MATH3075/3975 Financial Derivatives
Tutorial 4: Solutions
Exercise 1 We consider the elementary market model M = (B,S) with S0 > 0 and 0 < d <
1 + r < u.
(a) Our aim is to find the probability measure P̂ such that EP̂(B̂T ) = B̂0 where B̂t = Bt/St for
t = 0, 1. We will also compute the Radon-Nikodym density L of P̂ with respect to P̃ and we will
show that EP̃(L) = 1.
• We denote P̂(ω1) = p̂ and P̂(ω2) = q̂ = 1 − p̂. The postulated equality EP̂(B̂T ) = B̂0 means
that
EP̂
(
BT
ST
)
=
B0
S0
,
which can be expanded to the following equation for p̂
p̂
1 + r
Su
+ q̂
1 + r
Sd
=
1
S0
.
We obtain
p̂ =
(
1
Sd
−
1
(1 + r)S0
)
SuSd
Su − Sd
=
(1 + r)S0 − Sd
(1 + r)S0Sd
SuSd
Su − Sd
=
1 + r − d
u− d
u
1 + r
= p̃
u
1 + r
and
q̂ =
(
1
(1 + r)S0
−
1
Su
)
SuSd
Su − Sd
=
Su − (1 + r)S0
(1 + r)S0Su
SuSd
Su − Sd
=
u− (1 + r)
u− d
d
1 + r
= q̃
d
1 + r
where we denote q̃ = 1 − p̃. It is easy to see that p̂ > 0, q̂ > 0 and p̂ + q̂ = 1. Hence P̂ =
(P̂(ω1), P̂(ω2)) = (p̂, q̂) is a probability on Ω = (ω1, ω2) and it is equivalent to the risk-neutral
probability measure P̃ (hence also equivalent to the probability measure P).
The Radon-Nikodym density L of P̂ with respect to P̃ equals
L(ω1) =
dP̂
dP̃
(ω1) =
P̂(ω1)
P̃(ω1)
=
p̂
p̃
=
u
1 + r
, L(ω2) =
dP̂
dP̃
(ω2) =
P̂(ω2)
P̃(ω2)
=
q̂
q̃
=
d
1 + r
.
It is important to notice that
L =
dP̂
dP̃
=
STB0
S0BT
=
ŜT
Ŝ0
.
Furthermore,
EP̃(L) = p̃
u
1 + r
+ (1− p̃)
u
1 + r
= p̂+ q̂ = 1.
1
(b) We wish to show that the price π0(X) of any contingent claim X = g(ST ) satisfies
π0(X) = S0 EP̂
(
X
ST
)
= S0 EP̂
(
g(ST )
ST
)
.
• We already know from lectures that any contingent claimX can be replicated in the elementary
market modelM = (B,S) and its arbitrage price at time 0 can be computed using the risk-neutral
valuation formula
π0(X) = B0 EP̃
(
(1 + r)−1X
)
. (1)
First method. To show that π0(X) satisfies also the equality
π0(X) = S0 EP̂
(
S−1T X
)
, (2)
we may use Radon-Nikodym density L of P̂ with respect to P̃. It suffices to observe that
S0 EP̂
(
S−1T X
)
= S0 EP̃
(
LS−1T X
)
= S0 EP̃
(
ŜT Ŝ
−1
0 S
−1
T X
)
= B0 EP̃
(
B−1T X
)
= EP̃
(
(1 + r)−1X
)
.
Second method. Alternatively, we consider any portfolio ϕ = (ϕ0, ϕ1) where ϕ0 = x − ϕ1S0 =
V0(ϕ)− ϕ1S0 and
VT (ϕ) = ϕ
0(1 + r) + ϕ1ST = ϕ
0BT + S
−1
0 (V0(ϕ)− ϕ
0)ST .
Then (recall that B0 = 1)
VT (ϕ)
ST
=
V0(ϕ)
S0
+ ϕ0
(
BT
ST
−
B0
S0
)
,
so that, using the definition of the probability P̂, we obtain
EP̂
(
S−1T VT (ϕ)
)
=
V0(ϕ)
S0
+ EP̂
[
ϕ0
(
BT
ST
−
B0
S0
)]
=
V0(ϕ)
S0
+ ϕ0 EP̂(B̂T − B̂0) =
V0(ϕ)
S0
since EP̂(B̂T − B̂0) = 0. Hence if a portfolio ϕ replicates X so that VT (ϕ) = X, then
EP̂
(
S−1T X
)
=
V0(ϕ)
S0
=
π0(X)
S0
.
(c) We consider the put option with the payoff PT (K) = (K −ST )+ for some K > 0. We will show
that the arbitrage price P0(K) admits the following representation
P0(K) = K(1 + r)
−1 P̃(ST < K)− S0 P̂(ST < K).
