MATH3075/3975 Financial Mathematics
Tutorial 8: Solutions
Exercise 1 (a) We first focus on the conditional risk-neutral probabilities. To compute them, we
consider the three embedded single-period two-state models. In each of these models, we may use
the single-period formula
p̃ =
1 + r − d
u− d
.
Since r = 0.1, we obtain
p̃1 =
11
10
− 3
5
7
5
− 3
5
= 5
8
, p̃2 =
11
10
− 6
7
10
7
− 6
7
= 17
40
, p̃3 =
11
10
− 2
3
4
3
− 2
3
= 13
20
.
S2 = 10 Q(ω1) = 1764
S1 = 7
p̃2=
17
40
99tttttttttt
1−p̃2= 2340
%%K
KK
KK
KK
KK
K
S2 = 6 Q(ω2) = 2364
S0 = 5
p̃1=
5
8
CC�����������������
1−p̃1= 38
��7
77
77
77
77
77
77
77
77
S2 = 4 Q(ω3) = 39160
S1 = 3
p̃3=
13
20
99tttttttttt
1−p̃3= 720
%%K
KK
KK
KK
KK
K
S2 = 2 Q(ω4) = 21160
We conclude that the unique risk-neutral probability measure Q for the model M = (B,S) sa-
tisfies
Q = (q1, q2, q3, q4) =
(
17
64
, 23
64
, 39
160
, 21
160
)
.
(b) We start by noting that the digital call option has the following payoff at time T = 2
X = h(S2) =
(
X(ω1), X(ω2), X(ω3), X(ω4)
)
= (1, 0, 0, 0).
1
To compute the replicating strategy for the claim X, we proceed by the backward induction.
• We first consider the replicating portfolio for X at time 1 on the event
A1 = {S1 = 7} = {ω1, ω2}.
Let φ̃0 stand for the amount of cash in the savings account and let φ1 be the number of shares
held. Then we need to solve the linear equations{
1.1φ̃0 + 10φ1 = 1,
1.1φ̃0 + 6φ1 = 0.
We find that (φ̃0, φ1) =
(
−15
11
, 1
4
)
. The value of this portfolio equals, on the event A1,
V1(φ) = −1511 +
1
4
7 = 17
44
.
• We now consider the replicating portfolio for X at time 1 on the event
A2 = {S1 = 3} = {ω3, ω4}.
By solving the linear equations {
1.1φ̃0 + 4φ1 = 0,
1.1φ̃0 + 2φ1 = 0,
we get φ̃0 = φ1 = 0 and thus the value of the portfolio equals V1(φ) = 0 on the event ω ∈ A2.
• Finally consider the replicating portfolio for V1(φ) at time 0. We now solve the equations{
1.1φ̃0 + 7φ1 = 17
44
,
1.1φ̃0 + 3φ1 = 0,
and we obtain (φ̃0, φ1) =
(
−255
968
, 17
176
)
. The value of this portfolio equals, for all ω ∈ Ω,
V0(φ) = −255968 +
17
176
5 = 425
1936
.
To summarise, the replicating strategy for the claim X and its wealth process are given by:
t = 0 t = 1
ω1, ω2 ω4, ω4
φ0t −
255
968
−150
121
0
φ1t
17
176
1
4
0
Vt(φ)
425
1936
17
44
0
This means that we first buy 17
176
shares of the stock at time 0. If the price of the stock rises during
the first period, then we adjust our portfolio at time 1 by purchasing, in addition, 1
4
− 17
176
= 27
176
shares of the stock. However, if the price of the stock declines during the first period, then we sell
all 17
176
shares purchased at time 0 and we pay back our debt with interest; we then end up with
null portfolio at time 1 (and thus, obviously, also at time 2).
