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:

$$VaR_\alpha=\Phi^{-1}(\alpha) \cdot \sigma$$


$$SF_\alpha= \frac{1}{\alpha \sqrt{2 \pi}} \exp \left( - \frac{VaR_\alpha^2}{2} \right)$$

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)?

  • $\begingroup$ If samples are drawn from different distributions, I guess you can construct a counter-example e.g. by taking a standard normal distrbiton and skewed normal distribution. $\endgroup$ Feb 14, 2012 at 15:28
  • $\begingroup$ I'm not sure what you mean. The two sample have are drawn from a common population. $\endgroup$
    – SRKX
    Feb 14, 2012 at 15:43
  • $\begingroup$ Got it. Can you shed some light why you want to compare the sample std. dev. (which are directly observable from samples) looking at shortfalls? $\endgroup$ Feb 14, 2012 at 15:58
  • $\begingroup$ yeah, basically, I draw an efficient frontier using the $SF_\alpha$ as objective function. The thing is, I am asked to provide the best allocation for a given "volatility". Hence, I have to find the allocation on my frontier with the volatility the closest to some target $\bar{\sigma}$. I was wondering if looking "between" two points in the frontier to find an exact match for $\bar{\sigma}$ made sense. $\endgroup$
    – SRKX
    Feb 14, 2012 at 16:54

1 Answer 1


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.

  • $\begingroup$ Actually, that's what I'm doint you're right, I'm looking at different allocations out of a simple data sample. In that case, it seems fair to assume that shortfall is a monotonic function of volatility right? $\endgroup$
    – SRKX
    Feb 15, 2012 at 7:52
  • 1
    $\begingroup$ It still depends on how you do it but the assumption will fail only in fairly pathological cases (i.e. cases where your weighting effectively replicates the high-kurtosis effects cited above). $\endgroup$
    – Brian B
    Feb 15, 2012 at 13:32

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