Discrete time models
Derivatives 1, professor Alexei Zhdanov
Fall 2010
Handout 4
This handout has benefited from the materials provided by Tony Berrada.
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Convergence
Central Limit Theorem. Let Xn , n ∈ 0, 1, 2, ....N be a sequence
of independent, identically distributed random variables with finite
mean µ and variance σ 2 . Define
VN =
N
X
Xn
n=0
then
VN − µN
√
lim P
≤ z = N (z)
N →∞
σ N
where N (z) is the cumulative density function of the standard
normal distribution
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Convergence
In other words, the distribution of VN converges to
N ormal(µN, σ 2 N ).
Implication: Binomial distribution converges to normal
BN,p (j) =
N!
pj (1 − p)N −j → N (np, np(1 − p))
j!(N − j)!
Recall our binomial pricing formula
c0 = S0 Φ(a, N, qe) − K(1 + r)−T Φ(a, N, q)
where
T X
N
Φ(a, T, q) =
q n (1 − q)N −n
n
n=a
and
u
1+r
where do Φ(a, N, qe) and Φ(a, N, q) converge?
qe = q
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Convergence
Note that
N
X
Z
∞
BN,p (j) →
m
j=m
2
2
e−(x−µn,p ) /2σn,p
√
dx
2πσn,p
letting y = (x − µn,p )/σn,p , we have
N
X
j=m
Z
BN,p (j) →
∞
m−µn,p
σn,p
2
e−y /2
√
dy =
2π
m − np
m − np
1 − N (p
) = N (− p
)
np(1 − p)
np(1 − p)
So
a − Nq
)
Φ(a, N, q) → N (− p
N q(1 − q)
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Convergence
We get r∆t , u, and d using our calibration procedure (CRR):
1 + r∆t = eR∆t ≈ 1 + R∆t
so
√
1 √
u ≈ 1 + σ ∆t + (σ ∆t)2
2
√
1 √
d ≈ 1 − σ ∆t + (σ ∆t)2
2
√
1 (R − 12 σ 2 ) ∆t
q≈ +
2
2σ
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Convergence
Finally, Sua dN −a → K, so taking logs we have
log S + a log u + (N − a) log d → log K
and then solving for a gives
a→
log K − log S − n log d
=
log u − log d
√
− log(S/K) − n(−σ ∆t)
√
√
=
σ ∆t − (−σ ∆t)
√
− n log(S/K) N
√
+
2
2σ T
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Convergence
Taking all the pieces together we get
log(S/K) + (R − 12 σ 2 )T
−a + N q
p
√
→
≡ d2
σ T
N q(1 − q)
Then
Φ(a, N, q) → N (d2 )
Similarly, it can be shown that
Φ(a, N, qe) → N (d1 )
where
√
log(S/K) + (R + 21 σ 2 )T
√
d1 = d2 + σ T =
σ T
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Convergence
To summarize,
c0 = S0 N (d1 ) − Ke−RT N (d2 )
This is the famous Black-Scholes formula for the price of a
European call.
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Dividend paying stocks
We can use the binomial model to price options on dividend
paying stocks as long as either
We know the timing and dollar amount of each dividend to be
paid between today and the maturity date of the option
We know the timing and the proportion of the stock to be
distributed as dividends
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Example
IBM is currently trading at $100 and the simple quarterly
interest is 3.33%. Assume that every quarter the price of IBM
either rises by 30% or falls by 10%. IBM will pay a 5%
dividend in one quarter. What is the value of a European call
with strike K = $110 and six months to maturity?
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Example
The stock price tree looks like
So the price for a European call looks like
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Example
Risk-neutral probability
q=
1.033 − 0.9
= 1/3
1.3 − 0.9
so
.33 × 50.55 + .67 × 1.15
= 17.05
1.033
.33 × 1.15 + .67 × 0
= 0.37
c110 (down) =
1.033
So the day-call price is
c110 (up) =
c110 (100) =
.33 × 17.05 + .67 × 0.37
= 5.74
1.033
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Example
What about fixed dividends?
