250 Mathematics for Finance
when it has just one price independent of the state, namely the face value. We
are looking for a probabilityp∗such that
B(n, N;sn)=[p∗B(n+1,N;snu) + (1−p∗)B(n+1,N;snd)]
×exp{−τr(n;sn)}. (11.5)This equation can be solved forp∗, which in principle depends onn, Nandsn.
(As will turn out soon, see Proposition 11.2,p∗is in fact independent of ma-
turityN.) Recalling the definition of logarithmic returns, we have
B(n+1,N;sn+1)=B(n, N;sn)exp{k(n+1,N;sn+1)},which gives
p∗(n, N;sn)=exp{τr(n;sn)}−exp{k(n+1,N;snd)}
exp{k(n+1,N;snu)}−exp{k(n+1,N;snd)}. (11.6)
These numbers are called therisk-neutralormartingale probabilities. Condi-
tion (11.2) for the lack of arbitrage can now be written as
0 <p∗(n, N;sn)< 1.Example 11.7
We shall find the tree of risk-neutral probabilitiesp∗(n,3;sn)forn=0, 1 ,using
the data in Example 11.5 (the bond prices as shown in Figure 11.10).
First we compute the returns on the money market. The simplest way is
to use the yields (Figure 11.11). Withτ = 121 we haveτr(n;sn)=y(n, n+
1;sn)/ 12 ,n=0, 1 ,which gives the following values:
τr(1; u) = 0.521%
τr(0) = 0.995% <
τr(1; d) = 0.874%Next, we find the returnsk(1,3;s 1 )andk(2,3;s 2 ) on bonds. For example, if
s 2 = ud, thenk(2,3; ud) = ln(^00 ..^98759848 ).The results are collected below:
k(2,3; uu) = 0.58%
k(1,3; u) = 1.25% <
/k(2,3; ud) = 0.27%\ k(2,3; du) = 1.01%
k(1,3; d) = 0.84% <
k(2,3; dd) = 0.84%