2
The arbitrage price process of the digital call option coincides with the wealth process V (φ), so
it is given by:
π2(X) = 1 ω1
π1(X) =
17
44
17
40
77ooooooooooo
23
40
‘ ‘OO
OOO
OOO
OOO
π2(X) = 0 ω2
π0(X) =
425
1936
5
8
>>}}}}}}}}}}}}}}}}}}}
3
8
A
AA
AA
AA
AA
AA
AA
AA
AA
AA
A
π2(X) = 0 ω3
π1(X) = 0
13
20
77ooooooooooo
7
20
”OO
OOO
OOO
OOO
π2(X) = 0 ω4
(c) We now search for the arbitrage price process of an Asian option with the following payoff at
time 1
Y =
(
1
3
(
S0 + S1 + S2
)
− 4
)+
=
(
Y (ω1), Y (ω2), Y (ω3), Y (ω4)
)
=
(
10
3
, 2, 0, 0
)
.
To compute the arbitrage price of the Asian option, we may argue that the model is complete and
thus any contingent claim can be replicated. Hence the unique arbitrage price for the Asian option
can be computed by a direct application of the risk-neutral valuation formula
πt(Y ) = Bt EQ
(
Y
BT
∣∣∣Ft)
where Q is the unique risk-neutral probability measure, which was found in part (a). In fact, it
is better to rely on the backward induction, that is, the following relationship between πt(Y ) and
πt+1(Y )
πt(Y ) = Bt EQ
(
πt+1(Y )
Bt+1
∣∣∣Ft)
Within the two-period setup, it suffices to compute first the conditional expectation
π1(Y ) = (1 + r)
−1 EQ(Y | FS1 ) = (1 + r)
−1 EQ(Y |S1)
and subsequently the expected value
π0(Y ) = (1 + r)
−1 EQ(π1(Y )).
3
Using the conditional risk-neutral probabilities computed in part (a), we obtain the following re-
presentation for the arbitrage price of the claim Y
π2(Y ) =
10
3
ω1
π1(Y ) =
7
3
17
40
77ppppppppppp
23
40
”OO
OOO
OOO
OOO
π2(Y ) = 2 ω2
π0(Y ) =
175
132
5
8
??�������������������
3
8
��?
??
??
??
??
??
??
??
??
??
π2(Y ) = 0 ω3
π1(Y ) = 0
13
20
77ooooooooooo
7
20
”OO
OOO
OOO
OOO
π2(Y ) = 0 ω4
Exercise 2 (a) We first show explicitly that the contingent claim X is path-dependent. We have:
S0 = 80, S1 = (S
u
1 , S
d
1) = (104, 88) and
S2 = (S
uu
2 , S
ud
2 = S
du
2 , S
dd
2 ) = (135.2, 114.4, 96.8).
Hence for ω2 = (u, d) we get
X(ω2) =
(
S2(ω2)− S1(ω2)
)
1{S2(ω2)−S1(ω2)>20} =
(
114.4− 104
)
1{10.4>20} = 0
and for ω3 = (d, u) we get
X(ω3) =
(
S2(ω3)− S1(ω3)
)
1{S2(ω3)−S1(ω3)>20} =
(
114.4− 88
)
1{26.4>20} = 26.4.
Since X(ω2) ̸= X(ω3), we conclude that the claim is path-dependent.
(b) We will now compute the arbitrage price of the claim X using the risk-neutral valuation formula
πt(X) = Bt EP̃
(
XB−1T | Ft
)
, t = 0, 1, 2,
where P̃ is the unique equivalent martingale measure for the model M = (B,S). We have u =
104
80
= 1.3 and d = 88
80
= 1.1. Consequently,
p̃ =
1 + r − d
u− d
=
1.2− 1.1
1.3− 1.1
=
0.1
0.2
=
1
2
.
The claim X can be represented as follows:
X = (X(ω1), X(ω2), X(ω3), X(ω4)) = (31.2, 0, 26.4, 0).
4
Hence
πu1 (X) = π1(X)(ω1) = π1(X)(ω2) =
1
1.2
·
1
2
· 31.2 = 13,
πd1(X) = π1(X)(ω3) = π1(X)(ω4) =
1
1.2
·
1
2
· 26.4 = 11,
π0(X) =
1
1.2
·
1
2
· (13 + 11) = 10.
