We as- sume that the risk free rate r is constant. fF

: t 0 g is the …ltration generated by the two Brownian motion (with the usual regularity conditions.

The goal is to hedge the contingent claim C:

a) What is the risk neutral market model.

b) Explain why C is attainable.

c) Assume that the time t value of the contingent claim C is a function of the under- lying asset prices and time, that is, f (t; S

(t) ; S

(t)) : Express the discounted contingent claim value as a di¤usion process. More precisely,we search for the coe¢ cient a; b and c such that

t

f (t; S

(t) ; S

(t)) =

f (0; S

(0) ; S

(0)) +

Z

a (s) ds + Z

0

b (s) df W

(s) + Z

0

c (s) df W

(s)

where W f

, f W

are two independent ( ; F ; fF

: t 0 g ; Q ) Brownian motions. If you have to add some hypothesis, mention it.

d) Use the Martingale Representation Theorem to justify the existence of predictable processes and such that

t

f (t; S

(t) ; S

(t)) = f (0; S

(0) ; S

(0)) +

Z

u

d

S

(u) + Z

0

u

d

S

(u) : Why are you allowed to use this theorem?

e) Use the results obtained in c) and d) to get

and

in therms of partial derivatives of f and the market model components. More precisely, show that

t

= @f

@s

(t; S

(t) ; S

(t)) and

= @f

@s

(t; S

(t) ; S

(t))

f) Construct the hedging strategy for C, that is, the self-…nancing investment strategy with a time T value of C:

g) If the contingent claim C = S

(T ) S

(T ) ; then what is the hedging strategy?

Exercise 13.2. Let

X

= price of the underlying asset at time t and B

= bank account value at time t;

B

= 1:

The model is :

dX

= ( X

) dt + dW

dB

= r

B

dt

where the instantaneous risk free rate f r

: t 0 g is a time dependent deterministic variable.

a) What is the risk-neutral market model? Justify each step.

b) Find the replicating strategy with a time t value of f (X

; B

; t).

Exercise 13.3. The risky asset price dynamics satisfy

dS (t) = r (t) S (t) dt + (t) S (t) dW (t)

where f W (t) : t 0 g is a ( ; F ; fF

: t 0 g ; Q ) Brownian motion, Q is a martingale mea- sure and f r (t) : t 0 g and f (t) : t 0 g are fF

: t 0 g predictable processes (careful ! S is not a Geometrical Brownian motion). The time t value B (t) of the bank account satisfy

B (t) = exp Z

0

r (s) ds ; B (0) = 1:

a) Find an expression for the time t value P (t; T ) of a zero-coupon bond paying one dollar at time T in terms of the risk free rate f r (t) : t 0 g : (Easy question that do not require computation).

Forward contract

A forward contract set at time t and with a maturity of T determines the price at which

the underlying contract will be traded at maturity. The convention says that the delivery

price is the one that make nil the initial value of the forward contract. The delivery price

or T f orward price, determine at time t, is denoted F (t) : It is a F

measurable random

b) Determine F (t).

c) Determine the value V (t) at time t of the forward contract established at time 0:

d) Assuming that the interest rate r is a deterministic function of time, determine the hedging strategy of the forward contract set at time 0:

Change of numeraire

Any asset with positive price may be used as numeraire. The bank account can serve as numeraire. In this case, the risky asset worth

shares of the bank account.

The probability measure Q

is said to be risk neutral for the numeraire N if the price of any asset, divided by the numeraire value, is a Q

martingale. The risk neutral measure Q is a risk neutral measure for the bank account B.

Let

Q b (A) = 1

N (0) E

N (T )

B (T ) I

for each event A 2 F

where I

is the indicator function that worth 1 if ! 2 A and zero otherwise.

e) Show that Q b is a probability measure.

f) Knowing that for all F

measurable random variable X , E

[ X j F

] = N (0) B (t)

N (t) E

X N (T )

N (0) B (T ) F

; show that Q b is risk neutral for the numeraire N .

g) Consider the case where the numeraire is the zero-coupon bond of maturity T .

Determine the value of the risky asset S in terms of this numeraire and justify intuitively why

the risk neutral measure Q

for the numeraire P ( ; T ) is called the measure T forward.

