FRM Part II · FRM Exam Part II · Backtesting VaR
A risk manager backtests a 99% one-day VaR model over 250 trading days. Under the assumption that the model is correctly calibrated and exceptions are independent, what is the expected number of exceptions?
The expected number of exceptions is 2.5. With a 99% VaR the exception probability on each day is 1%, so across 250 independent trading days the expected count is 250 multiplied by 0.01, which equals 2.5.
- A0.25
- B1.0
- C2.5Correct
- D25
Explanation
Expected exceptions = T x p = 250 x 0.01 = 2.5. Choosing 25 uses a 10% tail probability, and 0.25 and 1.0 are arithmetic or scaling errors.
Did you get it right without looking?
One question tells you little. A timed set on Backtesting VaR shows your real accuracy, how long you take and where you lose marks.
More Backtesting VaR questions
- A bank moves from a 99% to a 95% confidence level for backtesting the same VaR model over 250 days, keeping the test on the number of except…
- Which limitation of the Christoffersen independence test, as typically specified, is most accurate?
- 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 ex…
- A bank's 99% one-day VaR model is backtested over 250 trading days and produces 1 exception. The head of risk argues that this proves the mo…
- A bank's 95% VaR backtest shows exactly 5% exceptions over a long sample, so the Kupiec test is not rejected. However, the independence test…
- A bank backtests a 99% one-day VaR over 500 days and observes 12 exceptions. The expected number is 5. A review finds that the model assumes…