A fixed dividend makes things slightly more complicated - we
will not have a ”recombining tree” anymore, but the
methodology is exactly the same.
To see this, let’s go back to our IBM example
Only this time, we are going to assume that IBM pays a
dividend of $5.17 at the end of the quarter - same PV as 5%
dividend:
2
1
1
5.17 + 23 5.17
3 6.5 + 3 4.5
= 3
1.033
1.033
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Example
The stock price tree looks like
What is the price of a European call struck at $110?
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Black-Scholes for dividend paying stocks
Suppose a firm pays a constant dividend yield δ
The payment of a dividend yield at rate δ then causes the
growth rate in the stock price to be less than it would
otherwise be.
If, with a dividend yield of q, the stock price grows from S0
today to ST at time T, then in the absence of dividends it
would grow from S0 e−δT today to ST at time T.
Then we can simply reduce the current stock price form S0 to
S0 e−δT and then value the option as though the stock pays
no dividends.
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Black-Scholes for dividend paying stocks
Implication 1 - Put-call parity. For a stock with a constant
dividend yield δ
c0 + Ke−RT = p0 + S0 e−δT
Implication 2 - Black-Scholes
c0 = S0 e−δT N (d+ ) − Ke−RT N (d− )
since
S0 e−δT
S0
= log
− δT,
log
K
K
the parameters d+ and d− are given by
d+ =
d− =
log(S0 /K) + (R − δ + 21 σ 2 )T
√
σ T
√
log(S0 /K) + (R − δ − 12 σ 2 )T
√
= d+ − σ T
σ T
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Adjusting the RN probability
In a binomial model we can price options on dividend paying
stocks by adjusting the RN probability
Suppose a stock pays a constant dividend yield δ. Then in the
risk-neutral world
q(uS + uSδ∆t) + (1 − q)(dS + dSδ∆t)
=S
1 + R∆t
then
q =
1+R∆t
1+δ∆t
−d
u−d
≈
1 + R∆t − d(1 + δ∆t)
=
u(1 + δ∆t) − d(1 + δ∆t)
e(R−δ)∆t − d
u−d
Then the binomial formula converges to the Black-Scholes
formula above
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Currency options
Mostly traded in the OTH market
Provide an alternative hedging tool to forward contracts
As we discussed before, a foreign currency is analogous to a
stock paying a known dividend yield - the owner of foreign
currency receives a yield equal to the foreign risk-free rate rf .
Then
c0 = S0 e−rf T N (d+ ) − Ke−rT N (d− )
where the parameters d+ and d− are given by
d+ =
d− =
log(S0 /K) + (r − rf + 21 σ 2 )T
√
σ T
√
log(S0 /K) + (r − rf − 21 σ 2 )T
√
= d+ − σ T
σ T
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Currency options
There is another way to write the same result
Recall that the forward rate F0 for a maturity T is given by
F0 = S0 e(r−rf )T
Then
c0 = e−rT [F0 N (d1 ) − KN (d2 )]
where
d1 =
log(F0 /K) + 12 σ 2 T
√
σ T
log(F0 /K) − 12 σ 2 T
√
σ T
we can also use a binomial tree to price currency options
d2 =
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Currency options
A currency is currently worth $1. Over each of the next 2
months it is expected to increase or decrease in value by 2%.
The domestic and foreign interest rates are 6% and 8%. What
is the value of a 2-month European call option with a strike of
$1?
