Computing default probability

[Kavita.bhangdia]

Hello David,

I am finding it little difficult to comprehend the way the words are drafted in problems related to computation of default probability.

1. The 10 yr bond pays annual coupons and probability of default is 2%. What is the probability that the bond pays three coupons and default at the end of three years.

I did 1-(.98*.98*.98) . I assumed that if the bond pays three coupons, it has to survvive for three years and then default.

but the solution was .98*.98*.02

2. The probability of default of a bond is 2 %. what is the probability that the bond defaults within 3 years.

That would be sum of

1. default in 1st year
2. survival in 1st year * default in 2nd year
3. Survival in 2 years and default in 3rd year..

Is that correct.?

Thanks
Kavita

 

[QuantMan2318]

Hi @Kavita.bhangdia , your solution answers Question #2, in fact the probability of Bond defaulting within 3 years, which is the three sums that you pointed above, is nothing but 1 - the probability of not at all defaulting in the three years (1-(0.98*0.98*0.98))

In fact, survival in two years and defaulting in the third year is 0.98*0.98*0.02. Lets see, Malz says that default is only the failure to redeem the principal amount of the Bond, I don't think he includes the payment of coupons in it. In fact, if I remember correctly, for him, default event is the value of Equity being eroded by the losses of the Bond issuer company and thus becoming zero/negative. For simplicity's sake, I think he assumes that the coupons are paid. ( I have to check this out )

 

[David Harper CFA FRM]

Hi @Kavita.bhangdia

This can be confusing. If we use your discrete (annual) probabilities and we assume the one-year conditional PD = 2.0%, then

• as @QuantMan2318 already noted, the 3-year cumulative pd = 1 - (1-0.02)^3 = 5.88%; i.e., the probability of default within three years
• The unconditional probability of default during the third year (as seen at time zero) is equal to 98%^2*2% = 1.92%; i.e., the probability of survival over the first two years then default in the third. Note this 3rd year unconditional pd can also be retrieved by: 5.88% - 3.96% = 1.92%, where 1.92% is the 2-year cumulative pd = 1-(1-0.02)^2=1.92%; i.e., 3rd year unconditional PD = (3-year cumulative pd) - (2-year cumulative pd). Why? Because both are from time zero, and if the bond doesn't not default in the first two years, but does default, it must be in the third year!
• The conditional pd remains 2.0% and can be retrieved (albeit this is circular but just to understand) by: (3rd year unconditional PD)/(1 - 2-year cumulative PD) = 1.92%/(1 - 3.96%) = 2.00%.

Another way to look at this, from time zero, is:

• The probability of the bond surviving all three years is 98%^3 = 94.12%
• The 3-year cumulative PD is simply the other outcome (a default), so it is 1-94.12%=5.88%
• The 5.88% can be parsed into the three unconditional PDs: 1st year 2.0% + 2nd year 1.96% + 3rd year 1.8824% = 5.88%. So that's a breakdown of our "time zero perspective" and the four possible outcomes which sum to 100%: default year 1, default year 2, default year 3 or survive all the way.

It's instructive to run these with continuous default probabilities, where the hazard rate (aka, default intensity) is the instantaneous conditional pd. I hope this helps!

 

[Delo]

Please also refer David's

 

[Karim_B]

Hi @David Harper CFA FRM
Your post and video above are very useful thanks.

The one piece I'd still need a hand with please is the difference between Marginal PD as found in the Excel sheet from here:
https://forum.bionicturtle.com/threads/p2-t6-710-merton-survival-time-and-z-spread.10598/#post-55132

Where:
Marginal PD = Hazard Rate * e^(-Hazard Rate * T)
Row 9 in my Excel sheet below.

As opposed to in the video above where:
Marginal PD is another name for Conditional PD = Unconditional PD(T) / Cumulative Survival(T-1)
Row 16 in my Excel sheet below.

I've attached an Excel sheet which is based on the one I linked above, and you can see the values in row 9 & 16 differ for the 2 Marginal PD calcs. (It's still a work in progress, so sorry about that.)

