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Let $V_\alpha(L)$ be the value at risk of a function $L$ at $\alpha\in[0,1]$, $$V_\alpha(L)=\inf\{x:P(L\le x)\ge\alpha\}.$$

Prove that for any $\alpha$, $$V_\alpha(-L) =-\lim_{\epsilon\rightarrow0^+}V_{(1-\alpha)+\epsilon}(L).$$ If $L$ has a continuous, strictly increasing CDF, then $V_\alpha(-L)=-V_{1-\alpha}(L)$

Doubt. Starting with the LHS, we have
$V_\alpha(-L)\\ =\inf\{x:P(-L\le x)\ge\alpha\}\\ =-\sup\{-x:P(-L\le x)\ge\alpha\}\\ =-\sup\{x:P(-L\le -x)\ge\alpha\}\\ =-\sup\{x:P(L\ge x)\ge\alpha\}\\ =-\sup\{x:P(L< x)\le1-\alpha\}$

Now, if $L$ had a continuous CDF, we can replace $P(L<x)$ by $P(L\le x)$. However, I'm not sure how to proceed from here for either non-continuous or continuous case. Hints are appreciated.

zaira
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2 Answers2

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The value at risk at $\alpha$ for the random variable $L$ is in fact the lowest $\alpha$-quantile of $L$. Let $F_L$ denote the cumulative distribution of $L$, then $V_L:\alpha\mapsto \inf\{x:F_L(x)\geq x\}$ is nondecreasing left continuous and $$F_L(x)\geq \alpha\quad\text{iff}\quad V_L(\alpha)\leq x \tag{0}\label{zero}$$ Similarly, there is $V^+_L:\alpha\mapsto\sup\{x: F_L(x-)\leq q\}$, is the largest $\alpha$-quantile of $L$ (where $F_L(x-)=\lim_{y\nearrow x}F(y)=P[L<x]$), it is nondecreasing right continuous and $$F_L(x-)=\lim_{y\nearrow x}F(y)\leq \alpha\quad\text{iff}\quad V^+_L(\alpha)\geq x\tag{1}\label{one} $$ Further more $$V_L(\alpha)=V^+_L(\alpha-)\leq V^+_L(\alpha)=V_L(\alpha+)=\lim_{\beta\searrow \alpha}V_L(\beta)\tag{2}\label{two}$$

This is not difficult to show by it take a few lines (see here).

Notice that $$F_{-L}(x)=P[L\geq -x]=1-P[L<-x]$$ Thus, from \eqref{one} and \eqref{two} \begin{align} V_{-L}(\alpha)&=\inf\{y:1-P[L<-y]\leq \alpha\}=\inf\{-y:P[L<y]\leq 1-\alpha\}\\ &=-\sup\{y:F_L(y-)\leq 1-\alpha\}=-V^+_L(1-\alpha) \end{align}

in particular, when $L$ is a continuous random variable $$V_{-L}(\alpha)=-V^+_L(1-\alpha)=-V_L(1-\alpha)$$

Mittens
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Let us start by making this claim (a picture always helps!).

Claim: for all $\alpha \in [0,1]$ and $x\in\mathbb{R}$, $$\displaystyle\sup\{x: \mathbb{P}(L<x)\le 1-\alpha\} = \displaystyle\sup\{x: \mathbb{P}(L\le x)\le 1-\alpha\}\,.$$

Proof: The CDF being monotone has at most countably many discontinuities, which means that the set of continuity points is dense in $\mathbb{R}$. The CDF being non-decreasing immediately gives that $$\displaystyle\sup\{x: \mathbb{P}(L<x)\le 1-\alpha\} \ge \displaystyle\sup\{x: \mathbb{P}(L\le x)\le 1-\alpha\}\,.$$ To show the inequality cannot be strict, argue by contradiction and show that if it were so, then there would be $y$ in between where $y$ is a continuity point of the CDF such that the following holds $$1-\alpha < \mathbb{P}(L\le y) = \mathbb{P}(L<y)\le 1-\alpha\,,$$ a contradiction.

Now, I claim (note that the limit exists by monotonicity of the CDF)

Claim: $\displaystyle\sup\{x: \mathbb{P}(L\le x) \le 1-\alpha\} = \displaystyle \lim_{\epsilon \to 0^+} \inf\{x: \mathbb{P}(L\le x) \ge 1-\alpha + \epsilon\}$.

Proof: to see this, the monotonicity of the CDF immediately gives the inequality $$\displaystyle\sup\{x: \mathbb{P}(L\le x) \le 1-\alpha\} \le \displaystyle \lim_{\epsilon \to 0^+} \inf\{x: \mathbb{P}(L\le x) \ge 1-\alpha + \epsilon\}\,.$$ Now, suppose for a contradiction that it is strict, that is, there exists some $y\in \mathbb{R}$ such that \begin{align}\displaystyle\sup\{x: \mathbb{P}(L\le x) \le 1-\alpha\} &< y < \displaystyle \lim_{\epsilon \to 0^+} \inf\{x: \mathbb{P}(L\le x) \ge 1-\alpha + \epsilon\}\\ & = \displaystyle \inf_{\epsilon \to 0^+} \inf\{x: \mathbb{P}(L\le x) \ge 1-\alpha + \epsilon\}\,. \end{align} where the las equality follows from the fact that $\alpha \mapsto V_\alpha (L)$ is non-decreasing. Then, it follows that $1-\alpha < \mathbb{P}(L\le y) \le 1- \alpha$, a contradiction.

Combining the above now gives the desired equality.

Pantelis Tassopoulos
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