Discrete Time Models Exercises
Exercise 6.1. The stochastic process f S
t1: t = 0; 1; 2 g represents the price evolution of one share of a riskless asset. The second stochastic process f S
t2: t = 0; 1; 2 g corresponds to the risky asset, S
t2being the price of one share at time t.
! (S
01(!) ; S
02(!)) (S
11(!) ; S
12(!)) (S
21(!) ; S
22(!))
!
1(1; 1) (1; 1; 1) (1; 2; 1)
!
2(1; 1) (1; 1; 1) (1; 2; 2)
!
3(1; 1) (1; 1; 2) (1; 3; 1)
!
4(1; 1) (1; 1; 2) (1; 3; 2)
!
5(1; 1) (1; 1; 2) (1; 3; 3)
The …rst investor holds at time t = 0 the portfolio = (3; 4) constituted of 3 shares of the riskless asset and 4 shares of the risky asset. The second investor has at the same time the portfolio = ( 2; 9). The second investor o¤ers to the …rst one the possibility to swap the portfolios at time t = 2, that is, if the value V (2) at time t = 2 of the …rst investor’s portfolio is smaller than the second investor’s portfolio, V (2), the …rst investor will swap the portfolios and realized a gain of V (2) V (2) ; if not, the …rst investor will keep is own portfolio and gain no advantage of the swap option.
a) What is the …ltered probability space describing this problem?
b) What are the martingale measures?
c) Justify why these measures are also called risk-neutral measures.
d) Verify that this market model admits no arbitrage opportunities.
e) Determine for each probability measure found in c), the option price.
f) Explain why the option price is not unique.
Exercise 6.2. We observe the price evolution of three assets during two periods. The
…rst stochastic process, S
(1), represents the price of the riskless asset while the two others
processes, S
(2)and S
(3)characterized the price evolution of two risky assets.
! S
0(1); S
0(2); S
0(3)S
1(1); S
1(2); S
1(3)S
2(1); S
2(2); S
2(3)!
1(10; 9; 12) (11; 9; 12) (12; 9; 11)
!
2(10; 9; 12) (11; 9; 12) 12; 11;
1459= 16; 111
!
3(10; 9; 12) (11; 12; 16) 13; 12;
122147= 25; 979
!
4(10; 9; 12) (11; 12; 16) (13; 12; 16)
!
5(10; 9; 12) (11; 12; 16) 13; 15;
122147!
6(10; 9; 12) (11; 12; 16) (13; 15; 16)
a) What is the …ltration representing the information available to all investors ?
b) Does a martingale measure exists in this model? If yes, specify all martingale measures.
If not, explain why.
c) Does the market model admits arbitrage opportunities? Why ? d) Are all contingent claims accessible? Justify.
Exercise 6.3. Do Exercise 5.2 again.
Solutions
1 Exercise 6.1
a) The fundamental set is = f !
1; !
2; !
3; !
4; !
5g ; the algebra is F = the set of all possible events of and the …ltration is F = fF
i: i = 0; 1; 2 g where
F
0= f? ; g ;
F
1= ff !
1; !
2g ; f !
3; !
4; !
5gg and F
2= F :
b) The martingale measures must satisfy for all t 2 f 1; 2 g ; E
QS
t2S
t1jF
t 1= S
t2 1S
t1 1which is equivalent to
8 t 2 f 1; 2 g ; E
QS
t2S
t1S
t2 1S
t1 1jF
t 1= 0:
Therefore, we search for ( Q (!
1) ; :::; Q (!
5)) = (q
1; :::; q
5) such that
0 = 1
1; 2 1 1; 1
q
1q
1+ q
2+ 2 1; 2
1 1; 1
q
2q
1+ q
2, 1
1; 2 1
1; 1 q
1+ 2 1; 2
1
1; 1 q
2= 0
0 = 1
1; 3 2 1; 1
q
3q
3+ q
4+ q
5+ 2 1; 3
2 1; 1
q
4q
3+ q
4+ q
5+ 3
1; 3 2 1; 1
q
5q
3+ q
4+ q
5, 1
1; 3 2
1; 1 q
3+ 2 1; 3
2
1; 1 q
4+ 3 1; 3
2
1; 1 q
5= 0
0 = 1
1; 1 1 (q
1+ q
2) + 2
1; 1 1 (q
3+ q
4+ q
5)
1 = q
1+ q
2+ q
3+ q
4+ q
5which is equivalent to solve the linear system 2
6 6 6 4
1 1:2
1 1:1
2 1:2
1
1:1
0 0 0
0 0
1:31 1:12 1:32 1:12 1:33 1:121
1:1
1
1:111
1:121
1:121
1:121
1 1 1 1 1
3 7 7 7 5
2 6 6 6 6 4
q
1q
2q
3q
4q
53 7 7 7 7 5
= 2 6 6 4
0 0 0 1
3 7 7 5
under the constraints that each probability q
imust be between 0 and 1.
