What is a coherent risk measure, and why do we care? Can you give a simple example of a coherent risk measure as opposed to a non-coherent one, and the problems that a coherent measure addresses in portfolio choice?
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$\begingroup$ There are 4 defining properties of coherent risk measures You can find them here as well as examples for coherent and counterexamples of those kind of risk measures. $\endgroup$– TheBridgeFeb 1, 2011 at 15:18
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$\begingroup$ Thanks. But I still don't get it: who defines these properties, where do they come from, is there any theoretical basis for imposing them, and why? $\endgroup$– DimitrisFeb 1, 2011 at 15:57
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$\begingroup$ I think these are just formal ways of describing informal "common-sense" ideas about risk. In the Wikipedia article, each axiom has a short sentence that descries the motivation -- such as "the risk of two portfolios together cannot get any worse than adding the two risks separately". $\endgroup$– Curt HagenlocherFeb 2, 2011 at 15:06
3 Answers
I'm just providing a global answer to the question, as I think it can be interesting for some beginners in quant finance.
The properties given by TheBridge:
Normalize
$\rho (\emptyset)=0$
This means you have no risk in taking no position.
Sub-addiitivity
$\rho(A_1+A_2) \leq \rho(A_1)+\rho(A_2)$
Having a position in two different can only decrease the risk of the portfolio (diversification)
Positive homogeneity
$\rho(\lambda A) = \lambda \rho(A)$
Doubling a position in an asset A doubles your risk.
And finally,
Translation invariance
$\rho(A + x) = \rho(A)-x$
That is, adding cash to a portfolio only diminishes the risk.
So a risk-measure is said to be coherent if and only if it has all these properties.
Note that this is just a convention, but it is motivated by the fact that all these properties are the ones an investor expects to hold for a risk measure.
Finally, notice that neither VaR nor Var are coherent risk measures, wherease the Expected Shortfall is.
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1$\begingroup$ What about monotonicity ? As per Klugman et al "it is useful to think of the random variables X and Y as the loss random variables for two divisions and X + Y as the loss random variable for the entity created by combining the two key divisions" ... and importantly "monotonicity means that if one risk always has greater losses than another risk, the risk measure should always be greater." $\endgroup$ Nov 10, 2015 at 1:46
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$\begingroup$ People claim VaR is not coherent is because VaR is not sub-additive. But based on you function in sub-addition part, it appears to me that VaR doesn't violate on this part. For example, using variance-covariance method to calculate VaR of a portfolio, the VaR result complies with the sub-addition function as portfolio VaR is always less than sum of sub-product VaR. Could you explain a little more about why VaR is not sub-additive or provide any example $\endgroup$– HuiJun 18, 2018 at 18:49
Coherent risk measures were created to address the problem that extant risk measures, like VaR, did not: namely that a risk measure should reward diversification.
I don't think that we should care if a risk measure is coherent.
The reason that VaR is not coherent is because it need not be sub-additive. I'm willing to stand corrected, but I doubt that VaR is very far from sub-additive in practical situations. And I don't see a great deal of harm if it were. I have several problems with VaR but non-coherent is not among them.
The homogeneity condition is wrong. I call this the Amaranth condition -- it turns out that being all of one side of a market is risky.
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$\begingroup$ I'm not sure I get what you don't "like" in the homogeneity. Could you please explain a bit further? $\endgroup$– SRKXSep 27, 2011 at 16:36
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$\begingroup$ The homogeneity condition claims that it is only 100 times more risky to own all of one side of a market than to share it equally with 99 others. I find that hard to believe. $\endgroup$ Sep 27, 2011 at 20:06
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$\begingroup$ I agree with you about homogeneity for large relative positions is not sensible, but it is worth noting that VaR doesn't address that, at least in implementations I have seen. $\endgroup$– Brian BOct 25, 2011 at 18:56