Also, from what point in time is the perspective for the row 9 Marginal PD?

Thanks!
Karim

 

[Karim_B]

Hi @David Harper CFA FRM
Your post and video above are very useful thanks.

The one piece I'd still need a hand with please is the difference between Marginal PD as found in the Excel sheet from here:
https://forum.bionicturtle.com/threads/p2-t6-710-merton-survival-time-and-z-spread.10598/#post-55132

Where:
Marginal PD = Hazard Rate * e^(-Hazard Rate * T)
Row 9 in my Excel sheet below.

As opposed to in the video above where:
Marginal PD is another name for Conditional PD = Unconditional PD(T) / Cumulative Survival(T-1)
Row 16 in my Excel sheet below.

I've attached an Excel sheet which is based on the one I linked above, and you can see the values in row 9 & 16 differ for the 2 Marginal PD calcs. (It's still a work in progress, so sorry about that.)

Also, from what point in time is the perspective for the row 9 Marginal PD?

Thanks!
Karim

Hi @David Harper CFA FRM
I'm feeling pretty dense, but I'm still not getting it

Aside from my question above, in the GARP 2018 P2 Question 6 we got another PD question and I thought "Unconditional Probability is a probability that does not take into account any other information, knowledge, or evidence" (from https://www.quora.com/Whats-the-difference-between-conditional-and-unconditional-probability [In my desperation I looked on the web too.]), so when they ask "What is the probability that the bond survives for 3 years and then defaults during Year 4?" in my mind it meant that you knew about the 3 year survival, so it was a Conditional probability. i.e. it's not asking for the probability of default in year 4 without knowing anything else (although I guess if you're defaulting in year 4, you must have survived the first 3 years. Argh! I don't get it).

I had a look in the forum but couldn't find a Conditional vs Unconditional Probability for Dummies post

I feel like I'm wasting your time now, but would appreciate your thoughts before I give up on the topic (at least for the exam on the 19th).

Thanks
Karim

Full question & answer below for ease of searching:

6. A risk analyst at a mid-size hedge fund is evaluating the credit risk of several trade positions. The hedge
fund specializes in corporate debt and runs a strategy that utilizes both relative value and long-only trades
using CDS and bonds. One of the new trades at the hedge fund is a B-rated long bond valued at
JPY 10 billion. Some of the hedge fund’s newest clients, including the B-rated bond holders, are restricted
from withdrawing their funds for four years. The analyst is currently evaluating the impact of various
default scenarios to estimate future asset liquidity. The analyst has estimated that the conditional
(marginal) probability of default of the B-rated bond is 7.7% in Year 1; 7.1% in Year 2; 6.6% in Year 3; and
6.1% in Year 4. What is the probability that the bond survives for 3 years and then defaults during Year 4?

A. 4.9%
B. 5.7%
C. 6.1%
D. 6.9%

Correct answer: A

Explanation:
A is correct. The probability that the bond survives for 3 years and then defaults in Year 4 can be
modeled as a Bernoulli trial given by the following equation, where MP stands for marginal probability:

P (Default at end of Year 4) = (1 – MP Year 1 default)*(1 – MP Year 2 default) * (1 – MP Year 3
default)* (MP Year 4 default) = (1 – 0.077)*(1 – 0.071)*(1 – 0.066)*(0.061) = 0.0489 = 4.9%.

B is incorrect. It is the probability that the bond defaults in Year 3.
C is incorrect. 6.1% is the marginal default probability in Year 4.
D is incorrect. The probability is incorrectly derived: 6.9% = [1 – (1-0.077)(1-0.071)(1-0.066)(1-
0.061)]1/4; or 6.9% = [(1+0.077)(1+0.071)(1+0.066)(1+0.061)]1/4 – 1.

garp18-p2-6

 

[David Harper CFA FRM]

Hi @Karim_B Right, I think I totally understand. Thank you for copying it wholesale. (We already identified GARP's previous error and the fixed the more basic flaw, but they didn't quite listen to me on how to best improve the question .... here is my original feedback complete with calculations/synonyms https://forum.bionicturtle.com/thre...s-2017-part-2-practice-exam.10510/#post-50919). I have already tried to explain to GARP why this question is strictly correct (after the fix) but remains open to misinterpretation and can be improved.

