2
$\begingroup$

I am able to see why Expected Shortfall will be subadditive for normal distribution or a uniform distribution. I am trying to prove the result for any generic distribution. I came across many proofs available on the internet, but the math involved is too complex in all of them. Is there any simple explanation why ES is subadditive for any generic distributions?

$\endgroup$
2
$\begingroup$

The expected shortfall is defined by \begin{align*} ES_{\alpha} = \frac{1}{1-\alpha}\int_{\alpha}^1 VaR_{p}(L) dp, \end{align*} where $L$ is the loss function. For the case with 500 scenarios, the $\alpha=99\,\%$ percentile VaR is approximately the $5^{\rm th}$ worst loss scenario. The expected shortfall can then be approximated by the average of the 5 worst losses, times $-1$ (we take $ES_{\alpha}$ to be positive). That is, \begin{align*} ES_{\alpha}(L) = -\frac{1}{5}\sum_{i=1}^5 L(i), \end{align*} where $L(i)$ is the $i^{\rm th}$ worst loss scenario. Assuming that the loss $L$ can be decomposed into the losses $L_1$ and $L_2$ from two sub-portfolios. That is $$L=L_1+L_2.$$ Then $$L(i) = L_1(i)+L_2(i).$$ However, it is easy to see that, though $L(1), \ldots, L(5)$ are the 5 worst loss scenarios of $L$, $L_j(1), \ldots, L_j(5)$, for $j=1, 2$, are not necessarily the 5 worst loss scenarios for $L_j$. In other words, for $j=1, 2$, \begin{align*} ES_{\alpha}(L_j) \ge -\frac{1}{5}\sum_{i=1}^5 L_j(i). \end{align*} Then \begin{align*} ES_{\alpha}(L) &= -\frac{1}{5}\sum_{i=1}^5 L(i)\\ &=-\frac{1}{5}\sum_{i=1}^5 L_1(i)-\frac{1}{5}\sum_{i=1}^5 L_2(i)\\ &\le ES_{\alpha}(L_1) + ES_{\alpha}(L_2). \end{align*}

$\endgroup$
  • $\begingroup$ Great answer. For others who are mathematically challenged like me, an example of the above proof is demonstrated @ [1]: medium.com/@sivarajesh.kasa/… $\endgroup$ – kasa Aug 14 '17 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.