Hi all.

I have one problem with calculate the probability of default. For calculation the probability of default I need of Default Point, but I don’t know how to calculate this point. I konw I using formula:

tDP = ttoday[date] + days_tDP

But I don’t know how I to calculate the Defult Point (number of days from today to Default Point).

I hope that someone help me!

Thank you very much!

hi david,
can u please tell me how to calculate PD with the help of Altman’s z score, i have some confusion relating it. can u please advice me whether i am calculating is correct or not, Firstly i am calculating z score from altman’s Z score method and then putting that Z value into recovery rate formula i.e.  recovery rate = 1/ (1+e-zi) here (-zi is an exponential function) watever value arrive from it can be subtracted from 1 to find PD. that’s how i calculate PD. Secondly please advice me what will be the range of PD’s showing this range is better and that range is bad. i’ll be really obliged if you answer my query by today cause i am in urgency. 

sabs

Hi Sabs,

If there is a standard translation of Altman’s-Z to a PD directly, I am not aware (I’ve read most of Altman’s new book and he himself does not offer this method). The Altman’s is a linear *discriminant:* it divides into classes without the accuracy of a PD. The classic “zone of ignorance” is 1.81 to 2.99; i.e., above that, predicts no default, below predicts default (But NOTE HE HAS PRIVATE FIRM and NON MANUFACTURER scales, so-called Z’ and Z’’).

It is common to map the Z-score to a rating. And of course you could map rating to PD. So, you could easily go Z > rating > PD. But i’ve never seen a direct map...?

the 1/(1+EXP(-PD)) is a logistic function. See Saunders Chapter 11. This is a common way to translate a linear combination into a bounded probabitliy [0,1]. Mathematically, nothing stops you from L = 1/[1+EXP(altman’s z)] and that will produce an L between [0,1] which happens to be higher if Z-score is higher. It is interesting and “directionally appropriate,” but I do not see the justification for this. I am not awares that Altman ought to be put to this service…

David

append: I mean, 1/[1+EXP(-altman’s Z)] would be “directionally appropriate” to (1-PD) or Saunders’ probability of repayment (p). For example, 1/[1+exp(-1.8)] = .85 so that PD = 15%. And, Z of 3 solves for PD of 5%. But, I can’t endorse this b/c I’ve not seen this done; i.e., logistic transform of altman’s Z

thanx alot david for your immediate reply i am really thankful to you, what do you think if i dont use Altman’s Z score then what should i do in calculating PD, if i dont have mean loss value what is the simple way to calculate PD, kindly let me know.

Sabs

Hi sabs,

I am not clear on the situation, what data do you have?
Have you tried just mapping to a rating? given a rating and a transition matrix, you can impute a PD
Or, per Saunders, if you have a credit spread, you can impute a PD this way…

David

hi david,

actually i want to calculate PD for obligor risk rating in the IRB approach of Basel II and the data i currently have is the financials of the obligor. Due to these reasons i was calculating PD through Altman’s Z score then Recovery rate and after that PD. Can you please tell me now how am i able to calculate PD in this circumstances.

Sabs

If you find a way to calculate you own PD and you can convince you regulator with it by using Altman Z it is allowed. But therefore you need data to proof that this is indeed a valid approach. Try to make an agreement with the regulators as soon as possible.

Hi david,

Can you please tell me whose PD model was discussed in chapter 4 of Servigny book title LGD in your notes. I want to know the practical implication of that model, the name of the person who has designed it and also the complete detail of this particular chapter of Servigny. kindly provie me this complete chapter I will be really grateful to you.

Sabs

Hi Sabs

The PD model in Chapter 3 is the Merton Model. The Black-Scholes approach was adapted by Robert Merton (in 1974). Now sees variety of applications under broad banner of contingent claims analysis. If you mean Ch 4, I don’t follow?

“kindly provide me this complete chapter” - I can’t, it would copyright violation. I have notes on the Chapter. Plus, I think the single chapter can be purchased at GARP

David

david,
I was talking about chapter 4 of servigny in your notes its on page no. 43 of your notes. i just want to know whose PD model was discussed there.

Sabs

Sabs,

Oh, sorry...GARP has several errors in the Credit risk AIMS, this among them. This first AIM should refer to Chapter 3 but it’s slotted in Chapter 4 (I debated placing in Ch 3, but I decided to match the Study Guide/AIMs)

So, it is still Chapter 3 and the PD model is still Merton.

Here is something that may help (and the exactly the kind of thing I will include in the cram, where ideas are especially linked):

The formula here for PD is N()
It is exactly the same as: N(d2) in the Black-Scholes(!)
The only difference, subtle but profound, is: the riskless rate in Black-Scholes (i.e., risk neutral pricing) is replaced by the expected return (> riskless rate) in the Merton.

Now you can see why it is Merton model?

PD = N() in merton model = N(d2) of B-S b/c
N(d2) is prob that the option will be struck; i.e., analogous to probabily firm value will breach default threshold.

David

hi david,

can you please tell me what does u(t-t) implies in your PD formula on page no 43 of your notes. It is understandable that T implies time but what does small t implies, and when it multiplies by t-t it gives 0 result in the end whereas you have used 5 as a value for small t in the eg given on your notes. for your information its on page 43 De servigny chapter 4: LGD of credit discipline notes. Please resolve this query soon as it’s urgent. 

sabs

sabs,

It should be (T-t). It is all just time, as shown in the example, 5 years is used for time. David

In an earlier post in this thread, David you say:

<....The formula here for PD is N()
It is exactly the same as: N(d2) in the Black-Scholes(!)

...>>

If I use d1 and d2 to refer to the expressions we countered in the Hull chapter on Mkt Risk, then the PD here is N(-d2) and not N(d2), correct?

--sridhar

sridhar,

Yes, but the thread happens to be referring to Stulz. They are the same; EITHER is correct.
You might look at this spreadsheet.
You’ll I showed both N(d2) and N(-d2). N(d2) is used by Stulz; N(-d2) by Hull.

But also, while the math is equivalent, the Merton uses an expected return (what is the firm’s expected growth?). The option pricing, per Hull, is using riskless rate (risk neutral pricing).

So to be careful,
“If I use d1 and d2 to refer to the expressions we countered in the Hull chapter on Mkt Risk, then the PD here is N(-d2) and not N(d2), correct?”

Mathematically, this is true, but the PD is using an expected return (r > riskless) and Hull because it is option pricing is using riskless rate.

David

hi david,

with refrence to my earlier posts regarding PD plz tell me how to calculate the value of meau in merton model in page 43 of your credit notes.

sabs