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.
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 $(X,t)$.
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: $LHS=t[w\ln M_X(t_1^{-1})+(1-w)\ln M_Y(t_2^{-1})]\geq t[\ln(E[e^{Xwt_1^{-1}}])+\ln(E[e^{Y(1-w)t_2^{-1}}])]$ $=t\ln(E[e^{Xwt_1^{-1}}]E[e^{Y(1-w)t_2^{-1}}])\geq t\ln(E[e^{Xwt_1^{-1}+Y(1-w)t_2^{-1}}])=t\ln(E[e^{X\lambda t^{-1}+Y(1-\lambda)t^{-1}}])=RHS$.
$\frac{1}{2}[\rho_{1-\alpha}(X) + \rho_{1-\alpha}(Y)]=\rho_{1-\alpha}(\frac{1}{2}X) + \rho_{1-\alpha}(\frac{1}{2}Y) = \inf_{t>0}\{\kappa_{\alpha}(\frac{1}{2}X,t)\}+\inf_{t>0}\{\kappa_{\alpha}(\frac{1}{2}Y,t)\} = \kappa_{\alpha}(\frac{1}{2}X,t_{X/2})+\kappa_{\alpha}(\frac{1}{2}Y,t_{Y/2})\geq \kappa_{\alpha}(\frac{1}{2}(X+Y),\frac{1}{2}(t_{X/2}+t_{X/2}))\geq \inf_{t>0}\{\kappa_{\alpha}(\frac{1}{2}(X+Y),t)\}=\rho_{1-\alpha}(\frac{1}{2}(X+Y))=\frac{1}{2}\rho_{1-\alpha}(X+Y)\Rightarrow \rho_{1-\alpha}(X) + \rho_{1-\alpha}(Y)\geq \rho_{1-\alpha}(X+Y)$
