108156.pdf

278 Mathematics for Finance

3.14Given the continuous risk-free rate of 14% and the time stepτ=1/12, we

find the one-step return

r=e^0.^14 /^12 − 1 ∼= 0. 0117.

By Condition 3.2,u>r∼= 0 .0117. This means that the middle valueS(0)(1 +

u)(1 +d)ofS(2) must be no less than about 22(1 + 0.0117)(1− 0 .01)∼= 22. 04

dollars.

3.15Consider the system of equations

S(0)(1 +u)^2 =32,

S(0)(1 +u)(1 +d)=28,

S(0)(1 +d)^2 =x.

It follows that

32

28 =

1+u

1+d=

28

x

andx=28^2 /32 = 24.50 dollars. However, the tree cannot be reconstructed

uniquely. For any valueS(0)>0 one can finduanddconsistent with the

data.

3.16The values ofuanddcan be found by solving the equations

S(0)(1 +u)^2 = 121,

S(0)(1 +d)^2 = 100,

and selecting those solutions that satisfy 1+u>0 and 1+d>0. IfS(0) = 100,

thenu=0.1andd=0.IfS(0) = 104, thenu∼= 0 .0786 andd∼=− 0 .0194.

3.17We only need to consider the values ofdbetween−1andr,thatis,− 1 <d<

1 /10. Asdincreases between these two bounds,p∗decreases from 11/13 to 0.

The dependence ofp∗ondis shown in Figure S.6.

Figure S.6 The risk-neural probabilityp∗as a function ofd

3.18By (3.4) the conditiond<r<uis equivalent tod<p∗u+(1−p∗)d<u.This,

in turn, can be written as 0<p∗(u−d)<u−dor, equivalently, 0<p∗<1.

3.19By Proposition 3.5

E∗(S(3)|S(2) = 110) = 110(1 +r) = 110(1 + 0.2) = 132

dollars.