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Some mathematics of the binomial model
The binomial model is a simple model which will be used as a
basis to develop finite probability spaces
A (finite) probability space is defined by a finite sample
space Ω and a probability measure P, i.e. a function assigning
a number in [0,1] to each element of ω ∈ Ω such that
X
P(ω) = 1
ω∈Ω
An event is a subset of Ω, the probability of the event A is
defined as
X
P(A) =
P(ω)
ω∈Ω
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Some mathematics of the binomial model
In the context of the binomial model with 3 periods an event
is a combination of stock price movements
Ω = {uuu, uud, udu, udd, duu, dud, ddu, ddd}
A random variable is a real-valued function defined on Ω
X(ω) : Ω → R
Binomial model: S2 (ω1 ω2 ) ∈ {Suu , Sud , Sdd } is a random
variable
Payoff of a derivative D(ST ) is also a random variable
A distribution assigns probability to the occurrence of each of
the possible values of the random variable
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Some mathematics of the binomial model
In the context of the binomial model with 3 periods an event
is a combination of stock price movements
Ω = {uuu, uud, udu, udd, duu, dud, ddu, ddd}
A random variable is a real-valued function defined on Ω
X(ω) : Ω → R
Binomial model: S2 (ω1 ω2 ) ∈ {Suu , Sud , Sdd } is a random
variable
Payoff of a derivative D(ST ) is also a random variable
A distribution assigns probability to the occurrence of each of
the possible values of the random variable
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Some mathematics of the binomial model
Changing the probability measure does not change the
random variable, it changes its distribution
Expectation of a random variable X is defined as
X
E(X) =
X(ω)P(ω)
ω∈Ω
Of course, expectation depends on the choice of probability
measure
X
X
e
X(ω)P(ω) 6=
X(ω)P(ω)
ω∈Ω
ω∈Ω
Expectations are linear
E(c1 X + c2 Y ) = c1 E(X) + c2 E(Y )
Jensen’s inequality. For a convex function ϕ(x)
E(ϕ(x)) ≥ ϕ(E(x))
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Conditional expectation
We can condition an expectation on the information available
at some time n, we denote by En [X] the conditional
expectation.
Recall that in the binomial model
Sn (ω1 ω2 ...ωn ) =
1
[qSn+1 (ω1 ω2 ...ωn U )+
1+r
(1 − q)Sn+1 (ω1 ω2 ...ωn D)
The conditional expectation is
EQ
n [Sn+1 ](ω1 ω2 ...ωn ) = qSn+1 (ω1 ω2 ...ωn U )+
(1 − q)Sn+1 (ω1 ω2 ...ωn D)
so
Sn =
1
EQ [Sn+1 ]
1+r n
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Conditional expectation
We can condition an expectation on the information available
at some time n, we denote by En [X] the conditional
expectation.
Recall that in the binomial model
Sn (ω1 ω2 ...ωn ) =
1
[qSn+1 (ω1 ω2 ...ωn U )+
1+r
(1 − q)Sn+1 (ω1 ω2 ...ωn D)
The conditional expectation is
EQ
n [Sn+1 ](ω1 ω2 ...ωn ) = qSn+1 (ω1 ω2 ...ωn U )+
(1 − q)Sn+1 (ω1 ω2 ...ωn D)
so
Sn =
1
EQ [Sn+1 ]
1+r n
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Conditional expectation
Q
Assume q = 1/2. What is EQ
1 [S2 ](U )? E1 [S2 ](D)?
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Conditional expectation
More generally, let X be a random variable depending on N
moves on the tree. Let n satisfy 1 ≤ n ≤ N, and let ω1 ...ωn
be given. There are 2N −n possible continuations ωn+1 ....ωN .
Denote by #U (ωn+1 ....ωN ) the number of up moves in the
continuation ωn+1 ....ωN and by #D(ωn+1 ....ωN ) the number
of down moves. Then
X
EQ
q #U (ωn+1 ....ωN ) ×
n [X] =
ωn+1 ....ωN
(1 − q)#D(ωn+1 ....ωN ) X(ω1 ...ωn ωn+1 ....ωN )
What is EQ
1 [S3 ](U )?
