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!
With a decreasing default intensity, the force with which the bond is compelled to default is reduced. So it's more likely that the bond will default during the first year (P(1) = 0.25) than during the 8th year (P(8)-P(7)=0.027273). It is in this sense that "the bond is more likely to default earlier rather than later" The numerical values you've listed are the cumulative probabilities, which by definition must increase. So the probability of defaulting by the end of the first year (P(1)=0.25) is less than the probability of defaulting by the end of the 8th year (P(8)=0.727273). Hope that helps.
