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?

  • $\begingroup$ The formula should be corrected. It should be ln (VaR/ initial value of the PRIIPS) and not ln(VaR) This is a typo. $\endgroup$
    – user22115
    Commented Jun 12, 2016 at 12:01
  • $\begingroup$ Do you have a source for that? $\endgroup$
    – Bob Jansen
    Commented Jun 12, 2016 at 14:53
  • $\begingroup$ @despieg I sort of guessed that, but still the formula is unclear. The number 3.842 is the 95% quantile of the chi-square distribution, but I don't see why it should play a role here. $\endgroup$ Commented Jun 12, 2016 at 16:26

2 Answers 2


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.


$$\ln(\mathrm{VaR}/S_0) = -1/2\mathrm{VEV}^2 - \mathrm{VEV} \times 1.96$$ 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.


To answer this question it might make sense to mention the VaR part and VEV part separately.

  1. VaR example using a parametric approach to VaR: assuming an investment of $V_0 = 10,000 $ into the financial asset and its daily log return following a normal distribution such that $r_t \backsim N(\mu, \sigma^2)$ with mean $\mu=0.02$ and standard deviation $\sigma = 0.7$, compute the VaR of the investment at $p = 2.5\%$ for 1 day. Since the VaR may be defined in several ways, e.g. as value $\bigtriangleup V_1 $ on which the investment can depreciate or as value $V_1$ ($V_1 < V_0$), which shows the new level of the investment, the later definition is used in the example. The solution would be:
    \begin{equation} VaR = V_t(exp(\Phi^{-1}(0.025)\sigma + \mu)). \end{equation} 2.5-quantile of standard Normal distribution is $\Phi^{-1}(0.025)$ is -1.96. \begin{equation} 10,000(exp(-1.96 * 0.7 + 0.02)) = 2587.22, \end{equation} meaning that with probability $2.5\%$ a one-day investment of 10,000 into the asset will be 2587.22 or less.

  2. As one may guess, given a Normal distributional assumption on daily log-return and already calculated VaR, for example by means of a Monte Carlo simulation, one may infer $\sigma$. It seems that a question being answered by VEV is: "what would be a scale parameter labelled $\sigma$ of a Normally distributed random variable, if one assumes "demeaned" distribution $N(-\dfrac{1}{2} \sigma T, \sigma^2T)$ and value of 2.5-quantile at the VaR level. $T$ denotes the period in years.
    [ see this link for more info on T and distributional assumption Is This A Viable Alternative Options Pricing Method? ]. \begin{equation} VaR = V_t(exp(\Phi^{-1}(0.025)\sigma \sqrt{T} +(-\dfrac{1}{2} \sigma T) )), \end{equation} which is rewritten in as a quadratic equation: \begin{equation} \dfrac{1}{2} T\sigma^2 + 1.96\sqrt{T}\sigma + \ln{\dfrac{VaR_0}{V_0}}=0, \end{equation} which can be solved for sigma via discriminant as $ax^2 + bx + c = 0$, when $\sigma$ replaces the $x$. Finally, a negative solution being disregarded one gets VEV formula: \begin{equation} \sigma = \dfrac{\sqrt{3.8416-2\ln{\dfrac{VaR_0}{V_0}}}-1.96}{\sqrt{T}}. \end{equation}


Not the answer you're looking for? Browse other questions tagged or ask your own question.