Partial Differential Equations with
Applications to Finance
Seminar 3: A reciprocal supershare option
Group 4: Bertan Yilmaz, Stephane Dumanois and Azza Alghamdi
May 21, 2015
-.
1
A reciprocal supershare option
In this seminar, we consider the following problem.
Exam 2012-08-22, Problem 4:
(a) Assume that the price process S(t) of a risky asset evolves as a geometric Brownian motion. Let L, U, M > 0 be positive constants such that L < U .
Suppose that T > 0 is the expiration date of the reciprocal supershare
option which at maturity pays the amount M U/S(T ) if S(T ) ∈ [L, U ], or
nothing if S(T ) ∈
/ [L, U ]. Using the standard Black-Scholes approach, derive
the price of such an option at time t ∈ [0, T ].
(b) Specify the partial differential equation that the pricing function for
the above option must satisfy.
2
Solution:
(a) Remark first of all that, although the similarity, we are not dealing
with any type of barrier option as the pay-off in that case would automatically
be zero if the price process S(t) ever hit L or U before expiration. We will
start our solution by stating some fundamental assumptions made in the
solution of the problem.
Assumption 1: We assume that:
1. The derivative instrument in question can be bought and sold on a market.
2. The market is free of arbitrage.
3. The price process of the derivative asset of the form:
Π(t; X ) = F (t, S(t))
where F is some smooth function.
We will in the solution of the problem employ the notation of the price
process of the asset according to S(t) = St . In order to solve the given
problem, we start by defining a Geometric Brownian Motion (GBM).
Definition 1 (Geometric Brownian Motion): A Geometric Brownian Motion
is a stochastic process Xt that satisfies the Stochastic Differential Equation
(SDE):
dXt = rXt dt + σXt dWt
(1)
together with some initial condition X0 = x where (r, σ) ∈ R2 , σ > 0 and Wt
is a Wiener process.
3
We are thereafter in the need of stating the Black-Scholes equation.
Theorem 1 (Black-Scholes equation): The price process of a European call
or put option in the framework of the Black–Scholes model on an underlying
stock paying no dividends is governed by the partial differential equation:
(
Ft + rsFs + 12 σ 2 s2 Fss − rF = 0, t ∈ (0, T ), x > 0
F (T, s) = Ψ(s)
where F = F (s, t) is the price of the option as a function of the stock price s
and time t. Moreover, σ is the volatility of the stock, r is the risk-free interest
rate and Ψ(s) is known as the payoff function.
From the problem formulation, we realize that our payoff function mathematically can be expressed as:
Ψ(ST ) = 1[L,U ] (ST )
MU
.
S(T )
(2)
Now, we know that the price process of the asset evolves as a GBM as written
in (1). Using the denotation for the price of the stock at time t as St we obtain
the SDE:
(
dSt = rSt dt + σSt dWt
St = s.
Define f (St ) = log St and apply Itô’s Lemma which we state before proceeding.
Theorem 2 (Itô’s Lemma): Let X = (Xt )t≥0 be an Itô process given by:
dXt = rdt + σdWt .
Let g(t, x) ∈ C 2 ([0, ∞) × R). Then Y = (Yt )t≥0 , where:
Yt = g(t, Xt )
is again an Itô process, and:
∂g
1 ∂ 2g
∂g
(t, Xt )dt +
(t, Xt )dXt +
(t, Xt )(dXt )2 .
dYt =
2
∂t
∂x
2 ∂x
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We employ Itô’s Lemma and hence obtain:
1
1
2
d log St0 = rSt0 f 0 (St0 )dt0 +σSt0 f 0 (St0 )dWt + σ 2 S 0t f 00 (St0 )dt = rdt0 +σdWt0 − σ 2 dt0 .
2
2
which after integration from t to T yields:
Z T
Z T
1 2 0
r − σ dt +
σdWt0 .
log ST − log St =
2
t
t
Note on the change of notation from t to t0 done in order to distinguish it
from the lower integration limit t. Hence, we finally obtain:
1 2
St = s exp r − σ (T − t) + σ(WT − Wt ) .
