Expected Shortfall
- Thread starter dwheats
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Hi @dwheats
Yes, that is exactly right, ES gives all tail losses (beyond the "conditional" quantile) equal weights. Dowd Chapter 2 introduces a generic risk-aversion function:
\({{M}_{\phi }}=\int_{0}^{1}{\phi \left( p \right){{q}_{p}}dp}\)
This is not ES (this generalizes ES but generalizes a lot!). ES is the case of this measure where the weighting function, phi(p), is constant at 1/(1-confidence); e.g., for 95% ES, phi(p) = 1/5% = 20 (by weighing the 5% tail densities by 20, the 5% tail is transformed into a probability function; it's the child's conditional probability within the overall "parent" probability).
Re: will this bias the expected value towards the lower end of the tail?
If you mean, bias toward the body rather than toward the extremities, then yes, given the normal is (by definition) not heavy tailed. For example, the 95% ES for a normal is 2.06 which is not too far from the 95% VaR of 1.645 (the 99% ES is 2.665 versus 99% VAR of 2.326). I hope that helps!
Yes, that is exactly right, ES gives all tail losses (beyond the "conditional" quantile) equal weights. Dowd Chapter 2 introduces a generic risk-aversion function:
\({{M}_{\phi }}=\int_{0}^{1}{\phi \left( p \right){{q}_{p}}dp}\)
This is not ES (this generalizes ES but generalizes a lot!). ES is the case of this measure where the weighting function, phi(p), is constant at 1/(1-confidence); e.g., for 95% ES, phi(p) = 1/5% = 20 (by weighing the 5% tail densities by 20, the 5% tail is transformed into a probability function; it's the child's conditional probability within the overall "parent" probability).
Re: will this bias the expected value towards the lower end of the tail?
If you mean, bias toward the body rather than toward the extremities, then yes, given the normal is (by definition) not heavy tailed. For example, the 95% ES for a normal is 2.06 which is not too far from the 95% VaR of 1.645 (the 99% ES is 2.665 versus 99% VAR of 2.326). I hope that helps!
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hi @dwheats
It's just to say mathematically that you are correct, in ES all the tail losses are given the same (equal) weights. Then, if you think about the normal distribution, it's tail is decreasing, so most of the tail mass is "bunched up" near the threshold so the conditional average (the ES) will be nearer to that threshold than way out in the extremity.
The fact that ES gives all tail losses the same weight is why Dowd thinks it is inferior to other measures, right? ES is better than VaR (according to him) yet it does not really capture true risk aversion; he says an even better measure should assign more weight to greater losses. Thanks,
It's just to say mathematically that you are correct, in ES all the tail losses are given the same (equal) weights. Then, if you think about the normal distribution, it's tail is decreasing, so most of the tail mass is "bunched up" near the threshold so the conditional average (the ES) will be nearer to that threshold than way out in the extremity.
The fact that ES gives all tail losses the same weight is why Dowd thinks it is inferior to other measures, right? ES is better than VaR (according to him) yet it does not really capture true risk aversion; he says an even better measure should assign more weight to greater losses. Thanks,
Yes @dwheats that's a great way to put it! (it's not easy....). If you look at the function above, there are really just two parts: the weight, phi(p), and the quantile, q(p). Each distribution will "give shape" to the quantile; e.g., the 95% quantile for a normal is 1.645, a heavier-tailed distribution will carry a higher quantile at 95%. But with respect to the weight of the quantile, the ES is just like an ordinary mean (hence it is a conditional mean): aside from being informed by the density ("factored into the probability density" is a great way to put it!), the losses all get the same weight. An extreme outlier, say 4 or 5 sigma, is extremely unlikely according to the density PDF; under ES, its extremely small probability is reflected in the density--the area of which is q(p)dp--but it gets no special weight treatment.
Dowd argues that a proper risk measure, unlike ES, should "assigned more weight ... to the more extreme losses." Dowd says that weighting function, phi(p), should instead be increasing such that an extremely unlikely loss receives a much higher weight. Thanks!
Dowd argues that a proper risk measure, unlike ES, should "assigned more weight ... to the more extreme losses." Dowd says that weighting function, phi(p), should instead be increasing such that an extremely unlikely loss receives a much higher weight. Thanks!
Pedro Cazorla
New Member
What do you mean "factored into the probability density"? To me, dp of q(p)dp is just notation, it does not mean anything. So, unless we are integrating with respect to some measure other than the Lebesgue measure, I don't know how the probabilities are included there.
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