Coherent risk measure

orit

Active Member
Hi David,
I would appreciate if you can clarify 2 points in respect to the coherent risk measure:
1. Can you please explain the Monotonocity and translation invariance?
2. ES is sub additive because the expected return will be higher in 3 bonds portfolio than in 2? Based on the example on the bonds?this is because the Bernoulli distribution is not normal and ES knows how to deal with non-normal distributions?
Thanks a lot,
 

skoh

Member
Hi David and Shakti,

What does it mean by empirical VS parametric non-normal VaR and its indication on heavy tails?

I'm looking at question 29.2 of Dowd Valuation and Risk Models.

Thanks
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
ShaktiRathore thanks for the help!

orit The sub-additive difference emanates from the fact that ES is a conditional mean, as opposed to VaR which is a quantile function (like median) and quantiles don't necessarily aggregate nicely (although VaR is subadditive if the distribution is normal [elliptical]. The 2-bond illustration is just Dowd's example of showing an easy violation; it is really easy to volatile sub-additive with discrete distributions like binomial.

skoh Parametric distributions are described by functions (e.g., normal, binomial, Poisson) where data (although informing parameters) is not displayed; also population, by definition. Empirical distributions (e.g., historical simulation) are data; e.g., histogram; also, tend to be samples. The point of that question is to give you a statement that is FALSE:
  • This is a parametric VaR and therefore cannot characterize a heavy-tailed distribution (FALSE)
Why?
  • For us, what is a heavy-tailed distribution? Answer: kurtosis> 3 or excess kurtosis > 0 (there are other defs, this is the FRMs)
  • A normal distribution, by def, is not heavy-tailed. Normal is most famous parametric, but hundreds of other parametric distributions exist.
  • But many (non-normal) parametric distributions are heavy-tailed; e.g., students-t
  • And a key advantage of an empirical distribution is that it can capture heavy tails.
It seems worth repeating b/c currently (in the wake of the JPM report), once again, mainstream financial journalism doesn't necessarily understand what VaR is. VaR doesn't presume normality. VaR doesn't say how the distribution is specified (param, simulation, empirical). VaR is just a (statistical) quantile function. Objecting to VaR is objecting to the idea of trying to quantify a possible future distribution (If we object to that, okay, but what else? ....) and employing statistics on the distribution, like median. thanks,
 

orit

Active Member
Thanks a lot David and Shakti.
One more question David regarding the calculation of the ES - I have noticed in some questions that sometimes you divide the average loss by alpha and sometimes you dont, can you please clarify.
Thanks
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi orit,

There is a good illustration of this in the answer to T4.31.1 here at http://forum.bionicturtle.com/threads/p1-t4-31-spectral-risk-measures.5709/
i.e., alpha is 1% due to 99% ES and the sum of the three (loss * pmf) is divided by 1% alpha

This is because the f(x) = pmf sum to 100% across the entire (unconditional) distribution; recall the definition of a discrete distribution is that the sum of f(x(i)) = 1.0
However, ES is a conditional mean: it wants the (conditional) average only of the 1% tail; in the T4.31.1 example, only the the three worst values. But their pmfs add up to only 1.0%. In order to compute their weighted average, we need divide by the sum of these "weights." It is to take a weighted average, where the pmfs are weights (in fact, they are weights!). Technically, the unconditional (total distribution) weights inform a conditional (tail) only average. It is the same for continuous distributions. Thanks,

P.S. it is maybe easier to use the example, in our 31.1 PQ, the three worst losses are:
$200 f(x) = 0.2%
$110, f(x) = 0.4%
$110, f(x) = 0.4%
... then the rest of the distribution such that sum of f(x) = 100% which is necessary to a probability distribution

ES is the conditional average; i.e., what is the weighted average of these three worst 1% losses?
$200 * 0.2% + 110 * 0.4% + 110 * 0.4% / (0.2% + 0.4% + 0.4%) =
$200 * 0.2% + 110 * 0.4% + 110 * 0.4% / 1.0% alpha
 

orit

Active Member
Hi David,
In the following practice question from 20112, the average loss which is 0.5*0+ 0.5*(0.2*10+0.5*18+0.3*25)=9.25M is not divided by the tail, can you please explain why?
30.4. Over the next year, a operational process model predicts an 95% probability of no loss occurrence and a 5% probability of a single loss occurrence. If the single loss occurs, the severity is characterized by three possible outcomes: $10.0 million loss with 20% probability, $18.0 million loss with 50% probability, and $25.0 million loss with 30% probability. What is the model's one-year 90% expected shortfall (ES)?

Thannks
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi orit,

It's only because the question gives you conditional probabilities, here is the question:
30.4. Over the next year, a operational process model predicts an 95% probability of no loss occurrence and a 5% probability of a single loss occurrence. If the single loss occurs, the severity is characterized by three possible outcomes: $10.0 million loss with 20% probability, $18.0 million loss with 50% probability, and $25.0 million loss with 30% probability. What is the model's one-year 90% expected shortfall (ES)?

We can re-phrase into unconditional probabilities, here is the worst 10% of the total (aka, parent, unconditional) distribution:
0 loss = 5% (i.e, the final 5% of the 95% no loss)
10 loss = 5%*20% = 1.0%
18 loss = 5% * 50% = 2.5%
25 loss = 5%* 30% = 1.50%

So that's the unconditional probabilities:
5% of 0
1% of 10
2.5% of 18, and
1.5% of 25
= sum to 10%

Expressed this way, you would use the "alpha" of 10% to compute the conditional mean:
10% ES = (0*5% + 10*1% + 18*2.5% + 25*1.5%)/10%

Thanks,
 
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