A flat distribution of defaults

Take a Credit basketA group of borrowers held together, each of which might default. The question is never about one borrower but about how many of them fail at once. of $n$ credits. Each one defaults with the same probability $p$ , and every pair of them has the same Asset correlationHow much the borrowers’ fortunes move together. At zero each fails on its own account; at one they share a single fate. $\rho$ . How many defaults should you expect, and how are they distributed?

Two cases are easy. If the names are independent, the count is Binomial distributionThe distribution of how many successes you get from a fixed number of independent tries. It is what the default count collapses to when the borrowers are unrelated.. If they are perfectly correlated they share one fate: either none default or all of them do. Everything between those two needs numerical work.

Except at one point, where the answer is as simple as an answer can be.

The model

This is the Gaussian copulaThe standard way of tying individual default chances together into a joint one, by giving each borrower a hidden bell-curve score and defaulting the ones whose score falls low enough., the standard model for a credit basket. Each name has a latent asset return, and the name defaults when that return falls below a threshold. Matching the threshold to the default probability $p$ puts it at $\Phi^{-1}(p)$ standard deviations.

Correlation enters through a single market factor $M$ :

R_k = \sqrt{\rho}\, M + \sqrt{1-\rho}\, Z_k

where $Z_k$ is noise particular to name $k$ . Conditional on the market — once $M$ is known — the names are independent, so the number of defaults is binomial with a shifted probability. Integrating that binomial over the market factor gives the distribution:

p(m) = \Phi\!\left( \frac{\Phi^{-1}(p) - m\sqrt{\rho}}{\sqrt{1-\rho}} \right)

\Pr[N = j] = \binom{n}{j} \int_{-\infty}^{\infty} p(m)^{j}\,\bigl(1 - p(m)\bigr)^{n-j} \varphi(m)\, dm

That integral normally wants a quadrature rule. Which is the point: closed-form answers are scarce here, and scarce enough that a single one is worth having, if only to check the numerical methods against.

The special case

Look at $p(m)$ . It would be much friendlier if $\Phi^{-1}(p)$ were zero — that is, if $p = \tfrac{1}{2}$ . And friendlier still if the $\sqrt{\rho}$ and $\sqrt{1-\rho}$ matched, which happens when $\rho = \tfrac{1}{2}$ . Put both together and the conditional probability collapses to

p(m) = \Phi(-m) = 1 - \Phi(m)

Substituting $u = \Phi(m)$ turns the integral into a beta integral, and out comes $B(n-j+1,\, j+1)$ . But there is a better route that needs no calculus at all.

With $p = \tfrac{1}{2}$ and $\rho = \tfrac{1}{2}$ the factor model is just

R_k = \sqrt{\tfrac{1}{2}}\,(M + Z_k)

and name $k$ defaults when $R_k < 0$ — equivalently when $M < -Z_k$ . Write $M^{*} = -M$ , and name $k$ defaults exactly when $M^{*} > Z_k$ .

So the number of defaults is the number of the $Z$ ‘s that $M^{*}$ exceeds. Sort $M^{*}$ and the $n$ noise terms into one list. They are identically distributed continuous random variables, so $M^{*}$ is equally likely to land in any of the $n+1$ positions. Therefore

\Pr[N = j] = \frac{1}{n+1}

Every outcome — no defaults, one default, all of them — is equally likely. The distribution is flat, and it stays flat for any $n$ .

Try it

Move the correlation and watch the binomial spread out, flatten, and then collapse toward the two-point distribution at $\rho = 1$ . The dashed line marks $1/(n+1)$ . It is only the answer at one setting, but you can see the shape pass through it.

Names in the basket10

Default probability50.0%

Asset correlation50.0%

E[defaults] 5.00σ 3.16Flat: every outcome from 0 to n is equally likely.

What to look for

The mean never moves. Slide the correlation from zero to one and the expected number of defaults stays at $n p$ throughout. Correlation does not change how likely any individual name is to default; it changes only how much they default together. Everything correlation does, it does to the shape.

The tails are what correlation buys. At low correlation the distribution concentrates near the mean and the extremes are negligible. Raise it and probability drains out of the middle into both ends — many defaults at once becomes far more likely, and so does none at all. This is the whole reason correlation matters for a tranche: a senior tranche only takes losses in the right tail, which barely exists at low correlation and is substantial at high.

Flat is not the same as random. A uniform distribution over $0 \ldots n$ defaults looks like the least informative outcome possible, but it comes from a very specific pair of parameters. Move either one and the flatness goes.

Notes

The distribution is computed by numerical integration over the market factor, using a composite Gauss-Legendre rule with panels placed on the transition region — as $\rho$ approaches 1 the conditional default probability becomes nearly a step function in $m$ , and a fixed grid misses it. At $p = \rho = \tfrac{1}{2}$ the computed values agree with $1/(n+1)$ to machine precision, which is the tightest check available: the code has to reproduce an exact identity, not merely come close.

This page follows Pete Benson, “Distribution of Defaults in a Credit Basket: An Interesting Special Case”, RiskMetrics Journal, Vol. 6, No. 1, Winter 2005, pp. 3–8. Per the editor’s note in that issue, the result became a standard interview question at RiskMetrics.