VaR model Unconditional Coverage Tests: Is this extension of Kupiec POF test correct?

Background: Kupiec P. in 1995, published paper "Techniques for Verifying the Accuracy of Risk Management Models" on Journal of Derivatives, v3, P73-84, it's a Unconditional Coverage Tests designe for VaR model.

This "proportion of failures" or POF test examines how many times a financial institution's VaR is violated over a given span of time. If the number of violations differs considerably from $\alpha * 100\%$ of the sample, then the accuracy of the underlying risk model is called into question. Kupiec's (1995) test statistic takes the form,

$$POF = 2 \cdot ln \left( \left(\frac{1-\hat\alpha}{1-\alpha}\right)^{T-I(\alpha)} \left(\frac{\hat\alpha}{\alpha}\right)^{I(\alpha)} \right)$$ $$\hat\alpha = \frac{1}{T}I(\alpha)$$ $$I(\alpha) = \sum_{t=1}^N I_t(\alpha)$$

, where $I_t(\alpha)$ is the violation of VaR.

Now, the problem is, if $I_t(\alpha) = 0$ , POF number is not defined.

However, I wonder if I could extend the POF test like this? --

First,

$$POF = 2 T \cdot ln \left( \left(\frac{1-\hat\alpha}{1-\alpha}\right)^{1-\hat \alpha} \left(\frac{\hat\alpha}{\alpha}\right)^{\hat \alpha } \right)$$

; then, $\lim_{x\to 0}x^x = 1$, for $x>0$

; so when $I_t(\alpha) = 0$, we have $\hat \alpha = 0$, and :

$$POF = 2 T \cdot ln \left( \left(\frac{1-\hat\alpha}{1-\alpha}\right)^{1-\hat \alpha} \right) = 2 T \cdot ln \left( \frac{1}{1-\alpha} \right)$$ , is this right?

• just to anser my own question. after study i think this is ok. Mar 20 '13 at 13:43