Solutions

1 Exercise 13.1

a) Under the risk neutral measure

dS

(t) = rS

(t) dt +

S

(t) df W

(t) ;

dS

(t) = rS

(t) dt +

S

(t) df W

(t) +

p

1

S

(t) df W

(t) :

b) The martingale measure is unique, son the market model is complete. All contin-

gent claims are attainable.

c) If the function f is twicely continuously di¤erentiable, then df (t; S

(t) ; S

(t))

= @f

@t (t; S

(t) ; S

(t)) dt + @f

@s

(t; S

(t) ; S

(t)) dS

(t) + @f

@s

(t; S

(t) ; S

(t)) dS

(t) + 1

2

@

f

@s

(t; S

(t) ; S

(t)) d h S

i (t) + 1 2

@

f

@s

(t; S

(t) ; S

(t)) d h S

i (t) + @

f

@s

@s

(t; S

(t) ; S

(t)) d h S

; S

i (t)

= @f

@t (t; S

(t) ; S

(t)) dt + @f

@s

(t; S

(t) ; S

(t)) rS

(t) dt + @f

@s

(t; S

(t) ; S

(t))

S

(t) df W

(t) + @f

@s

(t; S

(t) ; S

(t)) rS

(t) dt + @f

@s

(t; S

(t) ; S

(t))

S

(t) df W

(t) + @f

@s

(t; S

(t) ; S

(t))

p

1

S

(t) df W

(t) + 1

2

@

f

@s

(t; S

(t) ; S

(t))

S

(t) dt + 1 2

@

f

@s

(t; S

(t) ; S

(t))

S

(t) dt + @

f

@s

@s

(t; S

(t) ; S

(t))

S

(t) S

(t) dt

= 2 6 6 6 4

@f

@t

(t; S

(t) ; S

(t)) +

1

(t; S

(t) ; S

(t)) rS

(t) +

2

(t; S

(t) ; S

(t)) rS

(t) +

1

(t; S

(t) ; S

(t))

S

(t) +

2

(t; S

(t) ; S

(t))

S

(t) +

1@s2

(t; S

(t) ; S

(t))

S

(t) S

(t)

3 7 7 7 5 dt

+ @f

@s

(t; S

(t) ; S

(t))

S

(t) + @f

@s

(t; S

(t) ; S

(t))

S

(t) df W

(t) + @f

@s

(t; S

(t) ; S

(t))

p

1

S

(t) df W

(t) : The multiplication rule leads to

d

f (t; S

(t) ; S

(t)) =

df (t; S

(t) ; S

(t)) + f (t; S

(t) ; S

(t)) d

=

df (t; S

(t) ; S

(t)) rf (t; S

(t) ; S

(t))

dt

a (t) =

2 6 6 6 6 6 4

@f

@t

(t; S

(t) ; S

(t)) +

1

(t; S

(t) ; S

(t)) rS

(t) +

2

(t; S

(t) ; S

(t)) rS

(t) +

1

(t; S

(t) ; S

(t))

S

(t) +

(t; S

(t) ; S

(t))

S

(t) +

1@s2

(t; S

(t) ; S

(t))

S

(t) S

(t) rf (t; S

(t) ; S

(t))

3 7 7 7 7 7 5

b (t) =

@f

@s

(t; S

(t) ; S

(t))

S

(t) + @f

@s

(t; S

(t) ; S

(t))

S

(t) c (t) =

@f

@s

(t; S

(t) ; S

(t))

p

1

S

(t) :

d) The discounted value of the contingent claim is a Q martingale. Moreover, the discounted value of any primary asset are also Q martingales with positive di¤usion coe¢ cients, that is,

t

S

(t) = S

(0) + Z

0

1 u

S

(u)

| {z }

>0

df W

(u)

t

S

(t) = S

(0) + Z

0

2 u

S

(u)

| {z }

>0

df W

(u) + Z

0 2

p 1

S

(u)

| {z }

>0

df W

(u) :

Moreover,

E

Z

0 2 1

2

u

S

(u) du

=

Z

0

E

S

(u) du

=

Z

0

S

(0) E

"

exp ( rT ) exp r

2 1

2 u +

W

2

# du

=

Z

0

S

(0) E

exp

u + 2

W

du

=

S

(0) Z

0

exp

u + 2

u du

=

S

(0) Z

0

exp

u du = S

(0) e

1 < 1 :

