FRM Part I · FRM Exam Part I · Operational Risk
A bank models the number of fraud events in a year as Poisson with mean 4 and each loss as a fixed 50,000 for simplicity. What are the expected annual aggregate loss and the probability of exactly zero events in a year (e^-4 = 0.0183)?
Expected annual loss is 4 times 50,000, which equals 200,000. The probability of zero events under a Poisson distribution with mean 4 is e to the power minus 4, about 1.83%. The 98.17% figure would be the probability of at least one event.
- AExpected loss 200,000; P(zero events) = 1.83%Correct
- BExpected loss 200,000; P(zero events) = 98.17%
- CExpected loss 12,500; P(zero events) = 1.83%
- DExpected loss 50,000; P(zero events) = 4.00%
Explanation
Expected aggregate loss = E[N] x E[severity] = 4 x 50,000 = 200,000. For Poisson, P(N=0) = e^-4 = 0.0183, or 1.83%. The 98.17% option is the probability of at least one event, and 12,500 divides rather than multiplies.
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