2
which, using the properties of the Wiener process, we can rewrite as:
(3)
√
1 r − σ 2 (T − t) + σ T − tv
(4)
2
where v ∼ N (0, 1) and consequently:
√
√
1 1 r − σ 2 (T − t) + σ T − tZ ∼ N ( r − σ 2 (T − t), σ T − t).
2
2
St = s exp
Using the risk-neutral valuation, we have that the arbitrage free price is
given by:
F (t, s) = e−r(T −t) EQ [Ψ(ST )|St = s]
where the Q-dynamics is such that:
dSt = rSt dt + σSt dW t
where W t is a Q-Brownian motion.
Our problem is consequently reduced to solving the integral:
−r(T −t)
F (t, s) = e
Z∞ 1
√
v2
(r− 21 σ 2 )(T −t)+σ T −tv
√ e− 2 dv
Ψ se
2π
(5)
−∞
Using our payoff function as expressed in (2), we obtain:
√
√
1 2
1 2
Ψ(ST ) = Ψ se(r− 2 σ )(T −t)+σ T −tv = 1[L,U ] (se(r− 2 σ )(T −t)+σ T −tv )
5
MU
se
((r− 21 σ 2 )(T −t)+σ
(6)
√
T −tv)
Since we are dealing with the indicator function, we can reduce the integration limits from the whole of R by solving for v in the inequalities:
1
L ≤ se((r− 2 σ
2 )(T −t)+σ
√
T −tv)
≤U
By simple algebraic manipulations, we arrive at:
ln Us − r − 12 σ 2 (T − t)
ln Ls − r − 21 σ 2 (T − t)
√
√
≤v≤
σ T −t
σ T −t
|
|
{z
}
{z
}
= d1
= d2
and hence we obtain the integral:
−r(T −t)
d2
Z
F (t, s) = M U e
1
1
se((r− 2 σ
d1
2 )(T −t)+σ
√
v2
1
√ e− 2 dv
T −tv)
2π
(7)
which we can write as:
Z
d2
F (t, s) = a
e
d1
√
−σ T −tv
v2
1
√ e− 2 dv = a
2π
Z
d2
d1
√
v2
1
√ e− 2 −σ T −tv dv
2π
(8)
where
M U e−r(T −t)
(9)
1 2
se(r− 2 σ )(T −t)
We can now rewrite the integral in (8) by completing the square in the
exponential function in the integrand as:
a=
Z
d2
F (t, s) = a
d1
√
(v 2 +2σ T −tv+σ 2 (T −t)−σ 2 (T −t))
1
2
√ e−
dv = a
2π
Z
d2
d1
√
1
1
2 σ2
√ e− 2 (v+σ T −t) + 2 (T −t) dv
2π
(10)
We denote
b = ae
σ2
(T −t)
2
(11)
and rewrite (10) as
Z
d2
F (t, s) = b
d1
√
1
1
2
√ e− 2 (v+σ T −t) dv.
2π
6
(12)
We now perform the variable substitution
√
ξ = v + σ T − t ⇒ dv = dξ.
(13)
Consequently, we now integrate from
√
d˜1 = d1 + σ T − t
to
√
d˜2 = d2 + σ T − t
and our integral thus reduces to:
Z
d˜2
F (t, s) = b
d˜1
1 2
1
√ e− 2 ξ dξ
2π
(14)
which we recognize as the normalized form of the cumulative normal distribution function. Our final answer thus becomes:
F (t, s) = b(N (d˜2 ) − N (d˜1 ))
where
1 2
√
−
r
−
σ
(T − t)
2
√
d˜1 =
+ σ T − t,
σ T −t
U
1 2
√
σ
(T − t)
ln
−
r
−
x
√ 2
d˜2 =
+σ T −t
σ T −t
ln
L
x
and
b=
M U e−r(T −t)
se
e
(r− 1 σ 2 )(T −t)
σ2
(T −t)
2
2
7
=
M U (σ2 −2r)(T −t)
e
.
s
(15)
(b) We easily realize that the pricing function F (t, s) must satisfy the
Black-Scholes equation as:
(
Ft + rsFs + 21 σ 2 s2 Fss − rF = 0,
Ψ(s) = MsU 1{L≤s≤U } (s).
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t ∈ (0, T ),
x>0