Dear All
First of all, this is a great book.
My question is with the mu term in the expected excess loss. I found it strange that there would be a mu term at all, since the expected excess loss is derived from integrating from zero to infinite x*g(x) where g(x) is the GPD density function with beta and Xi parameters only.
In addition, when we set u to be VaR number, the expected loss conditional on loss greater than VaR looks different from Hull's result regarding expected shortfall.
I hope that I've made my point clear.
Joy
Expected Tail Loss
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 Posts: 815
 Joined: Sun Sep 28, 2008 10:30 pm
Re: Expected Tail Loss
Hi Joy
Thanks for your nice comments about MRA. I am not sure which volume you are referring to, but I suspect this thread should really be in Volume IV. In general questions should refer to the actual formulas numbers in the books. In your case, writing down Hull's formula would also be needed.
I can, however, already answer one question i.e. 'Why does the mean mu affect the ETL?' The answer is that as the mean changes the whole distribution shifts along the horizontal axis. So, the average taken along this axis will be translated: if you add mu to the mean, you also add mu to the ETL.
Cheers, Carol
Thanks for your nice comments about MRA. I am not sure which volume you are referring to, but I suspect this thread should really be in Volume IV. In general questions should refer to the actual formulas numbers in the books. In your case, writing down Hull's formula would also be needed.
I can, however, already answer one question i.e. 'Why does the mean mu affect the ETL?' The answer is that as the mean changes the whole distribution shifts along the horizontal axis. So, the average taken along this axis will be translated: if you add mu to the mean, you also add mu to the ETL.
Cheers, Carol
Re: Expected Tail Loss
Dear Carol
Thanks for your prompt response. I am referring to volume I page 104105, section 1.3.3.10. Equation 1.3.68 states that the mean excess loss e(u) equals to (beta+Xi*u)/(1Xi) where u is the threshhold. My question is that the mean excess loss is to integrate x*g(x) (x is the excess loss) from 0 to infinity and there is no u anywhere in the process. My derived result is (beta)/(1Xi).
Equation 1.3.69 states that the mean tail loss is VaR+e(VaR). This has no problem but when we plug in Equation 1.3.68. The result is (VaR+betaVaR*Xi+Xi*u)/(1Xi). This result is different from Hull's expected shortfall. In his third edition, it will be equation 14.10 Expected Shortfall=(VaR+betaXi*u)/(1Xi). Here if we set u=VaR, we have (VaR+betaVaR*Xi)/(1Xi).
If the formula for e(u) doesn't have the Xi*u in the numerator then the results would be the same.
I hope that I am not creating more confusions. This has troubled me these recent few days. Really look forward to your reply. Thanks a lot!
Joy
Thanks for your prompt response. I am referring to volume I page 104105, section 1.3.3.10. Equation 1.3.68 states that the mean excess loss e(u) equals to (beta+Xi*u)/(1Xi) where u is the threshhold. My question is that the mean excess loss is to integrate x*g(x) (x is the excess loss) from 0 to infinity and there is no u anywhere in the process. My derived result is (beta)/(1Xi).
Equation 1.3.69 states that the mean tail loss is VaR+e(VaR). This has no problem but when we plug in Equation 1.3.68. The result is (VaR+betaVaR*Xi+Xi*u)/(1Xi). This result is different from Hull's expected shortfall. In his third edition, it will be equation 14.10 Expected Shortfall=(VaR+betaXi*u)/(1Xi). Here if we set u=VaR, we have (VaR+betaVaR*Xi)/(1Xi).
If the formula for e(u) doesn't have the Xi*u in the numerator then the results would be the same.
I hope that I am not creating more confusions. This has troubled me these recent few days. Really look forward to your reply. Thanks a lot!
Joy

 Posts: 815
 Joined: Sun Sep 28, 2008 10:30 pm
Re: Expected Tail Loss
Hi Joy
Thanks for pointing out where the formula was. I had forgotten that I gave a brief overview of the GPD in Vol I. I drew heavily on my friend Alexander McNeil's work here, but also checked the formulas myself.
I believe formula (I.3.68) for the mean XS loss under the GPD is correct. I also checked the web and found the same formula here ftp://hubble21.math.ethz.ch/users/mcneil/cairns.pdf and here http://statisticalmodeling.wordpress.co ... tribution/ for example,
For the latter of the two references above, the parameters are different: alpha = Alex's beta/Xi and beta = Alex's 1/Xi....but on making that substitution you'll see that the two distributions are the same....and in the mean XS loss formula further down the page you'll again see the same result as (I.3.68).
I don't have Hull's book, but it is clear intuitively that the mean XS loss must depend upon the threshold u over which it is measured (and note: it appears in the excess loss function). If you get a different result to Hull using (I.3.68) perhaps you should check there isn't an error in Hull's equation?
