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\}.


The average value-at-risk is the expectation of the part of a distribution that sits above its (1-a)-quantile. The average value-at-risk interpolates between the expectation operator (alpha=1) and the essential maximum (alpha=0).

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<Y (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<Y (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<Y (a.s.). Then X\leq Y (a.s.). In that case, however, we may have X<Y 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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