Transcript Partial differential equation in option pricing

Partial differential
equations in option
pricing
Prof. Dr. Pairote Sattayatham
Suranaree Univ. of Technology
Feb. 25, 2010
Chiangmai University
Risk Lab Bangkok, SUT
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1. Review of the Black Scholes theory of
option pricing
• 1.1 Market model
• one riskless asset (saving account)
d t  r t dt
• The price St of other asset (the risky stock or
stock index evolve according to the stochastic
differential equation
dSt   St dt   St dWt
where  is a constant mean return rate.
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Or in the integral form
t
t
0
0
St  S0    S s ds    S s dWs
General class of stochastic diff eqn driven by
dSt   (t , St )dt   (t , St )dWt
Or in integral form
t
t
0
0
St  S0    ( s, S s )ds    ( s, S s )dWs
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Ito’s formula
Let Wt be a Brownian motion in [0, t ]. If we divide up
the time 0 to t in a partition with n+1 partition points
ti  it / n then
n
2
(
W

W
)
 ti ti1  t as n  
i 1
in the mean square limit, i.e.
n
E[ (Wti  Wti 1 ) 2  t ]  0 as n  
i 1
This is often written

t
0
(dWt )2  t , or equivalently (dWt ) 2  dt , dWt 2  dt
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If we were to do a Taylor series expansion of V (t ,Wt )
completely disregards the nature of Wt , and treat dWt as
a small increment of Wt we would get
V (t  dt ,Wt  dt )  V (t ,Wt )  V1 (t ,Wt )dt  V2 (t ,Wt )dWt
1
+ [V11 (t ,Wt )(dt ) 2  2V12 (t ,Wt )dtdWt  V22 (t ,Wt )(dWt ) 2 ]  ...
2
Delete higher order term, we get
V
V
1  2V
dV 
dt 
dW 
dt
2
t
W
2 W
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If V  V ( S , t ) where dSt  a (t )dt  b(t )dWt then
V
V
1 2  2V
dV 
dt 
dS  b
dt
2
t
S
2 S
This is the Ito’s formula
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Black Scholes Model
Let  to denote the portfolio of one long option position
and a short position in some quantity  ,i.e.
  V ( S , t )  S
We assume that the underlying follow a lognormal
random walk
dSt   dt   dWt
Then
d   dV  dS
From Ito’s formula
V
V
1 2  2V
d
dt  (
  )dS  
dt
2
t
S
2
S
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The random term is
V
(
 )dS
S
If we choose
V

S
Then the randomness is reduce to zero. Any reduction
of the randomness is termed to hedging.
After choosing the quantity  , we hold a portfolio
whose value changes by the amount
V
1 2  2V
d
dt  
dt
2
t
2
S
By principle of no arbitrage, we get
d   rdt
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Hence we find that
V
1 2  2V
V
dt  
dt  r (V  S
)dt
2
t
2
S
S
On dividing by dt and rearranging we get
V
1 2  2V
V
dt  
 rS
 rV  0
2
t
2
S
S
This is the Black-Scholes PDE for pricing European
call and put option (also binary option).
Initial conditions:
V (0, t )  0, V (S , t )  S as S   (European call option)
V (0, t )  0, V (S , t )  0 as S   (European put option)
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Final conditions
European call option: V (S , t )  max(S  E,0).
European put option: V (S , t )  max( E  S ,0).
Binary call option:V (S ,T )  H (S  E)
Binary put option: V (S ,T )  H ( E  S )
Where the Heaviside function is defined by
1 if x  0,
H ( x)  
0 if x  0.
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Method for solving B-S PDE for Europrean call
option
We set
1 2
S  Ee , t  T   /  , V  Ev( x, ).
2
x
The result equation is
v  2v
v
 2  (k  1)  kv
 x
v
1 2
where k  r /  . The initial condition become
2
v( x,0)  max(e x  1,0)
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Put
v  e x  u ( x, )
for some constants  and  to be found. After
differentiation and substituting in the above equation,
we get
ve
where
1
1
 ( k 1) x  ( k 1)2 
2
4
u ( x, )
v  2u
 2 , for    x  ,  0
 x
With
u ( x,0)  u0 ( x)  max(e
1
( k 1) x
2
e
1
( k 1) x
2
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,0)
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By method of Green, we have
u ( x , ) 
1
 
