# Cumulative vs marginal probability of default

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?

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.
• "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 $[1, 2]$. Nov 19, 2015 at 1:14
• Yes, That is correct. That is why the probability is given by $P(1 < \tau \le 2)$. Nov 19, 2015 at 2:05