In calculating of the average for ES, should I include the 100th loss?

[Steve Jobs]

In q 70.1, for calculation of ES at 95%, all losses beyond the 95th are included that is 96, 97, 98, 99, 100.
In Kaplan, there are 3s question related to same topic, in 2 of them the 100th is not provided and even in the answer provided for one of them, it's mentioned that for 95%, the number of slices would be n-1 that is 4 and hence the 100th will not be included.

I'm confused, is there any difference between the questions in which the 100th included and the others? and how can I know whether to include the 100th or not?

 

[David Harper CFA FRM]

@Steve Jobs If we observe a discrete distribution of 100 losses (n = 100) conveniently ordered (where losses are positive such that 100 is the worst loss) from 1, 2, ..., 96, 97, 98, 99, 100, the 95.0% ES is the average of the 5% tail, unequivocally. In this case, 95% ES is the average of 96, 97, 98, 99, and 100. (Implicitly the losses are equally weighted). ES does not throw out the worst. ES is a conditional average: weighted average of losses in excess of the VaR quantile. A key feature of its advantage over VaR is that it includes extreme losses.

I haven't seen the Kaplan questions but the two common sources of confusion are:

1. For the same dataset above (1, 2, ... , 99, 100), while the 95% ES is unambiguous, the 95% VaR has three valid definitions; specifically: 95 is best (Dowd), 96 is okay (Jorion), and 95.5 is understandable (assumes observations split the weight per T5 hybrid reading). But the ES, a conditional mean, does not depend on the VaR approach as it grapples with isolating on the quantile in a discrete series.
   
2. More likely in your case, per Dowd, we can estimate ES by taking an average of tail VaRs. So if we use n = 5 slices, there would be 4 tail VaRs. But why would we need to resort to this, in the case of a discrete distribution of (implicitly equally-weighted) loss observations? I hope that helps,

 

[Steve Jobs]

I wanted to send a private message to you, but your name has "," which is used by the forum system for separating names if sent to multiple receivers while choosing receivers; and it's the reason that we can't send a private message to you. We really appreciate if you could do something about it. Thanks a lot.

 

[David Harper CFA FRM]

@Steve Jobs thanks for the feedback, commas removed from my profile name

 

[David Harper CFA FRM]

@Steve Jobs thanks for sharing the specific examples privately

• The first Q&A (#2) is wrong. Notice the source is actually a 2010 FRM Practice Exam. I don't mean to lack humility, but that's one of the issues/questions about which I had to educate GARP. The average of the 1% tail clearly includes 10 worst losses; what is shown as the answer is actually the 99.10% expected shortfall, not the 99.00% ES, do you see why I say that? Current exam will not make that mistake
• The second Q&A (#17) is a good question, I like it. My only quibble is that "calculate" verb which is imprecise: It's a different question (see my 2nd point above). It is not asking for a calculation of ES, it is asking for an estimation of (the true) ES based on average VaRs. The answer given is not the exact ES (unlike Q #2 which is exactly the ES), which is unavailable with the the information given. (not only don't we have a list of equally weighted losses, but there is no reason to assume an "easy" distribution: this method is really handy). Thanks,

 

[David Harper CFA FRM]

... actually, I was too hasty: about Q #2, I cannot even say it is the 99.10% ES because that assumes those are the worst 9 out of 1,000 but if the worst is excluded then it's not the ES at all

 

[Steve Jobs]

So I understand that I should include the last loss, the 100th or the 1000th...for calculation of ES, thanks.

 

[David Harper CFA FRM]

Yes absolutely ES is an average (above a quantile) of the entire tail including the worst loss (the only rationale i can think of would be, by definition, an exclusion of the worst loss as a literal "outlier" in the dataset which should not inform the distribution)

 

[Steve Jobs]

After a second review of the materials, regarding including the worst loss, I see that Kaplan assumes the ES as part of coherent risk measures and that;s why it does not include the worst loss. By reading Kaplan/BT, I understand that ES is a coherent risk measure but BT differentiate between the two in calculations.

 

[David Harper CFA FRM]

Hi @Steve Jobs Where is the intuition even for omitting the worst loss in ES?