(c) We now find the replicating portfolio (φ0, φ1) for the claim X and check that the equality
Vt(φ) = πt(X) is satisfied for t = 0, 1, 2. For ω ∈ {ω1, ω2}, we need to solve
1.2 φ̃01 + 135.2φ
1
1 = 31.2,
1.2 φ̃01 + 114.4φ
1
1 = 0.
We get (φ̃01, φ
1
1) = (−143, 1.5) and thus V1(φ)(ω) = 13 for ω ∈ {ω1, ω2}. For ω ∈ {ω3, ω4}, we need
to solve
1.2 φ̃01 + 114.4φ
1
1 = 26.4,
1.2 φ̃01 + 96.8φ
1
1 = 0.
Here (φ̃01, φ
1
1) = (−121, 1.5) and thus V1(φ)(ω) = 11 for ω ∈ {ω3, ω4}. At t = 0, we solve
1.2φ00 + 104φ
1
0 = 13,
1.2φ00 + 88φ
1
0 = 11.
Hence (φ00, φ
1
0) = (0, 0.125). We check that V0(φ) = 0.125 · 80 = 10.
(d) We first show that in any CRR model
EP̃(S2 − S1) = r(1 + r)S0.
From the definition of P̃, we obtain
EP̃(S1) = (1 + r)S0, EP̃(S2) = (1 + r)
2S0.
Hence
EP̃(S2 − S1) = EP̃(S2)− EP̃(S1) = (1 + r)
2S0 − (1 + r)S0 = r(1 + r)S0.
Let us now consider the claim Y with maturity T = 2 given by
Y =
(
S2 − S1
)
1{S2−S1≤20}.
We wish to find the price of Y at time 0 using the additivity of arbitrage prices and the fact that
X + Y = S2 − S1. We have
π0(Y ) = EP̃
(
S2 − S1
(1 + r)2
)
− π0(X) =
r
(1 + r)
S0 − π0(X) =
0.2
1.2
80− 10 =
10
3
.
It remains to double-check this result by computing
π0(Y ) = B0 EP̃
(
Y
B2
)
.
5
We observe that Y = (0, 10.4, 0, 8.8) and thus we obtain
π0(Y ) = (1 + r)
−2 EP̃(Y ) = (1.2)
−2 0.25(10.4 + 8.8) =
10
3
.
(e) We need to find the unique probability measure P̂ on (Ω,F2) such that the process B̂t = Bt/St
for t = 0, 1, 2 is a martingale under P̂ with respect to the filtration F = (Ft)t=0,1,2. We may use the
following equalities
P̂(ω1, ω2) = p̂ =
p̃u
1 + r
=
13
24
, P̂(ω3, ω4) = 1− p̂ =
11
24
.
Hence
P̂ =
((
13
24
)2
,
13 · 11
(24)2
,
11 · 13
(24)2
,
(
11
24
)2)
.
We have
S0 EP̂
(
Y
S2
)
= 80
(
10.4
114.4
13 · 11
(24)2
+
8.8
96.8
(
11
24
)2)
=
10
3
.
Observe that this result is consistent with the computation performed in part (d).
Exercise 3 (MATH3975) (a) It is assumed that the process X is F-adapted has independent
increments with respect to F under P. We observe that since the process X is F-adapted, the process
Y is F-adapted as well since if Xt is Ft-measurable, then Xt + c is Ft-measurable for any constant
c. It thus suffices to consider the dates t and t + 1 for an arbitrary t = 0, 1, . . . and to show that
EP(Yt+1 | Ft) = Yt or, equivalently, EP(Yt+1 − Yt | Ft) = 0.
Since the random variable Xt+1 −Xt is independent of the σ-field Ft, it is clear that for any c ∈ R
the random variable Xt+1 −Xt + c is independent of Ft as well. Therefore, for any t = 0, 1, . . . ,
EP(Yt+1 − Yt | Ft) = EP
(
Xt+1 −Xt − EP(Xt+1) + EP(Xt) | Ft
)
= EP
(
Xt+1 −Xt − EP(Xt+1) + EP(Xt)
)
= EP(Xt+1)− EP(Xt)− EP(Xt+1) + EP(Xt) = 0.
This proves that the process Y is a martingale under P with respect to the filtration F.
