FRM Part II · FRM Exam Part II · Backtesting VaR
A bank backtests its one-day 99% VaR model using 250 trading days of P&L. Assuming the model is correctly calibrated and exceptions are independent, what is the expected number of exceptions over the period?
The expected number of exceptions is 2.5. A 99% VaR should be breached on 1% of days, so over 250 days the expected count is 250 times 0.01, which equals 2.5 exceptions if the model is correctly calibrated.
- A0.99
- B1.0
- C2.5Correct
- D25
Explanation
Under a correct model the exception probability each day is 1% (1 - 0.99). Expected exceptions = 250 x 0.01 = 2.5. Choosing 25 uses a 10% probability, and 1.0 is not derived from the data.
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's 99% one-day VaR model produced 4 exceptions in 250 days, all in the same two-week period following a market shock. Which conclusion…
- A risk manager backtests a 99% one-day VaR model and rejects the model whenever the number of exceptions exceeds a chosen cutoff. In this co…
- 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…
- A bank's 95% one-day VaR is backtested over 250 days with 20 exceptions. Using the normal approximation to the binomial under the null that …
- A risk manager notes that a 99% VaR model had 3 exceptions in 250 days, all in consecutive days during one stress week. What is the most app…
- 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…