• We denote A = {ST < K} so that
PT (K) = (K − ST )+ = (K − ST )1A = 1AK − 1AST = X1 −X2.
where, by definition, 1A = 1 on the event A and it equals 0 on the complement of A. Therefore, by
applying (1) to X1 and (2) to X2, we obtain
P0(K) = π0
(
X1
)
− π0
(
X2
)
= π0
(
1AK
)
− π0
(
1AST
)
= B0 EP̃
(
(1 + r)−11AK
)
− S0 EP̂
(
S−1T 1AST
)
= K(1 + r)−1 P̃(ST < K)− S0 P̂(ST < K).
2
Let CT (K) = (ST −K)+ for some K > 0. It is easy to show that the price C0(K) satisfies
C0(K) = S0 P̂(ST > K)−K(1 + r)−1 P̃(ST > K).
(d) Our goal is to show that the extended model Me = (B,S, P (K)) is arbitrage-free, in the sense
of Definition 2.2.3 from the course notes.
• For any trading strategy (x, ϕ1, ϕ2) ∈ R3, the wealth satisfies V0(x, ϕ1, ϕ2) = x and
V1(x, ϕ
1, ϕ2) =
(
x− ϕ1S0 − ϕ2P0(K)
)
(1 + r) + ϕ1S1 + ϕ
2P1(K).
If x = 0, then for every (ϕ1, ϕ2) ∈ R2
V1(0, ϕ
1, ϕ2) = ϕ1
(
ST − S0(1 + r)
)
+ ϕ2
(
PT (K)− P0(K)(1 + r)
)
= ϕ1
(
ST − S0(1 + r)
)
+ ϕ2
(
X − (1 + r)π0(X)
)
where we denote X = PT (K) and π0(X) = P0(K). Therefore,
EP̃(VT (0, ϕ
1, ϕ2)) = ϕ1 EP̃
(
ST − S0(1 + r)
)
+ ϕ2 EP̃
(
X − (1 + r)π0(X)
)
= 0.
If the wealth VT (0, ϕ
1, ϕ2) is non-negative and has the expected value equal to zero, then necessarily
VT (0, ϕ
1, ϕ2)(ωi) = 0 for i = 1, 2. We conclude that arbitrage opportunities do not exist in the
extended model Me = (B,S, P (K)).
(e) We take a fixed K such that S0d < K < S0u and we consider the modified market model
N = (B,P (K)). We will show that the price of an arbitrary claim X computed in N = (B,P (K))
coincides with its arbitrage price computed inM = (B,S). We will also find the arbitrage price at
time 0 for the claim X = ST .
• It suffices to observe that the probability measure Q = P̃ is the unique martingale measure
for the model N = (B,P (K)) since
EP̃(P1(K)) = (1 + r)P0(K).
Hence the arbitrage price of any contingent claim X computed in N = (B,P (K)) andM = (B,S)
are identical. In particular, the arbitrage price of the claim X = ST in the modelN can be computed
from the risk-neutral valuation
π0(X) = B0 EQ
(
B−1T X
)
= B0 EP̃
(
B−1T X
)
= B0 EP̃
(
B−1T ST
)
= S0
where the last equality is a consequence of the definition of P̃. One may also show directly that the
claim X = ST can be replicated by a portfolio composed of B and P (K).
Exercise 2 Consider a trading strategy (x, ϕ) = (x, ϕ1, . . . , ϕn). Its wealth satisfies V0(x, ϕ) = x
and
V1(x, ϕ) =
(
x−
n∑
j=1
ϕjS
j
0
)
(1 + r) +
n∑
j=1
ϕjS
j
1
and thus
V̂1(x, ϕ) :=
V1(x, ϕ)
B1
=
V1(x, ϕ)
1 + r
=
(
x−
n∑
j=1
ϕjS
j
0
)
+
n∑
j=1
ϕj(1 + r)−1S
j
1
=
(
x−
n∑
j=1
ϕjS
j
0
)
+
n∑
j=1
ϕjŜ
j
1
where Ŝ
j
1 := (1 + r)
−1S
j
1.
3
Consequently,
Ĝ1(x, ϕ) := V̂1(x, ϕ)− V̂0(x, ϕ) = V̂1(x, ϕ)− V0(x, ϕ) = V̂1(x, ϕ)− x
=
(
x−
n∑
j=1
ϕjS
j
0
)
+
n∑
j=1
ϕjŜ
j
1 − x =
n∑
j=1
ϕj(Ŝ
j
1 − S
j
0)
=
n∑
j=1
ϕj(Ŝ
j
1 − Ŝ
j
0) =
n∑
j=1
ϕj∆Ŝ
j
1.
