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VaR and CVaR Relationships for a Variable and Its Negation

Article Quant Q&A · Author: Xinyuan
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Summary

The answer derives how Value-at-Risk and Conditional Value-at-Risk change when a random variable is negated. Assuming a strictly increasing distribution function, it writes VaR as the inverse distribution function at the complementary tail probability. Applying the distribution of the negated variable yields a sign reversal and a complementary quantile: VaR at level α for −X equals negative VaR for X at level 1−α.

For CVaR, integrating the VaR identity gives a relation involving the mean of X and CVaR of X at the complementary level. The derivation uses the stated tail-integral definition and depends on the distributional regularity assumption; conventions for loss versus return, tail probability, and quantile definitions may differ across sources. It provides a mathematical identity, not an empirical risk comparison or trading application.

Key ideas

  • Under a strictly increasing distribution, VaR can be expressed using the inverse cumulative distribution function.
  • Negating a variable reverses the sign and switches VaR to the complementary probability level.
  • The CVaR of the negated variable depends on the mean and complementary-level CVaR of the original variable.
  • The formulas rely on the document’s stated VaR and CVaR conventions.

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Risk managementStatistics
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# Answer by Gordon (score 2, accepted)


# Questions about VaR and CVaR. Is there any relation between $VaR_{\alpha}(X)$ and $VaR_{\alpha}(-X)$, or $CVaR_{\alpha}(X)$ and $CVaR_{\alpha}(-X)$?












I have some questions when dealing with Value-at-Risk (VaR) and Conditional Value-at-Risk (CVaR).

Is there any relationship between $VaR_{\alpha}(X)$ and $VaR_{\alpha}(- X)$, or $CVaR_{\alpha}(X)$ and $CVaR_{\alpha}(-X)$ ?

Here, $VaR$ and $CVaR$ are defined as:

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

$$CVaR_{\alpha}(X) := \frac{1}{\alpha}\int_{0}^{\alpha}VaR_{s}(X)ds$$

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/36248

We consider the case where the distribution function $F$ of $X$ is strictly increasing. Then \begin{align*} VaR_{\alpha}(X) &= \inf\{x: P(X >x) \le \alpha \}\\ &=\inf\{x: F(x)\ge 1-\alpha \}\\ &=F^{-1}(1-\alpha). \end{align*} Moreover, we note that the distribution function $G$ of $-X$ is defined by \begin{align*} G(x) &= P(-X \le x) \\ &=1-F(-x), \end{align*} Then, \begin{align*} VaR_{\alpha}(-X) &= G^{-1}(1-\alpha)\\ &=-F^{-1}(\alpha)\\ &=-VaR_{1-\alpha}(X). \end{align*} Furthermore, \begin{align*} CVaR_{\alpha}(-X) &=\frac{1}{\alpha}\int_0^{\alpha}VaR_s(-X)ds\\ &=-\frac{1}{\alpha}\int_0^{\alpha}VaR_{1-s}(X)ds\\ &={\color{red}{-}}\frac{1}{\alpha}\int_{1-\alpha}^1 VaR_s(X)ds\\ &=-\frac{1}{\alpha}\left(\int_0^1 VaR_s(X)ds - \int_0^{1-\alpha} VaR_s(X)ds\right)\\ &=-\frac{1}{\alpha}\int_0^1F^{-1}(1-s)ds +\frac{1-\alpha}{\alpha}CVaR_{1-\alpha}(X)\\ &=-\frac{1}{\alpha}\int_0^1F^{-1}(s)ds +\frac{1-\alpha}{\alpha}CVaR_{1-\alpha}(X)\\ &=-\frac{1}{\alpha}\int_{-\infty}^{\infty} x\, dF(x) +\frac{1-\alpha}{\alpha}CVaR_{1-\alpha}(X)\\ &=-\frac{1}{\alpha} E(X) +\frac{1-\alpha}{\alpha}CVaR_{1-\alpha}(X). \end{align*}

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