MATH3075/3975 Financial Derivatives
Tutorial 6: Solutions
Exercise 1 (a) To show that the model is arbitrage-free, we need to show that there is no strategy
(x, ϕ) satisfying conditions of Definition 2.2.3 of an arbitrage opportunity:
(i) x = V0(x, ϕ) = 0, (ii) V1(x, ϕ) ≥ 0, (iii) EP(V1(x, ϕ)) > 0.
For x = 0, the wealth at time 1 of a strategy (0, ϕ) equals
V1(0, ϕ) = ϕ(S1 − S0(1 + r)).
Since S0 = 4 and r = 0.1, we have S0(1 + r) = 4.4 and thus
V1(x, ϕ)(ω1) = (8− 4.4)ϕ = 3.6ϕ,
V1(x, ϕ)(ω2) = (5− 4.4)ϕ = 0.6ϕ,
V1(x, ϕ)(ω3) = (3− 4.4)ϕ = −1.4ϕ.
There are now three cases to consider, namely,
ϕ > 0 =⇒ V1(x, ϕ)(ω3) = −1.4ϕ < 0,
ϕ < 0 =⇒ V1(x, ϕ)(ω1) = 3.6ϕ < 0,
ϕ = 0 =⇒ EP(V1(x, ϕ)) = 0.
In all three cases we have a contradiction, since either condition (ii) or condition (iii) in Definition
2.2.3 is not satisfied. We thus conclude that the model M = (B,S) is arbitrage-free.
(b) We will show that the call option with strike K = 4 is not attainable. To this end, we try to
compute the replicating strategy directly, that is, by solving the equation
1.1 81.1 5
1.1 3
[ϕ0
ϕ1
]
=
41
0
where we denote ϕ0 = x−ϕ1S0 and ϕ1 = ϕ. If we now augment the right-hand side and row reduce
the matrix then we find
1.1 8 41.1 5 1
1.1 3 0
=⇒ row operations =⇒
1.1 8 40 3 3
0 0 −1
.
We conclude that the above linear system has no solution and thus the call option with strike K = 4
is not attainable. Of course, the non-existence of a solution can be verified using other arguments,
for instance, by finding the unique solution to the first two equations and checking that it fails to
satisfy the last equation.
1
(c) Since Ω = {ω1, ω2, ω3} we may identify the class of all contingent claims with the vector space
R3. Let Y1 = (1, 1, 1) and Y2 = (8, 5, 3). Then the space of all attainable claims is given by{
X ∈ R3 |X = λ1Y1 + λ2Y2, λ1, λ2 ∈ R
}
,
meaning that it is the linear subspace of R3 spanned by the vectors Y1 and Y2. More explicitly, the
space of all attainable claims is the plane in R3 given by
X =
x1x2
x3
∈ R3 ∣∣∣λ1
11
1
+ λ2
85
3
, λ1, λ2 ∈ R
.
(d) We note that a martingale measure in this model is not unique. Indeed, we already know from
part (b) that the call option with strike K = 4 is not an attainable claim. Hence the model is
incomplete and thus the non-uniqueness of a martingale measure follows from Proposition 2.2.7 (or
Theorem 2.2.2). Our goal is now to compute explicitly the set of all martingale measures
Q = (Q(ω1),Q(ω2),Q(ω3)) = (q1, q2, q3) ∈M
for the model M.
First method. Recall that Q ∈ M if Q is equivalent to P and EQ(Ŝ1) = S0. More explicitly, a
probability measure Q = (q1, q2, q3) belongs to M whenever it satisfies the following conditions
S0 =
1
1 + r
(
q1S1(ω1) + q2S1(ω2) + q3S1(ω3)
)
,
q1 + q2 + q3 = 1, qi > 0, i = 1, 2, 3.
Equivalently, we need to solve the following system
[
8 5 3
1 1 1
]q1q2
q3
= [4.4
1
]
subject to the constraints 0 < qi < 1 for i = 1, 2, 3. By solving these equations, we conclude that
the set of all martingale measures M can be represented as follows (note that λ = q3 was chosen
here as a parameter)
M =
{
Q = (q1, q2, q3) =
(
2λ
3
− 1
5
,−5λ
3
+ 6
5
, λ
)
, λ ∈
(
3
10
, 18
25
)}
,
that is,
M =
{
Q = (q1, q2, q3) =
(
−1
5
, 6
5
, 0
)
+ λ
(
2
3
,−5
3
, 1
)
, λ ∈
(
3
10
, 18
25
)}
.