Note that
Q
EQ
0 [X] = E X
and
EQ
N [X] = X
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Conditional expectation
The following properties are important
Linearity
En [c1 X1 + c2 X2 ] = c1 En [X1 ] + c2 En [X2 ]
Taking out what is known. If at time n X is known then
En [XY ] = XEn [Y ]
Iterated conditioning. If 0 ≤ n ≤ m ≤ N then
En [Em [X]] = En [X]
In particular,
E[Em [X]] = E[X]
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Conditional expectation
Let us go back to our example. What is E1 [E2 [S3 ]]? E1 [S3 ]?
Another definition of conditional expectation:
E(X|A) =
E(X1A )
P(A)
where 1A (ω) = 1 if ω ∈ A and 1A (ω) = 0 otherwise.
According to this definition what is E1 [S3 ](U )?
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Martingales
Consider the N −period binomial model
A sequence of random variable M0 , M1 , M2 ...MN is a
stochastic process
If each Mn depends only on the first n movements, we call
this process adapted and
En [Mn ] = Mn
If
Mn = En [Mn+1 ]
then Mn is a martingale
If
Mn ≤ En [Mn+1 ]
then Mn is a submartingale
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Martingales
If
Mn ≥ En [Mn+1 ]
then Mn is a supermartingale
Note that the definition of a martingale implies
En [Mn+1 ] = En [En+1 [Mn+2 ]] = En [Mn+2 ]
Applying this argument iteratively
Mn = En [Mm ]
for 0 ≤ n ≤ m ≤ N
In particular
M0 = E[Mn ]
Under the risk-neutral measure, discounted stock prices are
martingales.
Sn+1
1
1
Sn
EQ
]=
EQ [Sn+1 ] =
n[
(1 + r)n+1
(1 + r)n 1 + r n
(1 + r)n
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Martingales
Imagine an investor who at each time n takes a position of
∆n shares of stock and holds this position until time n + 1,
when he takes a new position of ∆n+1 shares. This portfolio
is self-financing:
Xn+1 = ∆n Sn+1 + (1 + r)(Xn − ∆n Sn ),
n = 0, 1, ....N − 1
Note that Xn depends only on the first n movements, so the
wealth process is adapted.
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Martingales
Then
EQ
n[
Xn+1
∆n Sn+1
Xn − ∆n Sn
] = EQ
+
]=
n[
n+1
n+1
(1 + r)
(1 + r)
(1 + r)n
∆n Sn+1
Xn − ∆n Sn
EQ
+
]=
n[
(1 + r)n+1
(1 + r)n
Sn+1
Xn − ∆n Sn
Xn
∆n EQ
]+
=
n[
n+1
n
(1 + r)
(1 + r)
(1 + r)n
So self financing portfolios are martingales under the
risk-neutral measure. In particular,
EQ [
Xn
] = X0
(1 + r)n
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Martingales
But we have shown that given any derivative payoff DN , we
can construct a self-financing portfolio, such that XN = DN .
Therefore, the discounted derivative price is a martingale
under the risk-neutral measure:
D0 = EQ [
Dn
]
(1 + r)n
This has an important implication - Fundamental Theorem
of Asset Pricing. If there is an equivalent risk-neutral
measure (i.e. a measure that agrees with the actual measure
about which price paths have zero probability), then there can
be no arbitrage in the model.
XN
Indeed if X0 = 0 then EQ [ (1+r)
N ] = 0 and it cannot be that
Xn is always non-negative and strictly positive for some states.
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Markov processes
Let Xn be an adapted process. If for every n and every
function f (x) there exists another function g(x) such that
En [f (Xn+1 )] = g(Xn )
then Xn is a Markov process
En [f (Xn+1 )] is a random variable as it depends on the
evolution on the binomial tree up to date n
The Markov property means that this dependence is conveyed
through Xn only
Knowing Xn one may compute En [f (Xn+1 )]
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Markov processes
The stock price Sn+1 is a Markov process because
En [Sn+1 ] = Sn [pu + (1 − p)d]
and
En [f (Sn+1 )] = pf (uSn ) + (1 − p)f (dSn )
Are all processes in the binomial model Markov?
