MATH3075/3975 Financial Mathematics
Tutorial 9: Solutions
Exercise 1 Consider the CRR modelM = (B, S) with the horizon date T = 2, the risk-free
rate r = 0.1, and the following values of the stock price S at times t = 0 and t = 1:
S0 = 10, S
u
1 = 13.2, S
d
1 = 9.9.
Let X be a European contingent claim with the maturity date T = 2 and the payoff
X =
(
min(S1, S2)− 10
)+
.
(a) The stock price and the martingale measure are given by the following diagram
Suu2 = 17.424 P̃(ω1) = 1/9
Su1 = 13.2
1/3
77oooooooooooo
2/3
”OO
OOO
OOO
OOO
O
Sud2 = 13.068 P̃(ω2) = 2/9
S0 = 10
1/3
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2/3
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;;
;;
;;
;;
;;
;;
;;
;;
;;
Sdu2 = 13.068 P̃(ω3) = 2/9
Sd1 = 9.9
1/3
77oooooooooooo
2/3
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OOO
OOO
OOO
O
Sdd2 = 9.801 P̃(ω4) = 4/9
(b) To show that the claim X is a path-dependent, it suffices to observe that
X = (X(ω1), X(ω2), X(ω3), X(ω4)) = (3.2, 3.068, 0, 0).
Although we have the same value of the stock price S2(ω2) = 13.068 = S2(ω3), we have
different values of the payoff, depending on the level of the stock price at time 1 (note that
S1(ω2) = 13.2 6= 9.9 = S1(ω3)). Hence there is no function f : R→ R such that X = f(ST )
and thus X is a path-dependent claim (a special case of the lookback option).
1
(c) Our next goal is to compute the arbitrage price of X using the risk-neutral valuation
formula, for t = 0, 1, 2,
πt(X) = Bt EP̃
(
X
BT
∣∣∣Ft).
The price process πt(X) of the European claim X is given by
πuu2 (X) = 3.2 ω1
πu1 (X) = 2.8291
1/3
55kkkkkkkkkkkkkk
2/3
))SS
SSS
SSS
SSS
SSS
πud2 (X) = 3.068 ω2
π0(X) = 0.8573
1/3
;;xxxxxxxxxxxxxxxxxxxxxxx
2/3
##F
FF
FF
FF
FF
FF
FF
FF
FF
FF
FF
F
πdu2 (X) = 0 ω3
πd1(X) = 0
1/3
55kkkkkkkkkkkkkk
2/3
))SS
SSS
SSS
SSS
SSS
πdd2 (X) = 0 ω4
(d) We start by examining the replicating strategy (ϕ01, ϕ
1
1) at time t = 1.
• Assume first that the stock price has risen during the first period. Then we need to
solve
1.1ϕ̃01 + 17.424ϕ
1
1 = 3.2,
1.1ϕ̃01 + 13.068ϕ
1
1 = 3.068.
Hence (ϕ̃01, ϕ
1
1) = (2.4291, 0.0303) if the stock price has risen during the first period,
that is, for ω ∈ {ω1, ω2}. We check that
V u1 (ϕ) = ϕ̃
0
1 + ϕ
1
1S
u
1 = 2.4291 + 0.0303 · 13.2 = 2.8291 = π
u
1 (X).
• Let us now assume that the stock price has fallen during the first period. Then we need
to solve
1.1ϕ̃01 + 13.068ϕ
1
1 = 0,
1.1ϕ̃01 + 9.801ϕ
1
1 = 0.
Hence (ϕ̃01, ϕ
1
1) = (0, 0) if the stock price has fallen during the first period, that is, for
ω ∈ {ω3, ω4}. Obviously, V d1 (ϕ) = 0 = πd1(X).
2
Let us find the replicating portfolio at time t = 0. We need to solve
1.1ϕ00 + 13.2ϕ
1
0 = 2.8291,
1.1ϕ00 + 9.9ϕ
1
0 = 0.
Hence (ϕ00, ϕ
1
0) = (−7.7157, 0.8573) for all ωs. We check that
V0(ϕ) = ϕ
0
0 + ϕ
1
0S0 = −7.7157 + 0.8573 · 10 = 0.8573 = π0(X).
The dynamics of the portfolio are represented by the following diagram
V uu2 (ϕ) = 3.2
(ϕ̃01, ϕ
1
1) = (2.4291, 0.0303)
44iiiiiiiiiiiiiiii
**UUU
UUUU
UUUU
UUUU
U
V ud2 (ϕ) = 3.068
(ϕ00, ϕ
1
0) = (−7.7157, 0.8573)
77ppppppppppppppppppppppppppppp
”NN
NNN
NNN
NNN
NNN
NNN
NNN
NNN
NNN
NNN
V du2 (ϕ) = 0
(ϕ̃01, ϕ
1
1) = (0, 0)
44iiiiiiiiiiiiiiii
**UUU
UUUU
UUUU
UUUU
U
V dd2 (ϕ) = 0
Exercise 2 We apply the CRR call option pricing formula
C0 = S0
T∑
k=k̂
(
T
k
)
p̂k(1− p̂)T−k −
K
(1 + r)T
T∑
k=k̂
(
T
k
)
p̃k(1− p̃)T−k
where k̂ is the smallest integer k such that
k >
ln
(
K
S0dT
)
ln
(
u
d
) =: α0.
3
We consider the European call option with strike price K = 10 and maturity date T = 5
years. We assume that the initial stock price S0 = 9, the risk-free simple interest rate is
r = 0.01 and the stock price volatility equals σ = 0.1 per annum. The CRR parametrization
for u and d with ∆t = 1 gives
u = eσ
√
∆t = eσ = 1.105171, d =
1
u
= 0.904837.
