218 Mathematics for Finance
onebasis point, so here the rate drops by 100 basis points.) ThenB(6,12)∼=
0 .9692, which is more than in scenario 1. As a result, we are going to earn
more, achieving a logarithmic return of 4.13%.- The rate increases toy(6) = 8.26%.In this case the logarithmic return on
our investment will be 3.13%, which is lower than in scenario 1, the bond
price beingB(6,12)∼= 0. 9596.
We can see a pattern here: One is better off if the rate drops and worse off if
the rate increases. A general formula for the return on this kind of investment
is easy to find.
Suppose that the initial yieldy(0) changes randomly to becomey(n)=y(0)
at timen. Hence
B(0,N)=e−y(0)τN,B(n, N)=e−y(n)τ(N−n),and the return on an investment closed at timenwill be
k(0,n)=lnBB((0n,N,N))=lney(0)τN−y(n)τ(N−n)=y(0)τN−y(n)τ(N−n).We can see that the return decreases as the ratey(n)increases.The impact
of a rate change on the return depends on the timing. For example, ifτ= 121 ,
N=12andn=6,then a rate increase of 120 basis points will reduce the
return by 0.6% as compared to the case when the rate remains unchanged.
Exercise 10.2
Letτ = 121. Invest $100 in six-month zero-coupon bonds trading at
B(0,6) = 0.9400 dollars. After six months reinvest the proceeds in bonds
of the same kind, now trading atB(6,12) = 0.9368 dollars. Find the
implied interest rates and compute the number of bonds held at each
time. Compute the logarithmic return on the investment over one year.Exercise 10.3
Suppose thatB(0,12) = 0.8700 dollars. What is the interest rate after
6 months if an investment for 6 months in zero-coupon bonds gives a
logarithmic return of 14%?Exercise 10.4
In this exercise we take a finer time scale withτ = 3601 .(Ayearis
assumed to have 360 days here.) Suppose thatB(0,360) = 0.9200 dollars,
the rate remains unchanged for the first six months, goes up by 200 basis