Note. If you solve the linear system by hand, it is preferable to use a backward recursion in the tree. Indeed, in that case, we solve many smaller linear system.
There are in…nitely many solutions : depending on the choice of the free variable, the solutions are
! Q Q Q
!
1 119 119 119!
2 1109 1109 1109!
30 < q
3<
2207 2207 q24 1104+ q
5!
4 11072q
30 < q
4<
1107 110152q
5!
5q
3+
1104 22015 q24 1104< q
5<
22015c) Under any martingale measures found in b), The expected return of any risky asset is
the riskless interest rate. Indeed, E
QS
12S
02S
02= 0 (q
1+ q
2) + 1 (q
3+ q
4+ q
5)
= 1 q
3+ 7
110 2q
3+ q
3+ 4
110 = 11 110 = 1
10 E
QS
11S
01S
01= 1
10
8 ! 2 f !
1; !
2g ; E
QS
22S
12S
12jF
1(!) = 0 q
1q
1+ q
2+ q
2q
1+ q
2= 9 110
110 99 = 1
11 E
QS
21S
11S
11jF
1(!) = 1; 2 1; 1 1; 1 = 1
11 8 ! 2 f !
3; !
4; !
5g ; E
QS
22S
12S
12jF
1(!) = 1 2 2
q
3q
3+ q
4+ q
5+ 2 2 2
q
4q
3+ q
4+ q
5+ 3 2 2
q
5q
3+ q
4+ q
5= 1
2 q
3110 11 + 1
2 q
3+ 4 110
110 11 = 2
11 E
QS
21S
11S
11jF
1(!) = 1; 3 1; 1 1; 1 = 2
11
d) We have shown in b) that there are in…nitely many martingale measures. Theorem 2.7 of Harrison and Pliska says the there is no arbitrage opportunity if and only if there is at least one martingale measure.
e)
! (S
21; S
22) V (2) V (2) C =
max (V (2) V (2) ; 0)
!
1(1; 2; 1) 3 1; 2 + 4 1
= 7; 6
2 1; 2 + 9 1
= 6; 8 0
!
2(1; 2; 2) 3 1; 2 + 4 2
= 11; 6
2 1; 2 + 9 2
= 15; 6 4; 0
!
3(1; 3; 1) 3 1; 3 + 4 1
= 7; 9
2 1; 3 + 9 1
= 6; 4 0
!
4(1; 3; 2) 3 1; 3 + 4 2
= 11; 9
2 1; 3 + 9 2
= 15; 4 3; 5
!
5(1; 3; 3) 3 1; 3 + 4 3
= 15; 9
2 1; 3 + 9 9
= 24; 4 8; 5
E
QC
S
21= 0 1:2
9
11 + 4:0 1:2
9 110 + 0
1:3 q
3+ 3:5 1:3
7
110 2q
3+ 8:5
1:3 q
3+ 4 110
= 40 12
9
110 + 35 13
7
110 2q
3+ 85
13 q
3+ 4 110
= 15 22 + 15
13 q
3= 0; 68162 + 1; 1538q
3:
f ) We …nd in…nitely many values because the contingent claim is not attainable, that is, there is no self-…nancing investment strategy ! such that V
2! = C. Indeed,
0 = C (!
3)
= V
2!
; !
3=
12(!
3) S
21(!
3) +
22(!
3) S
22(!
3)
= 1; 3
12(!
3) +
22(!
3) )
22(!
3) = 1; 3
12(!
3) and
3; 5 = C (!
4)
= V
2! ; !
4=
12(!
4) S
21(!
4) +
22(!
4) S
22(!
4)
=
12(!
3) S
21(!
4) +
22(!
3) S
22(!
4)
= 1; 3
12(!
3) + 2
22(!
3)
= 1; 3
12(!
3) 2; 6
12(!
3)
= 1; 3
12(!
3) )
12(!
3) = 35
13 and
22(!
3) = 1; 3
12(!
3) = 35
10 :
But
V
2! ; !
5= 1; 3
12(!
3) + 3
22(!
3)
= 1; 3 35
13 + 3 35 10
= 2 35 10
= 7
6
= 8; 5:
Therefore, the result obtained is not in contradiction with the Corollary of page 228 of J.
M. Harrison and S. R. Pliska, ”Martingales, stochastic integrals and continuous trading”.
2 Exercise 6.2
a)
F
0= f? ; g
F
1= ff !
1; !
2g ; f !
3; !
4; !
5; !
6gg F
2= the set of all possible events of :
b) Let q
i= Q [!
i] ; i 2 f 1; 2; 3; 4; 5; 6 g : Since F
0= f? ; g ; E
Q"
S
1(2)S
1(1)jF
0#
= S
0(2)S
0(1), 9
11 (q
1+ q
2) + 12
11 (q
3+ q
4+ q
5+ q
6) = 9 10 : For the other risky asset,
E
Q"
S
1(3)S
1(1)jF
0#
= S
0(2)S
0(1), 12
11 (q
1+ q
2) + 16
11 (q
3+ q
4+ q
5+ q
6) = 12 10 : Since F
1= f? ; f !
