Exponential Hazard Rates and Conditional Default Risk

Summary

This note uses a constant hazard rate to model time to default with an exponential distribution. It gives the density and cumulative distribution, then derives the probability of default by a specified horizon. For the probability of default during the third year conditional on surviving the first two years, it subtracts cumulative default probabilities at the two endpoints and divides by survival through year two. Under this model, the conditional probability simplifies to 1 minus the exponential of negative one hazard-rate year.

The question also proposes multiplying the probability of surviving two years by an unconditional probability of default in the third year. That product represents a joint probability only if the second event is defined as default during years two to three unconditionally; it is not itself the conditional probability. The note poses the distinction but supplies no answer or broader credit-risk discussion. Its formulas rely on a constant hazard rate and continuous exponential timing.

Key ideas

• A constant hazard rate implies an exponential time-to-default distribution.
• Cumulative default probability by time t is one minus survival probability through t.
• Conditional default risk in a later interval divides the interval default probability by survival to its start.
• A joint probability and a conditional probability are different quantities.

Tags

# default probability


# default probability












Suppose the hazard rate is $\lambda$ the default probability density function follow exponential

$f(t) = \lambda e^{-\lambda t}$

and cumulative probability function is

$F(t) = 1 - e^{-\lambda t}$

the probability of default within 3 years is

$P(t<3) = F(3) = 1-e^{- 3 \lambda }$

and the conditional that it default in 3rd year given no default in the first 2 years is

$P(t<3|t>2) = \frac{P(t<3)-P(t<2)}{P(t>2)} = \frac{P(t<3)-P(t<2)}{1-P(t<2)} = \frac{e^{-2 \lambda}-e^{-3 \lambda}}{e^{-2 \lambda}} = 1 -e^{-\lambda} \hspace{0.05in} $ (1)

However, if I consider

- event A: no default in first 2 years

- event B: default in year 3

$P(A \cap B) = P(A) * P(B) ={[P(t<3)-P(t<2)]*}{P(t>2)} \hspace{0.65in}$ (2)

Which one is right for the default probability in year 3? (1) or (2), or neither

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.