Testing the Aumann–Serrano Risk Measure for Cash Invariance

Summary

The note examines whether the Aumann–Serrano risk measure satisfies a key condition for coherent risk measures: adding a deterministic cash return should reduce measured risk by exactly that amount. It defines the measure through an exponential expectation and tests the condition using a normally distributed gamble with mean and variance parameters.

Substituting a cash return into the normal case yields a measure that depends on the shifted mean through a ratio, rather than equaling the original measure minus the cash amount. The answer therefore concludes that the measure fails cash translation invariance and is not coherent under this criterion. The derivation is presented tentatively, with an explicit caveat that algebraic mistakes may have occurred; the document does not assess the other coherence axioms or establish whether the conclusion holds across all distributions.

Key ideas

• Cash invariance requires a deterministic return to shift measured risk by the opposite amount.
• The answer tests this property by applying the Aumann–Serrano definition to a normal gamble.
• The resulting expression does not match the cash-invariance condition, so the response concludes the measure is not coherent.
• The derivation is tentative and does not examine the remaining coherence properties.

Tags

# Is Aumann-Serrano risk measure coherent?


# Is Aumann-Serrano risk measure coherent?












Is the Aumann-Serrano risk measure (Robert J. Aumann, and Roberto Serrano: An Economic Index of Riskiness, JPE, Vol. 116 No. 5, October 2008. <link>) coherent? And why yes or no?

## Answer by Kermittfrog (score 3, accepted)

https://quant.stackexchange.com/a/69666

Let me give it a try: Afaik, the risk measure $R$ is defined such that $$R: \mathrm{E}_x\left(e^{-x/R}\right)\stackrel{}{=}1$$

One of the requirements for a coherent risk measure is that it is invariant to adding cash. Quoting wiki:

> If $A$ is a deterministic portfolio with guaranteed return $a$, $R(Z+A)=R(Z)-a$

Let's add a cash return $a$ to our normally distributed gamble with mean $\mu$ and variance $\sigma^2$:

$$ \begin{align} E(e^{-(x+a)/R})&=e^{-\frac{a+\mu}{R(x+a)}+\frac{1}{2}\frac{\sigma^2}{R(x+a)^2}}\stackrel{!}{=}1 \\ \Rightarrow R(x+a)&=\frac{\sigma^2}{2(a+\mu)}\neq \frac{\sigma^2}{2\mu}-a=R(x)-a \end{align} $$

Assuming I made no mistakes until here, the measure is not translation invariant.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.