The recent regulation (page 32) on PRIIPs requires to compute a VaR-equivalent volatility defined as
$$\mbox{VEV}=\frac{\sqrt{3.842-2\ln \mbox{VaR}}-1.96}{\sqrt{T}}$$
Does anyone have an idea how they came up with that formula?
Answer
Let's assume T=1 and let S be a geometric gaussian process with zero drift, i.e. $\ln(S_1/S_0)$ is normally distributed with mean $-1/2\times\mathrm{VEV}^2$ and volatility VEV.
Then
$$\ln(\mathrm{VaR}/S_0) = -1/2\mathrm{VEV}^2 - \mathrm{VEV} \times 1.96$$ with the VAR at $0.975$ quantile.
This is a quadratic equation in VEV, with solutions
$$\mathrm{VEV} = -1.96 \pm \sqrt{1.96^2 - 2\ln(\textrm{VaR}/S_0)}.$$
We take the positive solution and are done.
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