This confirms our claim that several martingale measures for the model M exist.
Second method. It is also possible to use two-state sub-models. Let Q1 (respectively, Q2) be the
unique martingale measure for the two-state sub-model with the elementary event ω2 (respectively,
ω1) discarded. Simple computations for two-state sub-models show that Q1 = (0.28, 0, 0.72) and
Q2 = (0, 0.7, 0.3). Hence this method gives the following representation for the class M
M =
{
Q = (q1, q2, q3) ∈ R3 |Q = αQ1 + (1− α)Q2, α ∈ (0, 1)
}
.
Let us stress that the probability measures Q1 and Q2 do not belong to the class M since they are
not equivalent to P (since q2 = 0 for Q1 and q1 = 0 for Q2).
2
(e) We search for the range of expected values
EQ
(
(S1 − 4)+
1 + r
)
= 4
1.1
q1 +
1
1.1
q2 +
0
1.1
q3
where Q = (q1, q2, q3) ∈M is any martingale measure.
First method. From part (d), we obtain
EQ
(
(S1 − 4)+
1 + r
)
= 4
1.1
(
2λ
3
− 1
5
)
+ 1
1.1
(
−5λ
3
+ 6
5
)
+ 0
1.1
λ.
We thus see that the range of arbitrage prices for the call option coincides with the range of the
linear function f(λ) = 4/11 + (10/11)λ where λ ∈ (3/10, 18/25). Since the range of values of f
equals (7/11, 56/55), we conclude that the arbitrage price of the option may take any value from
the open interval (7/11, 56/55).
Second method. This result can also be confirmed by noting that any Q ∈ M satisfies Q = αQ1 +
(1− α)Q2 and thus
EQ
(
(S1 − 4)+
1 + r
)
= αEQ1
(
(S1 − 4)+
1 + r
)
+ (1− α)EQ2
(
(S1 − 4)+
1 + r
)
where α ∈ (0, 1) and Q1 = (0.28, 0, 0.72) and Q2 = (0, 0.7, 0.3). Since
EQ1
(
(S1 − 4)+
1 + r
)
= 4
1.1
0.28 = 56
55
and
EQ2
(
(S1 − 4)+
1 + r
)
= 1
1.1
0.7 = 7
11
we conclude once again that the range of arbitrage prices for the call option is the open interval
(7/11, 56/55).
(f) (MATH3975) Generally speaking, when a contingent claim is not attainable, then by the su-
perhedging price for X we mean the minimal value of x ∈ R for which there exists ϕ ∈ R such that
V1(x, ϕ)(ω) ≥ X(ω) for every ω ∈ Ω. One can check that when a contingent claim X is attaina-
ble, then the superhedging price for X coincides with the arbitrage price for X, which is obtained
through replication.
In our case, we already know that the claim X = CT = (S1 − 4)+ is not attainable and the
superhedging conditions for X read:
1.1x+ 3.6ϕ1 ≥ 4, (1)
1.1x+ 0.6ϕ1 ≥ 1, (2)
1.1x− 1.4ϕ1 ≥ 0. (3)
It is now sufficient to sketch solutions (half-planes) for the above inequalities and find the corner
point in the feasible region with the lowest x. It appears to be the intersect of lines corresponding
to equations (1) and (3). Observe that the second inequality is strict in that case. This means that
an arbitrage opportunity will arise for the issuer of the option if she is able to sell the option at the
superhedging price computed below and the elementary events ω2 occurs at time T = 1.
3
We now proceed to explicit computations. By solving equations corresponding to inequalities
(1) and (3), that is,
1.1x+ 3.6ϕ1 = 4,
1.1x− 1.4ϕ1 = 0,
we obtain x = 56/55 and ϕ = 4/5. Then inequality (2) is strict, namely,
1.1x+ 0.6ϕ1 = 1.1
56
55
+ 0.6 4
5
> 1.
