Why Expected Shortfall Is Subadditive for Combined Losses

# Proof for expected shortfall sub additivity


# 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.

## Answer by Bob Jansen (score 2)

https://quant.stackexchange.com/a/71904

As per your previous question, please provide more details in your question. A quick hint:

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:

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

## Answer by Bennnn (score 0)

https://quant.stackexchange.com/a/71953

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:

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

For your problem:

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.

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