FRM Part II · FRM Exam Part II · Backtesting VaR
A risk analyst evaluates a bank's 99% VaR model with 250 observations. Assume the model is correct, so exceptions are binomial with p = 1%. The analyst needs the expected number of exceptions and the standard deviation of the count (rounded to two decimals). Which pair is correct, and how does the expected count compare with the green zone limit?
Under a correct model the exception count is binomial with mean 250 x 0.01 = 2.5 and standard deviation the square root of 250 x 0.01 x 0.99, about 1.57. The mean of 2.5 lies within the green zone, which runs from 0 to 4 exceptions.
- AExpected 2.5; standard deviation 1.57; the expected count lies inside the green zone, which ends at 4Correct
- BExpected 2.5; standard deviation 2.48; the expected count lies inside the green zone, which ends at 4
- CExpected 2.5; standard deviation 1.57; the expected count lies in the yellow zone
- DExpected 1.0; standard deviation 0.99; the expected count lies inside the green zone
Explanation
Expected exceptions = 250 x 0.01 = 2.5. Variance = 250 x 0.01 x 0.99 = 2.475, so the standard deviation = 1.57. The option with 2.48 uses the variance rather than the standard deviation. The expected count of 2.5 is below the green limit of 4, so it is green.
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