Sankar Krishna
Keen member
The status of a ten-year zero-coupon bond is to be modelled using the two-state model with a
default intensity at time t given by:
lambda(t) = 1/(3+t) for 0<=t<10
(i) Describe how the probability of the bond defaulting changes over the life of the bond
The answer given as part of the solution is:
The default intensity is a decreasing function of time, so the bond is more likely to default earlier rather than later.
My doubt:
P(default) = 1 - exp( - integral 0 to t lambda(s) ds)
integral 0 to t 1/(3+s) ds = ln(3+t) - ln(3)
Therefor, P(default) = 1 - exp(-ln(3+t) +ln(3)) = 1 - exp{ ln(3/(3+t) } = 1 - 3/(3+t) = t/(3+t)
Hence for 0<=t<10, we get the following result;
t P(default)
0 0
1 0.25
2 0.4
3 0.5
4 0.571429
5 0.625
6 0.666667
7 0.7
8 0.727273
9 0.75
This shows increasing probability of default contrary to the solution given.
Kindly clarify.
Thanks in advance!
default intensity at time t given by:
lambda(t) = 1/(3+t) for 0<=t<10
(i) Describe how the probability of the bond defaulting changes over the life of the bond
The answer given as part of the solution is:
The default intensity is a decreasing function of time, so the bond is more likely to default earlier rather than later.
My doubt:
P(default) = 1 - exp( - integral 0 to t lambda(s) ds)
integral 0 to t 1/(3+s) ds = ln(3+t) - ln(3)
Therefor, P(default) = 1 - exp(-ln(3+t) +ln(3)) = 1 - exp{ ln(3/(3+t) } = 1 - 3/(3+t) = t/(3+t)
Hence for 0<=t<10, we get the following result;
t P(default)
0 0
1 0.25
2 0.4
3 0.5
4 0.571429
5 0.625
6 0.666667
7 0.7
8 0.727273
9 0.75
This shows increasing probability of default contrary to the solution given.
Kindly clarify.
Thanks in advance!