Skip to content
All library documents

Deriving PRIIP VaR-Equivalent Volatility from a Normal VaR

Article Quant Q&A · Author: splinter123

Summary

The document derives PRIIP VaR-equivalent volatility (VEV) by assuming that an asset follows a geometric Brownian process with zero drift over the stated risk horizon. Under this model, log returns have a mean tied to minus half the variance and a standard deviation determined by volatility. Setting the lower-tail quantile of the modeled asset value equal to the reported VaR gives an equation for volatility.

Taking logarithms turns that relationship into a quadratic. Solving it and retaining the positive root yields the VEV expression, with the square-root-of-time adjustment for horizons other than one year. A second explanation connects the same calculation to normally distributed log returns and notes that VaR may be expressed as a loss or as the resulting asset value. The derivation depends on the assumed distribution, quantile convention, and consistent use of VaR relative to the initial value; it does not establish that these assumptions describe every PRIIP return distribution.

Key ideas

  • VEV is inferred by equating a modeled lower-tail asset value with the stated VaR.
  • The derivation assumes geometric Gaussian returns with a specific volatility-dependent mean.
  • Taking logarithms produces a quadratic equation in volatility.
  • The positive quadratic root gives the volatility estimate, adjusted for the horizon.
  • The calculation depends on the VaR definition and the model assumptions.

Tags

Full text
# How is the formula for the VEV (VaR-equivalent volatility) in the PRIIP document derived?


# How is the formula for the VEV (VaR-equivalent volatility) in the PRIIP document derived?












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 by Robert (score 12, accepted)

https://quant.stackexchange.com/a/27952

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.

## Answer by Олег Бойко (score 3)

https://quant.stackexchange.com/a/35705

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

- 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.

- 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}

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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