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Name the weaknesses of a credit portfolio framework built on correlation.

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

The idea

A credit portfolio framework built on correlation inherits several weaknesses. Correlation is a single linear summary, so it cannot describe tail dependence. Two portfolios with the same correlation can differ widely in the chance that many names default together. Defaults are rare, so default correlations are estimated from very few joint events and carry large sampling error. Correlations are not stable. They depend on the horizon and on the PD level, and they rise in downturns, so estimates taken in calm periods understate stress losses. With N obligors, the number of distinct pairwise correlations to calibrate is: Pairs = N × (N − 1) ÷ 2. That makes large portfolios hard to calibrate and the matrix hard to keep consistent. In an exam scenario, name the weakness it shows: instability, sparse data, nonlinearity or dimensionality.

Formulas

Number of distinct pairwise correlations

pairs = N × (N − 1) ÷ 2

N
the number of obligors

Each pair counted once, since the correlation of A with B equals that of B with A. Using N² or N × (N − 1) double counts.

A practice question

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

A bank wants to specify a pairwise default correlation for every pair of its 500 obligors. How many distinct pairwise correlations must it estimate?

  1. 500Why not: One parameter per obligor is the count for a single factor model, not for an unrestricted pairwise matrix.
  2. 124,750 - correct
  3. 250,000 (500²)Why not: Uses N² and counts every diagonal and mirrored entry of the matrix.
  4. 249,500Why not: Uses N(N-1) and double counts each pair, since the correlation of A with B equals that of B with A.

The number of distinct pairs is N(N-1)/2 = 500 x 499 / 2 = 124,750. This quadratic growth is one reason a pairwise correlation framework is hard to calibrate.

A harder one

A bank estimated default correlations from a calm five-year period. In a later crisis, many pairs of borrowers defaulted together and portfolio losses far exceeded the Credit VaR. The weakness of the correlation framework exposed here is best described as:

  1. Correlations are unstable and tend to rise in downturns, so calm-period estimates understate dependence in stress. - correct
  2. Correlation captures only linear dependence, so it cannot reflect any nonlinear link between the borrowers.Why not: A genuine weakness, but it does not explain why calm-period estimates failed once conditions changed.
  3. The correlation matrix grows with the square of the number of obligors and so is too large to estimate.Why not: Describes dimensionality, which is not what the change in behaviour between calm and crisis shows.
  4. Defaults are too rare to observe, so any default correlation estimate has a very large sampling error.Why not: A sparse-data weakness; the scenario instead shows estimates that were stable in the sample but did not hold afterwards.

The scenario is about stability over time. Correlations estimated in benign conditions typically rise in downturns, so Credit VaR built on them understates stress losses.

The hardest tier

A bank has 2,500 obligor-pair-years of history for two groups of obligors, each with a one-year PD of 0.5%. It wants to estimate their default correlation from the data. Which statement best describes what the data can show?

  1. About 12.5 joint defaults are expected under independence, so the sample is ample for estimating the correlation.Why not: Multiplies the number of pair-years by one PD only (2,500 x 0.005) instead of by the product of the two PDs.
  2. About 0.06 joint defaults are expected under independence and about 0.7 if the default correlation were 0.05, so the sample cannot reliably distinguish the two. - correct
  3. A default correlation of 0.05 implies 0.05 x 2,500 = 125 joint defaults, so the sample is ample.Why not: Treats the correlation as if it were a joint default probability.
  4. About 0.06 joint defaults under independence and 0.7 at correlation 0.05, so a correlation of 0.05 would almost always be detected.Why not: The expected counts are right, but a Poisson mean of 0.7 gives at least one joint default only about half the time, so detection is far from certain.

Independence gives 2,500 x 0.005² = 0.0625. With rho = 0.05, JDP = 0.000025 + 0.05 x 0.004975 = 0.000274, or 0.68 expected events. Such sparse events make default correlation estimates unreliable.

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