# Proof for expected shortfall sub additivity

I found on pag 5 https://faculty.washington.edu/ezivot/econ589/acertasc.pdf the proof about the sub additivity of expected shortfall. I understood the demonstration on the whole, but I would like to clear out this doubt: for which exact reason we can say that the first side of (9) formula is always <= than the second one? I thought this could be explained by the triangle inequality, but I think it doesn't work.

• It would be best if you could reproduce the derivation in summary at least up until the step you are asking about in your post, preferably using MathJax/LaTeX. If you are unable to do that, pls at least include in your post a picture of the page with the inequality you are enquiring about in addition to the link to the original paper. Commented Aug 12, 2022 at 15:16

It seems to me that on the LHS you have the worst $$\omega$$ realizations of $$X + Y$$ and on the RHS you have the worst $$\omega$$ realizations of them individually. If $$X$$ and $$Y$$ are not perfectly correlated the worst realizations of $$X$$ and $$Y$$ will not match so the worst of $$X$$ will be probably be compensated by a realization from $$Y$$ that is not the worst. Therefore, the value of the RHS will always at least be as high. For example, consider these realizations:

$$X = -10, 0, 10 \textrm{ and } Y = 10, 0, -10$$

Let random variables $$X,Y$$ correspond to the loss distributions of two assets. We invest in both and hence have loss distribution $$L = X + Y$$.

For $$n$$ realisations, denote the order statistics of $$L$$ as: $$L^{(1)} \leq L^{(2)} \leq ... \leq L^{(n-1)} \leq L^{(n)}$$

We're interested in the expectation given the $$\alpha$$ percentile case scenario is exceeded. Taking $$n$$ sufficiently large and setting $$m = \lfloor (1-\alpha) n \rfloor$$ as the number of observations exceeding this percentile level we have the estimator:

$$ES_{\alpha} = \frac{\sum_{i=0}^{m-1} L^{(n-i)}}{m} = \frac{\sum_{i=0}^{m-1} (X+Y)^{(n-i)}}{m}$$

The numerator, $$\sum_{i=0}^{m-1} L^{(n-i)}$$, is the sum of the $$m$$ largest values of $$(X+Y)_{1:n}$$

However the largest values of $$X+Y$$ do NOT correspond to the largest values of $$X$$ and $$Y$$.

We could have: $$X_{1:3} = (3, 0, 5)$$, $$Y_{1:3} = (3, 4, 0)$$ and so $$(X+Y)_{1:3} = (6, 4, 5)$$

Hence $$(X+Y)^{(3)} = 6$$.

However $$X^{(3)} + Y^{(3)} = 5 + 4 = 9$$

You're correct with the general idea of the triangle inequality, the paper just doesn't illustrate it very well.

$$\sum_{i=0}^{m-1} (X+Y)^{(n-i)} \leq \sum_{i=0}^{m-1} X^{(n-i)} + \sum_{i=0}^{m-1} Y^{(n-i)}$$