(b1) Recall that A0 = 0 and for all t = 0, 1, . . .
At+1 −At = EP
(
Xt+1 −Xt | Ft
)
. (1)
We wish to show the process Ỹ satisfying Ỹt = Xt − At for t = 0, 1, . . . is a martingale under P
with respect to F. We have, for all t = 0, 1, . . . ,
EP
(
Ỹt+1 − Ỹt | Ft
)
= EP
(
Xt+1 −Xt − (At+1 −At) | Ft
)
= EP
(
Xt+1 −Xt | Ft
)
−
(
At+1 −At
) (1)
= 0.
Hence the process Ỹ is a martingale. It is worth noting that for every t = 0, 1, . . . the random
variable At+1 is Ft-measurable (this can be proven by induction). Moreover, if X is a process of
independent increments then At = EP(Xt) − EP(X0). Hence this case can be seen as a natural
extension of part (a).
6
(b2) On the one hand, we assume that Ŷ is a martingale under P with respect to F so that
EP
(
Ŷt+1 − Ŷt | Ft
)
= 0 for all t = 0, 1, . . . . On the other hand, Ŷt = Xt − Ât for all t = 0, 1, . . . and
thus
EP
(
Ŷt+1 − Ŷt | Ft
)
= EP
(
Xt+1 −Xt − (Ât+1 − Ât) | Ft
)
= EP
(
Xt+1 −Xt | Ft
)
−
(
Ât+1 − Ât
)
where the second equality holds since  is assumed to be F-predictable. We conclude that for all
t = 0, 1, . . .
Ât+1 − Ât = EP
(
Xt+1 −Xt | Ft
)
= At+1 −At.
Since A0 = Â0 = 0, we conclude that At = Ât for all t = 0, 1, . . . .
(c1) We need to prove that: a game X is fair ⇔ X is a martingale.
(⇐) Let us first assume that the process X is a martingale under P with respect to F. We fix t ∈
{1, 2, . . . , T} and we consider an arbitrary gambling strategy H. Using the properties of conditional
expectation and the fact that H is F-adapted, we obtain
EP(Gt) = EP
( t−1∑
u=0
Hu(Xu+1 −Xu)
)
=
t−1∑
u=0
EP
(
EP
(
Hu(Xu+1 −Xu) | Fu
))
=
t−1∑
u=0
EP
(
Hu EP
(
Xu+1 −Xu | Fu
))
=
t−1∑
u=0
EP(Hu · 0) = 0
since the martingale property of X means that EP
(
Xu+1−Xu | Fu
)
= 0 for every u = 0, 1, . . . , t−1
(⇒) We now assume that the game is fair. Let us fix t and let us consider a gambling strategy,
which equals zero for all u ̸= t and Ht = 1A for an arbitrary event A from Ft. Since the game is
fair, we have that, for any event A ∈ Ft,
EP(Gt) = EP
(
Ht(Xt+1 −Xt)
)
= EP
(
1A(Xt+1 −Xt)
)
= 0.
From the definition of conditional expectation, the last equality in turn implies that
EP
(
Xt+1 −Xt | Ft
)
= 0.
Since the date t was arbitrary, we conclude that X is a martingale under P with respect to F.
(c2) Let A be an F-predictable process such that
At+1 −At = EP(Xt+1 −Xt | Ft).
We now assume that a player is required to ‘pay’ at time t an upfront fee At+1 − At per one unit
of the bet (note that At+1 − At is not necessarily positive). Then the cumulative profits/losses
generated by a gambling strategy H by time t are given by the following expression
Gt =
t−1∑
u=0
Hu
(
Xu+1 −Xu − (Au+1 −Au)
)
=
t−1∑
u=0
Hu
(
Ỹu+1 − Ỹu
)
where we denote Ỹ = X−A. From part (b1), we know that the process Ỹ is a martingale and thus,
by part (c1), we obtain EP(Gt) = 0 for all t = 1, 2, . . . , T . Put another way, if the fee At+1 −At is
charged at time t per one unit of the bet then a general game X becomes a fair game Ỹ .
7