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

A fund holds 200 independent bonds, each with a 2% annual default probability. A analyst approximates the number of defaults with a Poisson distribution. Which parameter value and justification is appropriate?

Lambda should be 4, because for a binomial with large n and small p the Poisson approximation uses the mean np, which is 200 times 0.02. The variance of 3.92 is nearly equal but is not the matching parameter, and 0.02 is only the single-bond probability.

  1. ALambda = 4, because the binomial with large n and small p has mean npCorrect
  2. BLambda = 4, because the Poisson variance must equal np(1-p)
  3. CLambda = 0.02, because it equals the individual default probability
  4. DLambda = 3.92, because the Poisson mean must equal the binomial variance

Explanation

With n=200 and p=0.02, np = 4. The Poisson approximation to a binomial matches the mean np when n is large and p is small. The variance np(1-p)=3.92 is close to 4 but is not used as the Poisson parameter.

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