Transcript Binomial Trees - Dublin City University | DCU

Binomial Option Pricing Model
(BOPM)
References:
Neftci, Chapter 11.6
Cuthbertson & Nitzsche, Chapter 8
1
Linear State Pricing


A 3-month call option on the stock has a strike price of 21.
Can we price this option?
◦ Can we find a complete set of traded securities to price the option
payoffs?
◦ If we make the simplifying assumption that there are only 2 states of
the world (up and down), then we only need the prices of two
independently distributed traded assets, e.g. the underlying stock
and the risk-free asset
Linear State Pricing

Algebraically,
s s
1
P  0  d 
1 R f
s 1  0
S

1

0 
 1 R f

S
s
(
1

R
)

d
 f s
s 1
qs
 S
   s
 s 1


If S = 2, this is a system of equations in two unknowns

To get a unique solution for it, we need at least 2 independent
equations
The Binomial Model


A stock price is currently S0 = $20
In three months it will be either S0u = $22
or S0d = $18
Stock Price = $22
Stock price = $20
Stock Price = $18
A One-Period Call Option

Option tree:
Stock Price = $22
Option Price = $1
Stock price = $20
Option Price=?
Stock Price = $18
Option Price = $0
Risk-Neutral Valuation
Reminder
Under the RN measure the stock price
earns the risk-free rate
 That is, the expected stock price at time T
is S0erT
 When we are valuing an option in terms
of the underlying the risk premium on the
underlying is irrelevant


See handout on RNV
6
Risk-Neutral Tree
f = [ q f u + (1 – q ) f d ]e-rT
 The variables q and (1 – q ) are the risk-neutral
probabilities of up and down movements
 The value of a derivative is its expected payoff in a riskneutral world discounted at the risk-free rate

S0
ƒ
S0u
ƒu
S0d
ƒd
Risk-Neutral Probabilities
S0u = 22
ƒu = 1
S0
ƒ

S0d = 18
ƒd = 0
Since q is a risk-neutral probability,
22q + 18(1 – q) = 20e0.12 (0.25)
 q = 0.6523
RN Binomial Probabilities
Formula

RN probability of up move:
e d
q
ud
rT

RN probability of down move:

1–q
With this probabilities, the underling grows
at the risk-free rate (check it out)
Using the RN Binomial
Probabilities Formula


In above example, u = 22/20 = 1.1
and d = 18/20 = 0.9
So, assuming r = 12% p.a.,
0.120.25
e d e
 0.9
q

 0.6523
ud
1.1  0.9
rT
Valuing the Option
S0u = 22
ƒu = 1
S0
ƒ
S0d = 18
ƒd = 0
The value of the option is
e–0.12(0.25) [0.6523  1 + 0.3477  0]
= 0.633
A Two-Step Example
24.2
22
19.8
20
18
16.2

Each time step is 3 months
Valuing a Call Option
D
22
20
1.2823
A
B
2.0257
18
E

19.8
0.0
C
0.0

24.2
3.2
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 A
= e–0.120.25(0.65232.0257 + 0.34770)
= 1.2823
A Put Option Example; K=52
D
60
50
4.1923
A
B
1.4147
40
72
0
48
4
E
C
9.4636
F
32
20
What Happens When an Option
is American (see spreadsheet)
D
60
50
5.0894
A
B
1.4147
40
72
0
48
4
E
C
12.0
F
32
20
15
And if we did not have u and d?
One way of matching the volatility of logreturns is to set
s t
ue
d  e s
t
where s is the volatility and Δt is the length of
the time step. This is the approach used by Cox,
Ross, and Rubinstein
◦ Handout on Asset Price Dynamics
The Probability of an Up Move
p
ad
ud
a  e rt for a nondividend paying stock
a  e ( r  q ) t for a stock index where q is the dividend yield on the index
ae
( r  r f ) t
for a currency where r f is the foreign risk - free rate
a  1 for a futures contract
17
EXOTICS PRICING (examples)
a) Average price ASIAN CALL
payoff = max {0, Sav – K}
Remember: cheaper than an ‘ordinary’ option
b) Barrier Options (e.g. up and out put)
pension fund holds stocks and is worried about fall in price
but does not think price will rise by a very large amount
◦ Ordinary put? - expensive
◦ Up and out put - cheaper
Pricing an Asian Option (BOPM)
Average price ASIAN CALL(T = 3)
1. Calculate stock price at each node of tree
2. Calculate the average stock price Sav,i at expiry, for each of
the 8 possible paths (i = 1, 2, …, 8).
3. Calculate the option payoff for each path, that is
max[Sav,i – K, 0] (for i = 1, 2, …, 8)
The risk neutral probability for a particular path is
qi* = qk(1 – q)n-k
q = risk neutral probability of an ‘up’ move
k = number of ‘up’ moves
(n – k) = the number of ‘down’ moves
Pricing an Asian Option (BOPM), cont’d
4. Weight each of the 8 outcomes for the call payoff
max[Sav,i – K, 0] by the qi* to give the expected payoff:
8
Eˆ ( S )   qi* max[Sav,i  K , 0]
i 1
5. The call premium is then the PV of ES*, discounted at
the risk free rate, hence:
C Asian  Eˆ ( S )e r 3
Pricing Barrier Options (BOPM)
Down-and-out call
S0 = 100. Choose K = 100 and H = 90 (barrier)
Construct lattice for S
Payoff at T is max {0, ST – K }
Follow every ‘path’ (ie DUU is different from UUD)
If on say path DUU we have any value of S < 90 , then the value
at T is set to ZERO (even if ST – K > 0).
Use BOPM risk neutral probabilities for each path and each
payoff at T
Example: Down-and-out call
S0 =100, K= 100,
q = 0.857,
(1 – q) = 0.143
H = 90
UUU ={115, 132.25, 152.09}
Payoff = 52.09 (q* = 0.8573, 0.629)
DUU ={80, 92,105.8}
Payoff = 0
NOT 5.08 (q* = 0.105)
C = e-rT  ‘Sum of [q*  payoffs at T]’
where qi* = qk(1 – q)n-k,