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Apr 18

## Expected shortfall is next after Value at Risk

A few years ago I took my American friend to Astana. It was the beginning of March, not the coldest time of the year, but it was impossible to walk in the street for more than a hundred yards. We used a taxi to move even for short distances. Before leaving for Astana we knew that the temperature would be below zero but didn't know the average temperature forecast. This is the difference between Value at Risk and expected shortfall: the VaR gives you the cut-off (your loss will be at least this much), while the expected shortfall provides the expected value of the loss conditional on the knowledge that the loss will be VaR or larger.

## Expected shortfall definition

There are two steps in determining expected shortfall: first you find Var and then calculate the expected loss conditioned on the cut-off VaR. As in case with VaR, there are two possible definitions. One is obtained if you are willing to work with negative quantities: the loss is a negative random variable and the associated quantities are negative. Thus, VaR is defined from

$P(L\le VaR^\alpha)=\alpha$

for a given (low) value of $\alpha$ and then the expected shortfall is defined by

(1) $ES^\alpha=E(L|L\le VaR^\alpha)$.

Alternatively, if you want positive quantities, for VaR use the definition

$P(L\le -VaR^\alpha)=\alpha$

and for expected shortfall use the equation

$ES^\alpha=-E(L|L\le -VaR^\alpha)$.

When reporting the results, be sure to use the right inequalities. In the first case, if, say, VaR is -\$1 mln, with probability $\alpha$ the loss satisfies $L\le -1,000,000$ and the expected shortfall is negative. In the second case the bound for the loss becomes  $L\ge 1,000,000$ and the expected shortfall is positive.

## Calculation example

This example is a part of the solution to one of the problems in the University of London exam 2016, Zone A. We work with negative quantities and suggest the reader to work out the calculations with definitions that give positive quantities.

Review the assumptions and calculations that give equation (7) in the post on VaR. For simplicity, as it is done in the UoL exam, we take $\mu_{t+1}=0$, so that

(2) $R_{t+1}=\sigma_{t+1} z_{t+1},$

and obtain

(3) $VaR^\alpha_{t+1}=\sigma_{t+1}\Phi^{-1}(\alpha)$.

We also need the result on conditional expectation for the standard normal:

(4) $E(z|z\le a)=-p_z(a)/\Phi(a)$.

To begin, plug (2) and (3) in (1) (using return in place of loss):

$ES^\alpha_{t+1}=E(R_{t+1}|R_{t+1}\le VaR^\alpha_{t+1})=E(\sigma_{t+1} z_{t+1}|\sigma_{t+1} z_{t+1}\le \sigma_{t+1}\Phi^{-1}(\alpha))$

(pulling out one sigma by homogeneity of means and cancelling out another)

$=\sigma_{t+1}E( z_{t+1}| z_{t+1}\le \Phi^{-1}(\alpha))$

(applying (4) with $a=\Phi^{-1}(\alpha)$)

$=-\sigma_{t+1}p_z(\Phi^{-1}(\alpha))/\Phi(\Phi^{-1}(\alpha))=-\frac{\sigma_{t+1}}{\alpha}p_z(\Phi^{-1}(\alpha))$

(by the property of an inverse function $\Phi(\Phi^{-1}(\alpha))=\alpha)$).