Exercise 3 In view of Definition 2.2.4, we need to find all probability measures Q on the space
Ω = {ω1, ω2, ω3} such that Q is equivalent to P (that is, Q(ωi) > 0 for i = 1, 2, 3) and
EQ(∆Ŝ1) = 0
where
∆Ŝ1 = Ŝ1 − Ŝ0 = 910
(
60
9
, 40
9
, 30
9
)
− (5, 5, 5) = (1,−1,−2).
Let us denote Q = (q1, q2, q3). Then we search for all solutions (q1, q2, q3) to the system
0 < qi < 1 for i = 1, 2, 3,
q1 − q2 − 2q3 = 0,
q1 + q2 + q3 = 1.
We obtain {
q1 =
1
2
q3 +
1
2
,
q2 = −32 q3 +
1
2
,
and we check that the condition qi ∈ (0, 1) is satisfied for every i = 1, 2, 3 whenever q3 ∈ (0, 1/3)
since it is clear that q3 should satisfy the following inequalities{
0 < 1
2
q3 +
1
2
< 1,
0 < −3
2
q3 +
1
2
< 1,
and this holds whenever q3 ∈ (0, 1/3). Hence the set of all risk-neutral probability measures forM
can be represented as follows
M =
{
Q =
(
1
2
, 1
2
, 0
)
+ q3
(
1
2
,−3
2
, 1
)
, q3 ∈
(
0, 1
3
)}
.
Exercise 4 (a) For any trading strategy (x, ϕ) = (x, ϕ1, ϕ2), we have
V1(x, ϕ) =
(
x−
2∑
j=1
ϕjS
j
0
)
(1 + r) +
2∑
j=1
ϕjS
j
1
or, more explicitly,
V1(x, ϕ)(ωi) =
10
9
x+ 10
9
ϕ1 + 20
9
ϕ2, i = 1,
10
9
x+ 10
9
ϕ1 − 20
9
ϕ2, i = 2,
10
9
x− 10
9
ϕ1 − 20
9
ϕ2, i = 3,
10
9
x− 30
9
ϕ1 + 20
9
ϕ2, i = 4.
Since
G1(x, ϕ) := V1(x, ϕ)− V0(x, ϕ) = V1(x, ϕ)− x,
4
we obtain
G1(x, ϕ)(ωi) =
1
9
x+ 10
9
ϕ1 + 20
9
ϕ2, i = 1,
1
9
x+ 10
9
ϕ1 − 20
9
ϕ2, i = 2,
1
9
x− 10
9
ϕ1 − 20
9
ϕ2, i = 3,
1
9
x− 30
9
ϕ1 + 20
9
ϕ2, i = 4.
Next
V̂1(x, ϕ) := (1 + r)
−1 V1(x, ϕ) =
9
10
V1(x, ϕ)
so that
V̂1(x, ϕ)(ωi) =
x+ ϕ1 + 2ϕ2, i = 1,
x+ ϕ1 − 2ϕ2, i = 2,
x− ϕ1 − 2ϕ2, i = 3,
x− 3ϕ1 + 2ϕ2, i = 4.
Finally,
Ĝ1(x, ϕ) := V̂1(x, ϕ)− V̂0(x, ϕ) = V̂1(x, ϕ)− x
and thus
Ĝ1(x, ϕ)(ωi) =
ϕ1 + 2ϕ2, i = 1,
ϕ1 − 2ϕ2, i = 2,
−ϕ1 − 2ϕ2, i = 3,
−3ϕ1 + 2ϕ2, i = 4.
We observe that
Ĝ1(x, ϕ) = ϕ
1∆Ŝ11 + ϕ
2∆Ŝ21 = ϕ
1(1, 1,−1,−3) + ϕ2(2,−2,−2, 2).
(b) It is clear that G1(x, ϕ) depends on the initial wealth x, but Ĝ1(x, ϕ) does not, so that for
any x, y ∈ R and arbitrary ϕ ∈ R2 we have Ĝ1(x, ϕ) = Ĝ1(y, ϕ). In particular, Ĝ1(x, ϕ) = Ĝ1(0, ϕ)
for every x ∈ R and ϕ ∈ R2.
Exercise 5 (MATH3975) (a) We have k = 4 and n = 2. Recall that
W :=
{
X ∈ R4 |X = Ĝ1(x, ϕ) for some (x, ϕ) ∈ R3
}
where in fact Ĝ1(x, ϕ) does not depend on x. More explicitly, X ∈ W if and only if X ∈ R4 and
X ∈ Ĝ1(x, ϕ) that is (from the previous exercise)
X =
ϕ1 + 2ϕ2, i = 1,
ϕ1 − 2ϕ2, i = 2,
−ϕ1 − 2ϕ2, i = 3,
−3ϕ1 + 2ϕ2, i = 4,
for some real numbers ϕ1 and ϕ2. This means that W is a plane given by
W =
{
X ∈ R4 |X = ϕ1(1, 1,−1,−3) + ϕ2(2,−2,−2, 2), ϕ1, ϕ2 ∈ R
}
or, more precisely, the two-dimensional linear subspace of R4 spanned by the vectors (1, 1,−1,−3)
and (2,−2,−2, 2).
