MATH3075/3975 Financial Mathematics
Tutorial 10: Solutions
Exercise 1 Assume the CRR modelM = (B,S) with T = 3, the stock price S0 = 100, Su1 = 120, Sd1 = 90,
and the risk-free interest rate r = 0.1. We consider the American put option on the stock S with the
expiration date T = 3 and the constant strike price K = 121. The option has the reward process g(St, t) =
(K − St)+ = (121− St)+ for t = 0, 1, 2, 3,
(a) We first compute the arbitrage price P at of this option for t = 0, 1, 2, 3. We start by noting that the
unique risk-neutral probability measure P̃ satisfies
p̃ =
1 + r − d
u− d
=
(1 + r)S0 − Sd1
Su1 − Sd1
=
1 + r − d
u− d
=
1.1− 0.9
1.2− 0.9
=
2
3
.
Since u = 1.2 and d = 0.9, the stock price process St is given by
Suuu3 = 172.8
Suu2 = 144
77nnnnnnnnnnn
”OO
OOO
OOO
OOO
Su1 = 120
88qqqqqqqqqqq
&&MM
MMM
MMM
MMM
Suud3 = 129.6
S0 = 100
p̃= 2
3
88rrrrrrrrrrr
1−p̃= 1
3 &&LL
LLL
LLL
LLL
Sud2 = 108
77ooooooooooo
”OO
OOO
OOO
OOO
Sd1 = 90
88qqqqqqqqqq
&&MM
MM
MM
MM
MM
Sudd3 = 97.2
Sdd2 = 81
77ooooooooooo
”OO
OOO
OOO
OOO
Sddd3 = 72.9
We first compute the terminal payoff P aT from the American put option at expiry date T = 3. We
use the following convention for sample paths of the stock price: ω1 = (u, u, u), ω2 = (u, u, d), ω3 =
(u, d, u), ω4 = (u, d, d), ω5 = (d, u, u), ω6 = (d, u, d), ω7 = (d, d, u), ω8 = (d, d, d). Hence the terminal
payoff P aT = (121− ST )
+ can be represented as follows (notice that ω4 = (u, d, d) and ω5 = (d, u, u) so
ST (ω5) = 129.6 > 97.2 = ST (ω4))(
P aT (ω1), P
a
T (ω2), P
a
T (ω3), P
a
T (ω4), P
a
T (ω5), P
a
T (ω6), P
a
T (ω7), P
a
T (ω8)
)
= (0, 0, 0, 23.8, 0, 23.8, 23.8, 48.1).
The above representation of the payoff is formally correct but not very convenient when we use the
backward induction since then we start by “splitting” the model into four submodels corresponding
to: {ω1, ω2}, {ω3, ω4}, {ω5, ω6} and {ω7, ω8}, that is, to the partition generating the σ-field F2.
1
To compute the price P at at times t = 0, 1, 2 through the risk-neutral valuation, we use the backward
induction
P at = max
{
(K − St)+, (1 + r)−1 EP̃
(
P at+1 | Ft
)}
with the terminal condition P a3 = (K−S3)+ = (121−S3)+. Easy computations show that price process
P at can be represented by the following diagram
g(S3, 3) = (121− 172.8)+ = 0
0
55kkkkkkkkkkkkkkkkk
))SS
SSS
SSS
SSS
SSS
SSS
3.9394
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1
(b) The rational exercise times for the holder of the American put option are: τ∗0 = 0,
τ∗1 (ω) = 2 for ω ∈ {ω1, ω2, ω3, ω4},
τ∗1 (ω) = 1 for ω ∈ {ω5, ω6, ω7, ω8},
τ∗2 = 2 and τ
∗
3 = 3.
2
(c) Suppose that the option was sold for the price P a0 = 21 and it was not immediately exercised by its
holder. Then the issuer may establish the replicating portfolio for the European claim X = (3.9394, 31)
with maturity 1 by solving the following equations
1.1ϕ00 + 120ϕ
1
0 = 3.9394,
1.1ϕ00 + 90ϕ
1
0 = 31.
We find that (ϕ00, ϕ
0
1) = (101.9835,−0.90202) and thus the initial wealth, which is needed to establish
this portfolio at time 0, equals
V0(ϕ) = 101.9835− 0.90202× 100 = 11.7815.
Hence the difference 21−11.7815 is the net profit of the issuer at time 0. This argument can be extended
to any date t.
Exercise 2 We consider the CRR binomial model with the risk-free rate r = 0 and the following values of
the stock price S at times t = 0 and t = 1:
S0 = 100, S
u
1 = 120, S
d
1 = 90.
We examine the American call option with maturity date T = 3 and the following reward process
g(St, t) = (St −Kt)+
where the variable strike Kt satisfies
K0 = K1 = 100, K2 = 105, K3 = 110.