If lack of an intuition (for excluding the worst loss when computing the average of a tail? ) is not enough, consider: if we give the worst loss no weight, then the risk-aversion function is no longer weakly increasing, which disqualifies the measure as a spectral measure (see below from Dowd: "Weakly increasing:If some probability p2 exceeds another probability p1, then p2 must have a weight bigger than or equal to that of p1."). This is related to VaR's lack of coherence: unlike ES, VaR is not weakly increasing, VaR is decreasing (degenerative actually) because it gives less weight (zero) to probabilities that exceeds the VaR confidence/significance (which receives a weight of 1.0). As Dowd says, "So one measure [i.e., ES] places equal weight on tail losses, and the other [i.e., VaR] places no weight at all on them."

So, I find the omission of the worst loss incompatible with the essential feature (according to Dowd) of a coherent risk measure. If that's not enough, I would add that I've simply never seen any ES omit the worst loss, including GARP's questions. True, GARP's historical ES questions do contain some flaws, now well documented, but none of this sort (GARP's older errors are only around computed different ES based on valid ambiguity of VaR in certain discrete distributions, which is a mistake b/c p% ES does not depend on the method of computing p% VaR, but GARP has never omitted the worst loss). Thanks,

Dowd page 39, on spectral (and coherent) risk measures: "However, we are concerned for the moment with the broader class of coherent risk measures. In particular, we want to know the conditions that φ(p) must satisfy in order to make Mφ coherent. The answer is the class of (non-singular) spectral risk measures, in whichφ(p) takes the following properties (Acerbi (2004, proposition 3.4)):

Non-negativity:φ(p)≥0 for allpin the range [0,1].
Normalization: φ(p)dp=1.
Weakly increasing:If some probability p2 exceeds another probability p1, then p2 must have a weight bigger than or equal to that of p1.

The first two conditions are fairly obvious as they require that weights should be positive and sum to 1. The critical condition is the third one. This condition reflects the risk-aversion, requiring that the weights attached to higher losses should be bigger than, or certainly no less than, the weights attached to lower losses. Given that it ensures coherence, this condition suggests that the key to coherence is that a risk measure must give higher losses at least the same weight as lower losses.

 

[Steve Jobs]

In q 72.1 in questions sets for Coherent Risk Measures, the 100th is not included. If I understood correctly, can we say that because ES is a coherent risk measure, then we should not include the 100th?

 

[David Harper CFA FRM]

Hi @Steve Jobs

• Question 72.1 does not compute the exact ES (which again, is a conditional mean of the entire tail about the quantile). Rather, it uses a method to approximate the ES, where the method is to slice the tail into (n) pieces; e.g., five slices requires 4 VaR quantiles. This is not an exclusion of the worst loss (in fact, the example uses a normal distribution: there is no identifiable worst loss), it is merely using selected VaR quintiles to approximate the ES
• Re: "can we say that because ES is a coherent risk measure, then we should not include the 100th?" No, the opposite: coherence requires including the worst loss, in my opinion, as above; to exclude the worst loss in an actual ES calculation is to disqualify coherence.

 

[Steve Jobs]

Hi DDavid,

"it is merely using selected VaR quintiles to approximate the ES", in the example questions of both BT and Kaplan for coherent Risk Measures, quantiles after the Var (say after 95, that is 96, 97, 98 , 99) included and 100th excluded. It seems like a standard rule to include all except the 100th, and not selected quantiles.

 

[David Harper CFA FRM]

Hi @Steve Jobs It confuses two things:

• Because calculating ES for a continuous distribution can be difficult/tedious, Dowd shows a way to estimate (approximate) the true ES by slicing the tail and taking an average of the VaR slices; for example, if 95% ES is estimated with only five slices, then the VaR quantiles used would be 96%, 97%, 98%, 99% and the ES ~ average of those tail VaRs. 95% to 96% is a slic,e, 96 to 97% is a slice ... and 99% to 100% is the "fifth" slice (n slices --> n -1 quantiles used). But that's just the one example where n = 5 slices. As n increases, the approximation tends toward the true ES, and higher quantiles of 99.x% also get included (consistent with an intent to take the average of all tail losses in the distribution). In this approximation method (this is a classic instance of numerical approximation in lieu of direct analytics), the 100% VaR will not be included, but I hope you can see why that does not mean that ES excludes the worst loss.
  