(b) Recall that the space W⊥ is the orthogonal complement of W, that is,
W⊥ :=
{
Z ∈ R4 | 〈X,Z〉 = 0 for all X ∈W
}
.
5
Hence a vector Z ∈ R4 belongs to W⊥ whenever it satisfies{
z1 + z2 − z3 − 3z4 = 0,
z1 − z2 − z3 + z4 = 0.
Therefore, z1 = z3 + z4 and z2 = 2z4 and thus
W⊥ =
{
Z ∈ R4 |Z = z3(1, 0, 1, 0) + z4(1, 2, 0, 1), z3, z4 ∈ R
}
which is the two-dimensional linear subspace of R4 spanned by the vectors (1, 0, 1, 0) and (1, 2, 0, 1).
We conclude that W⊥ is the plane orthogonal to the plane W.
(c) We will show that the model M = (B,S1, S2) is arbitrage-free using two methods.
• First method. In view of Remark 2.2.3, to check whether the model is arbitrage-free, it suffices
to show that W ∩ A = ∅ where
A =
{
X ∈ R4 |X 6= 0, xi ≥ 0, i = 1, . . . , 4
}
.
Since for all X ∈ W we have x3 = −x1 and for all X ∈ A we have xi ≥ 0, it is clear that if
X ∈W ∩A then x1 = x3 = 0. This in turn implies that ϕ1 = −2ϕ2 and thus x2 = −4ϕ2 and
x4 = 8ϕ
2. Condition xi ≥ 0 implies that ϕ2 = 0 and thus also x1 = x2 = x3 = x4 = 0. We
conclude that W ∩ A = ∅ and thus the market model is arbitrage-free.
• Second method. In view of Remark 2.2.4, in order to confirm that the model is arbitrage-free,
one may check that W⊥ ∩ P+ 6= ∅ where the set P+ is given by
P+ :=
{
Q ∈ R4
∣∣ 4∑
i=1
qi = 1, qi > 0
}
.
Assume that Z ∈W⊥∩P+. We observe that the condition qi > 0 for i = 1, . . . , 4 implies that
z3 > 0 and z4 > 0. Then z1 = z3 + z4 > 0 and z2 = 2z4 > 0 and thus it suffices to impose the
condition that Z is a probability measure so that
4∑
i=1
zi = 1 ⇒ 2z3 + 4z4 = 1.
It is now clear that W⊥ ∩ P+ 6= ∅ and thus the market model is arbitrage-free.
For instance, we may take z3 = 1/4 and z4 = 1/8. Then we obtain the following vector
Z = 1
4
(1, 0, 1, 0) + 1
8
(1, 2, 0, 1) =
(
3
8
, 2
8
, 2
8
, 1
8
)
∈W⊥ ∩ P+.
It is easy to check that for Q = Z we have that EQ
(
∆Ŝi1
)
= 0 for i = 1, 2 since ∆Ŝ11 =
(1, 1,−1,−3) and ∆Ŝ21 = (2,−2,−2, 2).
(d) From Lemma 2.2.1, we know that M = W⊥ ∩ P+. The class M of all risk-neutral probabilities
for this model is thus non-empty and non-uniqueness of a risk-neutral probability holds. We may
represent M as follows
M =
{
Q ∈ R4
∣∣Q = q3(1, 0, 1, 0) + q4(1, 2, 0, 1), q3 > 0, q4 > 0, 2q3 + 4q4 = 1}
or, equivalently,
M =
{
Q ∈ R4
∣∣Q = q3(1, 0, 1, 0) + 1−2q34 (1, 2, 0, 1), q3 ∈ (0, 12)}.
6
Exercise 6 (MATH3975) The proof of Proposition 2.2.1 is straightforward. It suffices to observe
that V1(x, ϕ) and V̂1(x, ϕ) have the same properties, since V̂1(x, ϕ) = cV1(x, ϕ) for a strictly positive
constant c (specifically, c = (1+r)−1). This argument shows that the first statement in Proposition
2.2.1 is valid.
Furthermore, we observe that in Definition 2.2.3 we only consider trading strategies with x = 0.
Therefore, we may use the equality V̂1(0, ϕ) = Ĝ1(0, ϕ) and thus the second part in Proposition
2.2.1 is true as well.
7