Distance to Default

[Yun chiang Tai]

Hello David:
The formula of Distance to Default (p 51 in the 2008 Formula)
DD=(V-D)/(V*sigma_a)
1. I try to think them as comparing (V-D)/V with the multiple of sigma_a, does it right?
2. Also, how can we connect it to the one given in the chapter of DeServigny
DD=[ln(V/X)-(mu-0.5*sigma^2)(T-t)]/(sigma*sqrt(T-t))??
Thank you so much....
chris

 

[David Harper CFA FRM]

Chris,

This post from last month should be directly to your point. (this thread may be helpful too)

1. Yes, I think this is a good interpretation. (V - D) is a distance from the mean of distribution to the tail (where default occurs). To divide it by V*sigma_a is to treat it like a dollar standard deviation. But, also (V-D)/D converts into into a return-based (%) standard deviation which can be then normalized (i.e., divided by sigma_a)
2. Please see the post above with < 10 min briefcast. I think it is good to see the intuition here. The DD is essentially similar to d2 of N(d2) in the Black-Scholes. I think you just need to see the numerator as composed of two pieces: the "built-in" growth already implied by LN(V/X) and the rest which is an expected return depressed by one-half variance (i.e., a geometric average)

David

 

[Yun chiang Tai]

Hello David:
As the screencast you give and the lecture note you provide...
May I think the topic as follows...
Take the Taylor's expansion of ln(x) around x=1
ln(x)= ln(1)+(1/x)(x-1)+remaining term
therefore we have
ln(x)=1-(1/x)....
Use this one for V/D
ln(V/D)= 1-(D/V)=(V-D)/V....
(The first approximation we use is proper since we assume the lognormal dist. of asset price movement)
as you have shown in the screencast or the lecture...
the distance to default is considered as the multiple of the volatility of asset we consider
it gives DD=ln(V/D)/sigma_asset=[(V-D)/V]/sigma_asset
Do you think it is a nice expression for the DD??
Thank you...
Chris

 

[Yun chiang Tai]

Hello David:
The issue is inspred from the lecture(the screencast) you give...
therefore it is simple to use the other one
DD=ln(V/Default value)/sigma_asset ...Right??
thanks...
Chris

 

[David Harper CFA FRM]

Chris,

I don't see the application of Taylor series (you could be right, I just don't see it).
Did you see the new Merton model XLS I uploaded to the member page?
http://www.bionicturtle.com/premium/editgrid/frm_credit_merton_model_for_pd_lgd_and_debt_equity/

The thing about the Merton model, IMO, is that it requires no special math; e.g., it isn't option pricing. The d2 is quite intuitive. d2 becomes complicated when inserted into Black-Scholes

If you consider the Stulz' example (2nd column in the above spreadsheet). Firm (V) = $120, Default = $100, Asset return = 20% with asset volatility = 20%. Time = 5 years.

The beauty of Merton is only two parts:
1. LN(F/V) = LN(120/100) = 18.2%
what is this? today's distance to default! (a.k.a., the continous return already in the capital structure)
the stock would drop 18.2%, continuously, to breach the default
So, this alone is like/analogous to (V0-D)/D

2. the expected return on the asset over five years:
(20%-20%^2/2)*5 years = 90%
This part changes (1) from today to the end of five years

Add them up, and = 108.2%. That is analogous to (V+5 - D)/D

I say analogous b/c Merton is operating with RETURNS. To use returns enables the use of NORMSINV(); i.e., returns are normal but the levels are lognormal. It occurs to me that a confusion could arise between DD in $ terms or DD in return (%) terms. I prefer to follow Merton model in "coping with" this in return (%) terms because price levels (i.e., V - D) are lognormal but the returns and standardized return (i.e., return/volatility%*SQRT[T]) can be treated with the normal (e.g., NORMSINV)

So, in this way, Merton is simply really:
[current implied DD% + expected return over period]/standardized by scaled volatility = a DD that is expressed in standard normal units

Because this is (eventually) intuitive, I don't see any need to invoke special math. As I've mentioned here often, this DD = d2 and Merton PD = N(d2) in Merton but it *does* become mathematically complicated we jump into the risk-neutral world of Black-Scholes (which none of the above needs to do!). That N(d2) is not DD because it uses the riskless rate instead of the expected return; a subtle difference with profound theoretical differences..

David

 

[Yun chiang Tai]

Hello David:
Thank you for providing me an intuitive way for this...
chris