(a) We first compute the arbitrage price Xat of this option for t = 0, 1, 2, 3. The unique risk-neutral
probability measure P̃ satisfies
p̃ =
1 + r − d
u− d
=
(1 + r)S0 − Sd1
Su1 − Sd1
=
1− d
u− d
=
1− 0.9
1.2− 0.9
=
1
3
.
Recall that we use the following notation for sample paths: for the “upper half” of the model
ω1 = (u, u, u), ω2 = (u, u, d), ω3 = (u, d, u), ω4 = (u, d, d)
and for the “lower half” of the model
ω5 = (d, u, u), ω6 = (d, u, d), ω7 = (d, d, u), ω8 = (d, d, d).
Hence the terminal payoff CaT from the American call option at time T = 3 equals (notice that
ω4 = (u, d, d) and ω5 = (d, u, u) so ST (ω5) = 129.6 > 97.2 = ST (ω4))(
CaT (ω1), C
a
T (ω2), C
a
T (ω3), C
a
T (ω4), C
a
T (ω5), C
a
T (ω6), C
a
T (ω7), C
a
T (ω8)
)
= (62.8, 19.6, 19.6, 0, 19.6, 0, 0, 0).
As in the previous exercise, when using the backward induction method we start by considering four
submodels corresponding to: {ω1, ω2}, {ω3, ω4}, {ω5, ω6} and {ω7, ω8}, that is, to the partition gene-
rating the σ-field F2.
To compute the price Cat at times t = 0, 1, 2 through the risk-neutral valuation, we use the recursive
formula
Cat = max
{
(St −Kt)+, (1 + r)−1 EP̃
(
Cat+1 | Ft
)}
with the terminal condition Ca3 = (S3 −K3)+ = (S3 − 110)+.
3
Since u = 1.2 and d = 0.9, the stock price process St satisfies
Suuu3 = 172.8
Suu2 = 144
77nnnnnnnnnnn
”OO
OOO
OOO
OOO
Su1 = 120
88qqqqqqqqqqq
&&MM
MMM
MMM
MMM
Suud3 = 129.6
S0 = 100
p̃= 1
3
88rrrrrrrrrrr
1−p̃= 2
3 &&LL
LLL
LLL
LLL
Sud2 = 108
77ooooooooooo
”OO
OOO
OOO
OOO
Sd1 = 90
88qqqqqqqqqq
&&MM
MM
MM
MM
MM
Sudd3 = 97.2
Sdd2 = 81
77ooooooooooo
”OO
OOO
OOO
OOO
Sddd3 = 72.9
and thus the price process Xat of the American call option is given by
g(S3, 3) = (172.8− 110)+ = 62.8
39
44jjjjjjjjjjjjjjjjjj
**TTT
TTTT
TTTT
TTTT
TTT
20
::tttttttttt
$$J
JJ
JJ
JJ
JJ
J g(S3, 3) = (129.6− 110)
+ = 19.6
8.1185
::tttttttttt
$$J
JJ
JJ
JJ
JJ
J 6.5333
44jjjjjjjjjjjjjjjj
**TTT
TTTT
TTTT
TTTT
T
2.1778
::tttttttttt
$$J
JJ
JJ
JJ
JJ
JJ
g(S3, 3) = (97.2− 110)+ = 0
0
44jjjjjjjjjjjjjjjjjj
**TTT
TTT
TTT
TTT
TTT
TTT
g(S3, 3) = (72.9− 110)+ = 0
4
Recall that r = 0 and thus Bt = 1 for all t. It is easy to check that the process X
a is a strict
supermartingale under P̃ since the inequality Xat ≥ p̃Xaut+1 + (1 − p̃)Xadt+1 is satisfied at all nodes and
it is strict at some nodes. For instance, at time 1 when S1 = uS0 we obtain
Xa1 = 20 > (1/3)39 + (2/3)6.5333 = p̃X
au
2 + (1− p̃)X
ad
2 .
(b) The rational exercise decisions of the holder are given by
1
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1
Hence the rational holder should exercise the American call option at time t = 1 whenever the stock
price rises during the first period. Otherwise, he should not exercise the option till time 2. Hence the
rational exercise time τ∗0 is a stopping time τ
∗
0 : Ω→ {0, 1, 2, 3} given by
τ∗0 (ω) = 1 for ω ∈ {ω1, ω2, ω3, ω4},
τ∗0 (ω) = 2 for ω ∈ {ω7, ω8},
τ∗0 (ω) = 3 for ω ∈ {ω5, ω6}.
(c) We now take the position of the issuer of the option:
– At t = 0, we need to solve
ϕ00 + 120ϕ
1
0 = 20,
ϕ00 + 90ϕ
1
0 = 2.1778.
