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)?
