Finance 1
Université d’Evry Val d’Essonne
Séance 3
BINOMIAL PRICING
Philippe PRIAULET
Introduction
• It is based on an arbitrage argument.
• The option can be replicated with the
underlying stock and cash.
• Objective:
1 Find the price of the option.
2 Derive the replicating portfolio: basis of
hedging and (therefore) investment banking.
Binomial Setting
• The price of the stock only can go up to a
given value or down to a given value.
uS
S
dS
• Besides, there is cash (bank account) that
will pay interest of r.
Binomial Setting (cont.)
• We assume u (up) > d (down).
• For consistency we also need u > (1+r) > d.
• Example: u = 1.25; d = 0.80; r = 10%.
S = 125
S=100
S = 80
Binomial Setting (cont.)
• Basic model that describes a simple world.
• As the number of steps increases, it
becomes more realistic.
• We will price and hedge an option: it
applies to any other derivative security.
• Key: we have the same number of states
and securities (complete markets)
⇒ Basis for arbitrage pricing.
Option Pricing
• Introduce a European call option:
– K = 110.
– It matures at the end of the period.
S
C (K=110)
uS = 125
Cu = 15
dS = 80
Cd = 0
S=100
Option Pricing (cont.)
• We can replicate the option with the stock
and the cash.
• Construct a portfolio that pays Cu in state u
and Cd in state d.
• The price of that portfolio has to be the
same as the price of the option.
• Otherwise there will be an arbitrage
opportunity.
Option Pricing (cont.)
• We buy Δ shares and invest B in the cash.
• They can be positive (buy or deposit) or
negative (shortsell or borrow).
• We want then,
ΔuS + B (1 + r ) = Cu ⎫
⎬
ΔdS + B (1 + r ) = Cd ⎭
• With solution,
Cu − C d
u × C d − d × Cu
;B =
Δ=
S (u − d )
(u − d )(1 + r )
Option Pricing (cont.)
• In our example, we get for stock:
Cu − C d
15 − 0
1
Δ=
=
=
S(u − d) 100 × ( 1.25 − 0.8 ) 3
• And, for cash:
u × C d − d × C u 1.25 × 0 − 0.8 × 15
=
= −24.24
B=
(u − d )(1 + r )
(1.25 − 0.8) × (1.1)
• The cost of the portfolio is,
1
ΔS + B = × 100 − 24.24 = 9.09
3
Option Pricing (cont.)
• The price of the European call must be 9.09.
• Otherwise, there is an arbitrage opportunity.
• If the price is lower than 9.09 we would buy
the call and shortsell the portfolio.
• If higher, the opposite.
• We have computed the price and the hedge
simultaneously:
– We can construct a call by buying the stock and
borrowing.
– Short call: the opposite.
Risk Neutral Pricing
• Remember that
Cu − C d
u × C d − d × Cu
;B =
Δ=
S (u − d )
(u − d )(1 + r )
• And
C = ΔS + B
• Substituting,
Cu − C d u × C d − d × Cu
C=
+
(u − d )
(u − d )(1 + r )
Risk Neutral Pricing (cont.)
• After some algebra,
⎡1 + r − d
1
u − (1 + r ) ⎤
×⎢
C=
Cu +
Cd ⎥
1 + r ⎣ (u − d )
(u − d )
⎦
• Observe the coefficients,
1 + r − d u − (1 + r )
,
(u − d ) (u − d )
• Positive.
• Smaller than one.
• Add up to one.
⇒Like a probability.
Risk Neutral Pricing (cont.)
• Rewrite
1
C=
× [ p × Cu + (1 − p) × Cd ]
1+ r
• Where
1+ r − d
u − (1 + r )
,1 − p =
p=
(u − d )
(u − d )
• This would be the pricing of:
– A risk neutral investor
– With subjective probabilities p and (1-p)
Multiperiod Setting
• Suppose the following economy,
u2 S
uS
udS
S
dS
d2S
• We introduce a European call with strike
price K that matures in the second period.
Multiperiod Setting (cont.)
• The price of the option will be:
1
2
2
C=
×
[
p
×
max(
0
,
u
S − X)
2
(1 + r )
+ (1 − p ) 2 × max(0, d 2 S − X )
+ 2 × p × (1 − p) × max(0, udS − X )]
• There are “two paths” that lead to the
intermediate state (that explains the “2”).
Multiperiod Setting (cont.)
• Consider now n periods.
n
1
n!
[
C=
n ∑
(1 + r ) j =0 j!(n − j )!
× p (1 − p)
j
n− j
j
max(0, u d
n− j
S − K )]
• Trick: count the minimum number of “up”
movements that puts the call in the money.
• Call that number a, substitute above and get
rid of 0.
Binomial Pricing
• The price of the European call becomes,
K
C = Sφ (a, n, p' ) −
φ (a, n, p)
n
(1 + r )
• Where,
p×u
p' =
(1 + r )
• And φ(a,n,p’) is the complementary
binomial distribution.
• Probability of getting at least a “up”
changes after n tosses with p’ the
probability of “up” at each toss.
Binomial Pricing (cont.)
• The binomial distribution is tabulated.
• φ(a,n,p’) is (approximately) the “delta” of the
call:
⇒Number of shares (smaller than one) we need to
replicate the European call.
• Suppose we know the volatility σ and the
time to maturity t.
• We can retrieve u and d (see B&S):
u=e
σ t/n
; d = 1/ u
American Options
• Objective: we want to value an American
call that matures in two periods.
• Strike price, K = 100.
• Interest rate: 5% (each period).
• The underlying will pay a dividend of 8
after the first period.
• Problem: should we exercise after the first
period or wait until maturity?
American Calls (no dividends)
• Consider that there will be no dividend
–
–
–
–
–
–
Remember, C > S - PV(X)
But, obviously, CA ≥ C
Therefore, CA > S - PV(X) > S - X
But if we exercise the option we get (S - X)
Therefore, we should never exercise it
It is like a European option
• American call with dividends, put with or without
dividends: not clear whether early exercise is
optimal
Back to the Example
Call Payoff
110
112.2
12.2
91.8
90.2
0
0
73.8
0
I
(102)
III
100
90
II
(82)
In parenthesis: ex-dividend
Back to the Example (cont.)
• Price of option at node II, 0: regardless of
what happens afterwards the call pays zero.
• Price of call option at node I:
a If we exercise it (before the dividend is
paid), 110-100 = 10.
b Unexercised: we compute the value of the
replicating portfolio.
⇒Then, we compare.
Back to the Example (cont.)
110
112.2
12.2
91.8
0
I
(102)
Replicating portfolio:
Δ × 112.2 + (1.05) B = 12.2⎫
⎬ ⇒ Δ = 0.6 ; B = −52.29
Δ × 91.8 + (1.05) B = 0
⎭
Value of call:
C = (0.6) × 102 − 52.29 = 8.91
⇒Solution: exercise and get 10.
Back to the Example (cont.)
• At node III:
110
10
90
0
III
100
• Replicating portfolio:
Δ × 110 + (1.05) B = 10 ⎫
⎬ ⇒ Δ = 0.5; B = −42.86
Δ × 90 + (1.05) B = 0 ⎭
• Value of call:
C = (0.5) × 100 − 42.86 = 7.14