Similarly,

E

Z

0 2 2 2 2

u

S

(u) du < 1 and E

Z

2

2

1

S

(u) du < 1 :

Consequently, the Martingale Representation Theorem implies the existence of predictable processes and such that

t

f (t; S

(t) ; S

(t))

=

f (0; S

(0) ; S

(0)) +

Z

u

d

S

(u) + Z

0

u

d

S

(u)

= f (0; S

(0) ; S

(0)) +

Z

u 1 u

S

(u) df W

(u) +

Z

u 2 u

S

(u) df W

(u) + Z

0 u 2

p 1

S

(u) df W

(u)

= f (0; S

(0) ; S

(0)) +

Z

[

S

(u) +

S

(u)] df W

(u) +

Z

u 2

p 1

S

(u) df W

(u) :

e) Since the equations in c) and in d) are both di¤usion processes for the discounted contingent claim value,

f (t; S

(t) ; S

(t)), then their di¤usion coe¢ cients must be equal.

Therefore, the coe¢ cients of W f

have to be equal :

t

@f

@s

(t; S

(t) ; S

(t))

p

1

S

(t) =

p

1

S

(t) which is veri…ed if and only if

t

= @f

@s

(t; S

(t) ; S

(t)) : The coe¢ cients of f W

have to be equal:

t

@f

@s

(t; S

(t) ; S

(t))

S

(t) + @f

@s

(t; S

(t) ; S

(t))

S

(t)

= [

S

(t) +

S

(t)] :

This equality is satis…ed if and only if

t

=

@f

@s1

(t; S

(t) ; S

(t))

S

(t) +

2

(t; S

(t) ; S

(t))

S

(t)

S

(t)

1

S

(t)

=

@f

@s1

(t; S

(t) ; S

(t))

S

(t) +

2

(t; S

(t) ; S

(t))

S

(t)

2

(t; S

(t) ; S

(t))

S

(t)

1

S

(t)

= @f

@s

(t; S

(t) ; S

(t))

f) Let

t

= the number of share of asset 1

t

= the number of share of asset 2

and

= the number of share of the riskless asset.

In our case, the riskless asset is B (t) = exp (rt) : We have shown that

t

= @f

@s

(t; S

(t) ; S

(t)) and

= @f

@s

(t; S

(t) ; S

(t)) : It su¢ ces to set

t

=

f (t; S

(t) ; S

(t))

S

(t)

S

(t)

=

f (t; S

(t) ; S

(t)) S

(t) @f

@s

(t; S

(t) ; S

(t)) S

(t) @f

@s

(t; S

(t) ; S

(t)) :

g) Since C = S

(T ) S

(T ) ; then its time t value is f (t; S

(t) ; S

(t))

= 1

t

E

[

S

(T ) S

(T ) jF

]

= e

E

2 6 6 4

e

S

(t) exp h

r

(T t) +

(W

(T ) W

(t)) i S

(t) exp

"

r

(T t) +

(W

(T ) W

(t)) +

p

1

(W

(T ) W

(t))

# jF

3 7 7 5

= exp [ r (T t)] S

(t) S

(t) E

2 6 6 4

exp h

2r

(T t) i exp [(

+

) (W

(T ) W

(t))]

exp h

2

p 1

(W

(T ) W

(t)) i jF

3 7 7 5

= exp r

2 1

2

2 2

2 (T t) S

(t) S

(t) E

[exp [(

+

) (W

(T ) W

(t))]] E

h

exp h

2

p 1

(W

(T ) W

(t)) ii

= exp r

2 1

2

2 2

2 (T t) S

(t) S

(t) exp

"

(

+

)

(T t) 2

# exp

2

2

(1

)

2 (T t)

= S

(t) S

(t) exp [(r +

) (T t)] : Consequently,

t

= @f

@s

(t; S

(t) ; S

(t)) = S

(t) exp [(r +

) (T t)]

t

= @f

@s

(t; S

(t) ; S

(t)) = S

(t) exp [(r +

) (T t)]

and

t

=

f (t; S

(t) ; S

(t))