The question is strictly correct because the "and" suggests it is looking for a joint probability; but I think the question would save many headaches if it simply were edited to the following:

• What is the joint (aka, unconditional) probability that the bond survives for 3 years and then defaults during Year 4?

i.e., the "and" is essential, but the "then" is actually subtly suggestive of conditional!

In the default probability context (and this seems to comport at least with the first answer I see on your quora thread):

• Unconditional PD = joint PD and is consistent with Hull's definition (emphasis mine): "We will refer to this as the unconditional default probability. It is the probability of default during the third year as seen today." ... the "as seen today" is meaningful! Unconditional is not conditional on the survival for 3 years (eg). In the PD context, it is not so much the lack of other information but the lack of an assumption about the default/survival in the initial periods.

 

[Karim_B]

Hi @Karim_B Right, I think I totally understand. Thank you for copying it wholesale. (We already identified GARP's previous error and the fixed the more basic flaw, but they didn't quite listen to me on how to best improve the question .... here is my original feedback complete with calculations/synonyms https://forum.bionicturtle.com/thre...s-2017-part-2-practice-exam.10510/#post-50919). I have already tried to explain to GARP why this question is strictly correct (after the fix) but remains open to misinterpretation and can be improved.

The question is strictly correct because the "and" suggests it is looking for a joint probability; but I think the question would save many headaches if it simply were edited to the following:

• What is the joint (aka, unconditional) probability that the bond survives for 3 years and then defaults during Year 4?

i.e., the "and" is essential, but the "then" is actually subtly suggestive of conditional!

In the default probability context (and this seems to comport at least with the first answer I see on your quora thread):

• Unconditional PD = joint PD and is consistent with Hull's definition (emphasis mine): "We will refer to this as the unconditional default probability. It is the probability of default during the third year as seen today." ... the "as seen today" is meaningful! Unconditional is not conditional on the survival for 3 years (eg). In the PD context, it is not so much the lack of other information but the lack of an assumption about the default/survival in the initial periods.

Thanks @David Harper CFA FRM
I think I finally understand the "as seen today".

So Conditional (aka Marginal) PD is a combination of assumed certainty and probability:

• e.g. We assume we are certain you haven't defaulted in the first 3 years, and are calculating the probability of you defaulting in the 4th given we know you didn't default in the first 3 years.
• Possible outcomes: You can default in year 4 or any subsequent year, but NOT in years 1-3.
• Words to look out for on the exam: Conditional (hopefully), Marginal (can be ambiguous)

Unconditional = Joint PD is a combination of different probabilities:

• e.g. We think you probably haven't defaulted in the first 3 years, and are calculating the probability of you defaulting in the 4th given we don't know what happened in the first 3 years.
• Possible outcomes: You can default in any year. i.e. Year 1, 2, 3, 4, ... , N
• Words to look out for on the exam: Unconditional, Joint, And

Edit: Maybe I should re-phrase my explanation above to say "Assumed Certainty" rather than Certainty. Fixed above.

Is that correct?

Thanks
Karim

 

[rana.nadeem]

Hi David

The below snapshot is a table given by Jon Gregory, citing example to calculate historical default probability from rating transition matrix,




He gave the following table as default probabilities of different rating categories for various years:



I understand that the table of PDs given shows cumulative default probabilities for the given years, but I could not grasp the calculation for probability of default, can you please help?

Thanks
Nadeem

 

[pascalb]

Perhaps I was not clear enough above -

What is the difference between the question mentioning " The analyst has estimated that the conditional (marginal) probability of default..." thus given the video above I would have thought C. 6.1% which is p(def4/surv3) as per video. Yet, the answer for marginal dist is different, it mentions:
A is correct.... where MP stands for marginal probability:

P (Default at end of Year 4) = (1 – MP Year 1 default)*(1 – MP Year 2 default) * (1 – MP Year 3

but in the video MP (or w call it cond prob), the math was different....