d1

u0 ( s ) e
 ( x  s ) 2 / 4
By the change of variable x '  (s  x) / 2 and with
some calculations, we get
V ( S , t )  SN (d1 )  Ee (T t ) N (d 2 )
where
1
log( S / E )  ( r   2 )(T t )
2
d1 
 ( T t )
1
log( S / E )  ( r   2 )(T t )
2
d2 
 ( T t )
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Early exercise and American put options
(paying no dividend)
1. The holder has the right to sell the asset for a specify
amount E.
2. The right to sell can be exercised any time up to
the expiry date t=T.
But as well as giving the holder more right, they
also give him more headaches; when should he
exercise ?
Part of valuation problem is deciding when is the
best time to exercise.
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This introduce the idea of optimal exercise price
into the modeling and convert the parabolic
equation model into a free boundary problem.
Free boundary problem involving in finding the solution
of PDE in domain that are unknown a priori: not only
the solution of the equations but also the domain of the
equation must be determined.
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Free boundary is the graph of the optimal exercise
price as a function of time t.
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V ( St , t )  the American put option value at asset price S
and time t.
Note that
V ( St , t )  max( E  St ,0), 0  t  T
[for if not, we have E  St  V ( St , t )  0 (arbitrage)]
Let us postulate that while the asset value is high we
won’t exercise the option. But if it fall too low we
immediately exercise the option, receiving E  St .
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Suppose that we decide at S  S * (t ) , i.e. as soon as S
reaches this value from above we exercise. [How do we
choose S  S * (t ) ?] When S  S * (t ) the option value
must be
V ( S * (t ), t )  E  S * (t )
It cannot be less, which would result in arbitrage
opportunity and it cannot be more or we would not
exercise
Hence, for each time t, if 0  S  S * (t ) then the option
should be exercised and in this range its value is E  S ,
while for S * (t )  S   the B-S equation still hold .
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That is, it must satisfies the equation
V
1 2  2V
V
*
dt  

rS

rV

0,
S
(t )  S  .
t
2 S 2
S
For simplicity, we assume that the above equation does
not depend on t, hence we have
1 2  2V
V
*


rS

rV

0,
S
 S  .
2
2
S
S
The solution of this equation is
V (S )  AS  BS
2 r /  2
, S*  S  
Since V (S )  0 as S   then the constant A must be
zero. The constant B, can be found by solving
* -2r/ 2
V ( S )  B( S )
*
 E  S *.
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Since both B and S * are unknown, we need one more
equation. We note that for S  S *, we have
* 2 r /  2
V (S )  ( E  S )(S / S )
*
(1)
We are going to choose S * to maximize the option’s
value at any time before exercise. We find this value by
differentiate (1) with respect to S * and setting to zero.

*
* 2 r /  2
( E  S )( S / S )
0
*
S
We find
S *  E /(1 
2
2r
)
This choice maximize V(S) for all S  S *. The solution
for this choice S * is
( / 2r ) /  E /(1   / 2r ) 
2
2
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1 2 r /  2
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Payoff
Option value
1
0.8
0.6
0.4
0.2
0
0.5 S*
S
1
1.5
The solution for the perpetual American put
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If the option depends on time we seek solution of
V
1 2  2V
V
*
dt  

rS

rV

0,
S
(t )  S  .
2
t
2 S
S
and S * (t ) an optimal exercise price. At expiry, if the
option still held, its payoff matches the European option,
so we have the final time condition:
V (S , T )  max( E  S (T ),0), S  0
V (S , t )  E as S  0, 0  t  T
V (S , t )  0 as S  , 0  t  T .
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V (S , t ) should satisfies the following free boundary
condition
V  E  S,
V
 1 at S  S * (t )
S
There is no closed form expression for V (S , t ) .
Can do only for some special cases
Research problem ?
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Method
1. construct linear complementarity problem for
American put.
2. Prove that linear complementarity formulation is
equivalent to the original free boundary problem.
3. Prove that there is a unique solution.
4. Techniques used are those of functional analysis
and variational inequalities.
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Transform the parabolic eqn. into heat eqns
1 2
1. S  Ee , t  T  ( /  ), V  Ev( x, )
2
2. S  S * (t )  x  x* ( )
3. S * (t )  E  x* ( )  0
4. Payoff fcn. max( E  S (t ),0)  g ( x, )
x
where g ( x, )  e
V
5.
t
u


