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FRM Part I · FRM Exam Part I · Common Univariate Random Variables

A credit analyst models the default of a single bond over one year as a Bernoulli random variable X, with X = 1 if default occurs, with probability p = 0.04. The bond has a loss of USD 500,000 if it defaults and nothing otherwise, so the loss is L = 500,000 X. What is the standard deviation of L?

The standard deviation of the loss is about USD 98,000. A Bernoulli variable has variance p(1-p) = 0.0384, so its standard deviation is about 0.196, and multiplying by the USD 500,000 loss gives roughly USD 97,980. USD 20,000 is only the expected loss.

  1. AUSD 20,000
  2. BUSD 98,000Correct
  3. CUSD 96,000
  4. DUSD 4,000

Explanation

Var(X) = p(1-p) = 0.04 x 0.96 = 0.0384. SD(X) = sqrt(0.0384) = 0.19596. Multiply by 500,000 to get about USD 97,980, roughly USD 98,000. USD 20,000 is the expected loss (0.04 x 500,000). USD 96,000 wrongly treats 0.96 x 0.04 x... ie. uses variance times the scale incorrectly as 0.192 SD approximations.

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