Why Capped-Loss CVaR Is Minimized at a Zero Cap
Summary
The note considers minimizing the conditional value at risk of a random variable capped above by a nonnegative decision level. It derives the value-at-risk of the capped variable from the distribution function, then evaluates CVaR in two cases depending on whether the relevant quantile lies below or above zero. In both cases, the argument concludes that setting the cap to zero attains the minimum.
The derivation assumes an absolutely continuous distribution with an invertible cumulative distribution function, while suggesting the result may hold more broadly. This setup also makes the result depend on the stated objective and the nonnegative domain for the cap; it is not a general prescription for choosing a trading risk limit, since a zero cap makes the capped quantity zero whenever the original variable is positive.
Key ideas
- The capped variable has a quantile equal to the smaller of the cap and the corresponding quantile of the original variable.
- The proof splits according to whether that original quantile is nonpositive or positive.
- Under the stated setup, the derivation finds the minimum CVaR at a cap of zero.
- The distributional assumptions in the proof are stronger than the author believes are necessary.
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Full text
# How to minimize $CVaR_{\alpha}(\min(X,d))$, where $X$ is a random variable and d is the decision variable?
# 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.
## Answer by David M. (score 2, accepted)
https://quant.stackexchange.com/a/41312
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
\begin{equation} (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} \end{equation}
By inspection, we clearly have that
\begin{equation} \min_{d\in\mathbb{R}_+}\text{CVaR}_\beta(\min\{d,X\})=0, \end{equation}
attained at $d=0$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.