The minimal superhedging price and the corresponding hedge ratio are thus given by (x, ϕ) =
(56/55, 4/5). It is worth noting that the minimal superhedging price is equal to the upper bound
for the arbitrage price of the option computed in part (e), that is,
min
{
x ∈ R |V1(x, ϕ) ≥ CT for some ϕ ∈ R
}
= sup
Q∈M
EQ
(
CT
1 + r
)
. (4)
In fact, it is possible to show that equality (4) is valid for an arbitrary contingent claim X and not
only for the claim X = CT = (S1 − 4)+.
Exercise 2 (a) A martingale measure Q satisfies: q1 + q2 + q3 = 1, 0 < qi < 1 and
EQ(Ŝ1) = EQ
(
S1
B1
)
= EQ
(
S1
1.1
)
= 7q1 + 5q2 + 4q3 = S0 = 5
or, equivalently, EQ(Ŝ1 − S0) = 2q1 − q3 = 0. Let q3 = α. Then
q1 =
α
2
=
1
3
(1− γ), q2 = 1−
3
2
α = γ, q3 = α =
2
3
(1− γ)
where 0 < α < 2/3 (or, equivalently, γ ∈ (0, 1)). We obtain
M =
{
(q1, q2, q3)
∣∣ q1 = α
2
, q2 = 1−
3
2
α, q3 = α, 0 < α < 2/3
}
and thus the market model M is not complete.
(b) It suffices to solve the following equations
1.1ϕ0 + 7.7ϕ1 = 5.5,
1.1ϕ0 + 5.5ϕ1 = 3.3,
1.1ϕ0 + 4.4ϕ1 = 2.2.
It is clear that the strategy (ϕ0, ϕ1) = (−2, 1) replicates X and thus X is attainable. The price of
X at time t = 0 thus equals π0(X) = V0(ϕ) = ϕ
0 + ϕ1S0 = −2 + 1 · 5 = 3.
(c) For an arbitrary 0 < α < 2/3, we obtain (recall that B1 = 1.1)
EQ
(
Y
B1
)
= (1.1)−1
(
3
α
2
+ 1
(
1−
3
2
α
)
+ 0α
)
= (1.1)−1.
Hence the claim Y is attainable.
4
(d) For 0 < α < 2/3, we obtain (recall that B1 = 1.1)
EQ
(
Z
B1
)
= 4
α
2
− 3α = −α.
Hence the range of prices is the interval (−2/3, 0) and thus the claim Z is not attainable.
(e) If π0(Y ) = −0.5 then α = 0.5 and thus the unique martingale measure for the extended model
M̃ = (B,S1, S2) equals.
Q̃ =
(
Q̃(ω1), Q̃(ω2), Q̃(ω3)
)
= (1/4, 1/4, 1/2).
The martingale measure for M̃ is unique and thus the model M̃ = (B,S1, S2) is complete.
Exercise 3 (MATH3975) (a) Consider the relative wealth V̂ (ϕ), which is given by
V̂t(ϕ) :=
Vt(ϕ)
S2t
, t = 0, 1.
For any trading strategy ϕ = (ϕ10, ϕ
2
0) ∈ R
2 with V0(ϕ) = 0, we obtain
V̂1(ϕ) = V̂1(ϕ)− V̂0(ϕ) =
V1(ϕ)
S21
−
V0(ϕ)
S20
=
ϕ10S
1
1 + ϕ
2
0S
2
1
S21
−
ϕ10S
1
0 + ϕ
2
0S
2
0
S20
= ϕ10
(
S11
S21
−
S10
S20
)
.
More explicitly
V̂1(ϕ)(ω1) = ϕ
1
0
(
s1
z1
−
s0
z0
)
,
V̂1(ϕ)(ω2) = ϕ
1
0
(
s2
z2
−
s0
z0
)
,
where ϕ10 is an arbitrary real number. By standard arguments, we obtain the following necessary
and sufficient conditions for the arbitrage-free property of the model M = (S1, S2)
(A)
s1
z1
<
s0
z0
<
s2
z2
or (B)
s2
z2
<
s0
z0
<
s1
z1
or (C)
s1
z1
=
s0
z0
=
s2
z2
.
(b) Let us now examine the completeness of the model. To this end, we ask whether the system of
equations
V1(ϕ)(ω1) = ϕ
1
0s1 + ϕ
2
0z1 = X1,
V1(ϕ)(ω2) = ϕ
1
0s2 + ϕ
2
0z2 = X2,
has a solution ϕ = (ϕ10, ϕ
2
0) ∈ R
2 for any claim X = (X1, X2) ∈ R2.