S

(t)

S

(t)

= exp ( rt) S

(t) S

(t) exp [(r +

) (T t)]

S

(t) exp [(r +

) (T t)] exp ( rt) S

(t) S

(t) exp [(r +

) (T t)] exp ( rt) S

(t)

= exp ( rt) S

(t) S

(t) exp [(r +

) (T t)] :

Veri…cation. At time T , we hold S

(T ) shares of asset 1, S

(T ) shares of asset 2 and we are short of exp ( rt) S

(T ) S

(T ) shares of the bank account. The time t value of this investment strategy is

V

=

S

(t) +

S

(t) +

B (t)

= S

(t) exp [(r +

) (T t)] S

(t) + S

(t) exp [(r +

) (T t)] S

(t) exp ( rt) S

(t) S

(t) exp [(r +

) (T t)] exp ( rt)

= S

(t) S

(t) exp [(r +

) (T t)] : In particular,

V

= S

(T ) S

(T ) :

Other veri…cation. a (t) has to be equal to 0. But 2

6 6 6 6 6 4

@f

@t

(t; S

(t) ; S

(t)) +

1

(t; S

(t) ; S

(t)) rS

(t) +

2

(t; S

(t) ; S

(t)) rS

(t) +

1

(t; S

(t) ; S

(t))

S

(t) +

2

(t; S

(t) ; S

(t))

S

(t) +

1@s2

(t; S

(t) ; S

(t))

S

(t) S

(t) rf (t; S

(t) ; S

(t))

3 7 7 7 7 7 5

= 2 6 6 6 6 4

(r +

) s

s

exp [(r +

) (T t)]

+s

exp [(r +

) (T t)] rs

+ s

exp [(r +

) (T t)] rs

0 + 0

+ exp [(r +

) (T t)]

s

s

rs

s

exp [(r +

) (T t)]

3 7 7 7 7 5

= 2 6 6 4

(r +

) s

s

exp [(r +

) (T t)]

+rs

s

exp [(r +

) (T t)] + rs

s

exp [(r +

) (T t)]

+

s

s

exp [(r +

) (T t)]

rs

s

exp [(r +

) (T t)]

3 7 7 5

= 0:

2 Exercise 13.2

a) The multiplication rule implies that

dB

X

= B

dX

+ X

dB

+ d X; B

= B

( ( X

) dt + dW

) + X

r

B

dt

= B

(r

+ ) B

X

dt + dW

Let us introduce the process : W

= W

+ R

0 s

ds and

dB

X

= (

) B

(r

+ ) B

X

dt + B

d W

+ Z

0 s

ds

= (

) B

(r

+ ) B

X

dt + dW

: To cancel the drift term,

(

) B

(r

+ ) B

X

= 0 ,

= (r

+ ) X

+ :

Assuming that the Novikov condition

E

h

exp

R

2 t

dt i

is satis…ed, we can apply Girsanov theorem and state that there exists a measure Q under which W is a Brownian

1

E^P

"

exp 1 2

Z T 0

2 tdt

!#

= E^P

"

exp 1 2

Z T 0

(rt+ )Xt

+

2

dt

!#

= E^P

"

exp 1 2

Z T 0

(rt+ )Xt 2

2(rt+ )Xt

+

2! dt

!#

= exp 1 2

2

T

! E^P

"

exp 1 2

Z T 0

(r_t+ )X_t ²

2(r_t+ )X_t ! dt

!#

exp 1 2

2

T

! vuutE^P

"

exp Z T

0

(rt+ )Xt 2

dt

!#vuutE^P

"

exp 2 Z T

0

(rt+ )Xt

dt

!#

(Cauchy-Schwartz inequality)

= exp 1 2

2

T

! vuutE^P

"

exp Z T

0

(rt+ )Xt 2

dt

!#vuutE^P

"

exp 2 Z T

0

(rt+ )Xt

dt

!#

TO BE COMPLETED

motion. Let …nd the SDE of X in function of the Q Brownian motion:

dX

= ( X

) dt + dW

= ( X

) dt + d W

Z

0 s

ds

= (

X

) dt + dW

= (r

+ ) X

+ X

dt + dW

= r

X

dt + dW

: b) According to Itô’s lemma

df (X

; B

; t)

