If the probability distribution function $f(x)$ is not symmetric, is there any relationship between $VaR_{\alpha}(X)$ and $VaR_{1-\alpha}(X)$?

Here, $VaR$ is defined as $$ VaR_{\alpha}(X) := \inf\left\{x \in \mathbb{R}| Pr(X>x)\leq \alpha\right\}, \alpha \in [0, 1]. $$

  • $\begingroup$ If the distribution is not symmetric but fully general (discrete, continuous, mixed) then anything can happen ... $\endgroup$
    – Richi Wa
    Oct 13, 2017 at 7:38
  • $\begingroup$ In my answer to this question, I didn't assume any symmetric property. That is, the same relationship should be hold. What kind of symmetric property you are referring to? Can you revise your question to make it more specific? $\endgroup$
    – Gordon
    Oct 13, 2017 at 13:04

1 Answer 1


No, because the VaR is defined as a quantil. For example, you have the loss-vector l=(-1,-2,3,4,5,6,7,8,9,10). The VaR(90%) is 9. And it is also VaR(90%)=9, if you have l=(8,8,8,8,8,8,8,8,9,10). The VaR is independent of the values before and after his value. This is also a disadvantage of the VaR and one reason to take also the expected shortfall (mean of the losses that are bigger than the VaR).

  • $\begingroup$ Thank you for your help. There seems no relationship between these two for a nonsymmetric distribution. $\endgroup$
    – Xinyuan
    Oct 13, 2017 at 13:00
  • 1
    $\begingroup$ The relationship $VaR_{\alpha}(-X) =-VaR_{1-\alpha}(X)$ should always hold. You can check with you examples. $\endgroup$
    – Gordon
    Oct 13, 2017 at 13:10
  • $\begingroup$ @Gordon: This hold if first f(X) is symmetric (not true) and second X has mean zero. $\endgroup$
    – Klaus
    Oct 13, 2017 at 13:41
  • $\begingroup$ It holds without assuming any symmetric property and we do not need to assume that it have zero mean. Please check with your example. $\endgroup$
    – Gordon
    Oct 13, 2017 at 14:08
  • $\begingroup$ I did not see the minus in the first VaR. So, my comment to your special definition is not right. But an example: L=( VaR90% (-x)=2 and -VaR10% (x)=-2. It is not equal. But mostly, you are right. $\endgroup$
    – Klaus
    Oct 13, 2017 at 14:33

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