Solutions 279
3.20By Condition 3.3 the random variablesS(1)/S(0) = 1+K(1) andS(2)/S(1) =
1+K(2) are independent, each taking three values 1 +d,1+nand 1 +uwith
probabilitiesp,qand 1−p−q,respectively. ThereforeS(2), which is the
product of these two random variables and the numberS(0), takes up to nine
values. Among these nine values there are only six different ones,S(2) =
S(0)(1 +u)^2 with probabilityp^2 ,
S(0)(1 +n)^2 with probabilityq^2 ,
S(0)(1 +d)^2 with probability (1−p−q)^2 ,
S(0)(1 +u)(1 +n) with probability 2pq,
S(0)(1 +u)(1 +d) with probability 2p(1−p−q),
S(0)(1 +n)(1 +d) with probability 2q(1−p−q).
3.21Letp∗,q∗, 1 −p∗−q∗be the probabilities of upward, middle and downward price
movements, respectively. Condition (3.6) implies that 0. 2 p∗− 0 .1(1−p∗−q∗)=
0, that is,q∗=1− 3 p∗and 1−p∗−q∗=2p∗. Observe thatp∗, 1 − 3 p∗, 2 p∗∈[0,1]
if and only ifp∗∈[0, 1 /3]. It follows thatp∗,q∗, 1 −p∗−q∗are risk-neutral
probabilities if and only ifq∗=1− 3 p∗andp∗∈[0, 1 /3].
3.22Solving the system of equations
ln(1 +u)=mτ+σ√
τ,
ln(1 +d)=mτ−σ√
τ,
we findσ∼= 0 .052 andm∼= 0 .059.
3.23Sincep=1/2andξ(n)^2 =τ,E(ξ(n)) =^12√
τ−^12√
τ=0,Var(ξ(n)) =E(ξ(n)^2 )−E(ξ(n))^2 =^12 τ+^12 τ=τ,
E(k(n)) =mτ+σE(ξ(n)) =mτ,
Var(k(n)) =σ^2 Var(ξ(n)) =σ^2 τ.
3.24By (3.2)S(1) =S(0)ek(1)=S(0)emτ+σξ(1),
S(2) =S(0)ek(1)+k(2)=S(0)e^2 mτ+σ(ξ(1)+ξ(2)).
3.25Lett=Nn. Because theξN(i) are independent,E(ξN(i)) = 0 and Var(ξN(i)) =
1
Nfor eachi=1,^2 ,..., it follows that
E(wN(t)) =E(ξN(1) +···+ξN(n))
=E(ξN(1)) +···+E(ξN(n)) = 0,
Var(wN(t)) = Var(ξN(1) +···+ξN(n))
=Var(ξN(1)) +···+Var(ξN(1)) =Nn=t.