CS计算机代考程序代写 MATH3075/3975 Financial Derivatives

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)

=


=

u

1 + r
, L(ω2) =

dP̂
dP̃

(ω2) =
P̂(ω2)
P̃(ω2)

=


=

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