Handbook example 10.5

ckyeh

New Member
Hi David:

On Handbook 5th example 10.5 page 258:
Given the following 30 ordered percentage returns of an asset, calculate
the VAR and expected shortfall at a 90% confidence level: −16,−14,
−10,−7,−7,−5,−4,−4,−4,−3,−1,−1, 0, 0, 0, 1, 2, 2, 4, 6, 7, 8, 9, 11,
12, 12, 14, 18, 21, 23.

the answer is
b. VAR (90%) = 10, expected shortfall = 15

However, acoording to the equation: P(|L|>Var)≦1-c, I get P(|L|>7)≦10%.
Which one is the correct answer?

If VaR is 7, how to calculate expected shortfall?

Thanks
 

ckyeh

New Member
One more relative issue, on your webnair 「2010-5-d-Market-Risk」,page 19:

ES(Expected shortfall) is
6027 (CL=0.995, Var= 2988)
4276 (CL=0.990, Var= 2524)
3685 (CL=0.985, Var= 2503)
3380 (CL=0.980, Var= 2466)
……

However, according to the equation on webnair 「2010-5-d-Market-Risk」page 10,
I get the ES:
(1/(1-CL))*(1/200)*3039 = (1/0.005)*0.005*3039=3039~CL=0.995, Var= 2988
(1/0.01)*{0.005*3039+0.005*2988}=3014 ~CL=0.990, Var= 2524
(1/0.015)*{0.005*3039+0.005*2988+0.005*2524}=2850~CL=0.990, Var= 2524
……

Which one is correct? Thanks!
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi ckyeh,

Great observation on the handbook question; I will include this item in my list of issues for GARP's attention (I have already alerted them to the general issue, of their methodology problem in regard to HS, but I wasn't aware of this exact example which really highlights the confusion).

This handbook answer is wrong in regard to ES: 15 is the average of two 3.33% outcomes, therefore 15 is the 1-1/30*2 = 93.33% ES. The 90% ES cannot be 15, regardless of the VaR.. Or to put another way, 90% ES means "average loss in the 10% tail, or conditional on the 90% loss threshold: unlike HS, it has only one answer!. For example, 92% "chops into" the 3rd worst loss, but as a mean, there is a decisive 92% ES.

I agree with your 10%. Just as 10% is consistent with the formula, it is also consistent with Dowd; i.e., the three worst losses occupy the 10% tail, such that the 4th worst loss is the 90% VaR.
However, there is no right/wrong answer in the DISCRETE distribution: 90% VaR can be 3rd worst (-10; Jorion), 4th worst (-7; per Dowd) or interpolation between the two
…. My request to GARP is, obviously, the exam should standardize (!)

The ES, however, is not debatable. Per a discrete distribution, the 90% ES is the average of the worst 10% and, in this case, is therefore: (1/30*16 + 1/30*14 + 1/30*10)*1/10% = 13.3333; i.e., average(L | Losses in 10% tail) = 13.333

In regard to p 19, that is actually not mine but a snapshot from Dowd … I will try to retrieve the original to settle that, tomorrow. Thanks

David
 
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