NuclearPhynance Archive

As a complementary risk measure to historical VaR, I would like to use either the expected shortfall or the conditional VaR. What is the best practise? As far as I understood, expected shortfall=E(X/X<-VaR) whereas Conditional VaR=E(X/X<=-VaR). So ES and VaR looks mutually exclusive, is it the good reason to use it as a complementary measure? Or should I prefer the CVaR?



Thx!

It's kinda a crapsoot with either/forecasting risk anyways.  I generally send our company's risk managers VaR, parametric, and Monte Carlo calcs.  Probably CVaR in your case.

The two are basically synonyms, although "Expected Shortfall" has become the standard term and "Conditional VaR" has fallen into disuse. Some people distinguish by subtracting the VaR number from Expected Shortfall, but that has also fallen into disuse. The term "Conditional VaR" is used for other things, and CVaR is used for even more things, so ES is the best choice for clarity and up-to-date usage.



Both measures always gave the same information, and there was more divergence of definition within each one than between the two.

So I should call it ES and calculate it including the VaR, is that correct? What is the reason why some people substract it or not? Is it a different point of view or a different information that is required?

Yes.



If you use ES or CVaR as a complement to VaR, it makes sense to subtract the VaR. That is "our VaR loss is $10 million, and if we exceed that, our expected shortfall is an additional $6 million." As a standalone risk measure to replace VaR, it obviously makes more sense to report the "if we get into the 1% tail of the distribution, we expect to lose $16 million." It's the same information either way, but it's always wise to ask which number you've been given.

Thx a lot!

I'd like to estimate a confidence interval for HS VaR and ES using the statistics order approach.



The distribution of the VaR is binomial so I can easily get a confidence interval for the VaR, but how does it work exactly for the ES? The precision will probably fall but how can I deduce a distribution for ES using this order statistics approach?



Thx

Take a look at Variance of the CTE Estimator by Manistre and Hancock.

You have put your finger on the reason I don't like Expected Shortfall. You cannot put a confidence interval on it without a strong distributional assumption. The VaR interval requires only that the returns are i.i.d.



Suppose you have 1,000 observations. As long as they are i.i.d. (and tomorrow is an i.i.d. draw from the same distribution) you know the 95% VaR point will be between the 37th and 63rd worst days 95% of the time. The shape of the P&L distribution doesn't matter.



Let V be the 95% VaR point, that is Pr{X < V} = 0.05. Suppose that for C < V, Pr{X < C} = 0.05*(V/C)^a for some a > 0. The expected shortfall point is at the 0.05*(1 - 1/a)^a fractile point of the distribution. For large a, this is near the 1.8% point, so you have enough data to estimate it with some reliability. But the smaller the a, the lower the fractile point. For a < 1.022, the expected shortfall point will be worse than the worst of your 1,000 observations.



Things are even worse that this. If a is constant in the VaR interval, you can hope to guess its value from the 50 points you do observe. You won't get a good estimate, but you have something to go on. But what if a decreases beyond the range of your observations? In other words, what if there is the possibility of a large loss that you don't observe in your data? It's easy to look at 4 years of bond returns without guessing there is a probability of default.

"You have put your finger on the reason I don't like Expected Shortfall. You cannot put a confidence interval on it without a strong distributional assumption."



I thought Extreme Value Theory guaranteed that you can model Expected Shortfall with a generalized Pareto distribution (Pickands-Balkema-deHaan theorem).  So ES has a known limiting distribution, and you can estimate its variance from the data (per Manistre and Hancock).  Isn't that enough to get a confidence interval?



IIRC, all EVT assumes is that the underlying loss distribution is continuous and invertible.  Maybe you mean that's already too strong an assumption?