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FRM Part II · FRM Exam Part II · Beyond Exceedance-Based Backtesting of Value-at-Risk Models

Two banks each report 99% one-day VaR of USD 10 million and each records exactly 3 exceptions in 250 days. Bank A's exception losses averaged USD 11 million; Bank B's averaged USD 40 million. Which conclusion is most consistent with the limitations of exceedance-based backtesting?

Identical exception counts make the two banks indistinguishable to count-based tests, yet Bank B's exceedances average USD 40 million versus 11 million for Bank A. The test is blind to severity, so Bank B's worse tail risk goes undetected.

  1. ABank B's model is better because its VaR was exceeded by larger amounts, showing higher sensitivity
  2. BThe banks' models are equivalent because exception counts are identical and the test cannot separate them, though Bank B's tail risk is clearly worse than the test indicatesCorrect
  3. CBank A's model should be rejected because its exceptions are smaller
  4. DOnly Bank B passes the Basel traffic-light test because its losses are larger

Explanation

Both have 3 exceptions, so any count-based test (Kupiec or traffic light) gives identical results. The severity differs greatly (11 vs 40 million), which exceedance tests ignore. Bank B's heavy tail is hidden, so magnitude-based approaches such as expected shortfall backtests or loss-function measures are needed.

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