4.1
Option Prices:
numerical approach
Lecture 4
4.2
Pricing:
1.Binomial Trees
Binomial Trees
• Binomial trees are frequently used to
approximate the movements in the price
of a stock or other asset
• In each small interval of time the stock
price is assumed to move up
by a proportional amount u or
to move down by a proportional
amount d
4.4
A Simple Binomial Model
• A stock price is currently $20
• In three months it will be either $22 or
$18
Stock Price = $22
Stock price = $20
Stock Price = $18
probabilities can’t be 50%50%, unless you are riskneutral
4.5
A Call Option
A 3-month call option on the stock has a strike price of
21.
Stock Price = $22
Option Price = $1
Stock price = $20
Option Price=?
if you were risk-neutral (and r=0),
you could say that the option is
worth: 0.5=50%*1$+50%*$0
Stock Price = $18
Option Price = $0
• prob(22) has to be > prob (18), because otherwise
Utility function is linear
4.6
U(22)
U(20)
Expcted U
(50% prob)
U(18)
22
18
20
• Then, in order to know prob we need to know the Utility
function. But this is an impossible task, and we have to find a
shortcut .... i.e. we have to find a way of “linearizing” the world
Setting Up a Riskless Portfolio
• Consider the Portfolio: long D shares
short 1 call option
22D – 1
18D
• Portfolio is riskless when 22D – 1 = 18D or
D = 0.25
4.7
4.8
Valuing the Portfolio
• risk-fre rate=12% p.a. ---> 3% quarterly ---> disc.
factor=exp(-0.12*0.25)=0.970446
• The riskless portfolio is:
long 0.25 shares
short 1 call option
• The value of the portfolio in 3 months is
220.25 – 1 = 4.50
• Note that this pay-off is deterministic, so
its PV is obtained by simple discounting
4.9
Valuing the Option
• The value of the portfolio today is
4.5e – 0.120.25 = 4.3670
• The portfolio that is
long 0.25 shares
short 1 option
is worth 4.367
• The value of the shares is
5.000 (= 0.2520 )
• The value of the option is therefore
0.633 (= 5.000 – 4.367 )
4.10
Valuing the Option
• note that the value of the option has been
obtained without knowing the shape of
the utility function
• but if the solution is independent of
preferences functional form, then it is
valid also for all utility function
• Then, it is valid also for risk-neutral
preferences .....
• ... eureka !!! let’s imagine a risk-neutral
world ---> derive risk-neutral probabilities
Summing up
... Movements in Time Dt
Su
S
Sd
Risk-neutral Evaluation
hyp: risk-free rate=12% p.a.; t = 3m
Su = 22
ƒu = 1
S
ƒ
Sd = 18
ƒd = 0
• Since p is a risk-neutral probability
20e0.12 0.25 = 22p + 18(1 – p ); p =
0.6523
• p is called the risk-neutral probability
• show simple_example.xls
4.12
Tree Parameters for a
Nondividend Paying Stock
• We choose the tree parameters p, u, and d so
that the tree gives correct values for the mean
& standard deviation of the stock price
changes in a risk-neutral world
er Dt = pu + (1– p )d
s2Dt = pu 2 + (1– p )d 2 – [pu + (1– p )d ]2
• A further condition often imposed is u = 1/ d
Tree Parameters for a
Nondividend Paying Stock
• When Dt is small a solution to the
equations is
ue
s Dt
 s Dt
d e
ad
p
ud
a  e r Dt
The Complete Tree
S0u 3
S0u 4
S0u
S0u 2
S0u 2
S0u
S0
S0d
S0
S0d
S0d 2
S0d 3
S0
S 0d 2
S 0d 4
Backwards Induction
• We know the value of the option
at the final nodes
• We work back through the tree
using risk-neutral valuation to
calculate the value of the option
at each node, testing for early
exercise when appropriate
4.17
Valuing the Option
Su = 22
ƒu = 1
S
ƒ
Sd = 18
ƒd = 0
The value of the option is
0.120.25 [0.65231 + 0.34770]
= 0.633
e–
A Two-Step Example
24.2
22
19.8
20
18
16.2
• Each time step is 3 months
4.18
Valuing a Call Option
D
22
20
1.2823
A
B
2.0257
18
24.2
3.2
E
19.8
0.0
C
0.0
F
16.2
0.0
• Value at node B
= e–0.120.25(0.65233.2 + 0.34770) = 2.0257
• Value at node C =0
• Value at node A
= e–0.120.25(0.65232.0257 + 0.34770)
= 1.2823
4.19
4.20
Pricing:
2.Monte Carlo
An Ito Process for Stock Prices
(See pages 225-6)
dS  mSdt  sSdz
where m is the expected return s
is the volatility.
The discrete time equivalent is
DS  mSDt  sS Dt
Monte Carlo Simulation
• We can sample random paths for the
stock price by sampling values for 
• Suppose m= 0.14, s= 0.20, and Dt =
0.01, then
DS  0.0014 S  0.02 S
see simple_example.xls
Monte Carlo Simulation – One Path
(continued. See Table 10.1)
Period
Stock Price at
Random
Start of Period Sample for 
Change in Stock
Price, DS
0
20.000
0.52
0.236
1
20.236
1.44
0.611
2
20.847
-0.86
-0.329
3
20.518
1.46
0.628
4
21.146
-0.69
-0.262
Monte Carlo Simulation
When used to value European stock options, this
involves the following steps:
1. Simulate 1 path for the stock price in a risk neutral
world
2. Calculate the payoff from the stock option
3. Repeat steps 1 and 2 many times to get many sample
payoff
4. Calculate mean payoff
5. Discount mean payoff at risk free rate to get an
estimate of the value of the option
A More Accurate Approach
(Equation 16.15, page 407)
Use


d ln S  m  s 2 / 2 dt  s dz
The discrete version of this is


ln S  DS   ln( S )  m  s 2 / 2 Dt  s
or
m  s / 2  Dt  s 

S  DS  S e
2
Dt
Dt
Extensions
When a derivative depends on several
underlying variables we can simulate
paths for each of them in a risk-neutral
world to calculate the values for the
derivative
To Obtain 2 Correlated Normal
Samples
Obtain independent normal samples
1 and  2 and set
1  x1
 2  x1  x2 1  2
A procedure known a Cholesky's
decomposition can be used when
samples are required from more than
two normal variables
Standard Errors
The standard error of the estimate of
the option price is the standard
deviation of the discounted payoffs
given by the simulation trials divided by
the square root of the number of
observations.
Application of Monte Carlo
Simulation
• Monte Carlo simulation can deal with
path dependent options, options
dependent on several underlying
state variables, & options with
complex payoffs
• It cannot easily deal with Americanstyle options