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 FORUMS > Technical Forum < refresh >
 Topic Title: Expected shortfall calculation Created On Sat Jan 18, 03 04:21 AM Topic View: Branch View Threaded (All Messages) Threaded (Single Messages) Linear

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abhandari
Junior Member

Posts: 13
Joined: Sep 2002

Sat Jan 18, 03 04:21 AM

Hi,

I have the variance covariance matrix which I use to compute the VAR of a portfolio. I would like to use the same matrix and the weights to compute the expected shortfall for the portfolio. Can someone explain me how to go about doing this. I want to compute the expected shortfall at the 5% VAR levels.

If someone has an excel sheet with a model calculation of both the VAR statistic and expected shortfall, it would be very helpful.

Best Regards
Arjun Bhandari

-------------------------
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To make ourselves sublime,

And leave behind us when we depart,

Footprints on the sands of time.

Anthis
Senior Member

Posts: 4301
Joined: Oct 2001

Sat Jan 18, 03 11:13 AM

I have the impression that ES can be calculated only with simulation (historical or MC) methods. And you need enough tail observations so that the results can be reliable.

If i am wrong, someone can correct me

MRNAXOS
Junior Member

Posts: 1
Joined: Jan 2003

Sat Jan 18, 03 05:56 PM

I think that you can calculate ES only with simulation methods.

Of course you may suppose a P/L distribution and use the parametric VaR figure to estimate ES (using statistics).

mrnaxos

acabrol
Junior Member

Posts: 16
Joined: Jul 2002

Tue Jan 21, 03 04:53 PM

simulation seems to be the only practical way, however I found somme answer to your question in papers on this web site (sorry in french, however the bibliography can help !!!)

GRO credit lyonnais
Paper in french that can help

The idea is to use a Generalised Pareto Distribution, which is parametric, as a proxy of the excess return distribution (F(u,y)=Pr(X>y+u | X>u))).
The expected shortfall is then a function of the VaR(5%), and the GPD parameters (beta and epsilon).
In a multidimensional framework, you have to use extreme copulae and it seems to be quite hard beyond dimension 2.

hope it can help

Axel

Aaron
Senior Member

Posts: 6435
Joined: Jul 2001

Wed Jan 22, 03 12:51 AM

Anthis and MRNAXOS are giving you conservative empirical answers, acabrol is giving you a conservative theoretical one.

If you assume a Normal distribution, the expected shortfall is easy to compute. If your standard deviation is S then your 95% VaR is 1.64*S and the expected shortfall is 2.06*S. I use the definition of Expected Shortfall as expected loss conditional on loss being greater than VaR. Sometimes people use variants of this, such as my definition minus VaR.

The general Normal distribution formula for VaR with confidence a is -NORMSINV(a)*S. Expected Shortfall is exp(-(VaR^2)/2)/(a*(2*pi)^0.5). Be sure to put in the extra parenthesis around VaR^2 or Excel will take the negative first and then square, in defiance of standard operator precedence rules.

For a non-Normal distribution you should have no trouble computing VaR and Expected Shortfall either analytically or numerically.

People seldom use non-Normal distributions to compute Expected Shortfall in practice, it's too hard to simulate. However, it is very important to include security payoffs that are non-linear in their market factors. For many purposes it's not too bad to assume the underlying price changes follow a normal distribution, but don't apply that assumption to option prices. However, since you're using a covariance matrix for the computation, you can't include non-linear effects.

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Aaron Brown

Junior Member

Posts: 2
Joined: Jul 2002

Thu Nov 03, 05 01:50 PM

Apologies for resuscutating a long-closed thread, but what is the general formula for expected shortfall assuming a normal distribution? The definition of expected shortfall I'm using is E(x|x<a). This is easy to calculate from simulations: run your scenarios; define a shortfall level (a); take all of the observations that fall short of (a) and add up the total difference between these obs and (a); divide the total by the total number of observations (including those that are greater than or equal to (a)).

I think the formula to calculate this is the integral between negative infinity and (a) of x.f(x), which you might be able to integrate by parts, but my A-level stats is ropey and my A-level maths is ropier. Can anyone shed any light on this?

Cheers

RR

Aaron
Senior Member

Posts: 6435
Joined: Jul 2001

Sun Nov 13, 05 04:53 PM

Quote

Originally posted by: Aaron
The general Normal distribution formula for VaR with confidence a is -NORMSINV(a)*S. Expected Shortfall is exp(-(VaR^2)/2)/(a*(2*pi)^0.5). Be sure to put in the extra parenthesis around VaR^2 or Excel will take the negative first and then square, in defiance of standard operator precedence rules.

-------------------------
Aaron Brown

JakeManson
Junior Member

Posts: 14
Joined: Mar 2002

Sun Feb 15, 09 05:52 PM

Hey Aaron,

I know this formula must be in several books (Jorion for one), but I was lazy and googled this, and this forum came up. I am simulating loss distributions on huge dollar amounts, on the order of about a trillion, and S = 7.94E+10. I am scaling this to be in units of trillions so as to avoid overflow. VaR(99.97%) = 9.49e+11=\$ 0.949 Trillion (I have losses as positive). Using your formula I am getting something nonsensical: ES999 = exp(((0.949)^2)/2)/(.001*(2*3.14)^0.5) = 626.07. Is it obvious to you where I am going wrong?

