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Proving Subadditivity of Entropic Value at Risk

Article Quant Q&A · Author: Amy Zhang
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Summary

The document addresses how to prove subadditivity for an entropic risk measure, after monotonicity, positive homogeneity, and translation invariance have been established. The response introduces an auxiliary function of a random variable and a positive parameter, then relies on its joint convexity. A convexity inequality is sketched using Jensen’s inequality and log-convexity of moment-generating functions, under a stated condition that the relevant exponential moments exist.

The response then combines parameter choices and positive homogeneity to derive the subadditivity inequality. This provides a route for establishing coherence, but the proof as written has notation and indexing inconsistencies in the final parameter substitutions, and it assumes minimizers exist where infima may not be attained. The derivation should therefore be checked and formalized before relying on it, including the domain of the random variables and the treatment of nonattained infima.

Key ideas

  • The proof strategy is to establish joint convexity of an auxiliary function in the loss and a positive parameter.
  • Jensen’s inequality and log-convexity of moment-generating functions support the convexity argument.
  • Combining convexity with positive homogeneity can yield subadditivity of the entropic risk measure.
  • The argument assumes exponential moments exist for all positive parameters in its stated domain.
  • The displayed derivation contains notation issues and may assume infima are attained, so formal verification is needed.

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Risk managementStatistics
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# Prove Subadditivity - Entropic Value at Risk


# Prove Subadditivity - Entropic Value at Risk












Any insight in how to prove the following risk measure is subadditive? $\rho_{1-\alpha}(X) = \inf_{z>0}\{z^{-1}\ln(\frac{E[e^{zX}]}{\alpha})\}$, with $\alpha \in ]0,1]$

I want to prove it is a coherent risk measure and already proved monotonicity, positive-homogeneity and translation invariance.

## Answer by David Nguyen (score 2)

https://quant.stackexchange.com/a/45313

If you can prove that $\kappa_{\alpha}(X,t)=a_X(\alpha,t^{-1})=t\ln(\frac{E[e^{t^{-1}X}]}{\alpha})$ is convex and apply the property of positive homogeneity, then the sub-additivity follows.

In original paper, authors show that $\kappa_{\alpha}(X,t)=a_X(\alpha,t^{-1})$ is convex in $(X,t)$.

- Lemma:

For fixed $\alpha$, all $\lambda\in[0,1],X,Y\in L_{M^+}$ and $t_1,t_2>0$, where $L_{M^+}$ is the space of random variables such that moments $M_X(z)$ exist for all $z>0$, then $\lambda\kappa_{\alpha}(X,t_1)+(1-\lambda)\kappa_{\alpha}(Y,t_2)\geq \kappa_{\alpha}(\lambda X+(1-\lambda)Y,\lambda t_1+(1-\lambda)t_2)$.

Proof:

$\lambda\kappa_{\alpha}(X,t_1)+(1-\lambda)\kappa_{\alpha}(Y,t_2)\geq \kappa_{\alpha}(\lambda X+(1-\lambda)Y,\lambda t_1+(1-\lambda)t_2)$

$\Leftrightarrow\lambda t_1\ln M_X(t_1^{-1})+(1-\lambda) t_2\ln M_Y(t_2^{-1})\geq (\lambda t_1+(1-\lambda)t_2)\ln M_{\lambda X+(1-\lambda)Y}[(\lambda t_1+(1-\lambda)t_2)^{-1}]$

Let $t=\lambda t_1+(1-\lambda)t_2$ and $w=\frac{\lambda t_1}{t}$, then LHS: $\lambda t_1\ln M_X(t_1^{-1})+(1-\lambda) t_2\ln M_Y(t_2^{-1})=t[w\ln M_X(t_1^{-1})+(1-w)\ln M_Y(t_2^{-1})]$

Recall the Jensen's Inequality in Probabilistic Form for the concave function $x^w$ for $x>0;w\in[0,1]$ and replace $x$ by $e^{X/t}$: $\phi(E[X])\geq E(\phi(X))\Rightarrow (E[e^{X/t}])^w\geq E[(e^{X/t})^w]\Rightarrow w\ln(E[e^{X/t}])\geq \ln(E[(e^{X/t})^w])$ or $w\ln(M_X(t^{-1}))\geq ln(M_X(wt^{-1}))$.

So $w\ln(E[e^{Xt_1^{-1}}])\geq \ln(E[e^{Xwt_1^{-1}}]);(1-w)\ln(E[e^{Yt_2^{-1}}])\geq \ln(E[e^{Y(1-w)t_2^{-1}}])$.

Hence, remind that moment-generating function is log-convex: $LHS=t[w\ln M_X(t_1^{-1})+(1-w)\ln M_Y(t_2^{-1})]\geq t[\ln(E[e^{Xwt_1^{-1}}])+\ln(E[e^{Y(1-w)t_2^{-1}}])]$ $=t\ln(E[e^{Xwt_1^{-1}}]E[e^{Y(1-w)t_2^{-1}}])\geq t\ln(E[e^{Xwt_1^{-1}+Y(1-w)t_2^{-1}}])=t\ln(E[e^{X\lambda t^{-1}+Y(1-\lambda)t^{-1}}])=RHS$.

- Sub-additivity:

$\frac{1}{2}[\rho_{1-\alpha}(X) + \rho_{1-\alpha}(Y)]=\rho_{1-\alpha}(\frac{1}{2}X) + \rho_{1-\alpha}(\frac{1}{2}Y) = \inf_{t>0}\{\kappa_{\alpha}(\frac{1}{2}X,t)\}+\inf_{t>0}\{\kappa_{\alpha}(\frac{1}{2}Y,t)\} = \kappa_{\alpha}(\frac{1}{2}X,t_{X/2})+\kappa_{\alpha}(\frac{1}{2}Y,t_{Y/2})\geq \kappa_{\alpha}(\frac{1}{2}(X+Y),\frac{1}{2}(t_{X/2}+t_{X/2}))\geq \inf_{t>0}\{\kappa_{\alpha}(\frac{1}{2}(X+Y),t)\}=\rho_{1-\alpha}(\frac{1}{2}(X+Y))=\frac{1}{2}\rho_{1-\alpha}(X+Y)\Rightarrow \rho_{1-\alpha}(X) + \rho_{1-\alpha}(Y)\geq \rho_{1-\alpha}(X+Y)$

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.