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P2.T6.309. Default correlation, Malz sections 8.1 and 8.2

Thread starter #1
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Hi @Kavita.bhangdia You are right about 98.32% but let's list them out. To re-cap, for question 309.3, N = 20, pd = 0.01, and we are assuming independence which justifies the binomial:
  • Prob(exactly 0 defaults) = 0.99^20 = 81.79%
  • Prob(exactly 1 default) = 0.99^19*.01*20 = 16.52%, so the cumulative Prob[1 default or less] = Prob[0 or 1 default] = 81.79%+16.52%=98.31%
  • Prob(exactly 2 defaults) = 0.99^18*01^2*190= 1.59%, so the cumulative Prob[2 defaults or less] = Prob[0 or 1 or 2] = 98.31% + 1.59% = 99.90%
If we want the 95.0% VaR, we only go 95.0% "into the tail:" 0 defaults only gets us to 81.79% so we go to the "next worse" outcome which is 98.31%. We've gone far enough, the 95.0% quantile is one default. In fact, the 98.0% VaR is still one default; and so is the 99.0% VaR still one default, because 98.0% < 99.90% and 99.0% < 99.90%. The precise Prob[2 defaults or less] = 99.8996%, so the 99.9% VaR here is two (2) defaults, barely :) I hope that helps!
Hi, how do we do this in case of 2 bonds with default correlation of 30% and joint default probability of 1.5%

Thanks
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
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#2
Hi @taunk I don't see how that's related. Question 309.3 assumes 20 independent positions such that default correlation is zero. As a totally separate question, if you have two identical bonds, you can use joint PD =pd^2 + covariance such that 1.5% = pd^2 + pd*(1-pd)*30% which is just E[XY]= E(X)*E(Y) + covariance[X,Y], so pd would be about 1.50% in your case.
 
#4
Hi David....in page no. 20 of bionic notes pertaining to malz readings....in Table 8.1 I am unable to gauge how did u arrive at 28 defaults for total credits of 1000 as according to me 95th percentile should be 50 credits....
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
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#5
Hi @meenalbaheti That table assumes the credits are independent (independence --> uncorrelated). Each credit either defaults or survives (ie, does not default), which is a Bernoulli (0 or 1, yes or no). A series or set of i.i.d. Bernoullis is characterized by a binomial distribution (this is one of the archetypal credit risk distributions). It's not realistic, but it's the easiest way to characterize a portfolio of credits: by assuming they all have the same default probability (satisfying the "identical" part of i.i.d.) and that they are uncorrelated (satisfying the "independent" part of i.i.d.; although don't forget that independence implies correlation but correlation does not imply independence, but in this context they are used synonymous).

So, the table presumes a binomial. And in the scenario where N = 1,000 and PD = 2.0%, we have the binomial below. The answer is 28.0 because that is the quantile associated with 95.0% probability (VaR is a quantile); i.e., the 5.0% loss tail includes 29 losses and worse, or more accurately part of the 28th loss and worse. Put another way, only 5.0% of the time, we'd expect this portfolio to experience 28 or more defaults. In recreating the table, I got 28.0 in a manner analogous to the maybe more familiar = NORM.S.INV (95%) = 1.645 which is generically =NORM.S.INV(probability) = quantile. VaR is the quantile associated with a selected probability. That's the inverse cumulative distribution, but for the binomial distribution, such that in Excel it is given by =BINOM.INV(1000, 2%, 95% ) so it's the reverse of computing the cumulative distribution: =BINOM.DIST(28, 1000, 2%, true) = 96.71% which is greater that 95.0% but =BINOM.DIST(27, 1000, 2%, true) = 94.93% so we can see why the inverse cumulative function returned 28 defaults. I hope that helps!

 
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