• In a discrete distribution, were we have actual loses where the worst three are x, y, z; e.g., 98, 99, 100 is unrealistic but let's say the actual worst losses where 98, 99, 100 ... then the 100 gets included. In a discrete distribution, the worst x% losses are averaged which be definition includes the worst loss. It matters because, in a discrete distribution (like a simple historical simulation, which is the most testable) there is no need to do the approximation: the actual (conditional; i.e., tail) average can be calculated by averaging the (1 - confidence)% worst losses. thanks,

 

[Steve Jobs]

So the key here is whether its discrete or continuous...I have to review these samples questions again to confirm. Thanks David.

 

[David Harper CFA FRM]

In an indirect way, i think that has merit. I think it's important to have a grasp on the definition of ES as a probability weighted averages of all losses (including the worst) in the tail above the threshold (i.e., the conditional average). Then i think the key difference is: (i) exact ES or (ii) approximation via slicing the tail into (n) slices.

However, it is true that approximations will tend to be needed for only continuous distributions; while (i think) it's generally easy to be exact with discrete distributions. So, yes to the extent approximations connote continuous, it has some merit (but it misses the point, in my opinion; to be argumentative: we could use the approximation method on a discrete distribution, which shows why it is not the key).

ES can seem complicated in continuous distributions, but i don't think we want to lose sight of the forest for the trees. The most likely exam question re: ES is much like my 70.1 (if only b/c it can be quickly calculated):

70.1 Assume an empirical loss/profit distribution is (most conveniently!) a uniform distribution of 100 losses where the highest seven losses are $94, $95, $96, $97, $98, $99, and $100. What is the 95% expected shortfall (ES)?

Unlike 95.0% VaR, which has three good, valid answers (and they are?), the 95.0% ES has only one answer: it is the average of the worst 5 (any ES is going to include the $100 here).

 

[thomas.hood]

1. For the same dataset above (1, 2, ... , 99, 100), while the 95% ES is unambiguous, the 95% VaR has three valid definitions; specifically: 95 is best (Dowd), 96 is okay (Jorion), and 95.5 is understandable (assumes observations split the weight per T5 hybrid reading). But the ES, a conditional mean, does not depend on the VaR approach as it grapples with isolating on the quantile in a discrete series.

...so what would be the 97.5% VaR and ES in this example?

Thanks,

Tom

 

[David Harper CFA FRM]

Hi Tom, that's a great question!

• The 97.5% ES is the weighted average of the 2.5% tail, so it's the worst loss of $100 (weighted its full 1.0%), the next worst $99 (weighted its full 1.0%) and the next worst $98 (but weighted only 0.5%, since halfway through this value gets us the 2.5% tail = 1% + 1% + 0.5). So what is the 97.5% ES if the worst 2.5% includes $100 @ 1.0%, $99 @ 1.0% and 98 @ 0.5%?
• As discussed above, unlike ES, we have some discretion. However, under Hull's technical approach (i.e., 2014 Reading 33), it turns out that the 97.5% VaR is conveniently, exactly $98.00 (i.e., the third worst loss without modification). Why is that?

Thanks

 

[Steve Jobs]

Hi! I passed P2 but honestly still find these confusing

If someone asked me this q in an interview, I'll say:
1. Imagine you have 1000 days returns with no 1 being the best return and no 1000 the worst.
2. Var 97.5 is the return no 975.
3. ES 97.5 is the [(976 + 977 + ...+1000) / 25]

I'm sure David is now thinking what's this bs! but that's what comes to my mind....

 

[rahul.goyl]

I come across numerous question about 99% VaR is equal to 97.5%, some research has also put up the below

View attachment 374
or
View attachment 375

However when using excel functionality I am not able to produce the same 2.34 value for 97.5% ES.
The below are results which excel 2010 produce, please correct me why am not able to produce 2.34 as shown in the above 2 screen-shot.

View attachment 376

David any help will be appreciated.