Derivative Securities: Lecture 5 American Options and ...

[Pages:30]Derivative Securities: Lecture 5 American Options and Black Scholes

PDE

Sources: J. Hull Avellaneda and Laurence

The Black Scholes PDE

? The hedging argument for assets with normal returns presented at the end of Lecture 4 gave rise to the Black Scholes PDE

CS,t 2S 2 2CS,t (r q)S CS,t rCS,t 0

t

2 S 2

S

r=interest rate, q=dividend yield, volatility. The volatility is the

annualized standard deviation of returns (it is not a market price or, rate, but rather a model input).

? We introduce a method for solving this PDE numerically on a grid.

Finite-difference scheme, or ``trinomial tree''

Mx

2x x 0

t 2t

S

j n

S0e jx,

M j M

Cnj C Snj , nt , 0 n N

Mx Nt

S

Finite-difference template

n

n+1 t

Change of variables

BS equation in log-price

S S0ex

S C S C x S C 1 C S x S x S x

S2

2C S 2

S2

S

1 S

S

C S

S2

S

1 S

C x

S 1 C x S x

2C C x2 x

C t

r

q

1 2

2

C x

1 2

2

2C x 2

rC

0

Taylor expansion & symmetric finite-difference approximations for derivatives

f x f 0 f '0x 1 f ''0x2 ...

2

f x f 0 f '0x 1 f ''0x2 ...

2

f (x) f (x) 2 f '0x o x2 f (x) f (x) 2 f 0 f ''0x2 o x3

f '0 f (x) f (x) ox

2x

f ''0

f (x)

f

(x) x2

2

f

0

ox

Symmetric finite difference approximations for first and second derivatives

Discretization of the PDE

C S , t

Cj n1

Cnj

t

t

C S,t

C j1 n1

C j1 n1

x

2x

2C S,t x2

C j1 n1

C j1 n1

x 2

2Cnj1

Here we do not use symmetric differences

Here use symmetric differences

C j n1

Cnj

t

(r

q

2

)

C j1 n1

C j1 n1

2 2x

2

2

C j1 n1

C j1 n1

2Cnj1

x 2

rCnj

0

From PDE to recursive scheme

C j n1

Cnj

t

(r

q

2

)

C j1 n1

C j1 n1

2 2x

2 2

C j1 n1

C j1 n1

2Cnj1

x 2

rCnj

0

Cnj

Cj n1

2t

2x2

(r

q

2 2x

/

2)t

Cnj11

1

2t

x2

Cnj1

2t

2x2

(r

q

2 2x

/

2)t

Cnj11

rtCnj

Cnj

1 1 rt

pUCnj11

pM

Cj n1

pDCnj11

pU

2t

2x2

(r

q

2 2x

/

2)t

pM

1

2t

x2

pD

2t

2x2

(r

q

2 2x

/

2)t

Interpreting the weights

? Notice that

pU pM pD 1

? Set

x max t

rq2 2

p

2t

2x2

2

2

2 max

? The weights become

pU

p

t 2 max

pM 1 2 p

t pD p 2 max

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