Different approaches to caluculating Probability of Defaut

hsuwang

Member
Hello David,
I think there are different approaches to calculating PD in terms of formulas to use,
PD = CS/(1-RR) and 1-((1+risk free rate)/(1+yield)
but would they give the same value of PD? because when I tried to calculate the PDs with the following assumptions, I get different values:

Risk free rate: 5%
Prob. of Default: 2%
Recovery rate: 40%

I don't know if I'm doing this right, but I used the first formula to get the CS (spread) first, which turns out to be 1.2% (0.02*(1-.4)), and I add the spread onto the risk free rate to get the yield (6.2%), but when I plug this number into the second formula (1-(1.05/1.062)), I get a different value for PD. Maybe I'm thinking in the wrong direction?

Thanks!
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi Jack

You bring up a great point, the cirriculum indeed approaches (implied) PD estimation from at least two angles...I see two things straightaway:

1. Hull's approximation (i.e., PD = spread / (1 - recovery)) includes recovery whereas the Saunder's formula assumes zero recovery. So the apples-to-apples comparison, I think, would be to assume zero recovery. So, something like: 3% riskless & 5% risky = 2% spread and:
Hull's approximation = 2% / (1 - zero recovery) = 2% PD
and Saunders' = 1 - (1.03/1.05) = 1.9%

2. Hull's is an approximation (as evidenced by the fact that subsequently he shows a more precise method)

...but okay it still, for me, raises the question of "how to reconcile?" b/c I think there is an additional difference: the Saunders' method uses annual compounding and, I think or i would assume, that Hull uses continous compounding.

Saunders does give a formula that includes recovery (p 323), where he solves for spread
k - i = (1+ i) / (recovery + prob[repay] - recovery*prob[repay]) - (1+i), where i riskless and k = risky

...so I am glad you prompted me to thing about Hull's approximation...assuming it uses continuous compounding, then the continuous equivalent to Saunders would be, I think, set no-arbitrage:

riskless payoff = weighted average exp risky payoff
EXP(i) = p*EXP(k) + (1-p)*lambda*EXP(k); lambda = recovery
EXP(i) = EXP(k) * (p + (1-p) * lambda)
EXP(i-k) = p + (1-p)*recovery
EXP(i-k) = p + recovery - (p)(recovery)
EXP(i-k) = p * (1-recovery) + recovery
EXP(i-k) - recovery = p * (1-recovery)
p = (EXP(i-k) - recovery) / (1 - recovery)
...but I want PD so,

1-p = 1 - (EXP(i-k) - recovery) / (1 - recovery)
1 - p = (1-recovery)/(1-recovery) - (EXP(i-k) - recovery) / (1 - recovery)
1-p = (1 - recovery - EXP(i-k) + recovery) / (1-y)
1-p = (1-EXP(i-k)) / (1-y)

...that is what I get for the continuous solution to PD ... assuming I did that right, then indeed Hull's spread/(1-recovery) looks like an approximation because 1-EXP(i-k) is an approximation for spread; e.g., 1 - EXP(-2%) = about 2%.

...okay, sorry for detour (although I had never derived Hull's approximation, so I am grateful for the chance to understand this!), I think the Hull and Saunder's can be reconciled if:

1. we use recovery on both sides
2. we recognize difference btwn continous (Hull) and annual (Saunders), and
3. we see that spread/(1-recovery) is an (imprecise) approximation

hope that helps? (I will definitely spreadsheet this...)

David
 
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