FRM Part I · FRM Exam Part I · Common Univariate Random Variables
Defaults in a portfolio follow a Poisson distribution with a rate of 2 per year. Assuming a constant rate and independence across periods, what is the probability of exactly 2 defaults over a 6-month period (nearest value)?
The probability is about 0.184. The rate must be scaled to the six-month horizon, giving a mean of 1. Then the probability of exactly two defaults is e^-1 times 1 squared divided by 2 factorial, which is 0.1839. Using the annual rate of 2 gives 0.271 incorrectly.
- A0.184Correct
- B0.271
- C0.368
- D0.092
Explanation
The 6-month mean is lambda = 2 x 0.5 = 1. P(2) = e^-1 x 1^2 / 2! = 0.3679/2 = 0.1839. The value 0.271 uses the annual lambda of 2 (e^-2 x 2^2/2), which is the wrong time base.
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