# How to prove the following relation of Conditional Value-at-Risk and Value-at-Risk?

How to prove the following relation of Conditional Value-at-Risk $$\text{CVaR}_{\alpha}(X)$$ and Value-at-Risk $$\text{VaR}_{\alpha}(X)$$, $$\begin{equation} \text{CVaR}_{\alpha}(X) = \text{VaR}_{\alpha}(X)+\frac{1}{\alpha}E[(X-\text{VaR}_{\alpha}(X))^{+}]? \end{equation}$$ Here are the definations of Value-at-Risk and Conditional Value-at-Risk.

Value-at-Risk

Suppose $$X$$ is a random variable, the value-at-risk (VaR) of $$X$$ at a confidence level $$1-\alpha$$ where $$0<\alpha<1$$ is defined as $$\begin{equation} \text{VaR}_{\alpha}(X) := \inf\left\{x :Pr\{X>x\}\leq\alpha\right\}. \end{equation}$$

Conditional Value-at-Risk

Based on the definition of Value-at-Risk, the Donditional Value-at-Risk (CVaR) of $$X$$ at a confidence level $$1-\alpha$$ (namely, significance level $$\alpha$$) is defined to be $$\begin{equation} \mathrm{CVaR}_{\alpha}(X) = \frac{1}{\alpha}\int_{0}^{\alpha}\mathrm{VaR}_{s}(X)ds. \end{equation}$$ Let $$F$$ be the cumulative distribution function of $$X$$. We assume that $$F$$ is continuous. Then, for $$x\ge 0$$, \begin{align*} F^{-1}(x) = \inf\{s: F(s) \ge x \}. \end{align*} Moreover, \begin{align*} \text{VaR}_{\alpha}(X) &= \inf\left\{x :1-F(x) \le \alpha\right\}\\ &=F^{-1}(1-\alpha). \end{align*} Consequently \begin{align*} E\Big(\big(X-\text{VaR}_{\alpha}(X)\big)^+\Big) &= \int_{-\infty}^{\infty} \Big(x-\text{VaR}_{\alpha}(X)\Big)^+ dF(x)\\ &=\int_{\text{VaR}_{\alpha}(X)}^{\infty} \Big(x-\text{VaR}_{\alpha}(X)\Big) dF(x)\\ &=\int_{1-\alpha}^1 \Big(F^{-1}(y)-\text{VaR}_{\alpha}(X)\Big) dy\\ &=\int_{1-\alpha}^1 F^{-1}(y) dy - \alpha \text{VaR}_{\alpha}(X) \\ &=\int_{1-\alpha}^1 \text{VaR}_{1-y}(X) dy - \alpha \text{VaR}_{\alpha}(X) \\ &=\int_0^{\alpha} \text{VaR}_{s}(X) ds - \alpha \text{VaR}_{\alpha}(X). \end{align*} That is, \begin{align*} \text{VaR}_{\alpha}(X)+\frac{1}{\alpha}E\Big(\big(X-\text{VaR}_{\alpha}(X)\big)^+\Big) &= \frac{1}{\alpha}\int_{0}^{\alpha}\text{VaR}_{s}(X)ds. \end{align*}