I am trying to build a model based on Gibson's paper. In particular I would like to replicate the results he gets in Table 1. I have been able to derive the conditional default distribution, but can get the Unconditional default distribution. I wonder if anyone can kindly give me a breakdown of what I need to do. My understanding is that he applied eqn (6) on page 6 to the results, but when I try that I get very different values.Grateful for any assistance.

The formula you use is the right one. You need to integrate the unconditional default probabilities over M. g(M) is the standard normal density. If you apply this correctly, you should get the same results.

That's odd, DarkQuant. I could match Gibson's conditional probabilities exactly, but not his unconditional probabilities. There is something funny going on with the way he normalizes the factor density before he integrates the factor out of the distribution.

At least for zero defaults (first line of table 1), I could get his results. This is my Matlab code:>> p_0_given_M=inline('(1-(normcdf((x-sqrt(rho)*m)/sqrt(1-rho)))).^100')p_0_given_M = Inline function: p_0_given_M(m,rho,x) = (1-(normcdf((x-sqrt(rho)*m)/sqrt(1-rho)))).^100>> rho=.3rho = 0.3000>> x=norminv(1-exp(-.01))x = -2.3282>> m=-1m = -1>> q_i(m,rho,x)ans = 0.1863>> m=0m = 0>> q_i(m,rho,x)ans = 0.7635>> m=1m = 1>> q_i(m,rho,x)ans = 0.9711So, the conditional zero default probabilities are correct.For the unconditional ones:>> int_fct_for_p_0=inline('(1-(normcdf((x-sqrt(rho)*m)/sqrt(1-rho)))).^100.*normpdf(m)')int_fct_for_p_0 = Inline function: int_fct_for_p_0(m,rho,x) = (1-(normcdf((x-sqrt(rho)*m)/sqrt(1-rho)))).^100.*normpdf(m)>> quad(int_fct_for_p_0,-5,5,[],.0001,rho,x)ans = 0.6439So, this should be okay.

Mosi, Anuj Thanks. I was doing something stupid, but I have sorted it out and I am able to get the results as in Gibson's paper. I am using EXCEL-VBA so it isn't a perfect match but I just need it for illustrative purposes. These are the results I get for the conditional default probabilities using the recursive method he suggested in the paper.-2.000 -1.000 0.000 1.000 2.0000.001 0.189 0.766 0.971 0.9980.005 0.318 0.205 0.028 0.0020.020 0.264 0.027 0.000 0.0000.050 0.145 0.002 0.000 0.0000.090 0.059 0.000 0.000 0.0000.130 0.019 0.000 0.000 0.0000.154 0.005 0.000 0.000 0.0000.155 0.001 0.000 0.000 0.0000.134 0.000 0.000 0.000 0.0000.103 0.000 0.000 0.000 0.0000.070 0.000 0.000 0.000 0.0000.043 0.000 0.000 0.000 0.0000.024 0.000 0.000 0.000 0.0000.012 0.000 0.000 0.000 0.0000.006 0.000 0.000 0.000 0.000... and so on

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