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American Derivatives
An option whose owner can choose to exercise at any date
prior to maturity (including the maturity as a European
contract)
At least as valuable as a European contract
Intermediate between European and American is :
Bermudean option
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American Derivatives
An American option is always worth at least as much as the
immediate payoff from exercise
Americanoption ≥ intrinsicvalue
We will use a hedging argument to price this type of option
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American Derivatives
Starting with a short position in the option, how do we choose
a strategy providing a perfect hedge both
1
2
at the terminal date
at any other time if the holder of the long position decides to
exercise
Simultaneously we will obtain the optimal exercise time
Consider an N −period binomial tree
d<1+r <u
⇒
no arbitrage
A European derivative with payoff at maturity f (SN )
40 / 71
American Derivatives
The value of the self-financing replicating strategy is given by
Vn
From the Markov property of the stock price we know that at
time n,
Vn = vn (Sn )
a function v() of S at time n only
We have shown that vn can be obtained recursively
backward
vN = f (SN )
vn (Sn ) =
qvn+1 (Sn u) + (1 − q)vn+1 (Sn d)
1+r
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American Derivatives
Consider now an American derivative, with potential early
exercise with payoff f (Sn )
Then the ”replicating” strategy must satisfy
Vn ≥ f (Sn )
n = 0, 1, 2.....N
It is then natural to guess the following algorithm
vN (SN ) = f (SN )
vn (Sn ) =
qvn+1 (Sn u) + (1 − q)vn+1 (Sn d)
max f (Sn ),
1+r
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American Derivatives
We try this algorithm on the following example
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American Derivatives
We have u = 2 and d = 0.5
We further assume that r =
1
4
and thus q =
1
2
We consider an American Put option with strike price K = 5
Given a stock price level x the algorithm becomes
v2 (x) = [5 − x]+
qvn+1 (2x) + (1 − q)vn+1 ( x2 )
vn (x) = max (5 − x),
1+r
n = 0, 1
44 / 71
American Derivatives
We have u = 2 and d = 0.5
We further assume that r =
1
4
and thus q =
1
2
We consider an American Put option with strike price K = 5
Given a stock price level x the algorithm becomes
v2 (x) = [5 − x]+
qvn+1 (2x) + (1 − q)vn+1 ( x2 )
vn (x) = max (5 − x),
1+r
n = 0, 1
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American Derivatives
v1 (2) differs from the European Derivative value
1
[q1 + (1 − q)4] = 2 < 3 = K − 2
1+r
at this node, keeping the option alive does not seem to be
optimal for the holder of the long position
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American Derivatives
The price of a European option with exercise price K at this
node would be 2
The holder of the long position can thus exercise his option,
then buy a European option with similar characteristic for 2 $
and keep the remaining 1 $. Let us consider now the
replicating portfolio
We already know the value of the portfolio, vn , so we just
need to compute the ∆n
At time 0 we must have
0.4 = Su ∆0 + (1 + r)(v0 − ∆0 S0 )
3 = Sd ∆0 + (1 + r)(v0 − ∆0 S0 )
⇒
∆0 = −0.433
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American Derivatives
where (v0 − ∆0 S0 ) = γ0 = 3.093 is the number of units of the
risk free bond held at time 0. After an initial up-move we have
∆0 Su + (1 + r)γ0 = −0.43 × 8 + 3.093 × 1.25 = 0.4
If the option is exercised at this node, the payoff is
5 − 8 = −3 and clearly the replicating portfolio is sufficient to
cover the cost
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American Derivatives
If the option is not exercised, then the amount needed to
replicate the following 2 possible payoffs is exactly 0.4
1
[qv2 (Suu ) + (1 − q)v2 (Sud )] =
1+r
1
1
1
2
1 2 × 0 + 2 × 1 = 5 = 0.4
1+ 4
After an initial down-move we have
∆0 Sd + (1 + r)γ0 = −0.43 × 2 + 3.093 × 1.25 = 3
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American Derivatives
If the option is exercised at this node, the payoff is 5 − 2 = 3
which is precisely the replicating portfolio’s value
If the option is not exercised, then the amount needed to
replicate the following 2 possible payoffs is only 2
1
[qv2 (Suu ) + (1 − q)v2 (Sud )] =
1+r
1
1
1
×1+ ×4 =2
2
1 + 41 2
If the holder of the long position chooses not to exercise, then
the replicating strategy generates a 1$ cash outflow !