= @f

@x (X

; B

; t) dX

+ @f

@b (X

; B

; t) dB

+ @f

@t (X

; B

; t) dt + 1

2

@

f

@x

(X

; B

; t) d h X i

+ 1 2

@

f

@b

(X

; B

; t) d h B i

+ @

f

@x@b (X

; B

; t) d h X; B i

= @f

@x (X

; B

; t) (r

X

dt + dW

) + @f

@b (X

; B

; t) r

B

dt + @f

@t (X

; B

; t) dt + 1

2

@

f

@x

(X

; B

; t)

dt

= r

X

@f

@x (X

; B

; t) + r

B

@f

@b (X

; B

; t) + @f

@t (X

; B

; t) +

2

2

@

f

@x

(X

; B

; t) dt + @f

@x (X

; B

; t) dW

dB

f (X

; B

; t)

= r

B

X

(X

; B

; t) + r

(X

; B

; t) + B

(X

; B

; t)

+

B

(X

; B

; t) r

B

f (X

; B

; t) dt + B

@f

@x (X

; B

; t) dW

According the Martingale Representation Theorem,there is a predictable process such that

dB

f (X

; B

; t) =

dB

X

=

B

dW

1

Therefore

0 = r

B

X

@f

@x (X

; B

; t) + r

@f

@b (X

; B

; t) + B

@f

@t (X

; B

; t) +

2

2 B

@

f

@x

(X

; B

; t) r

B

f (X

; B

; t) ; and

t

B

= B

@f

@x (X

; B

; t) :

The last equation implies that the number of shares of the risky asset is

t

= @f

@x (X

; B

; t) : The number of shares of the riskless asset is,

t

= B

f (X

; B

; t) @f

@x (X

; B

; t) B

X

:

3 Exercise 13.3

3.1 Solution a)

P (t; T ) = E

B (t) B (T ) F

= E

exp

Z

r (s) ds F

:

3.2 Solution b)

The delivery price F (t) satis…es E

B (t)

B (T ) (S (T ) F (t)) F

= 0:

But,

0 = E

B (t)

B (T ) (S (T ) F (t)) F

= B (t) E

S (T )

B (T ) F

B (t) F (t) E

1 B (T ) F

= B (t) S (t)

B (t) B (t) F (t) E

1

B (T ) F

which implies that

F (t) = S (t) E

h

B(T)

F

i = S (t) P (t; T ) :

3.3 Solution c)

V (t) = B(t)E

S (T ) F (0) B (T ) F

= B(t)E

"

S (T )

B (T ) F

#

= B(t) E

S (T )

B (T ) F

S (0)

P (0; T ) E

1 B (T ) F

= B(t) S (t) B (t)

S (0) P (0; T )

P (t; T )

B (t) since S

B is a Q martingale

= S (t) S (0)

P (0; T ) P (t; T ) :

Therefore, V (0) = S (0)

P (0; T ) = 0 (the initial contract value is nil) and V (T ) = S (T )

= S (T ) F (0) (At maturity, the contract value corresponds to its cash ‡ow).

3.4 Solution d)

Assuming deterministic interest rate, the time t value of the forward contract is V (t) = S (t) S (0)

P (0; T ) P (t; T )

= S (t) S (0) exp Z

0

r (s) ds

= S (t) S (0) B (t) :

It is a twicely continuously di¤erentiable function of S and the bank account B. To replicates

this contract, it su¢ ces to hold one share of the risky asset and be short of S (0) shares of

the bank account.

3.5 Solution e)

First, note that Q b (A) 0 since N (T ) > 0, B (T ) > 0, N (0) > 0 and I

0: Second, Q b ( ) = 1

N (0) E

N (T )

B (T ) = 1 N (0)

N (0) B (0) = 1

where the second equality is justify by the fact that Q is risk neutral for the bank account, which implies that N=B is a Q martingale. Finally, if A

; A

; ::: are disjoint events

Q b [

A

!