This holds if and only if the determinant ∆ := s1z2 − z1s2 6= 0, that is, whenever the vectors
(s1, s2) and (z1, z2) are not collinear. Observe that conditions (A) and (B) imply that ∆ 6= 0. Hence
under either (A) or (B) the model is arbitrage-free and complete.
5
It is clear that if (C) holds then ∆ = 0 and thus the model is still arbitrage-free, but it is incomplete.
We also observe that if ∆ = 0 then the model is incomplete and it is arbitrage-free whenever
s1
z1
=
s0
z0
=
s2
z2
.
(c) We need to find a vector ϕ = (ϕ10, ϕ
2
0) ∈ R
2 such that
ϕ10s1 + ϕ
2
0z1 = (s1 − z1)
+,
ϕ10s2 + ϕ
2
0z2 = (s2 − z2)
+.
Complete case. We first assume that either (A) or (B) holds, so that the determinant ∆ :=
s1z2 − z1s2 6= 0. Hence the model is arbitrage-free and complete. Then the unique replicating
strategy (ϕ10, ϕ
2
0) can be easily computed for any contingent claim X = (X1, X2). Specifically, the
unique solution to equations
ϕ10s1 + ϕ
2
0z1 = X1,
ϕ10s2 + ϕ
2
0z2 = X2,
reads
ϕ10 =
X1z2 −X2z1
s1z2 − z1s2
, ϕ20 =
X1s2 −X2s1
s1z2 − z1s2
.
The arbitrage price of X at time 0 equals
π0(X) = ϕ
1
0s0 + ϕ
2
0z0
or, more explicitly,
π0(X) =
X1z2 −X2z1
s1z2 − z1s2
s0 +
X1s2 −X2s1
s1z2 − z1s2
z0
Of course, this formula can be applied to the claim
X = (S11 − S
2
1)
+ =
(
(s1 − z1)+, (s2 − z2)+
)
.
Incomplete case. Assume now that condition (C) is satisfied so that the determinant ∆ := s1z2−
z1s2 = 0. Hence the model is arbitrage-free and incomplete. It is now clear that (s1, s2) = λ(z1, z2)
for some strictly positive real number λ. Consequently, the system of equations for the replicating
strategy ϕ = (ϕ10, ϕ
2
0) becomes
ϕ10λz1 + ϕ
2
0z1 = (λz1 − z1)
+ = z1(λ− 1)+,
ϕ10λz2 + ϕ
2
0z2 = (λz2 − z2)
+ = z2(λ− 1)+.
• Let us first assume that 0 < λ ≤ 1. Then ϕ = (ϕ10, ϕ
2
0) solves
ϕ10λ+ ϕ
2
0 = 0.
Hence ϕ = (ϕ10,−λϕ
1
0) for any real number ϕ
1
0. Note that, in view of condition (C), we have
that s0 = λz0 as well. We conclude that for 0 < λ ≤ 1 the price equals
π0(X) = ϕ
1
0s0 − λϕ
1
0z0 = 0.
6
• Assume now that λ > 1. Then ϕ = (ϕ10, ϕ
2
0) solves
ϕ10λ+ ϕ
2
0 = λ− 1 > 0.
Hence ϕ = (ϕ10, λ − 1 − λϕ
1
0) for any real number ϕ
1
0. We conclude that for λ > 1 the price
satisfies
π0(X) = ϕ
1
0s0 − (λ− 1− λϕ
1
0)z0 = (λ− 1)z0.
To summarise, the unique price of X in an arbitrage-free and incomplete model equals π0(X) =
(λ− 1)+z0 where λ = s0/z0 or, simply, π0(X) = (s0 − z0)+.
(d) We note that
X − Y = (S11 − S
2
1)
+ − (S21 − S
1
1)
+ = S11 − S
2
1
and thus, by the linearity of the arbitrage price map, we obtain
π0(X)− π0(Y ) = π0(X − Y ) = π0(S11 − S
2
1) = π0(S
1
1)− π0(S
2
1) = S
1
0 − S
2
0 = s0 − z0.
Hence the put-call parity relationship reads
π0(X)− π0(Y ) = s0 − z0.
7