It follows that the replicating portfolio does not satisfy the
self-financing condition if the holder of the long position in
the derivative does not behave optimally
Nevertheless the suggested algorithm guarantees that the
replicating strategy is a perfect hedge
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American Derivatives
Let us now consider the evolution of the discounted
American put price
51 / 71
American Derivatives
Recall that the discounted price of any European derivative is
a martingale under Q
The discounted American put price is a supermartingale
vn
vn+1
Q
≥ En
(1 + r)n
(1 + r)n+1
From the pricing algorithm we get
vn+1
Q
vn = max (K − Sn ), En
(1 + r)
and the supermartingale property follows
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Stopping time
Can we model the optimal exercise decision ?
In the example we have argued that when S1 = Sd the holder
of the long position should exercise
The exercise date is thus not only a function of time but also
a function of the underlying random stock price
We call this random date a Stopping Time and usually write
it as τ (ω)
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Stopping time
In the 2 period example there are 4 possible trajectories and
thus
τ ∈ {τ (uu), τ (ud), τ (du), τ (dd)}
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Stopping time
Definition - Stopping time. In an N-period binomial model,
a stopping time is a random variable τ ∈ [{0, 1, 2, ...N }, ∞]
which satisfies the condition that if
τ (ω1 ω2 ...ωn ωn+1 ωn+2 ...ωN ) = n, then
0
0
0 ) = n for all ω 0
0
0
τ (ω1 ω2 ...ωn ωn+1
ωn+2
...ωN
n+1 ωn+2 ...ωN .
Let τ define the optimal exercise policy as given in the
previous graph
Consider the discounted put price Yn , then Yn∧τ is a stopped
process
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Stopping time
Let us see graphically how Yn∧τ behaves
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Stopping time
The stopped process Yn∧τ is a Q−martingale
This property holds for any American derivative
If an optimal exercise time is passed, the martingale property
disappears and the discounted price process becomes a
supermartingale
We have just seen that we can stop a process that is not a
martingale and thereby obtain a martingale. This is a
consequence of a more general theorem
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Stopping time
Theorem - Optional Sampling Part I. A martingale stopped
at a stopping time is a martingale. A submartingale stopped
at a stopping time is a submartingale. A supermartingale
stopped at a stopping time is a supermartingale.
In general the value of an American derivative can be
presented as
f (Sτ )
Q
Vn = max E 1τ ≤N
τ
(1 + r)τ −n
where 1{τ ≤N } = 1 if τ ≤ N and 1{τ ≤N } = 0 if τ = ∞ (the
option is left unexercised).
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Stopping time
Let us see if this makes sense. Let us go back to our example.
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American Call
It is never optimal to exercise an American call option Cn on
a non-dividend paying stock prior to maturity
Setting f (Sn ) = (Sn − K)+ , we get from the pricing
algorithm
vN = f (SN )
Q vn+1
vn = max f (Sn ), E
(1 + r)
Can it be optimal to exercise at date N − 1
Is it possible that
(SN −1 − K)+ > EQ
N −1
(SN − K)+
(1 + r)
?
