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Why Expected Shortfall Does Not Transform Directly from Log to Arithmetic Returns

Article Quant Q&A · Author: shenflow
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Summary

The document asks how to convert expected shortfall (ES) calculated from log returns into ES for arithmetic returns. It contrasts this with value at risk (VaR), for which a stated exponential conversion relates the log-return threshold to the arithmetic-return threshold. The question suggests that an ES conversion may depend on the assumed return distribution.

The accepted answer confirms that dependence: ES involves a conditional tail expectation, and applying a logarithm versus using the return itself changes the quantity being averaged. Consequently, the VaR conversion cannot simply be applied to ES. The discussion offers no general conversion formula or worked numerical example, and its brief answer does not specify distributional assumptions or conventions for defining the tail. Researchers need those details to derive a usable ES transformation for a particular return model.

Key ideas

  • A VaR threshold for log returns can be mapped to an arithmetic-return threshold using an exponential transformation.
  • Expected shortfall averages outcomes in a tail, so transforming it depends on the return distribution.
  • A threshold conversion alone does not provide a general formula for converting expected shortfall.

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StatisticsRisk management
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# Transforming log-return expected shortfall to arithmetic-return expected shortfall


# Transforming log-return expected shortfall to arithmetic-return expected shortfall












When deriving the Value at Risk (VaR) for log returns, one can easily transform the log-return VaR to an arithmetic-return VaR via $VaR_{arithmetic} = e^{VaR_{log}}-1$

However, how is the log-return Expected Shortfall (ES) transformed to an arithmetic-return ES? I guess the exact method would depend on the respective (assumed) distribution, right?

## Answer by Wei (score 2, accepted)

https://quant.stackexchange.com/a/80796

Correct. Effectively you are comparing $\mathbb E[R\cdot 1_{\{R\ge r\}}]$ to $\mathbb E[\log R\cdot 1_{\{R\ge r\}}]$ which will depend on the distribution of $R$.

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