= 1

N (0) E

N (T )

B (T ) I

= X

i=1

1

N (0) E

N (T )

B (T ) I

= X

i=1

Q b (A

) :

3.6 Solution f )

Let f Y (t) : t 0 g be the value of an asset of the model. We have that Y =B is a Q martingale.

We want to show that Y =N is a Q b martingale. First, since Y and N are fF

: t 0 g adapted, Y =N is. Second, we need to check the integrability condition : E

h

Y(t) N(t)

i

< 1 ... Finally, for all 0 s t, we have

E

Y (t) N (t) F

= N (0) B (s)

N (s) E

Y (t) N (t)

N (T )

N (0) B (T ) F

= B (s)

N (s) E

Y (t)

N (t) E

N (T )

B (T ) F

F

(law of iterated conditional expectation)

= B (s)

N (s) E

Y (t) N (t)

N (t)

B (t) F

since N

B is a Q martingale

= B (s)

N (s) E

Y (t) B (t) F

= B (s) N (s)

Y (s)

B (s) since Y

B is a Q martingale

= Y (s) N (s) :

3.7 Solution g)

In this case, the risky asset worth

shares of the zero-coupon bond for all 0 t T ,

that is, the risky asset worth F (t)shares of the zero-coupon bond where F (t) is the delivery

price.

E

S (t)

P (t; T ) F

= S (s)

P (s; T ) = F (s) :

4 Exercise 13.4

The model consist of a bank account B (t), a stochastic interest rate r, a risky asset S (t) with a stochastic volatility (t):

dS (t) = r (t) S (t) dt + (t) S (t) dW

(t) ; dB (t) = r (t) B (t) dt;

dr (t) =

(t) dt +

(t) dW

(t) ; d (t) = (t) dt + (t) dW (t) :

The Q Brownian motions W

, W

and W are independent. We assume that the drift and di¤usion coe¢ cients are such that the system of SDE admits a unique solution. Intuitively, this market model is incomplete because there is less tradable assets than there are sources of randomness. To complete the market, we add a contingent claim of maturity T whose time t value is C

= f (t; S

; B

;

) : Itô’s lemma imply that

dC = df

= f

dt + f

dS + f

dB + f d + 1

2 f

d h S i + 1

2 f

d h B i + 1

2 f d h i + f

d h S; B i + f

d h S; i + f

d h B; i

= f

+ f

rS + f

rB + f + 1

2 f

S

+ 1

2 f

dt + f

SdW

+ f (t) dW : the discounted value of C satis…es

dB

C

= B

dC + CdB

= B

f

+ f

rS + f

rB + f + 1

2 f

S

+ 1

2 f

rf dt +f

B

SdW

+ f B

(t) dW :

For B

C a Q martingale, the drift term as to be nil, that is, f

+ f

rS + f

rB + f + 1

2 f

S

+ 1

2 f

rf = 0:

Therefore,

dC = rf dt + f

SdW

+ f dW :

We consider a second contingent claim T < T whose time t value is g (t; S

; B

; C

) : Itô’s

lemma imply that

dg = g

dt + g

dS + g

dB + g

dC + 1

2 g

d h S i + 1

2 g

d h B i + 1

2 g

d h C i + f

d h S; B i + f

d h S; C i + f

d h B; C i

= g

+ g

rS + g

rB + g

rf

+

g

S

+

g

f

S

+ f

+ f

f

S

dt + (g

+ g

f

) SdW

+ g

f dW :

The discounted value satisfy dB

g

= g

+ g

rS + g

rB + g

rf

+

g

S

+

g

f

S

+ f

+ f

f

S

rg dt + (g

+ g

f

) B

SdW

+ g

B

f dW

On the other hand, the Martingale Representation Theorem implies the existence of a predictable processes and such that

dB

g = dB

S + dB

f

= ( + f

) B

SdW

+ f B

dW : We conclude that

f B

= g

B

f

( + f

) B

S = (g

+ g

f

) B

S 0 = g

+ g

rS + g

rB + g

rf

+ 1

2 g

S

+ 1

2 g

f

S

+ f

+ f

f

S

rg which is equivalent

= g

;

= g

+ g

f

f

= g

and

0 = g

+ g

rS + g

rB + g

rf + 1

2 g

S

+ 1

2 g

f

S

+ f

+ f

f

S

rg: