Conditional PD - GARP practice question # 10 (2015)

[afterworkguinness]

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.

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.

 

[Matthew Graves]

I think the difference is that your approach gives the probability of surviving for 2 years THEN defaulting but the question asks for the probability of defaulting at any point in year 2, given that you've survived year 1.

 

[afterworkguinness]

Thanks Matthew, I can see that now. These default probabilities might just be my undoing. I thought I had them locked in but I'm still struggling with them. Especially marginal PD .

 

[afterworkguinness]

@David Harper CFA FRM CIPM Looking at this again, I think there is a mistake in Garp's answer.
Assuming they are asking for the 2 year conditional PD, should they not divide by the 1 year probability of survival?

The question asks for the probability of survival in the first year followed by default in the second. They solve it as PD_2 year CUMULATIVE - PD_1 year. The answer given states "Then, the cumulative probability that the firm defaults in the second year is equal to 1-exp(-2*lambda) or 0.18127, and the conditional one year default probability given the firm survived the first year is the difference between the two year cumulative probability of default and the one year probability: 0.18127 - 0.09516 = .08611"

But in Malz (pg 241)...

"The difference between the two and one year default probabilities - the probability of the joint event of survival through the first year and default in the second - is 0.11989 (value from Malz example 7.4), The conditional one year default probability given survival through the first year, is the difference between the two probabilities divided by the one-year survival probability "

Secondly,
I thought the wording for the conditional probability had to be tighter?

Is it unreasonable to interpret "probability of survival in the first year followed by default in the second" as the cumulative (aka unconditional) probability?

 

[David Harper CFA FRM]

HI @afterworkguinness I agree with you that the answer is imprecise. The only assumption we are given is the hazard rate, which can safely be interpreted as the instantaneous (conditional) default probability such that:

• One year cumulative PD = 1 - exp(-0.10*1) = 9.516%, which under a constant hazard rate will equal each year's conditional PD; Two year cumulative PD = 1 - exp(-0.10*2) = 18.127%
• The unconditional PD in the second year = 18.127% - 9.516% = 8.611%. This is the same as (using Malz terms) the joint probability of survival in the first year followed by default in the second year; i.e., where survival in first year = exp(-0.10*1) = 90.4837%, this joint probability = 90.4837% * conditional PD of 9.516% = 8.611%. Another way to look at this is (per Hull) "the probability of default in the 2nd year as seen at time 0"; i.e., not assuming survival in the first year
• Then the conditional PD in the second year (and any year under constant hazard) = 8.611%/ (1 - 9.516%) = 9.516%; i.e., the probability of default in 2nd year conditional on survival through the end of the first year.

The question asks for "The probability of survival in the first year followed by a default in the second year is closest to:" which would be more semantically helpful if they wrote "The joint probability of survival in the first year followed by a default in the second year is closest to:" but, nonetheless in my opinion, is clearly asking for an unconditional probability such that the given answer of 8.61% is correct.

However, in the explanation, this sentence is incorrect: "the conditional one year default probability given that the firm survived the first year is the difference between the two year cumulative probability of default and the one year probability: 0.18127 - 0.09516 = .08611" as the conditional one year default probability given that the firm survived the first year is 9.516%. To answer your question, the solution text is definitely unreasonable (incorrect!). I hope that clarifies!

 

[Karim_B]

HI @afterworkguinness I agree with you that the answer is imprecise. The only assumption we are given is the hazard rate, which can safely be interpreted as the instantaneous (conditional) default probability such that:

• One year cumulative PD = 1 - exp(-0.10*1) = 9.516%, which under a constant hazard rate will equal each year's conditional PD; Two year cumulative PD = 1 - exp(-0.10*2) = 18.127%
• The unconditional PD in the second year = 18.127% - 9.516% = 8.611%. This is the same as (using Malz terms) the joint probability of survival in the first year followed by default in the second year; i.e., where survival in first year = exp(-0.10*1) = 90.4837%, this joint probability = 90.4837% * conditional PD of 9.516% = 8.611%. Another way to look at this is (per Hull) "the probability of default in the 2nd year as seen at time 0"; i.e., not assuming survival in the first year
• Then the conditional PD in the second year (and any year under constant hazard) = 8.611%/ (1 - 9.516%) = 9.516%; i.e., the probability of default in 2nd year conditional on survival through the end of the first year.

The question asks for "The probability of survival in the first year followed by a default in the second year is closest to:" which would be more semantically helpful if they wrote "The joint probability of survival in the first year followed by a default in the second year is closest to:" but, nonetheless in my opinion, is clearly asking for an unconditional probability such that the given answer of 8.61% is correct.

However, in the explanation, this sentence is incorrect: "the conditional one year default probability given that the firm survived the first year is the difference between the two year cumulative probability of default and the one year probability: 0.18127 - 0.09516 = .08611" as the conditional one year default probability given that the firm survived the first year is 9.516%. To answer your question, the solution text is definitely unreasonable (incorrect!). I hope that clarifies!

Hi @David Harper CFA FRM
I've also got these various PD definitions swimming around in my head, and after looking at our beloved De Laurentis I'm even more confused than when I started

This post is the closest I've found to explanations of what the various flavors of PD mean, and how to calculate them.

Is there another post of yours where you explain what all the various PDs listed in the Excel sheet here represent https://forum.bionicturtle.com/threads/p2-t6-710-merton-survival-time-and-z-spread.10598/#post-55132

Screenshot:


In the excel sheet I can see the calculations, but I'm trying to understand what they represent, and which words to look out for in the questions so I know which flavor of PD to use depending on what GARP throws our way on exam day.

Thanks for all your help!
Karim

 

[David Harper CFA FRM]

Hi @Karim_B Howabout https://forum.bionicturtle.com/threads/computing-default-probability.9504/ (per the tag)? We do have a lot of these threads ... @Nicole Seaman can you see what we've got, ayc, thanks?

 

[Sitemaze]

Trying to get the intuition behind conditional and unconditional PD.

Seems to me conditional PD is using the remaining surviving total for that year as the base whereas unconditional is using the initial starting total.