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Outline the Gaussian copula model of time to default and use the one-factor Gaussian copula to compute default probability.

FRM® Part II · Credit Risk Measurement and Management · Credit Risk · FRM2.CR.12.f

The idea

The Gaussian copula links individually estimated distributions of time to default into one joint distribution. Each firm's cumulative default probability Qᵢ(t) is mapped to a standard normal variable by matching percentiles. Firm i defaults before t when its normal variable xᵢ is below N⁻¹(Qᵢ(t)). The dependence between the xᵢ is multivariate normal. The one-factor version builds each firm's variable from two pieces. x = √ρ × M + √(1 − ρ) × Z. Here x is the variable for firm i, M is a common factor, and Z is an independent idiosyncratic factor for that firm. Both M and Z are standard normal, so every pair has correlation ρ. Conditional on the common factor taking the value F, the probability of default by T is PD = N[(N⁻¹(Q(T)) − √ρ × F) ÷ √(1 − ρ)]. A negative F is a bad economy and raises the probability. Exam point: the square root of ρ, not ρ itself, multiplies F, and the sign of F matters.

Formulas

Firm variable in the one-factor Gaussian copula

x = √ρ × M + √(1 − ρ) × Z

ρ
the copula correlation between any two firms
M
the common factor, standard normal
Z
the firm's idiosyncratic factor, standard normal and independent of M

The loading on M is √ρ, not ρ. The firm defaults by T when x is below N⁻¹ of its cumulative default probability Q(T).

Default probability by T conditional on the common factor

PD = N[(N⁻¹(Q(T)) − √ρ × F) ÷ √(1 − ρ)]

N
the standard normal cumulative distribution function, with N⁻¹ its inverse
Q(T)
the firm's unconditional cumulative default probability by T
ρ
the copula correlation
F
the value taken by the common factor

A negative F is a bad economy and raises the probability. Q(T) must be cumulative to T, such as 1 − e^(−λT) from a hazard rate, not a one-year figure.

A practice question

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

What is the role of the Gaussian copula in modelling time to default for several firms?

  1. It estimates each firm's default probability from the firm's equity price and debt level, treating equity as a call option.Why not: Describes the Merton structural approach; the copula takes the marginal distributions as given.
  2. It combines each firm's own distribution of time to default into a joint distribution, using correlated standard normal variables. - correct
  3. It assumes that each firm's time to default is itself normally distributed.Why not: Only the transformed variables are normal, via matching percentiles; default times are not assumed normal.
  4. It sets recovery rates so that portfolio losses are fixed in advance.Why not: The copula describes dependence, not recovery rates.

Each marginal default time is mapped percentile by percentile to a standard normal variable, and the dependence among these normals defines the joint distribution of default times.

A harder one

In a one-factor Gaussian copula, a firm has a one-year default probability of 2% and a copula correlation of 0.15. If the common factor takes the value -2 (a poor economy), the conditional default probability is closest to:

  1. 0.11%.Why not: Reverses the sign of the common factor, treating -2 as a strong economy.
  2. 2.86%.Why not: Multiplies the factor by rho instead of the square root of rho.
  3. 10.04%.Why not: Omits the division by sqrt(1 - rho).
  4. 8.27%. - correct

Conditional probability = N[(N^-1(0.02) - sqrt(0.15) x (-2)) / sqrt(0.85)] = N[(-2.054 + 0.775)/0.922] = N(-1.387) = 8.27%.

The hardest tier

In a one-factor model each firm's variable is x = a M + sqrt(1 - a²) Z, with M and Z independent standard normals and a = 0.4. A firm's one-year default probability is 5%. If M = -2, the conditional default probability is closest to:

  1. 17.8%. - correct
  2. 31.2%.Why not: Treats the loading a = 0.4 as the correlation rho, so it uses sqrt(0.4) and sqrt(0.6).
  3. 7.4%.Why not: Multiplies the factor by a² (the pairwise correlation) instead of the loading a.
  4. 13.8%.Why not: Uses sqrt(1 - a) in the denominator instead of sqrt(1 - a²).

Here the loading is a, so the pairwise correlation is a² = 0.16. Conditional probability = N[(N^-1(0.05) - a x M)/sqrt(1 - a²)] = N[(-1.645 + 0.8)/0.9165] = N(-0.922) = 17.8%.

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