Ch4 Q13: first approach

```Ch4 Q13: first approach
I
u = 3/2, d = 2/3, interest rate factor= β = 13/12, P0 = 1
I
The question doesn’t specify the real-world probabilities of the up- and
down-move. But for now let’s suppose they are both 1/2
I
Idea: find an adapted process Ln such that Ln Pn and Ln Rn are martingales,
where Pn and Rn are the processes of the stock and bond price
respectively. Note that Rn = β n R0
γ Rn
Try Ln = PPn0
= PPn0 β nγ , with L0 = 1
R0
I
I
Exercise: establish the below equations from the martingale properties
E(Ln+1 Rn+1 |Fn ) = Ln Rn and E(Ln+1 Pn+1 |Fn ) = Ln Pn
γ+1 1 2
13
1 3
+
=1
(1)
12
2 2
2 3
!
γ
+1
+1
13
1 3
1 2
+
=1
(2)
12
2 2
2 3
Ch4 Q13: first approach (cont)
I
Solving the equations would give = 0 and γ = −1. Hence Ln =
I
The price of an ATM call option maturing at n = 2 with payoff
V2 = (P2 − 1)+ is given by
2
1
12
E((P2 − 1)+ )
E(L2 V2 ) =
L0
13
144
1
9
=
&times; &times;
−1
169
4
4
45
=
169
12 n
13
Ch4 Q13: first approach (cont)
I
What if the real-world probability of the up-move now becomes 1/3??
I
The new set of equations derived from the martingale properties of Ln Rn
and Ln Pn now become (check!)
γ+1 1 3
2 2
13
=1
(3)
+
12
3 2
3 3
!
γ
+1
+1
2 2
13
1 3
+
=1
(4)
12
3 2
3 3
I
Exercise: from these equations, verify that and γ will satisfy
√
γ
= 9132
and 13
12
I
The price of the same call option is given by
P2
1
(P2 − 1)+
E(L2 V2 ) = E β 2γ
L0
P0
2γ
13
1
9
9
=
&times; &times;
&times;
−1
12
9
4
4
√ 2
√ 2
9 2
1
9
45
&times; &times;
−1 =
=
2 &times;
13
9
4
169
3 2
=
√
2
Ch4 Q13: second approach
I
Find the probability q of the up-move under which the discounted stock
price process is a martingale:
3
1
2
1
&times; + (1 − q) &times; &times; = 1
2
β
3
β
β − 2/3
13/12 − 2/3
q=
=
= 1/2
3/2 − 2/3
3/2 − 2/3
q&times;
I
Option price is given by the expected value of the discounted payoff under
this q-probability
V2
1
9
2
EQ
= 2 &times;q &times;
−1
β2
β
4
2 2 12
1
9
45
=
&times;
&times;
−1 =
13
2
4
169