0
$\begingroup$

A coherent risk measure is:

$\rho(\lambda X_1+(1-\lambda X_2))$

How can it be shown that everey convex risk measure is indeed a coherent risk measure?

I assume that it is enough to show that a convex risk measure is coherent by using, subadditivity, positive homogeniety. So we get: $\rho(\lambda X_1+(1-\lambda X_2))=\rho(\lambda X_1)+\rho((1-\lambda)X_2)=\lambda \rho(X_1)+(1-\lambda)\rho(X_2))$ right?

$\endgroup$
  • $\begingroup$ What does your first formula say? This is $\rho$ of something .. this is no property not theorem .. just nothing. $\endgroup$ – Richard Dec 30 '15 at 14:29
  • $\begingroup$ $\rho$ is a risk measure $\endgroup$ – Elekko Dec 30 '15 at 15:03
  • $\begingroup$ the first formula is still not a statement ... $\endgroup$ – Richard Dec 30 '15 at 15:40
2
$\begingroup$

We define a convex risk measure as $$ \rho( \lambda X_1 + (1-\lambda) X_2) \le \lambda \rho( X_1 ) + (1-\lambda) \rho(X_2), $$ for $\lambda \in(0,1) $.

A coherent risk measure is subadditive and homogeneous thus for coherent $\rho$ we get: $$ \rho( \lambda X_1 + (1-\lambda) X_2) \le \rho( \lambda X_1) + \rho( (1-\lambda) X_2) $$ by subadditivity and $$ \rho( \lambda X_1) + \rho( (1-\lambda) = \lambda \rho(X_1) + (1-\lambda)\rho(X_2) $$ by homogeneity. Thus a coherent risk measure is convex. The reverse is not true in general.

$\endgroup$
  • $\begingroup$ So a convex risk measure satisfies 3 axioms: convexity, subadditivity and homogeniety? $\endgroup$ – Elekko Dec 30 '15 at 15:05
  • $\begingroup$ no, but if you have subadditivity and homogeneity then you automatically have a convex risk measure. This is what "implied" means. Coherent implies convex. Thus convex is more general. $\endgroup$ – Richard Dec 30 '15 at 15:41
  • $\begingroup$ Every coherent measure is convex. The reverse is not true (but maybe in a lot of cases ... still not in general). $\endgroup$ – Richard Dec 30 '15 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.