Expected Shortfall and Variance Are Not Generally Monotonic
Summary
The document examines whether a lower estimated expected shortfall implies lower sample standard deviation. Under a normal distribution, it describes a relationship in which scaling volatility changes expected shortfall in the same direction, while the sign of value at risk depends on the confidence-tail convention used. The question then asks whether that relationship extends to samples from an unknown population.
The answer says no general inference follows from comparing separate samples: sampling variation and differences in distribution shape, such as kurtosis, can change the ordering of estimated shortfall and standard deviation. It suggests a more controlled comparison for an efficient-frontier-like analysis: vary portfolio weights while keeping the underlying observations fixed, which may make expected shortfall monotonic in volatility. This is a proposed setup, not a proof for arbitrary distributions or a guarantee when samples or portfolio composition change.
Key ideas
- For normally distributed returns, expected shortfall scales with standard deviation under the stated tail convention.
- Comparisons of expected shortfall from separate samples do not determine the ordering of their standard deviations.
- Sampling variation and distribution shape can cause expected shortfall and volatility rankings to diverge.
- Holding observations fixed while varying portfolio weights may support a monotonic risk comparison.
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# How are Expected Shortfall and Variance related?
# How are Expected Shortfall and Variance related?
I would like to know how Expected Shortfall $SF_\alpha$ and variance $\sigma^2$ are related.
If I follow what Aaron Brown answered in this post, when the underlying distribution is Normal with standard deviation $\sigma$, then:
$$VaR_\alpha=\Phi^{-1}(\alpha) \cdot \sigma$$
and
$$SF_\alpha= \frac{1}{\alpha \sqrt{2 \pi}} \exp \left( - \frac{VaR_\alpha^2}{2} \right)$$
Hence, if I take an $\alpha < 0.5$, I know that $\Phi^{-1}(\alpha) < 0$ and hence that:
$$\frac{d(VaR_\alpha)}{d \sigma} = \Phi^{-1}(\alpha) < 0$$
and that
$$\frac{d(ES_\alpha)}{d \sigma} = SF_\alpha \frac{-1}{2} \Phi^{-1}(\alpha) 2 \sigma = \underbrace{SF_\alpha}_{>0} \underbrace{ (-1) \Phi^{-1}(\alpha)}_{>0} \sigma > 0$$
Hence, I know that if my volatilty grows, my shortfall will grow along with it, under the normality assumption.
However, I wonder if we can say the same in general.
Assume I have two large samples from an unknown population, and that I estimate the $ES_\alpha$ of the samples which give me values $r_1$ for sample 1 and and $r_2$ for sample 2. Assume that $s_1$ and $s_2$ are the sample standard deviations.
Finally, assume that $r_1<r_2$.
Can I say, looking only at the shortfall measures, that $s_1<s_2$?
Mathematically, with no prior assumptions, does $r_1<r_2 \Rightarrow s_1<s_2$ ? Or even $r_1<r_2 \Leftrightarrow s_1<s_2$ (that seems unlikely)?
## Answer by Brian B (score 5, accepted)
https://quant.stackexchange.com/a/2972
Even if everything is truly normally distributed, once you are taking separate samples anything can happen. In particular, your first sample may have gotten a small standard deviation but (by chance) a huge kurtosis, so you will have $s_1<s_2$ but $r_1>r_2$.
Given your goal of finding an analog to the efficient frontier, you could try making your volatility dependent on weight/position sizes and forcing your sample populations to always have the same elements, just with different weights. You should then be able to attain expected shortfall as a monotonic function of volatility.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.