Answer: 5

## Explanation For a 1-day 99% VaR model, we expect losses to exceed VaR on 1% of trading days. With 225 days in a year, the expected number of exceedances is: \[\text{Expected exceedances} = 225 \times 1\% = 2.25\] To determine the acceptable number of exceedances at a 95% confidence level, we use the binomial test. The null hypothesis is that the model is correctly calibrated (true probability of exceedance = 1%). Using the binomial distribution with: - n = 225 (number of trials) - p = 0.01 (probability of success) - Confidence level = 95% We need to find the maximum number of exceedances k such that: \[P(X \leq k) \geq 0.95\] Calculating the cumulative probabilities: - P(X ≤ 3) ≈ 0.757 (too low) - P(X ≤ 4) ≈ 0.900 (too low) - P(X ≤ 5) ≈ 0.967 (≥ 0.95) - P(X ≤ 6) ≈ 0.990 (≥ 0.95) The first value that gives us at least 95% confidence is k = 5. This means that if we observe 5 or fewer exceedances, we cannot reject the null hypothesis that the model is correctly calibrated at the 95% confidence level. Therefore, **5 exceedances** is the maximum acceptable number to conclude the model is calibrated correctly.

Author: LeetQuiz .