# How to minimize $CVaR_{\alpha}(\min(X,d))$, where $X$ is a random variable and d is the decision variable?

How to solve the following problem, $$\min_{d \in \mathbb{R}^{+}} \text{CVaR}_{\alpha}(\min(X,d))$$, where, X is a random variable whose distribution function $f_{X}(x)$ is given and $d$ is the decision variable.

• Given $X$ is a random variable, can I say that $\text{CVaR}_{\alpha}(\min(X,d))$ is a increasing function of $d$? – Xinyuan Wei Jul 16 '18 at 2:50

The minimum value is always attained at $d=0$.

In this proof, I will assume that the distribution of the random variable $X$ is absolutely continuous and monotonically increasing, and thus the CDF of $X$ is invertible (though I believe the result holds generally).

Fix $\beta\in(0,1)$. We have that

$$\Psi(d,\alpha)\equiv\int_{\min\{d,x\}\leqslant\alpha}p(x)dx=\begin{cases}1,&d\leqslant\alpha,\\F(\alpha),&d>\alpha,\end{cases}$$

where $p$ and $F$ are the PDF and CDF of the random variable $X$, respectively.

By definition, $\text{VaR}_\beta(\min\{X,d\})=\min\{\alpha\in\mathbb{R}\ |\ \Psi(d,\alpha)\geqslant\beta\}$. By inspection, we have that

$$\text{VaR}_\beta\big(\min\{X,d\}\big)=\min\{d,F^{-1}(\beta)\}.$$

Then by definition of CVaR:

$$\text{CVaR}_\beta\big(\min\{X,d\}\big)=\frac{1}{1-\beta}\int_{\min\{d,x\}\geqslant\min\{d,F^{-1}(\beta)\}}\min\{d,x\}p(x)dx.$$

There are two cases:

Case 1: $F^{-1}(\beta)\leqslant0$. In this case, $\text{VaR}_\beta(\min\{d,X\})=F^{-1}(\beta)$ for all $d\in\mathbb{R}_+$. Hence

\begin{align} (1-\beta)\text{CVaR}_\beta(\min\{d,X\})&=\int_{F^{-1}(\beta)}^\infty\min\{d,x\}p(x)\ dx\\ &=\underbrace{\int_{F^{-1}(\beta)}^0xp(x)\ dx}_{\displaystyle\text{Constant w.r.t. }d}+\underbrace{\int_0^dxp(x)\ dx}_{\displaystyle\geqslant0\ \forall\ d\in\mathbb{R}_+}+\underbrace{\int_d^\infty dp(x)\ dx}_{\displaystyle=d(1-F(d))} \end{align} Since $d(1-F(d))\geqslant0$ for all $d\in\mathbb{R}_+$, we have that

$$\min_{d\in\mathbb{R}_+}\text{CVaR}_\beta(\min\{d,X\})=\frac{1}{1-\beta}\int_{F^{-1}(\beta)}^0xp(x)\ dx,$$

attained at $d=0$.

Case 2: $F^{-1}(\beta)>0$. A direct calculation similar to the above shows that

$$(1-\beta)\text{CVaR}_\beta(\min\{d,X\})= \begin{cases} d(1-F(d)),&0\leqslant{d}\leqslant{F^{-1}(\beta)},\\ \displaystyle d(1-F(d))+\int_{F^{-1}(\beta)}^dxp(x)\ dx,&F^{-1}(\beta)<d. \end{cases}$$

By inspection, we clearly have that

$$\min_{d\in\mathbb{R}_+}\text{CVaR}_\beta(\min\{d,X\})=0,$$

attained at $d=0$.