1
$\begingroup$

I understood the cumulative (aka unconditional) probability of default to be the probability of defaulting in a given period eg: between years 1 and 5. Further $\pi_{cumulative} = 1-e^{-\lambda*t}$ where lambda is a hazard rate.

I understood the marginal (aka conditional) probability of default to be the probability of defaulting at time $T$ given survival up to that point. Further $\pi_{marginal} = \lambda e^{-\lambda*t}$ where lambda is a hazard rate.

Attempting to solve the following problem, I came up with a close but off value.

Problem

1 year hazard rate = 0.1. What is the probability of surviving in the first year followed by defaulting in the second?

My solution was to calculate the marginal probability of default = $0.1\lambda e^{0.1*2}$ = 8.19%

But the given answer was 8.61% arrived at by:

1 year cumulative (also called unconditional) PD = 1 - e^(- hazard*time) = 9.516%

2 year cumulative (also called unconditional) PD = 1 - e^(- hazard*time) = 18.127%

solution - 18.127% - 9.516% = 8.611%

Is my approach incorrect or merely an approximation?

$\endgroup$
3
$\begingroup$

The question sounds like a conditional probability problem. However, note that, for conditional probability, people will generally say if survived to or conditional on. Here it says that survived in year one and (i.e., followed by) will default in year two. Then we should not treat this as a conditional or marginal probability.

Based on the above understanding, the probability can be computed as follows: \begin{align*} P(\tau >1 \ and \ \tau \le 2) &= P(1 < \tau \le 2)\\ &=P\big((\tau \le 2) \setminus(\tau \le 1) \big)\\ &=P(\tau \le 2) - P(\tau \le 1)\\ &= \big(1- e^{-2\lambda}\big) - \big(1- e^{-\lambda}\big)\\ &= 18.127\,\% - 9.516\,\% \\ &= 8.611\,\%. \end{align*} Here, $\tau$ is the default time.

$\endgroup$
  • $\begingroup$ Interesting observation. So "survived in year 1 and defaulted in year 2" is not the same as "survived to year 2"? If so, is it because the default, in this scenario, is not necessarily at the end of year 2? $\endgroup$ – AfterWorkGuinness Nov 19 '15 at 0:54
  • 1
    $\begingroup$ No, defaulted in year 2 does not mean to survive to year 2; it means that default happens some time from year 1 to year 2. $\endgroup$ – Gordon Nov 19 '15 at 1:06
  • 1
    $\begingroup$ "defaulted in year 2" means that the default happens at a time $t$, where $1<t \le 2$. Here, first year refers to the time interval $[0, 1]$, while the second year refers to the time interval $[1, 2]$. $\endgroup$ – Gordon Nov 19 '15 at 1:14
  • 1
    $\begingroup$ No, it didn't say anything that might happened already, that is, it doesn't guarantee a default didn't happen already at some point. $\endgroup$ – Gordon Nov 19 '15 at 1:30
  • 1
    $\begingroup$ Yes, That is correct. That is why the probability is given by $P(1 < \tau \le 2)$. $\endgroup$ – Gordon Nov 19 '15 at 2:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.