$\rho_\gamma(X)=\frac{1}{\gamma} \log \mathbb{E}[e^{-\gamma X}]$

Question

$\rho_\gamma(X)=\frac{1}{\gamma} \log \mathbb{E}[e^{-\gamma X}]$

is a convex risk measure, but it fails the subadditivity property in order to be called coherent.

A mapping $\rho:L^\infty(\Omega,\mathcal{F},\mathbb{P})\rightarrow \mathbb{R}$ is called a coherent risk measure if the following properties hold

If $X \geq 0$ then $\rho(X)\leq 0$.

Subadditivity: $\rho(X_1+X_2)\leq \rho(X_1)+\rho(X_2)$

Positive homogeneity: for $\lambda \geq 0$ we have $\rho(\lambda X)=\lambda\rho(X) $

Translation Invariance: For all $ c\in \mathbb{R}$ $\rho(X+c)=\rho(X)-c$

Im trying to investigate why it fails, or better, under which condition it is not true. \begin{align*} \rho_\gamma(X_1+X_2) & = \frac{1}{\gamma}\log\mathbb{E}[e^{-\gamma(X_1 + X_2)}]\\ & = \frac{1}{\gamma}\log\mathbb{E}[e^{-\gamma X_1}e^{-\gamma X_2}] \end{align*} at this stage I am a bit stuck, becasue I think it depends on how $X_1$ and $X_2$ are correlated to continue.. hm any help would be great!

Answer 1

Cauchy-Schwarz inequality shows that $\rho_\gamma(2X)\geqslant2\rho_\gamma(X)$ for every random variable $X$ and every positive $\gamma$, and that the inequality is strict as soon as $X$ is not almost surely constant.

Thus, subadditivity and positive homogeneity both fail.