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FRM Part II · FRM Exam Part II · Backtesting VaR

Assume a bank's VaR model is truly accurate at 99% over 250 days, so exceptions are binomial with p = 0.01. The probability of 4 or fewer exceptions is about 89.2%. A competitor's true exceedance probability is 3%, and its probability of 4 or fewer exceptions is about 13.2%. Which conclusion is correct?

The green zone captures the accurate model about 89% of the time, so it lands in yellow or red about 11% of the time. The 3% exceedance model still lands in green about 13% of the time, which is a Type 2 error of wrongly accepting a bad model.

  1. AThe green zone accepts the accurate model about 89% of the time, and wrongly accepts the 3% model about 13% of the timeCorrect
  2. BThe green zone rejects the accurate model 89% of the time
  3. CThe 3% model is in the red zone 13% of the time
  4. DThe accurate model lands in the yellow or red zone with probability 13.2%

Explanation

Green means 0-4 exceptions. P(green | accurate) is 89.2%, so the accurate model is in yellow or red 10.8% of the time. The 3% model is in green 13.2% of the time, which is a Type 2 error (accepting a bad model). The other options misread these probabilities.

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