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American Call
f (x) is a convex function and f (0) = 0, we therefore get the
property (*)
f (λx) ≤ λf (x)
∀ λ ∈ [0, 1], x ≥ 0
From the martingale property of the discounted price we have
SN −1 =
EQ
N −1
SN
(1 + r)
⇒
f (SN −1 ) = f
EQ
N −1
SN
(1 + r)
From Jensen’s inequality we get
SN
SN
Q
Q
f EN −1
≤ EN −1 f
(1 + r)
(1 + r)
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American Call
using (*) it follows that
SN
1
Q
Q
EN −1 f
≤ EN −1
f (SN )
(1 + r)
(1 + r)
Finally we obtain that
f (SN −1 ) ≤
EQ
N −1
f (SN )
(1 + r)
It can never be optimal to exercise at N − 1
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American Call
f (Sn )
More generally the process (1+r)
n is a submartingale under Q
as the previous discussion holds for any n:
1
Q
f (Sn−1 ) ≤ En−1
f (Sn )
(1 + r)
Using this argument iteratively, we see that
1
Q
f (SN −1 ) =
f (SN −2 ) ≤ EN −2
(1 + r)
1
1
Q
Q
EN −2
E
f (SN ) =
(1 + r) N −1 (1 + r)
1
Q
EN −2
f (SN ) = cN −2
(1 + r)2
and so on, so for any n ≤ N
f (Sn ) ≤ cn
63 / 71
American Call
Therefore, it is never optimal to exercise and American call on
a non-dividend paying stock before maturity
This result can also be obtained from
64 / 71
American Call
Theorem - Optional Sampling Part II. Let Xn be a
Q-submartingale and τ be a stopping time, then
EQ Xn∧τ ≤ EQ Xn
From the submartingale property we obtain from
f (SN ∧τ )
f (SN )
Q
Q
E
≤E
(1 + r)N ∧τ
(1 + r)N
But the right hand side of this inequality is the value of a
European call option
Any early exercise policy yields a value for the American call
option which is lower or equal to that of a European call
option
65 / 71
Dividends
It is only optimal to exercise an American call
At maturity
Just before a dividend payment
Intuitive proof: consider the picture below
66 / 71
Dividends
By exercising at 1 instead of 2 we
Lose the interest, lose the right not to exercise (option to wait)
Gain nothing (no dividends between 1 and 2)
Therefore, we are always better off exercising at 2, or at
maturity. Exercising at 1 cannot be optimal
67 / 71
Dividends
Likewise, consider the picture below
By exercising at 3 instead of 2 we
Lose the dividend, lose the right not to exercise (option to
wait)
Gain almost nothing (very little interest between 2 and 3)
Therefore, we are always better off exercising at 2, or at
maturity. Exercising at 3 cannot be optimal
68 / 71
Dividends
No-arbitrage proof. An American call is always worth at
least as much as its intrinsic value
C ≥S−K
We can always sell an option.
So we’ll never exercise the call if C > S − K
We’ll only consider exercising if C = S − K
So we only need to show that C > S − K, except possibly at
maturity or a cum-dividend date (the day before the
ex-dividend date)
Suppose t < T is not a cum-dividend date. Then if
C = S − K you can buy a call, short the stock, and buy a
bond with a face value K maturing on the next cum-dividend
date. (or, if no more dividends, the call’s expiration)
This provides positive cashflow today and non-negative cash
flow later - arbitrage.
69 / 71
Example
ABC is trading at $52 and will pay a $5 dividend in 2 months.
The simple 2-month interest rate is 1%. You find a 2-month
call with a strike of $50 selling at $2. Is there an arbitrage
opportunity?
70 / 71
Example
In each of the next 2 months the price of the stock will either
go up by u = 1.25 or down by d = 0.8. The current price
S0 = 100. The simple 1-month risk-free interest rate is 2.5%.
The stock will pay a 10% dividend at the end of the first
month. What is the price of a two-month at-the-money
American call?
71 / 71