Merton Model - PD and LGD

[RomanS]

Hi David,
I saw your video on the Merton Model where you explain how PD can be derived.
A very good one! I have 2 questions relating to that video:

(1) Does the same logic carry over to the LGD formula? (I wasn't to transfer)
(2) mu - 0.5 x sigma² = drift but what is mu + 0.5 x sigma²?

Best Roman

 

[David Harper CFA FRM]

Hi Roman,
Thanks much appreciated!

(1) I myself cannot intuit Stulz' Merton-based LGD. He explains it as if it were an expected shorfall (loss condition on default) and the formula looks suspiciously like a BSM put. But, as paid members can verify (http://www.bionicturtle.com/how-to/spreadsheet/2011.t6.d.1.-stultz-merton/), the practical issue with Stulz' Merton-based LGD is that the outputs are too low (near to zero, often). GARP kindly removed the LGD from the AIMs at my request a couple/few years ago because the answers simply are not realistic (the lognormal/normal assumption of course doesn't help but it seems to be worse that that would explain)

(2) Right, you can use either due to the symmetry of the normal! (keep in mind, the d1 or DD is here a standardized RETURN which is normal, the asset price/firm value is lognormal).

So, Stulz happens to use in the numerator: mu + 0.5 x sigma² Then he finds PD with N(DD)
Whereas de Servigny more naturally uses: mu - 0.5 x sigma² in the numeration, and gets to the same PD with N(-DD)
... again paid members can see the verification in learning XLS http://www.bionicturtle.com/how-to/spreadsheet/2011.t6.d.1.-stultz-merton

I MUCH PREFER de Servigny (mu - 0.5 x sigma²) because you can memorize along with the BSM:
distance to default (DD) = d2 in BSM!
... only except we replace riskfree rate with firm asset drift (mu)
and PD = N(-d2)

so:
N(d2) in BSM uses a risk free rate is the probability the option will expire in the money; i.e., will be exercised in the risk-neutral world

N(-d2) as PD = 1 - N(d2), with mu replaced Rf, is the probably the "option" will expire out of the money (ie, firm will default)

Hope that is helpful, thanks,
David

 

[RomanS]

Hi David,
thanks for the comprehensive answer. Not for the first time...
I also prefer de Servigny (mu - 0.5 x sigma²) and will use that.
Thanks and regards, Roman