1
( k 1)2 
2
1
( k 1) x
2
1
( k 1) x
2
max(e
e
,0)
1 2  2V
V
*
dt  

rS

rV

0,
S
(t )  S  .
2
2 S
S
 2u
 2 for x* ( )  x
x
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Heat equation.
u  2u
 2 for x* ( )  x
 x
u ( x, )  g ( x, ) for x  x* ( )
Initial conditions
1
( k 1)2 
2
g ( x, )  e
max(e
u( x,0)  g ( x,0)
lim u ( x, )  0
1
( k 1) x
2
e
1
( k 1) x
2
,0)
x 
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Linear complementarity for American put
 u  2u 
   x 2  .  u ( x, )  g ( x, )   0,


 u  2u 
   x 2   0,  u ( x, )  g ( x, )   0.


with initial and boundary condition,
u ( x,0)  g ( x,0),
u ( x, ) and g ( x, ) are cont.
A solution to the linear complementarity problem can be found
by using technique of functional analysis and variational method.
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Example: Concept of variational method
original problem
variational problem
1
min  | u |2 dx
u f
2
u  0
free boundary cond.
u f
u  0, u  f , (u)(u  f )  0
 u  (v  u )dx  0, v  f .
n.u  n.f

(See article, page 23)
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We must use numerical method.
Concept: The value of an American option can be
define as the maximum over all the stopping time of the
discounted, risk-neutral payoff of the option evaluate at
the stopping time.
Vn i  max  max( E  Sn i ,0),  pVni11  (1  p)Vni 1  
0  n  i, 0  i  M  1
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Recombining binary tree
21
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American call option no dividend payment
Since it is never optimal to exercise an American call
option before expiry date, an American call option must
have the same value as a European call option.
1. The holder exercise the call option at time t  T
2. The holder receive a payoff of St  E  0
1’. Instead short the asset at the market price at time t.
(receiving St )
2’. Exercising the option at time t  T or buying the
asset at the market price at time T. Hence the holder
paid out and amount less than or equal to E at time T.
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American call option with dividend
American put option with dividend
Bermuda option
(A type of option that can only be exercised on predetermined
dates, usually every month.)
Canary option is an option whose exercise style lies
somewhere between European options and Bermudan options.
Typically, the holder can exercise the option at quarterly dates.
Exotic (path-dependent)
Asian option
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The relationship between SDE and PDE
Theorem (Feynman-Kac). Consider the SDE
dS (u )   (u , S (u ))du   (u , S (u ))dW (u ).
Let h(y) be a Borel measurable function. Fix T  0,
and let t  [0, T ] be given. Define the function
f (t , x)  E t , x [e  r (T t ) h( S (T ))].
We assume that E t , x | h( S (T )) |  for all t , x.) Then f (t , x)
satisfy the PDE
1 2
ft (t , x)   (t.x) f x (t , x)   (t , x) f xx (t , x)  rf (t , x)
2
and the terminal condition
f (T , x)  h( x) for all x.
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Example: SDE for interest rate is
dR(t )   (t , R(t ))dt   (t , R(t )dWˆ (t )
T
  R ( s ) ds
B (t , T )  Eˆ [e t
| F (t )]
= f (t , R (t ))
To find f(t,R(t)), we must solve PDE
1 2
f x (t , r )   (t , r ) f r (t , r )   (t , r ) f rr R(t ))  rf (t , r ).
2
with terminal condition
f (T , r )  1 for all r.
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End of presentation
Thank for your attention
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