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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

share|improve this question
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. – David Nehme Dec 10 '13 at 6:56
Thanks. So, is (1) is the correct answer? – Jay Dec 13 '13 at 0:12
The exponential distribution has the memoryless property, (1) is correct. – jensa Dec 14 '13 at 19:20

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