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Using a binomial distribution for credit losses, estimate their mean and their standard deviation.

FRM® Part I · Valuation and Risk Models · Measuring Credit Risk · FRM1.VR.6.e

The idea

The binomial model treats a portfolio of n loans as n independent trials, each with the same default probability p. The number of defaults then has mean np and variance np(1 − p). The standard deviation of the number of defaults is SD = √(np(1 − p)). If every default costs the same amount L, which is exposure times LGD, the loss has mean L × np. The standard deviation of the loss is SD = L × √(np(1 − p)). The probability of at least one default is 1 − (1 − p)ⁿ. For large n the distribution is close to normal. The loss at a confidence level is then roughly the mean plus z times the standard deviation, where z is 2.326 at 99 percent. The independence assumption is the weakness. Real defaults are positively correlated, so the model understates both the spread and the tail. The usual errors are forgetting to multiply by the loss per default and reporting the variance instead of the standard deviation.

Formulas

Mean credit loss in the binomial model

mean loss = L × n × p

L
the loss per default, exposure times LGD
n
the number of loans
p
the default probability of each loan

Holds for identical loans whatever the correlation, since means add. Forgetting to multiply the expected number of defaults by L is the usual slip.

Standard deviation of credit loss in the binomial model

standard deviation of loss = L × √(n × p × (1 − p))

L
the loss per default, exposure times LGD
n
the number of loans
p
the default probability of each loan

Assumes independent defaults, so it understates the spread and tail when defaults are correlated. Take the square root: np(1 − p) is the variance of the count.

Probability of at least one default

P(at least one) = 1 − (1 − p)ⁿ

p
the default probability of each loan
n
the number of independent loans

Requires independence. Multiplying n by p overstates it and can exceed one.

A practice question

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

Which set of assumptions underlies a simple binomial model of credit losses on a portfolio of loans?

  1. All loans share one default probability and one loss given default, and defaults are independent. - correct
  2. Every loan has the same default probability, and defaults are perfectly correlated.Why not: Perfect correlation would make the portfolio all-or-nothing; the binomial model requires independence.
  3. Default probabilities change with the economic cycle, so each year's defaults are drawn from a different binomial.Why not: A time-varying PD is not part of the basic binomial model, which uses one constant p.
  4. Loss amounts are random and independent of whether a loan defaults, with a different default probability for each loan.Why not: The binomial model needs a common default probability and a fixed loss per default.

The binomial model counts defaults among n identical, independent loans each with probability p. Loss is the number of defaults times a fixed loss per default.

A harder one

A bank holds 500 loans, each with a default probability of 2%, defaults independent, and a loss of USD 0.8 million per default (exposure net of recovery). The mean and standard deviation of the portfolio's loss, in USD million, are closest to:

  1. mean 10.0, standard deviation 3.13Why not: Reports the number of defaults rather than the loss amounts, so both figures are in the wrong unit.
  2. mean 8.0, standard deviation 2.50 - correct
  3. mean 8.0, standard deviation 3.13Why not: Gives the standard deviation of the number of defaults and forgets to scale it by the USD 0.8 million loss per default.
  4. mean 8.0, standard deviation 7.84Why not: Multiplies the variance of the count by the loss, mixing units; the loss scales the standard deviation, not the variance.

Mean loss = 500 x 0.02 x 0.8 = 8.0. Standard deviation = 0.8 x sqrt(500 x 0.02 x 0.98) = 0.8 x 3.13 = 2.50.

The hardest tier

A portfolio has 1,000 independent loans, each with a default probability of 1% and a loss of USD 1 million per default. Using a normal approximation to the binomial, the 99th percentile of the loss (z = 2.326) is closest to:

  1. USD 33.0 millionWhy not: Uses the variance of the count (9.9) in place of the standard deviation (3.15).
  2. USD 7.3 millionWhy not: Reports only the unexpected part, z x sigma, and leaves out the mean loss of USD 10 million.
  3. USD 17.3 million - correct
  4. USD 15.2 millionWhy not: Uses z = 1.645, the 95% quantile, instead of 2.326 for 99%.

Mean = 10. Standard deviation = sqrt(1000 x 0.01 x 0.99) = 3.146. 99th percentile = 10 + 2.326 x 3.146 = USD 17.3 million.

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