In https://sites.math.washington.edu/~rtr/papers/rtr206-RiskTutorial_INFORMS2007.pdf, Rockafellar defines a coherent risk measure in the following way:
$\rho : \mathcal{L}^2(\Omega, \mathcal{F}, P) \rightarrow (-\infty, \infty]$ is a coherent measure of risk in the basic sense if
- Constantness: $\rho(C) = C $ for all constants $C$,
- Convexity: $\rho( \lambda X + (1-\lambda)X') \leq \lambda \rho(X) + (1-\lambda)\rho(X')\quad$ for $0\le\lambda\le 1$,
- Monotonicity: $\rho(X) \leq \rho(X')$ when $X \leq X'$,
- Closedness: $\rho(X) \leq 0$ when $\|X_k - X \|_ 2 \rightarrow 0 $ with $\rho(X_k) \leq 0$.
- Homogeneity: $\rho(\lambda X) = \lambda \rho(X),$ when $ \lambda >0 $
These axioms are slightly different than other definitions, and in particular the axiom of translation invariance is not included (but closedness is), cf. for instance https://en.wikipedia.org/wiki/Coherent_risk_measure#Translation_invariance.
I wonder: Can we really derive translation invariance (ie., $\rho(X + m) = \rho(X)+m$, when $m \in \mathbb{R}$ ) from the axioms above? It is clear that: \begin{align} \rho(X + m) \leq \rho(X) + m, \end{align} for $m \in \mathbb{R}$ by axiom 1 and 2 and 3. But this does not show equality. Is it possible to derive the opposite inequality $\rho(X) + m \leq \rho(X+m)$, or in some other way show $\rho(X + m) = \rho(X) + m$, or is simply not translation invariance a feature of these axioms?
Reviewing this some more, I wonder if the following argument is simply enough: Let $X$ be a random variable, $m > 0$ a constant. Then, \begin{align} \rho(X) &= \rho(\frac{1}{2}( 2X + 2m) - \frac{1}{2}2m ) \\ &\leq \rho(\frac{1}{2}( 2X + 2m)) - m \\ & \leq \rho(X) +m -m \\ &= \rho(X), \end{align} giving \begin{align} \rho(X) +m \leq \rho(X +m) \leq + \rho(X) +m. \end{align}