Hence (ϕ00, ϕ
1
0) = (−51.2888, 0.5941) for all ωs.
– If the stock price has risen during the first period, the option is exercised by its holder. Hence we
do not need to compute the strategy at time 1 for ω ∈ {ω1, ω2}.
– If the stock price has fallen during the first period, we need to solve
ϕ01 + 108ϕ
1
1 = 6.5333,
ϕ01 + 81ϕ
1
1 = 0.
Hence (ϕ01, ϕ
1
1) = (−19.599, 0.2420).
– If the stock price has fallen twice then the option is exercised and has value 0.
5
– If the stock price has fallen during the first period and has risen during the second period then
we need to solve
ϕ02 + 129.6ϕ
1
2 = 19.6,
ϕ02 + 97.2ϕ
1
2 = 0.
Hence (ϕ01, ϕ
1
1) = (−58.8, 0.6050).
We conclude that the replicating strategy ϕ = (ϕ0, ϕ1) is represented by the following diagram
V u1 (ϕ) = 20
(ϕ00, ϕ
1
0) = (−51.29, 0.59)
88ppppppppppppppppppppppp
%%K
KK
KK
KK
KK
KK
KK
KK
KK
KK
KK
KK
KK
(ϕ02, ϕ
1
2) = (−58.8, 0.61)
(ϕ01, ϕ
1
1) = (−19.60, 0.24)
44iiiiiiiiiiiiiiiii
**UUU
UUUU
UUUU
UUUU
UU
V dd2 (ϕ) = 0
Exercise 3 (MATH3975) 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 interest rate is r = 0.01 and the
stock price volatility equals σ = 0.1 per annum.
We use the CRR parametrization for u and d with ∆t = 1, that is, we set
u = eσ
√
∆t = 1.105171, d =
1
u
= 0.904837.
Consequently,
p̃ =
1 + r − d
u− d
= 0.524938.
Using the backward induction method, we obtain the following results.
(a) The price at time 0 of the European call option equals C0 = 0.5522. The detailed computations are
summarised on the next page.
(b) The price at time 0 of the European put option equals P0 = 1.0669.
(c) The put-call parity at time t = 0 reads
C0 − P0 = 0.5522− 1.0669 = −0.5147 = 9−
10
(1.01)5
= S0 −
K
(1 + r)T
.
(d) The price at time 0 of the American put option equals P a0 = 1.2112. The option should not be exercised
at time 0, but it should be exercised by its holder at time 1 if the stock price falls during the first
period. For the full description of the rational exercise decisions of the holder, see the foregoing pages.
6
Exercise 4 (MATH3975) Numerical results for Exercise 4 are given on the foregoing pages.
(a) Notice that the computations of the price X
g
t were done using the backward induction
X
g
t = min
{
h(St, t), max
{
`(St, t), (1 + r)
−1(p̃Xgut+1 + (1− p̃)Xgdt+1)}}
with the terminal condition X
g
T = πT (X
g) = `(ST , T ) = (K − ST )+.
(b) Rational exercise times τ∗0 and σ
∗
0 can be found either using the respective expressions
τ∗0 = inf
{
t ∈ {0, 1, . . . , T} |Xgt = `(St)
}
and
σ∗0 = inf
{
t ∈ {0, 1, . . . , T} |Xgt = h(St)
}
or when computing the price process by comparing the continuation value with respective cancellati-
on/exercise values for the issuer and holder.
