# Strict monotonicity of expected shortfall

The expected shortfall, also known as average value-at-risk or conditional value-at-risk, is a coherent risk measure defined as

$\mathrm{AV@R}_{\alpha}[Z]=\inf_{t\in\mathbb{R}} \{t+\alpha^{-1}\mathbb{E}[Z-t]_+\}$

for $Z\in\mathcal{Z}:=\mathcal{L}_p(\Omega,\mathcal{F},\mathrm{P})$ for some $p\in[1,+\infty]$.

The average value-at-risk can also be computed by the following useful formula

$\mathrm{AV@R}_{\alpha} = \mathbb{E}[X\mid X\geq \mathrm{V@R}_{\alpha}[X]],$

where $\mathrm{V@R}_{\alpha}[X]$ is the value-at-risk of $X$, that is

$\mathrm{V@R}_{\alpha}[X]=\inf\{t\mid \mathrm{P}[X\leq t]\geq 1-\alpha\}.$

We know that $\mathrm{AV@R}_{\alpha}$ is monotone, i.e., $\mathrm{AV@R}_{\alpha}[X]\leq \mathrm{AV@R}_{\alpha}[Y]$ whenever $X\leq Y$ (a.s.).

Is it, however, true that $\mathrm{AV@R}_{\alpha}[X] <\mathrm{AV@R}_{\alpha}[Y]$ whenever $X (a.s)? This property is known as strict monotonicity.

We know that this holds when $\mathrm{AV@R}_{\alpha}$ is replaced with the expectation operation, whereas, for $\|\cdot\|_{\infty}$ this not true.

\begin{aligned} \mathrm{AV@R}_{\alpha}[X]&=\mathbb{E}[X\mid X\geq \mathrm{V@R}_{\alpha}[X]]\\ &= \alpha^{-1}\int_{\mathrm{V@R}_{\alpha}[X]}^{+\infty} X\mathrm{dP}\\ &< \alpha^{-1}\int_{\mathrm{V@R}_{\alpha}[X]}^{+\infty}Y\mathrm{dP}\\ &\leq \alpha^{-1}\int_{\mathrm{V@R}_{\alpha}[Y]}^{+\infty}Y\mathrm{dP}\\ &= \mathrm{AV@R}_{\alpha}[Y] \end{aligned}

The strict inequality above is because of the strict monotonicity property of the integral and the second inequality is because $X (a.s.) implies $\mathrm{V@R}_{\alpha}[X]\leq \mathrm{V@R}_{\alpha}[Y]$.

Note: $\alpha^{-1}\int_{\mathrm{V@R}_{\alpha}[X]}^{+\infty}Y\mathrm{dP} = \mathbb{E}[Y \mid Y \geq \mathrm{V@R}_{\alpha}[X]] \leq \mathbb{E}[Y \mid Y \geq \mathrm{V@R}_{\alpha}[Y]]$ and this is because $\mathrm{V@R}_{\alpha}[X]\leq \mathrm{V@R}_{\alpha}[Y]$.

Note: We have denoted by $\mathbb{E}[X\mid H]$ the expectation of the random variable $X$ conditioned by the event $H\in\mathcal{F}$.

Now note that $\mathrm{AV@R}_0[\cdot]=\mathbb{E}[\cdot]$ by convention; this is because $\mathrm{AV@R}_{\alpha}[X]=\mathbb{E}[X\mid X\geq \mathrm{V@R}_{\alpha}[X]]$ and $\mathrm{V@R}_{0}[X]=\mathrm{esssup}[X]=\|X\|_{\infty}$. Take $X,Y\in\mathcal{L}_{\infty}(\Omega, \mathcal{F}, \mathrm{P})$ such that $X (a.s.). Then $X\leq Y$ (a.s.). In that case, however, we may have $X and $\|X\|_{\infty}=\|Y\|_{\infty}$; take for instance $\mathcal{Z}=\mathcal{L}_{\infty}(\Re_+,\mathcal{B}_{\Re_+}, \mathrm{P})$ and

\begin{aligned} X(\omega)&=\frac{\omega}{\sqrt{1+\omega^2}}\\ Y(\omega)&=\mathrm{erf}\left(\frac{\sqrt{\pi}}{2}\omega\right) . \end{aligned}

over the domain $\omega \in [0,\infty)$.

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