2
$\begingroup$

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:

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

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

Improve this question
$\endgroup$
4
  • $\begingroup$ Down-voted, I'd like to know what have you tried? Did you write down the definitions? $\endgroup$ Commented Sep 28, 2017 at 14:07
  • $\begingroup$ I believe you should improve your question to get a better response. $\endgroup$ Commented Sep 29, 2017 at 5:45
  • $\begingroup$ So, given those formula's, would you say that there is a relationship? $\endgroup$ Commented Sep 29, 2017 at 8:35
  • $\begingroup$ Actually, I am not sure. Since $VaR_{\alpha}(-X) = \sup{x \in \mathbb{R}| F_{X}(x) \leq 1-\alpha}$}, I was wondering whether there could be some relation between $VaR_{\alpha}(-X)$ and $VaR_{\alpha}(X)$. It does not seem apparently. Maybe some relation holds by applying some trick. $\endgroup$ Commented Sep 29, 2017 at 9:07

1 Answer 1

Reset to default
2
$\begingroup$

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*}

Improve this answer
$\endgroup$
3
  • $\begingroup$ Hi Gordon. Thank you again for your help. Your answer is really helpful. But I still doubt about a sign. Could you help me check it ? \begin{equation} \begin{split} CVaR_{\alpha}(-X) &= \frac{1}{\alpha}\int_{0}^{\alpha}VaR_{s}(-X)ds \\ &= -\frac{1}{\alpha}\int_{0}^{\alpha}VaR_{1-s}(X)ds\\ &= -\frac{1}{\alpha}\int_{1}^{1-\alpha}VaR_{t}(X)d(1-t)\\ &= -\frac{1}{\alpha}\int_{1-\alpha}^{1}VaR_{t}(X)dt)\\ &= -\frac{1}{\alpha}\left(\int_{1-\alpha}^{0}VaR_{t}(X)dt)+\int_{0}^{1}VaR_{t}(X)dt)\right)\\ \end{split} \end{equation} $\endgroup$ Commented Oct 11, 2017 at 10:59
  • $\begingroup$ \begin{equation} \begin{split} &= -\frac{1}{\alpha}\left(\int_{0}^{1}VaR_{t}(X)dt)-\int_{0}^{1-\alpha}VaR_{t}(X)dt)\right)\\ &= -\frac{1}{\alpha}\left(\int_{0}^{1}F^{-1}(1-t)dt - (1-\alpha)CVaR_{1-\alpha}(X)\right)\\ &= \frac{1}{\alpha}\int_{1}^{0}F^{-1}(s)ds + \frac{1-\alpha}{\alpha}CVaR_{1-\alpha}(X)\\ &= -\frac{1}{\alpha}\int_{0}^{1}F^{-1}(s)ds + \frac{1-\alpha}{\alpha}CVaR_{1-\alpha}(X)\\ &=-\frac{1}{\alpha}E[X] + \frac{1-\alpha}{\alpha}CVaR_{1-\alpha}(X). \end{split} \end{equation} $\endgroup$ Commented Oct 11, 2017 at 11:00
  • $\begingroup$ @Xinyuan: Thanks for pointing this out. See the correction. $\endgroup$ Commented Oct 11, 2017 at 12:51

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.