1; !
2g ; f !
3; !
4; !
5; !
6g ; g
E
Q"
S
2(2)S
2(1)jF
1#
(!
1) = S
1(2)(!
1) S
1(1)(!
1) , 9
12 q
1q
1+ q
2+ 11 12
q
2q
1+ q
2= 9 11
, 9
12 9
11 q
1+ 11 12
9
11 q
2= 0:
and
E
Q"
S
2(2)S
2(1)jF
1#
(!
3) = S
1(2)(!
3) S
1(1)(!
3) , 12
13
q
3+ q
4q
3+ q
4+ q
5+ q
6+ 15 13
q
5+ q
6q
3+ q
4+ q
5+ q
6= 12 11 , 12
13 12
11 q
3+ 12 13
12
11 q
4+ 15 13
12
11 q
5+ 15 13
12
11 q
6= 0:
For the second asset, we get
E
Q"
S
2(3)S
2(1)jF
1#
(!
1) = S
1(3)(!
1) S
1(1)(!
1) , 11
12 q
1q
1+ q
2+
145 9
12 q
2q
1+ q
2= 12 11 , 11
12 12 11 q
1+
145 9
12 12
11 q
2= 0 and
E
Q"
S
2(3)S
2(1)jF
1#
(!
3) = S
1(3)(!
3) S
1(1)(!
3) ,
1221 47
13
q
3+ q
5q
3+ q
4+ q
5+ q
6+ 16 13
q
4+ q
6q
3+ q
4+ q
5+ q
6= 16 11 ,
1221 47
13 16
11 q
3+ 16 13
16 11 q
4+
1221 47
13 16
11 q
5+ 16 13
16
11 q
6= 0:
Since P
6i=1
q
i= 1, it su¢ ces to solve the linear system
M 2 6 6 6 6 6 6 4
q
1q
2q
3q
4q
5q
63 7 7 7 7 7 7 5
= 2 6 6 6 6 6 6 6 6 6 4
9 10 12 10
0 0 0 0 1
3
7 7
7 7
7 7
7 7
7 5
where
M = 2 6 6 6 6 6 6 6 6 6 6 6 4
9 11
9 11
12 11
12 11
12 11
12 11 12
11
12 11
16 11
16 11
16 11
16 11 9
12 9 11
11 12
9
11
0 0 0 0
0 0
1213 1211 1213 1211 1513 1211 1513 121111 12
12 11
145 9
1 12
12
11
0 0 0 0
0 0
122147 131 1611 1613 1611 122147 131 1611 1613 16111 1 1 1 1 1
3 7 7 7 7 7 7 7 7 7 7 7 5
According to the chosen free variable, the solution is,
Q (!
1)
22091 22091 22091 22091Q (!
2)
22063 22063 22063 22063Q (!
3) q
32 0;
1109 1109q
4 25 7952256q
5q
6 25 7953372Q (!
4)
1109q
3q 2 0;
1109q
5 51 590291 10 3182193q
6Q (!
5)
25 7952256q
3 51 590291+ q
4q 2
51 590291;
25 7952256 1255q
6Q (!
6)
25 7953372+ q
3 219310 318
q
4 1255
q
5q
62
25 7953372;
10 3182193Therefore,
Q (!
1) = 0; 41364 = 0; 41364 Q (!
2) = 0; 28636 = 0; 28636 Q (!
3) q
32 (0; 0; 081818) 0; 081818 q
4Q (!
4) 0; 081818 q
3q
42 (0; 0; 081818) Q (!
5) 0; 087459 q
30; 0056406 + q
4Q (!
6) 0; 13072 + q
30; 21254 q
4Q (!
1) = 0; 41364 = 0; 41364
Q (!
2) = 0; 28636 = 0; 28636
Q (!
3) 0; 087459 q
5q
60; 130 72
Q (!
4) q
50; 0056406 0; 212 54 q
6Q (!
5) q
52 (0; 0056406; 0; 087459) 0; 21818 q
6Q (!
6) 0; 21818 q
5q
62 (0; 130 72; 0; 212 54)
Graphs showing the constraint on q
iq
3q
40.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 -0.5
0.0 0.5 1.0
x y
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 -0.5
0.0 0.5 1.0
x y
0.02 0.04 0.06 0.08 0.10
0.0 0.1 0.2
x y
0.05 0.10 0.15 0.20
-0.1 0.0 0.1 0.2
x y
q
5q
60.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 -0.5
0.0 0.5 1.0
x y
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 -0.5
0.0 0.5 1.0
x y
0.1 0.2 0.3 0.4
-0.2 0.0 0.2
x y
0.1 0.2 0.3
-0.1 0.0 0.1 0.2
