What's new

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

Taunk

Member
H
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
Subscriber
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.

Taunk

Member
I am sorry. I goofed up. I had this question no. 74 in mind when I raised the query.

[email protected]

Member
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
Subscriber
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!

[email protected]

Member
Yes David....that surely helped.....thanks so much...hopefully the question in exam shall provide the number of defaults