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Large-Sample Variance Formula for an Expected Shortfall Estimator

Article Quant Q&A · Author: Kumar
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Summary

The response offers a large-sample variance expression as a starting point for estimating uncertainty in an empirical Expected Shortfall measure. The formula expresses the estimator’s asymptotic variance through a double integral over the distribution’s quantile range. Its integrand is built from the minimum of two cumulative probabilities minus their product, with scaling determined by the sample size and tail probability.

The question asks specifically about confidence intervals or standard errors under normality, but the response does not work through a normal distribution or derive a confidence interval. Instead, it points to a paper and a cited reference for the formula, suggesting that the reader use it to develop an interval. The result is therefore a general asymptotic starting point; applying it requires a specified distribution and careful attention to the Expected Shortfall convention and tail level used.

Key ideas

  • The response provides an asymptotic variance formula for an Expected Shortfall estimator.
  • The formula uses a double integral involving the cumulative distribution function.
  • Sample size and the selected tail probability determine the variance scaling.
  • The response does not derive a normal-specific standard error or confidence interval.
  • Applying the expression requires matching the distribution and tail convention to the estimator.

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Risk managementStatistics
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# Variability in the Expected Shortfall estimator


# Variability in the Expected Shortfall estimator












Are there any results for calculating the variability in the Expected Shortfall measure. I am looking for Large sample confidence intervals under Normality for Expected Shortfall or calculation of standard error for the same.

## Answer by SRKX (score 1)

https://quant.stackexchange.com/a/15183

I wrote this paper a couple of years ago where we discuss this kind of topic.

On page 6, you see a formula that comes from a paper from Acerbi available in Szego's book:

$$\sigma^2(ES^{(N)}_\alpha(X)) \overset{N>>1}{=} \frac{1}{N(1-\alpha)^2} \int_0^{F^{-1}(1-\alpha)} dx \int_0^{F^{-1}(1-\alpha)} dy \{ \min( F(x), F(y) ) - F(x)F(y) \}$$

This should be a good starting point and reference for you to derive the confidence interval.

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