Level 2: Post what your remember here...

[David Harper CFA FRM]

Hi Tom,

If we continue to assume your losses are a simple historical simulation (i.e., sorted and each has the same weight of 1.0%), then you can't use 96*1.5%; unless that reflects the parent distribution, in which case it would have to reflect 99, 98, 97, 96, 96. Then your formula works, but only b/c the 5th worst is also 96. To get the 95.5 ES, you need the 4.5% tail, which means that you need the worst four losses plus 0.5%*the 5th worst. (so, you'd never get this question). Your numerator is retrieving the actual "parent" x*f(x) values; or in continuous terms, x*f(x)dx, so something should feel wrong about 96*1.5%.

While we are here, this does just happen to be a sorted list were each loss is given the same weight of 1/n; i.e., simple historical simulation. In the (Dowd) variations, the 1% weights can vary according to some rule (e.g., EWMA), but still if we want the X% ES, we do need to retrieve the parent's (1-x)% probabilities "as they actually are." Thanks,

 

[Tom77]

Hi David,

hmm so what would be your answer to the question of "post what you can remember here":

5) Confidence VAR
94 a
95 b
96 c
97 d
98 e
99 f

What is the Expected shortfall at 95.5% confidence
I guess the answer is Answer: Average of c,d,e,and f​

According to this they just gave 4 losses over 95.5 ? If each loss had the same weight of 1/n then it would just be the average i guess and much less complicated than i thought

 

[David Harper CFA FRM]

I can't say that i precisely understand the actual-exam-question posted, I just assume something was lost in translation. But, okay, aren't the worst five losses listed, aren't they: 99, 98, 97, 96, and 95? In which case, the 95.5% ES is 97.22. The average of the worst four losses, per our discussion, is necessarily the (1 - 4/n)ES or 96.0% ES if n = 100. The VaR assumption, or outcome for that matter, is irrelevant to ES; ES (a conditional mean) doesn't use the VaR (a quantile). It's possible the question used VaR as a red herring for ES. Thanks,

 

[Tom77]

Finally i got it. Thanks a lot I always was missing the 100, but if n was indeed 100 the 100th loss could also have been any number between 0 and 94. Thanks.

 

[cqbzxk]

Hi David,

it seems like GARP is trying to confuse people during the exam with questions about stuff you though easy while learning. But then in the exam it turns out that the questions are very different from the ones you were working on while preparing. One specific question on Expected shortfall. We learned like ES 95% is just the average of tail losses above the 95% VaR. What if they question you 95.5% or 96.5% ES like the question below, which i took from the section "Post what you can remember here"... ?

5) Confidence VAR
94 a
95 b
96 c
97 d
98 e
99 f

What is the Expected shortfall at 95.5% confidence
I guess the answer is Answer: Average of c,d,e,and f

it's very easy from my view, you just simply calculate ES at 95% and choose the answer which lightly lager than ES at 95%.

 

[cqbzxk]

I can't say that i precisely understand the actual-exam-question posted, I just assume something was lost in translation. But, okay, aren't the worst five losses listed, aren't they: 99, 98, 97, 96, and 95? In which case, the 95.5% ES is 97.22. The average of the worst four losses, per our discussion, is necessarily the (1 - 4/n)ES or 96.0% ES if n = 100. The VaR assumption, or outcome for that matter, is irrelevant to ES; ES (a conditional mean) doesn't use the VaR (a quantile). It's possible the question used VaR as a red herring for ES. Thanks,

hi david, for this question, can I simply calculate ES at 95% and ES at 96% and find the answer between these two answers? thanks

 

[cqbzxk]

I can't say that i precisely understand the actual-exam-question posted, I just assume something was lost in translation. But, okay, aren't the worst five losses listed, aren't they: 99, 98, 97, 96, and 95? In which case, the 95.5% ES is 97.22. The average of the worst four losses, per our discussion, is necessarily the (1 - 4/n)ES or 96.0% ES if n = 100. The VaR assumption, or outcome for that matter, is irrelevant to ES; ES (a conditional mean) doesn't use the VaR (a quantile). It's possible the question used VaR as a red herring for ES. Thanks,

Hi, David I was totally confusing at this concept, since ES at 95.5% , why should we calculate 95% VaR since ES suppose to exceed VaR at 95.5% ? thanks

 

[David Harper CFA FRM]

Hi cqbzxk, but i never said that (i have not seen the source exam question, i suggested that if it asked for ES and included VaR, the VaR is a red herring). In my experience, the #1 confusion in regard to ES is the mistaken idea that it has some relationship to VaR. It does not. A 95% ES is the (conditional) average of the worst 5% (of the distributional tail), period. A 95.5% ES is the average of the worst 4.5%. We identify the distribution, we find the 1-confidence = signficance% tail, and we translate that tail into its own probability distribution such that its mean (weighted within the tail distribution only) is the ES. The 95% VaR is a quantile, it plays no role is this calculation. The confusion typically reduces to a confusion about the meaning of a conditional mean; it does not mean conditional on VaR! We may simultaneously want both the 95% VaR and the 95% ES, but the ES does not need anything from the VaR. Thanks,

 

[David Harper CFA FRM]

I forget (is another reason i regret trying to decipher these partial question snippets) that we can estimate ES by taking an average of the tail VaRs, per Dowd Chapter 3 (Dowd 3.4.1). That is an instance of using easier-to-access VaRs (quantiles) as a means to approximate the conditional (tail) mean, but I have no clue if this question was looking for that approach, having yet to see the actual question thanks,