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

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