MA370 Study Guide - Quiz Guide: Compound Interest, Standard Deviation, Random Variable
MA 370 - Summer 2013
Assignment #4 (Optional) Solutions
1. In this problem we investigate continuously compounded stock returns
in the binomial model. To this end consider the N-period binomial
model with parameters u,dand p.
(a) Determine the mean and standard deviation of the one-period
continuously compounded return log(Sn+1/Sn).
The key here is to recall (July 15 lecture) that
Sn+1 =SnuZn+1 d1−Zn+1 ,
where
Zn+1 =(1 with probability p ,
0 with probability 1 −p .
Observe that Zn+1 is a binomial with parameters n= 1 and p, so
that its mean and variance are pand p(1 −p), respectively. Now
write the continuously compounded single-period return in terms
of Zn+1 as follows
log(Sn+1/Sn) = log(uZn+1 d1−Zn+1 )
= log(uZn+1 ) + log(d1−Zn+1 )
=Zn+1 log(u) + (1 −Zn+1) log(d)
= log(d) + [log(u)−log(d)]Zn+1 .
So the continuously compounded return is a linear function of the
random variable Zn+1, which allows us to calculate its mean and
variance as follows
E[log(Sn+1/Sn)] = E[log(d) + [log(u)−log(d)]Zn+1]
= log(d) + [log(u)−log(d)]E[Zn+1]
= log(d) + [log(u)−log(d)]p
=plog(u) + (1 −p) log(d),
and
Var(log(Sn+1/Sn)) = Var(log(d) + [log(u)−log(d)]Zn+1)
= [log(u)−log(d)]2Var(Zn+1)
= [log(u)−log(d)]2p(1 −p).
The standard deviation of the single-period return is therefore
SD(log(Sn+1/Sn)) = [log(u)−log(d)]pp(1 −p).
(b) Determine the mean and standard deviation of the m-period con-
tinuously compounded return log(Sn+m/Sn).
The trick here is to recall (July 15 again) that
Sn=S0uXndn−Xn,
where Xn=Pn
k=1 Zkand Z1, Z2, . . . are independent random vari-
ables such that
Zk=(1 with probability p ,
0 with probability 1 −p .
Recall that we (i) interpreted Zkas an indicator random variable,
telling us whether the stock moved up or down from period k−1to
period k, and (ii) noted that Xnhas a binomial distribution with
parameters nand p. Now observe that
Sn+m/Sn=S0uXn+md(n+m)−Xn+m/S0uXndn−Xn
=uXn+m−Xndm−(Xn+m−Xn).
Now Xn+m−Xn=Pm
k=n+1 Zkcounts the number of times the
stock moves up from period nto period n+m(over which there
are mtotal movements) and therefore has a binomial distribution
with parameters mand p. In particular the mean and variance
of Xn+m−Xnare mp and mp(1 −p), respectively. As in (a) the
trick is to write the continuously compounded m-period return as
a linear function of Xn+m−Xn, as follows
log(Sn+m/Sn) = log(uXn+m−Xndm−(Xn+m−Xn))
= (Xn+m−Xn) log(u) + [m−(Xn+m−Xn)] log(d)
=mlog(d) + [log(u)−log(d)](Xn+m−Xn).
We can now get the mean and variance of the return from that of
Xn+m−Xnas follows
E[log(Sn+m/Sn)] = E[mlog(d) + [log(u)−log(d)](Xn+m−Xn)]
=mlog(d) + [log(u)−log(d)]E[Xn+m−Xn]
=mlog(d) + [log(u)−log(d)]mp
=mp log(u) + m(1 −p) log(d)
=m[plog(u) + (1 −p) log(d)] ,
and
Var(log(Sn+m/Sn)) = Var(mlog(d) + [log(u)−log(d)](Xn+m−Xn))
= [log(u)−log(d)]2Var(Xn+m−Xn)
= [log(u)−log(d)]2mp(1 −p).
The standard deviation of the m-period return is therefore
SD(log(Sn+m/Sn)) = [log(u)−log(d)]pmp(1 −p).
It is interesting to note that
E[log(Sn+m/Sn)] = m·E[log(Sn+1/Sn)] ,
SD(log(Sn+m/Sn)) = √m·SD[log(Sn+1/Sn)] .
So the expected return over m= 10 days is just ten times the
expected daily return, while the standard deviation of the ten-day
return is √10 ≈3.16 times the daily standard deviation. Or, the
