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Default time in structured product simulation

Thread starter #1
Goodmorning,

this is my first post on BT, i hope not to violate any of the forum rules.
As you can see in the post description, my question regard the modeling of default time laid out in the book FRM PART II Credit Risk Measurement and Management, Chapter 8 page 172.
What I'm struggling to understand is the definition of default time used in the book.
The author simulate N standard normal variables Z, find the CDF corresponding value and use it to derive the time (t) element from the cumulative default time distribution t = log(1- CDF(z))/Lambda. The meaning that the book attribute to this t is the EXACT time when the default occurs of the nth security, confirmed by the fact that based on this value the security default is allocated to one of the 5 years of the securitization life to calculate the cashflow waterfall.
The meaning that I attribute to this t is that the nth security has a 1-CDF(Z) probability of defaulting BEFORE time t, thus default could occur in any of the 5 years, leading to 5 possible different cashflows structures.
I'm struggling to understand this example.
Could anyone help me?
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
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#2
Hi @giuseppe.dattis Welcome! That part is interesting, I have too struggled with it. But isn't he just using inverse transformation to infer any default time in years? (although I get a bit flummoxed on the cumulative aspect ...).

The cumulative PD = 1 - exp(-λT); he's using λ = 2.25%. So the 10-year cumulative PD profile is:
2.22% (1 year), 4.40% (2 years), 6.53%, 8.61%, 10.64%, 12.63%, 14.57%, 16.47%, 18.33% (9 years), 20.15% (10 years)

As cumulative PD is a probability, he randomizes the cumulative PD as a normal CDF such that cum PD = N(Z) = 1 - exp(-λT); solving for T:
exp(-λT) = 1 - N(Z); take LN(.) both sides:
-λT = ln[1 - N(Z)];
T = -ln[1 - N(Z)]/λ, but call Z* a random Z deviate, so maybe best is:
T = -ln[1 - N(Z*)]/λ

So we randomize the Z values; e.g., his first row of random correlated Z values (is why they are suspiciously negative..) is:
−1.2625, −0.3968, −0.4285, −1.0258

so that the first simulated default time, with his λ = 2.25%, is:
T = -ln[1 - N(−1.2625)]/0.025 = 4.85 years cumulative. Except ... hmmm ... Malz seems to be translating between instantaneous λ and the 1-year conditional PD, which is constant every year; i.e., PD(1-yr conditional) = 1 - exp(-λ) so that λ = -ln(1 - PD)... I'm not immediately sure why that translation step, but it has minor effect. In any case he's got:
T = -ln[1 - N(−1.2625)]/-ln(1 - 0.025) = 4.795 years (versus the 4.85 years i might get)

But, maybe to your point, on reflection it doesn't trouble me that it's a function of a cumulative PD: consider the 10-year profile above and imagine 100 credits. Only 2.25% default within the first years, cumulatively 4.40% within two years, 20.15% within 10 years. If we randomize the cumulative normal, N(Z*), and we get 4.40%, that corresponds to default in 2 years. It's inverse transform, we're mapping, I think (?!), the cumulative probability to the cumulative time. Let me know ... thanks,
 
Thread starter #3
Hi David,
Thanks for your answer. Unfortunately I keep not understanding the process or better how the author can say default happens on year 4.85 instead of default will happen before 4.85 years from now. Thanks for your support
 
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