3
$\begingroup$

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

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ P(A∩B)=P(A)∗P(B) is not correct. It's P(A)*P(B|A). Suppose A is a coin toss of heads and B is a coin toss of tails. In that case $P(A \cap B) = 0$, but P(A)*P(B) = 0.25. $\endgroup$
    – David Nehme
    Dec 10, 2013 at 6:56
  • $\begingroup$ Thanks. So, is (1) is the correct answer? $\endgroup$
    – Jay
    Dec 13, 2013 at 0:12
  • 1
    $\begingroup$ The exponential distribution has the memoryless property, (1) is correct. $\endgroup$
    – jensa
    Dec 14, 2013 at 19:20

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.