7
Week 10
Exercise 3 Interest Volatility Stock at 0
0.01 0.1 9
up down tilde p
1.105171 0.904837 0.524938
Year 0 1 2
Stock 9 9.946538 10.99262
price 8.143537 9
7.368577
European 0.552247 0.920649 1.498351
call price 0.156793 0.301676
0
European 1.066904 0.583914 0.211627
put price 1.62306 1.007577
2.337325
Put-call parity -0.51466
American 1.211179 0.650276 0.233532
put price 1.856463 1.124461
2.631423
Exercise No No No
decision Yes No
Yes
Strike price
10
1 – tilde p
0.475062
3
12.14873
9.946538
8.143537
6.667364
2.357597
0.580436
0
0
0.011828
0.436858
1.659424
3.135597
0.011828
0.483428
1.856463
3.332636
No
No
Yes
Yes
Maturity
5
4 5
13.42642 14.83849
10.99262 12.14873
9 9.946538
7.368577 8.143537
6.03288 6.667364
5.458776
3.525432 4.838491
1.116781 2.148729
0 0
0 0
0 0
0
0 0
0.025146 0
0.90099 0.053462
2.532413 1.856463
3.86811 3.332636
4.541224
0 0
0.025146 0
1 0.053462
2.631423 1.856463
3.96712 3.332636
4.541224
No Yes
No Yes
Yes Yes
Yes Yes
Yes Yes
Yes
Week 10
Exercise 4 Interest Stock at 0 Strike K
0.05 25 27
up down tilde p 1 – tilde p alpha
1.1 0.9 0.75 0.25 0.02
Year 0 1 2 3 4 5 6
Stock 25 27.5 30.25 33.275 36.6025 40.26275 44.28903
price 22.5 24.75 27.225 29.9475 32.94225 36.23648
20.25 22.275 24.5025 26.95275 29.64803
18.225 20.0475 22.05225 24.25748
16.4025 18.04275 19.84703
14.76225 16.23848
13.28603
Upper 2.02 0.02 0.02 0.02 0.02 0.02 0.02
payoff 4.52 2.27 0.02 0.02 0.02 0.02
H_t 6.77 4.745 2.5175 0.06725 0.02
8.795 6.9725 4.96775 2.762525
10.6175 8.97725 7.172975
12.25775 10.78153
13.73398
Lower 2 0 0 0 0 0 0
payoff 4.5 2.25 0 0 0 0
L_t 6.75 4.725 2.4975 0.04725 0
8.775 6.9525 4.94775 2.742525
10.5975 8.95725 7.152975
12.23775 10.76153
13.71398
Game 2 0.02 0.009022 0.005964 0.001683 0.000408 6.43E-05
option 4.5 2.25 0.02 0.02 0.005847 0.001519
price 6.75 4.725 2.4975 0.06725 0.02
X^g_t 8.775 6.9525 4.94775 2.742525
10.5975 8.95725 7.152975
12.23775 10.76153
13.71398
Exercise L H N N N N N
decision L L H H N N
L L L H H
L L L L
L L L
L L
L
7 8 9 10 11 12
48.71793 53.58972 58.94869 64.84356 71.32792 78.46071
39.86012 43.84613 48.23075 53.05382 58.35921 64.19513
32.61283 35.87411 39.46152 43.40767 47.74844 52.52328
26.68322 29.35154 32.2867 35.51537 39.06691 42.9736
21.83173 24.0149 26.41639 29.05803 31.96383 35.16022
17.86232 19.64855 21.61341 23.77475 26.15223 28.76745
14.61463 16.07609 17.6837 19.45207 21.39728 23.537
11.95742 13.15316 14.46848 15.91533 17.50686 19.25755
10.76168 11.83785 13.02163 14.3238 15.75618
9.685512 10.65406 11.71947 12.89142
8.716961 9.588657 10.54752
7.845265 8.629791
7.060738
0.02 0.02 0.02 0.02 0.02 0.02
0.02 0.02 0.02 0.02 0.02 0.02
0.02 0.02 0.02 0.02 0.02 0.02
0.336777 0.02 0.02 0.02 0.02 0.02
5.188272 3.0051 0.60361 0.02 0.02 0.02
9.157677 7.371445 5.40659 3.245249 0.867774 0.02
12.40537 10.94391 9.336301 7.567931 5.622724 3.482996
15.06258 13.86684 12.55152 11.10467 9.513138 7.762451
16.25832 15.18215 13.99837 12.6962 11.26382
17.33449 16.36594 15.30053 14.12858
18.30304 17.43134 16.47248
19.17474 18.39021
19.95926
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0.316777 0 0 0 0 0
5.168272 2.9851 0.58361 0 0 0
9.137677 7.351445 5.38659 3.225249 0.847774 0
12.38537 10.92391 9.316301 7.547931 5.602724 3.462996
15.04258 13.84684 12.53152 11.08467 9.493138 7.742451
16.23832 15.16215 13.97837 12.6762 11.24382
17.31449 16.34594 15.28053 14.10858
18.28304 17.41134 16.45248
19.15474 18.37021
19.93926
0 0 0 0 0 0
0.00027 0 0 0 0 0
0.005572 0.001134 0 0 0 0
0.336777 0.02 0.004762 0 0 0
5.168272 2.9851 0.60361 0.02 0 0
9.137677 7.351445 5.38659 3.225249 0.847774 0
12.38537 10.92391 9.316301 7.547931 5.602724 3.462996
15.04258 13.84684 12.53152 11.08467 9.493138 7.742451
16.23832 15.16215 13.97837 12.6762 11.24382
17.31449 16.34594 15.28053 14.10858
18.28304 17.41134 16.45248
19.15474 18.37021
19.93926
L L L L L L
N L L L L L
N N L L L L
H H N L L L
L L H H L L
L L L L L L
L L L L L L
L L L L L L
L L L L L
L L L L
L L L
L L
L
MATH_3075_3975_sol_10.pdf
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MATH_3075_3975_sol_10n
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MATH3075_3975_s10.pdf