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.