#### afterworkguinness

##### Active Member

Hi,

I thought the conditional probability of default (the probability of default in year T given survival up to that point) was hazard*e^(- hazard * time). Putting that to the test in GARPs practice question below yields a close but incorrect value.

Is the problem asking for conditional PD?

Is this method valid to solve for it, but yields only an approximation?

Hazard rate of 0.1. What is the probability of survival in the first year followed by default in the second?

(phrasing mine)

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%

P(default in year 2 given survival in year 1) = 18.127% - 9.516% = 8.61%

Trying my solution gives 0.1(e^(-0.1*2)) = 8.19%

Malz page 241 seems to support my approach.

I thought the conditional probability of default (the probability of default in year T given survival up to that point) was hazard*e^(- hazard * time). Putting that to the test in GARPs practice question below yields a close but incorrect value.

**Question:**Is the problem asking for conditional PD?

Is this method valid to solve for it, but yields only an approximation?

**Problem (Question # 10 GARP's 2015 practice exam):**Hazard rate of 0.1. What is the probability of survival in the first year followed by default in the second?

**Given solution:**(phrasing mine)

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%

P(default in year 2 given survival in year 1) = 18.127% - 9.516% = 8.61%

Trying my solution gives 0.1(e^(-0.1*2)) = 8.19%

Malz page 241 seems to support my approach.

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