By way of context, I am not doing a VaR for trading P/L, but for time series of different types of banks (these #'s are from a hypothetical mega-bank) losses for an economic capital calculation. I am fitting different copulas, and want to benchmark my result against assuming joint normality. By the way, I am not having much trouble numerically getting ES for simulated distribution, due to the simple nature of this data (100 quarters of losses for 5 risk types - can't get more primitive than that!)

Thanks!

Jake

lemair93
Junior Member

Posts: 1
Joined: Feb 2009

Thu Feb 19, 09 01:43 PM

Hi,

My integration may be off, but here is the general equation for ES N(mu,sigma) ES at confidence of 1-a is:

ES = (sigma)/(a Root(2Pi)) exp[-(1/2)z^2]

Where z is the cumulative normal at a (1.645, 2.33, ...)

Hope that helps,

P.

raul07
Junior Member

Posts: 1
Joined: Mar 2009

Sun Mar 08, 09 09:58 PM

Lemair and Aaron,

thanks for the general formula.. could you guys possibly give me a citation/link for this please.

thanks,

raul

Edited: Sun Mar 08, 09 at 09:59 PM by raul07

cm27874
Member

Posts: 69
Joined: Jul 2007

Wed Mar 11, 09 02:09 PM

McNeil / Frey / Embrechts: Quantitative Risk Management (Princeton Series in Finance), p. 45

daveyale
Junior Member

Posts: 1
Joined: Oct 2009

Fri Oct 23, 09 01:07 AM

Thanks Lemair I've seen plenty of supposed analytic solutions to this problem for normal distribution and yours is the one I'm now using to test a simulation suite before letting it loose on the world of asymmetric distributions! For anyone else needing this, a useful shortcut in addition to Aaron's note below from years back is that for 99% conf the expected shortfall is approx 2.67*StdDev.

Cheers.

MHill
Senior Member

Posts: 435
Joined: Feb 2010

Mon Mar 01, 10 11:42 AM

Having used the info you have all kindly provided, I thought it only fair to (finally) join & share:
If you look at Aaron's formula, it looks very like the formula for the Standard Normal Density Function. So in Excel, you should be able to write:
=(Volatilty)*NORMDIST(NORMSINV(Confidence),0,1,FALSE)/(1-Confidence).

Cheers,

iamorlando
Member

Posts: 85
Joined: Oct 2009

Mon Mar 01, 10 06:41 PM

The results I calculate from Aaron, Mhill, and lemair93's methods are vastly difrent. Aaron's formula gives a CVaR of about 50% portfolio size, Lemair's gives a CVaR that is only marginally greater than VaR, and Mhill's is way smaller than VaR. Whats going on here?

-------------------------
The O-Factor

MHill
Senior Member

Posts: 435
Joined: Feb 2010

Thu Mar 11, 10 10:29 AM

I'd guess that either you've an error in translating the formula, or the 'a' 'z' and 'Confidence' are getting confused.
When I put in 'Confidence', I'd have 0.99 from using a 99% VaR. Lemair & Aaron have 'a' as 0.01 for 99% VaR, and 'z' as NORMSINV(1-a), equivalent to -NORMSINV(a).
I'd have Lemair's formula as =(Volatility)/((1-Confidence)*SQRT(2*PI()))*EXP(-(0.5*NORMSINV(Confidence)^2)) in Excel,
and Aaron's as =Volatility*EXP(-(NORMSINV(Confidence)^2)/2)/((1-Confidence)*(2*PI())^0.5).

Hope that helps,

mcgrat07
Junior Member

Posts: 2
Joined: May 2010

Wed May 26, 10 03:56 AM

This is a question/response to a very, very old answer from Aaron--hopefully, it is still relevant.

Quote

If you assume a Normal distribution, the expected shortfall is easy to compute. If your standard deviation is S then your 95% VaR is 1.64*S and the expected shortfall is 2.06*S.

Of course, I understand the VaR calculation here--that is by the book. But, give that and the 95% level of confidence, how do you derive the critical value for CVaR? I expect this is really simple, and I apologize for missing something obvious.

Thanks,
James

acastaldo
Senior Member

Posts: 1369
Joined: Oct 2002

Wed May 26, 10 11:09 PM

Evaluate Aaron Brown's formula

exp(-(VaR^2)/2)/(a*(2*pi)^0.5)

with VaR = 1.644854, a= 0.05 and pi=3.14159

and you will get 2.062713 as advertised.

acastaldo
Senior Member

Posts: 1369
Joined: Oct 2002

Wed May 26, 10 11:45 PM

...but if the question is: where does Aaron Brown's formula come from, a derivation occurs at the bottom of Page 61 (section III.A) in

Yamai, Y. and T. Yoshiba. (2002) ?On the Validity of Value-at-Risk: Comparative Analyses with Expected Shortfall.? Monetary and Economic Studies, V.20, 57-85.

a copy of which is available here [BOJ web site]

mcgrat07
Junior Member

Posts: 2
Joined: May 2010

Thu May 27, 10 11:36 PM

Excellent! That was exactly what I was looking for. Thanks for the reference.

Tedypendah
Member

Posts: 69
Joined: May 2013

Thu Mar 19, 15 07:23 AM

Quote

Originally posted by: Aaron
...
The general Normal distribution formula for VaR with confidence a is -NORMSINV(a)*S

Can someone please confirm for me that the miu is missing here because we looking at market returns and therefore the assumption of iid as well as a short time period allow us to assume a zero mean?

Otherwise for a different time series we will have to fit a normal and get the miu and standard dev

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