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When the risk is defined by a discrete random variable, is CVaR a coherent risk measure? I stick to the following definition of CVaR:

$$ CVaR_\alpha(R) = \min_v \quad \left\{ v + \frac{1}{1-\alpha} \mathbb{E}[R-v]^+ \right \}$$

where $R$ is the DISCRETE random variable for the loss and $\alpha$ is the confidence level.

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I'm almost sure CVaR has been created to satisfy the sub-additivity property which isn't valid for normal VaR. The whole point of this new measure was to "enhance" VaR and have a coherent measure. I must say that I am unsure of what the fact that it is DISCRETE changes? – SRKX Jun 24 '11 at 10:51

3 Answers 3

Conditional VaR (CVaR), which is also called Expected Shortfall, is a coherent risk measure (although being derived from a non-coherent one, namely VaR).

See this paper:

Expected Shortfall: a natural coherent alternative to Value at Risk
from Carlo Acerbi and Dirk Tasche

EDIT: I just saw that you emphasized discrete but that shouldn't change the general situation.

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[2002 - Acerbi] Spectral measures of risk: a coherent representation of subjective risk aversion This paper actually says that CVaR is not a coherent measure in general. But, now I think the author was confused with different names for similar concepts. – FEQ Jun 24 '11 at 19:46
I agree to @vonjd's answer. In his link in equation (12) that's the thing to do if you have atoms in the distribution (could be continuous with points of mass - for example at 0 in an insurance/operational loss example). Formula (12) applies for discrete distributions and the "correction term" vanishes for continuous ones (or when there is no atom). – Richard Jul 24 '12 at 12:04

$VaR^\alpha$ is not a coherent risk measure because it fails sub-additivity (a coherent risk measure is monotonic, sub-additive, positive homogenous, and translation invariant). The expectation operator $E[\cdot]$ is linear, so it meets sub-additivity, as well as the other three properties, so $CVaR$ is a coherent risk measure.

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Why can't use "minimum"? I can't think of a good example without a minimum. Could you suggest one? – FEQ Jun 24 '11 at 20:01
@chang -- Good catch! You're right. I read too quickly and misinterpreted discrete $R$ as $R$ in a finite set, which would require infimum. – Richard Herron Jun 25 '11 at 15:58
Thanks. I thought you thought in that way. – FEQ Jun 27 '11 at 17:31
up vote 8 down vote accepted

I found this paper: Conditional value-at-risk for general loss distributions by Rockafellar and Uraysev

which says CVaR is coherent for general loss distributions, including discrete distributions.

I think that I was confused by other authors who were also confused with the definitions of CVaR. In particular, in the following paper, the author mistakenly stated that Tail Conditional Expectation (TCE) is same as CVaR, and they are not coherent.

However, TCE is not same as CVaR in general. If the underlying distribution is continuous, they are same.

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