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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}

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A slightly different take here:

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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*}

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