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Describe a simulation of joint defaults as a way to calculate Credit VaR.

FRM® Part II · Credit Risk Measurement and Management · Portfolio Credit Risk · FRM2.CR.11.g

The idea

A simulation of joint defaults estimates Credit VaR when no closed form exists. In each trial, draw one common factor M and an independent idiosyncratic shock e for every obligor. Using each obligor's factor loading β, form its asset return in that trial as: R = β × M + √(1 − β²) × e. If the return is below the threshold N⁻¹(PD), the obligor defaults and loses exposure × LGD. Add the losses to get the portfolio loss for that trial. Repeat many thousands of times, sort the losses and read off the quantile at the chosen confidence level. Credit VaR is that quantile less the average loss. With 10,000 trials, the 99.9% quantile is about the 10th largest loss, so tail estimates are noisy unless trials are plentiful. Dropping the common factor removes correlation and understates Credit VaR.

Formulas

Trials needed to place a given number of losses in the tail

trials = k ÷ (1 − q)

k
the number of simulated losses wanted beyond the quantile
q
the confidence level

The tail holds a share 1 − q of the trials. At 99.9% the tail is 0.001, so the 99% figure is ten times too few.

A practice question

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

Which description best matches a simulation of joint defaults used to compute Credit VaR?

  1. Draw an independent default indicator for each obligor from its PD, sum the losses, and take the quantile over many repeats.Why not: Omits the common factor, so defaults are independent and Credit VaR is understated.
  2. Draw the factor, compute each firm's conditional PD, and report exposure x LGD x conditional PD as Credit VaR.Why not: Gives a conditional expected loss for one factor value and does not simulate default events or a loss distribution.
  3. Draw a common factor and idiosyncratic shocks, form asset returns, compare them with default thresholds, sum the losses, repeat many times and take the loss quantile less the mean loss. - correct
  4. Run one trial with the factor set at its worst plausible value and report the resulting loss as Credit VaR.Why not: This is a single stress scenario, not a simulated distribution with a confidence level.

Correlated defaults come from the common factor. Repeating the trial many times builds a loss distribution whose upper quantile, less the mean, is Credit VaR.

A harder one

In one trial of a joint default simulation, a firm has beta 0.6 and PD 3% (default threshold N^-1(3%) = -1.88). The trial draws M = -1.5 and e = -0.8. What are the firm's asset return and default outcome?

  1. Asset return -1.54, no default. - correct
  2. Asset return -1.54, default.Why not: Compares the asset return with the PD (0.03) instead of with N^-1(PD) = -1.88.
  3. Asset return -2.30, default.Why not: Adds the two shocks without the weights beta and sqrt(1 - beta²).
  4. Asset return -1.41, no default.Why not: Uses 1 - beta² = 0.64 as the weight on e instead of its square root 0.8.

R = 0.6 x (-1.5) + sqrt(1 - 0.36) x (-0.8) = -0.90 - 0.64 = -1.54. Since -1.54 is above the threshold -1.88, the firm does not default in this trial.

The hardest tier

A simulation estimates a joint default probability of about 0.1% as the fraction of trials in which both obligors default. The standard error of the fraction is sqrt(p(1-p)/n) for n independent trials. To bring the standard error down to 0.0001, the number of trials needed is closest to:

  1. 10,000Why not: Rule of thumb of ten times 1 / p, which does not use the standard error formula.
  2. 316Why not: Uses sqrt(p(1-p)) / se instead of squaring the ratio.
  3. 99,900 - correct
  4. 9,990,000Why not: Targets a standard error 100 times smaller, 0.000001, instead of 0.0001.

Set sqrt(p(1-p)/n) = 0.0001, so n = p(1-p) / 0.0001² = 0.000999 / 0.00000001 = 99,900.

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