MATH3075/3975 Financial Derivatives
Tutorial 5: Solutions
Exercise 1 For any trading strategy (x, ϕ), the wealth at time 1 equals
V1(x, ϕ) = (x− ϕS0)(1 + r) + ϕS1.
Hence the class of all attainable contingent claims is the two-dimensional subspace of the linear
space R3 spanned by the vectors (1, 1, 1) and (6, 4, 3). This means that the considered model is
incomplete since the space R3 of all contingent claims is three-dimensional.
We wish to find out for which values of the strike K the call option with the payoff CT =
(S1 −K)+ is an attainable claim. To this end, we need to examine four subcases:
• We first assume that K ≤ 30
9
. Then CT = S1 −K and thus it is an attainable claim with the
unique arbitrage price at time 0 given by C0 = S0 − 910 K.
• Next, we assume that 30
9
< K < 40
9
. Then
CT = (S1 −K)+ =
(60
9
−K,
40
9
−K, 0
)
,
so that we now search for α, β ∈ R satisfying
α+ 6β = 60
9
−K,
α+ 4β = 40
9
−K,
α+ 3β = 0.
From the last two equations, we obtain β = 40
9
−K and α = −3β. Then the first equation
becomes
3β = 120
9
− 3K = 60
9
−K,
which yields K = 30
9
. Hence the option is not attainable when 30
9
< K < 40
9
.
• We now assume that 40
9
≤ K < 60
9
. Then
CT = (S1 −K)+ =
(60
9
−K, 0, 0
)
and thus we search for α, β ∈ R such that
α+ 6β = 60
9
−K,
α+ 4β = 0,
α+ 3β = 0.
The last two equations give α = β = 0 and thus the first equation is not satisfied. Hence the
option is not attainable when 40
9
≤ K < 60
9
.
• Finally, we assume that K ≥ 60
9
. Then CT = 0 and thus it is an attainable claim with the
unique arbitrage price at time 0 given by C0 = 0.
We conclude that the call option is attainable only when either K ≤ 30
9
or K ≥ 60
9
. However, in
the former case CT = S1 −K and thus we deal with the forward contract, and in the latter case
CT = 0 so that the contract is trivial. In contrast, if we take any K ∈ (309 ,
60
9
), then the call option
cannot be replicated in our model since the claim CT = (ST −K)+ is not attainable. This confirms
that the model is incomplete, as was observed before.
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Exercise 2 The model M = (B,S) introduced in Example 2.2.3 postulates that the stock price
S1 satisfies
S1(ω) =
(1 + h)S0 if ω = ω1,
(1 + l)S0 if ω = ω2,
(1− l)S0 if ω = ω3,
(1− h)S0 if ω = ω4,
where 0 < l < h < 1. The savings account B satisfies B0 = 1, B1 = 1 + r where, by assumption,
0 ≤ r < h.
(a) For a trading strategy (x, ϕ) we have
V1(x, ϕ) = (x− ϕS0)(1 + r) + ϕS1.
Hence the class of all attainable contingent claims is the two-dimensional subspace of R4 spanned
by the vectors (1, 1, 1, 1) and (1+h, 1+ l, 1− l, 1−h). It is thus clear that the model is not complete.
(b) By Definition 2.2.4 of the martingale measure, a probability measure Q = (q1, q2, q3, q4) belongs
to M when 0 < qi < 1 for i = 1, 2, 3, 4 and EQ(Ŝ1) = S0. More explicitly,
S0 =
1
1 + r
4∑
i=1
qiS1(ωi),
that is,
(1 + r)S0 = q1(1 + h)S0 + q2(1 + l)S0 + q3(1− l)S0 + q4(1− h)S0.
After simplifications, we obtain the following system:{
q1 + q2 + q3 + q4 = 1,
q1h+ q2l − q3l − q4h = r,
with the constraints 0 < qi < 1 for i = 1, 2, 3, 4. By multiplying the first equation by h, we obtain{
q1h+ q2h+ q3h+ q4h = h,
q1h+ q2l − q3l − q4h = r.
Hence q1 and q4 can be expressed in terms of q2 and q3, specifically,
q1 =
h+ r
2h
−
h+ l
2h
q2 −
h− l
2h
q3,
and
q4 =
h− r
2h
−
h− l
2h
q2 −
h+ l
2h
q3.
Let us write q1 = f(q2, q3) and q4 = g(q2, q3). We denote by D the following domain in R2
D :=
{
(q2, q3) ∈ (0, 1)2
∣∣ 0 < f(q2, q3) < 1, 0 < g(q2, q3) < 1}.
After sketching this domain, we realise that it is non-empty. We conclude that the class of all
martingale measures for M is a non-empty set, which can be represented as follows
M =
{
(q2, q3) ∈ D
∣∣ (h+r
2h
, 0, 0, h−r
2h
)
+ q2
(
−h+l
2h
, 1, 0,−h−l
2h
)
+ q3
(
−h−l
2h
, 0, 1,−h+l
2h
)}
.
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(c) (MATH3975) We assume that S0(1 + l) < K < S0(1 + h) and thus the call option can be
identified in our model with the following payoff
CT = ((1 + h)S0 −K, 0, 0, 0).
According to Proposition 2.2.5, an arbitrage price of any contingent claim X is given by the equality
π0(X) = EQ
(
(1 + r)−1X
)
where Q is an arbitrary martingale measure forM. Recall that we denote q1 = f(q2, q3). Therefore,
the set of arbitrage prices at time 0 for the call option is given by{
f(q2, q3)(1 + r)
−1((1 + h)S0 −K), (q2, q3) ∈ D}
or, more explicitly,{(
h+r
2h
− h+l
2h
q2 − h−l2h q3
)
(1 + r)−1
(
(1 + h)S0 −K
)
, (q2, q3) ∈ D
}
.
(d) (MATH3975) We now assume that r = 0. As before, we have that
CT = ((1 + h)S0 −K, 0, 0, 0)
and thus we search for α, β ∈ R such that
α+ β(1 + h) = (1 + h)S0 −K,
α+ β(1 + l) = 0,
α+ β(1− l) = 0,
α+ β(1− h) = 0.
It is obvious that no solution (α, β) exists since (1 + h)S0 − K > 0 and thus the option is not
attainable. Because any arbitrage price for CT at time 0 is equal to ((1 + h)S0 − K)q1 for some
value of q1, it suffices to find the lower and upper bounds for q1 when Q ranges over the class M.
• The lower bound for q1 equals 0, since for an arbitrarily small value of q1 there exists a
risk-neutral probability Q ∈M.
• The upper bound for q1 can be found by considering the situation when q2 and q3 are arbitra-
rily small. It is then easy to verify that the upper bound equals 0.5. Finally, one may check
directly that there is no martingale measure Q such that q1 ≥ 0.5. Indeed, for q1 = 0.5, we
obtain q2 = q3 = 0 and q4 = 0.5, and thus Q is not equivalent to P. If q1 > 0.5 then we get
q4 < 0 and thus Q is not a probability measure.
We conclude that q1 ∈ (0, 0.5) when Q ranges over the class M of all martingale measures. Hence
the set of all possible arbitrage prices for the call option in an extended arbitrage-free market model
is the open interval (0, c) where c = 0.5((1 + h)S0 −K).
Remark. For a slightly different approach, you may consult Example 2.2.4 from the course notes.
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