A case study in a exam material goes like this:

"Assume that the bank reports a daily VAR of \\$100 million at the 99% level of confidence. Under the null hypothesis that the VAR model is correctly calibrated, the number of exceptions should follow a binomial distribution with expected value of $$E[X] = np = 250(1 − 0.99) = 2.5$$. The risk manager then has to pick a cutoff number of exceptions above which the model would be rejected. The type 1 error rate is the probability of observing higher numbers than the cutoff point. Say the risk manager chooses $$n = 4$$, which corresponds to a type 1 error rate or significance level of $$10.8\%$$. Above 4, the risk model is rejected."

I calculated the significance level corresponding to observing 4 exceptions to be 17.11%, using standard normal distribution as follows:

$$z= (4 - np)/\sqrt{p(1-p)n} = (4-250(1-0.99))/\sqrt{0.01(1-0.01)*250}=1.5/1.573=0.95$$

$$P(-\infty < Z <0.95)=0.8289$$ $$\alpha = (1-P)*2 = 0.1711 = 0.1811=17.11\%$$

Was there any error in my calculation?

• Try the continuity correction-change 4 to 4.5-and you will get close to the figures you have in the case study. Sep 11, 2020 at 16:53
• That willl yield 20.4% Sep 11, 2020 at 18:43
• z=(4.5−250×0.01)÷SQRT(250×0.01×0.99)=1.27128; N(1.27128)=0.8982; 1- N(1.27128)=0.1018 Sep 11, 2020 at 18:53
• Thanks. your calculation is right and closer. but it turned out that the caculation used the exact Binominal model as I posted below. Sep 11, 2020 at 19:00
• thanks, yes binomial is the best in this case, thought you wanted to use normal approximation! Sep 11, 2020 at 19:02

$$\alpha=1-(P(4)+P(3)+P(2)+P(1)+P(0))=1-(0.134+0.214+0.257+0.205+0.081)=0.108$$