Use the Exact Binomial Tail for a VaR Exception Test

Summary

The note examines the significance level for testing whether a daily value-at-risk model is calibrated. Under the stated null hypothesis, exceptions over the observation period follow a binomial distribution, with the exception probability set by the VaR confidence level. The risk manager rejects the model when the observed exception count exceeds a chosen cutoff, and the significance level is the probability of landing in that rejection region under the null.

The questioner applies a normal approximation to the exception count and obtains a different result. The correction is to calculate the discrete binomial tail directly: add the probabilities for counts at or below the cutoff, then take the complement for counts above it. In the example, the source reports a 10.8% tail probability for rejection above four exceptions. The note illustrates why a normal approximation can misstate a tail probability in this setting; it does not discuss alternative test designs or adjustments for repeated testing.

Key ideas

• Under the calibration null, the number of VaR exceptions is modeled with a binomial distribution.
• The significance level is the probability of observing an exception count in the rejection region.
• For the stated cutoff, calculate the exact discrete binomial tail rather than relying on a normal approximation.
• The example defines rejection as counts above four and reports a 10.8% significance level.

Tags

# question about significance level


# question about significance level












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?

## Answer by techie11 (score 1)

https://quant.stackexchange.com/a/57942

I figured it out now.

When calculate the probability use Binominal itself, the result matches:

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

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.