Consequently,
p̃ =
1 + r − d
u− d
= 0.524938, p̂ =
p̃u
1 + r
= 0.574402.
(a) We first compute the call price at time 0. It is easy to check that α0 = 3.02680… and
thus k̂ = k̂(S0, T ) = 4 and thus
C0 = 9
5∑
k=4
(
5
k
)
p̂k(1− p̂)5−k −
10
(1.01)5
5∑
k=4
(
5
k
)
p̃k(1− p̃)5−k = 0.552247.
Hence the price of the option at time 0 equals C0 = 0.552247.
(b) We will now compute the prices of the option at time t = 1. Notice that the time to
maturity is now equal to T − 1 = 4.
• If S1 = uS0 = 9.9465 then we obtain the new value of k̂(uS0, T − 1) = 3 since αu1 =
2.02680… and thus in that case
Cu1 = 9.9465
4∑
k=3
(
4
k
)
p̂k(1− p̂)4−k −
10
(1.01)4
4∑
k=3
(
4
k
)
p̃k(1− p̃)4−k = 0.920649.
• If S1 = dS0 = 8.1435 then we obtain the value of k̂(dS0, T − 1) = 4 since αd1 = 3.02680…
(you may check that αd1 = α0) and thus
Cd1 = 8.1435
4∑
k=4
(
4
k
)
p̂k(1− p̂)4−k −
10
(1.01)4
4∑
k=4
(
4
k
)
p̃k(1− p̃)4−k
= 8.1435 (0.574402)4 −
10
(1.01)4
(0.524938)4 = 0.156793.
In Week 10, the price of the same option at any time t (in particular, for t = 1) will be
computed using the backward induction method.
(c) The hedge ratio at time 0 equals
Cu1 − Cd1
Su1 − Sd1
=
0.920649− 0.156793
9.9465− 8.1435
= 0.423659 = ϕ10
and thus ϕ00 can be computed from the equality
C0 = 0.552247 = ϕ
0
0 + ϕ
1
0S0 = ϕ
0
0 + 0.423659× 9.
We find that ϕ00 = −3.260657, that is, we need to borrow 3.260657 units of cash at time 0.
4
Exercise 3 (MATH3975) (a) It is clear that L0 = 1 and L is strictly positive. The mar-
tingale property of L under under Q̃ follows from the definition of Q̃ since L = cS/B where
c is a constant and, by assumption, the process S/B is a martingale respect to the filtration
F under the probability measure Q̃
(b) Let us denote M = B/S. It suffices to observe that the product LM = L(B/S) = B0/S0
is constant and thus the process LM is clearly is a martingale with respect to the filtration
F under the probability measure Q̃. From part (c) in Exercise 4 in Week 7, we deduce that
the process M = B/S is a martingale with respect to the filtration F under Q̂.
(c) Once again, from part (b) in Exercise 4 in Week 7, we know that the equality
EQ̃(Y | Ft) = (Lt)
−1 EQ̂(Y Ls | Ft). (1)
holds for any Fs-measurable random variable Y . Recall that the Radon-Nikodým density L
satisfies
Lt =
B0
S0
St
Bt
, Ls =
B0
S0
Ss
Bs
.
Since Ss is Fs-measurable, if X is any Fs-measurable random variable X, then the random
variable X/Ss is Fs-measurable as well.
Therefore, by applying (1) to the random variable Y = X/Ss, we obtain for
EQ̂
(
X
Ss
∣∣∣Ft) = (Lt)−1EQ̃
(
LsX
Ss
∣∣∣Ft)
or, more explicitly,
EQ̂
(
X
Ss
∣∣∣Ft) = S0
B0
Bt
St
EQ̃
(
B0
S0
Ss
Bs
X
Ss
∣∣∣Ft).
This in turn yields
St EQ̂
(
X
Ss
∣∣∣Ft) = Bt EQ̃
(
X
Bs
∣∣∣Ft). (2)
as was required to show.
(d) The payoff CT of the call option can be decomposed as follows
CT = ST1D −K1D = X1 −X2.
Since CT is assumed to be an attainable claim, its arbitrage price can be computed using
any martingale measure for the process S/B, that is, for the model M. Hence we may take
the martingale measure Q̃ ∈ M and compute the arbitrage price of the option using the
risk-neutral valuation formula
Ct = Bt EQ̃
(
CT
BT
∣∣∣Ft).
5
Using the additivity property of conditional expectation and part (c) (specifically, equation
(2) applied to X1), we obtain
Ct = Bt EQ̃
(
CT
BT
∣∣∣Ft) = Bt EQ̃
(
X1
BT
∣∣∣Ft)−Bt EQ̃
(
X2
BT
∣∣∣Ft)
(c)
= St EQ̂
(
X1
ST
∣∣∣Ft)−Bt EQ̃
(
X2
BT
∣∣∣Ft)
= St EQ̂
(
ST1D
ST
∣∣∣Ft)−Bt EQ̃
(
K1D
BT
∣∣∣Ft)
= St Q̂(D | Ft)−KBt(BT )−1 Q̃(D | Ft)
= St Q̂(D | Ft)−KB(t, T ) Q̃(D | Ft)
since K and BT are deterministic and, obviously,
EQ̂(1D | Ft) = Q̂(D | Ft), EQ̃(1D | Ft) = Q̃(D | Ft).
(e) It suffices to argue that the payoff X = 1 at time T can be replicated by investing
α = Bt/BT units of cash at time t in the asset B and keeping the portfolio constant till time
T . Then the wealth of the portfolio at time T will be αBT/Bt = 1. This means that the
arbitrage price B(t, T ) of the unit zero-coupon bond at time t satisfies B(t, T ) = Bt/BT .
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