Let $X$ be a position with a probability distribution $F$ that is strictly increasing and smooth. Let $f(x)=F^{\prime}(x)$ be the associated probability density.

(a) Verify that

\[\begin{equation*}\operatorname{CVaR}_{h}(X)=-\frac{1}{h} \int_{-\infty}^{-\operatorname{VaR}_{h}(X)} x f(x) \mathrm{d} x \tag{10.13}\end{equation*}\]

(b) For any $u \in(0,1)$ let $x=F^{-1}(u)$ be the value of $X$ that defines the $u$-quantile of $X$. Conversely, for any specific value $x$ of $X$, we have $u=F(x)$ as the quantile value associated with $x$. Using the change of variable $u=F(x)$ in equation (10.13), show that

\[\begin{equation*}\operatorname{CVaR}_{h}(X)=-\frac{1}{h} \int_{0}^{h} F^{-1}(u) \mathrm{d} u . \tag{10.14}\end{equation*}\]

(c) Interpret the right-hand side of equation (10.14) to obtain

\[\begin{equation*}\operatorname{AVaR}_{h}(X)=-\frac{1}{h} \int_{0}^{h} F^{-1}(u) \mathrm{d} u . \tag{10.15}\end{equation*}\]

and hence conclude that $\mathrm{CVaR}_{h}(X)=\operatorname{AVaR}_{h}(X)$. [Equation (10.15) can be extended to be coherent in general (e.g., with jumps in $F$ ) because the upper limit of the integral will force atoms to be split if necessary so that the integration is strictly over the range zero to $h$.]