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Summarise how the Gaussian copula produces a default time for an asset whose defaults are correlated with all the other names held in a portfolio.

FRM® Part II · Market Risk Measurement and Management · Financial Correlation Modeling: Bottom-Up Approaches · FRM2.MR.10.c

The idea

To model many names at once, the Gaussian copula needs a correlation structure across all of them. The usual simulation draws correlated standard normals, for instance by a Cholesky decomposition of the correlation matrix applied to independent normals. The one-factor version is simpler. For name i, the draw is written Xi. Xi = √ρ × M + √(1 − ρ) × Zi. Here M is the common systematic factor shared by every name, and Zi is the independent idiosyncratic term. Each Xi becomes a uniform number Ui through the normal distribution function N. Ui = N(Xi). That uniform number is then turned into a default time by inverting the name's cumulative default distribution. With a constant hazard rate λ, the default time of name i is written Time. Time = −ln(1 − Ui) ÷ λ. A low normal draw gives an early default. Repeating the trial many times gives the portfolio loss distribution. A frequent error is using ρ instead of the square root of ρ on the common factor.

Formulas

One-factor Gaussian copula draw for name i

Xᵢ = √ρ × M + √(1 − ρ) × Zᵢ

ρ
the common correlation between any two names' draws
M
the common systematic factor, a standard normal shared by every name
Zᵢ
the independent idiosyncratic standard normal for name i

The square-root weights keep Xᵢ standard normal and make the pairwise correlation ρ. Using ρ and 1 − ρ as the weights is the usual error.

Default time from a uniform draw with a constant hazard rate

τᵢ = −ln(1 − Uᵢ) ÷ λ

Uᵢ
the uniform number N(Xᵢ) from the name's correlated normal draw
λ
the name's constant hazard rate

Invert the cumulative default distribution, so use 1 − Uᵢ inside the logarithm, not Uᵢ. A low normal draw gives an early default.

A practice question

From the question bank, with the reason each wrong option is wrong.

In the Gaussian copula simulation of a default time for one name in a multi-name portfolio, which sequence is correct?

  1. Draw a uniform number, apply N^-1 to get a normal, then multiply by the portfolio correlation to get the default time.Why not: Reverses the mapping and treats the correlation as a scaling to time; the uniform is mapped through the inverse default distribution.
  2. Draw independent uniform numbers for each name and set each default time equal to its recovery-adjusted loss.Why not: Skips the correlation step entirely, so the simulated defaults would be independent, and confuses default time with loss.
  3. Draw a correlated normal, convert it to a uniform with N(), then invert the name's default time distribution. - correct
  4. Draw a normal for each name, then set the default time to the smaller of the name's normal draw and the average draw of the other names.Why not: Invents an averaging rule; correlation enters through the joint normal draw, not through a minimum of draws.

The correlated normal X is converted to a uniform U = N(X), and the default time is the value tau with cumulative default probability U, that is tau = Q^-1(U). Correlation is carried entirely by the joint normal draws.

A harder one

In a one-factor Gaussian copula, X_i = sqrt(rho) x M + sqrt(1 - rho) x Z_i with rho = 0.36. A trial draws M = -1.5 and Z_i = 0.4. The simulated X_i is:

  1. -0.58. - correct
  2. -0.28.Why not: Uses rho and 1 - rho as the weights instead of their square roots (0.36 x -1.5 + 0.64 x 0.4).
  3. -0.96.Why not: Swaps the weights, putting sqrt(1 - rho) on the common factor and sqrt(rho) on the idiosyncratic term.
  4. -1.10.Why not: Adds the two draws with no weights, ignoring the factor loadings that make X_i standard normal.

sqrt(0.36) = 0.6 and sqrt(0.64) = 0.8, so X_i = 0.6 x -1.5 + 0.8 x 0.4 = -0.90 + 0.32 = -0.58. The weights keep X_i standard normal and make the pairwise correlation between names rho.

The hardest tier

In a one-factor Gaussian copula with rho = 0.25, a name has a 5% cumulative default probability to the horizon. The common factor takes the value M = -2. Given M, the probability that the name has defaulted is N((N^-1(0.05) - sqrt(rho) x M) / sqrt(1 - rho)). With N(-3.00) = 0.0013, N(-1.50) = 0.0668, N(-1.00) = 0.1587, N(-0.75) = 0.2266, the conditional probability is closest to:

  1. 0.1%.Why not: Adds sqrt(rho) x M instead of subtracting it, so the factor pushes the wrong way: (-1.645 - 1.0) / 0.866 = -3.05.
  2. 9%.Why not: Uses rho instead of sqrt(rho) on the factor: (-1.645 + 0.5) / 0.866 = -1.32.
  3. 23%. - correct
  4. 19%.Why not: Divides by 1 - rho instead of sqrt(1 - rho): -0.645 / 0.75 = -0.86.

N^-1(0.05) = -1.645. The argument is (-1.645 - 0.5 x -2) / sqrt(0.75) = -0.645 / 0.866 = -0.745, and N(-0.745) is about 0.23. A bad systematic outcome raises the default probability from 5% to roughly 23%.

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