282 Mathematics for Finance
4.8The put option gives the right (but no obligation) to sell one share of stock
for the strike priceX= 110 dollars at time 2. We consider an extended model
with three assets, the stock, the money market, and the option. The unit prices
of these assets areS(n),A(n),PE(n), wherePE(n) is the market price of the
put option at timen=0, 1 ,2. The time 2 price of the put option is
PE(2) = max{X−S(2), 0 }.
According to the Fundamental Theorem of Asset Pricing, the discounted stock
and option pricesS ̃(n)=S(n)/A(n) andP ̃E(n)=PE(n)/A(n) should form a
martingale under some probability measureP∗, or else an arbitrage opportu-
nity would arise. From Example 4.5 we know that there is only one probability
P∗turningS ̃(n) into a martingale. Because of this,P ̃E(n) must be a martin-
gale under the same probabilityP∗. It follows that
PE(1) =A(1)
A(2)
E∗(PE(2)|S(1)) and PE(0) =A(0)
A(1)
E∗(PE(1)).
Using the values ofP∗for each scenario found in Example 4.5, we can compute
PE(1) andPE(0).For example
PE(1,ω 3 )=PE(1,ω 4 )=AA(2)(1)P∗(ω^3 )P
E(2,ω 3 )+P∗(ω 4 )PE(2,ω 4 )
P∗(ω 3 )+P∗(ω 4 )
=^110121
1
25 ×20 +
1
1 100 ×^30
25 +
1
100
∼= 20. 00
dollars. In this manner, we obtain
Scenario PE(0) PE(1) PE(2)
ω 1 1. 96 1. 21 0. 00
ω 2 1. 96 1. 21 4. 00
ω 3 1. 96 20. 00 20. 00
ω 4 1. 96 20. 00 30. 00
Chapter 5
5.1In the first investment project
E(K 1 )=0. 12 × 0 .25 + 0. 12 × 0 .75 = 0. 12 ,
Var(K 1 )=(0. 12 − 0 .12)^2 × 0 .25 + (0. 12 − 0 .12)^2 × 0 .75 = 0.
In the second project
E(K 2 )=0. 11 × 0 .25 + 0. 13 × 0 .75 = 0. 125 ,
Var(K 2 )=(0. 11 − 0 .125)^2 × 0 .25 + (0. 13 − 0 .125)^2 × 0 .75 = 0. 000075.
Finally, in the third project
E(K 3 )=0. 02 × 0 .25 + 0. 22 × 0 .75 = 0. 17 ,
Var(K 3 )=(0. 02 − 0 .17)^2 × 0 .25 + (0. 22 − 0 .17)^2 × 0 .75 = 0. 0075.
The first project is the least risky one, in fact, it is risk-free. The third project
involves the highest risk.