Cheers, Carol
Thanks for pointing out where the formula was. I had forgotten that I gave a brief overview of the GPD in Vol I. I drew heavily on my friend Alexander McNeil's work here, but also checked the formulas myself.
I believe formula (I.3.68) for the mean XS loss under the GPD is correct. I also checked the web and found the same formula here ftp://hubble21.math.ethz.ch/users/mcneil/cairns.pdf and here http://statisticalmodeling.wordpress.co ... tribution/ for example,
For the latter of the two references above, the parameters are different: alpha = Alex's beta/Xi and beta = Alex's 1/Xi....but on making that substitution you'll see that the two distributions are the same....and in the mean XS loss formula further down the page you'll again see the same result as (I.3.68).
I don't have Hull's book, but it is clear intuitively that the mean XS loss must depend upon the threshold u over which it is measured (and note: it appears in the excess loss function). If you get a different result to Hull using (I.3.68) perhaps you should check there isn't an error in Hull's equation?
Cheers, Carol
Re: Expected Tail Loss
Dear Carol
Thanks very much for your reply.
I checked the websites you provided but they don't have the derivation. I am enclosing my proof in this post and please help me check whether the first four lines are right. If they are, then the equation doesn't have u in it and the result won't contain u.
I am not sure of the intuition too. Even if the threshhold increases, it doesn't necessarily mean that mean excess loss would be greater. In addition, when we choose different threshhold, beta and Xi are reestimated, so even if we don't have u directly in the formula, the mean excess loss could be larger (as the parameters themselves would change).
Sorry to keep bothering you. But I really want to figure it out. Thanks!
Joy
Thanks very much for your reply.
I checked the websites you provided but they don't have the derivation. I am enclosing my proof in this post and please help me check whether the first four lines are right. If they are, then the equation doesn't have u in it and the result won't contain u.
I am not sure of the intuition too. Even if the threshhold increases, it doesn't necessarily mean that mean excess loss would be greater. In addition, when we choose different threshhold, beta and Xi are reestimated, so even if we don't have u directly in the formula, the mean excess loss could be larger (as the parameters themselves would change).
Sorry to keep bothering you. But I really want to figure it out. Thanks!
Joy
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 Posts: 815
 Joined: Sun Sep 28, 2008 10:30 pm
Re: Expected Tail Loss
Suppose u is a very large positive number. Then the mean excess loss is more or less the mean of the distribution, yes? Now set u to be far below the mean, how can the mean excess loss still be the mean?
You need to integrate (x  u) g(x) where g(x) is the density function for the GPD from  infty to u, and u appears as the upper limit of integration. See the definition here http://statisticalmodeling.wordpress.co ... function/
You need to integrate (x  u) g(x) where g(x) is the density function for the GPD from  infty to u, and u appears as the upper limit of integration. See the definition here http://statisticalmodeling.wordpress.co ... function/
Re: Expected Tail Loss
Dear Carol
Thanks for answering my questions very patiently. I still disagree on the intuition.
When u is very large, mean loss is more or less the same as the mean of the distribution. But mean excess los should be more or less mean of the distribution  u. Therefore if we change u to be very small, mean loss is much smaller. But it doesn't imply that the mean excess loss is much smaller.
But I think the core of the problem lies in how to interpret the GPD. I think it is defined on excess loss already. As the CDF when x= 0 is 0 and when x approach infinity CDF is 1. x here then should be the excess loss. If the g(x) is defined on excess loss, shouldn't we integrate x instead of xu from 0 to infinity?
Best
Joy
Thanks for answering my questions very patiently. I still disagree on the intuition.
When u is very large, mean loss is more or less the same as the mean of the distribution. But mean excess los should be more or less mean of the distribution  u. Therefore if we change u to be very small, mean loss is much smaller. But it doesn't imply that the mean excess loss is much smaller.
But I think the core of the problem lies in how to interpret the GPD. I think it is defined on excess loss already. As the CDF when x= 0 is 0 and when x approach infinity CDF is 1. x here then should be the excess loss. If the g(x) is defined on excess loss, shouldn't we integrate x instead of xu from 0 to infinity?
Best
Joy
Re: Expected Tail Loss
Dear Carol
After checking the website, I finally am able to get a consistent result. The implicit assumption in the website is that the cutoff threshhold for using the GPD is zero. And we calculate an excess loss when loss is greater than deductible. The expected tail loss now in 1.3.69 is consistent with Hull's expected shortfall. We only need to set his cuoff threshhold to be 0.
And if the imlicit threshhold is set to be zero, certainly the mean excess loss (greater than an arbitrary deductible u) is going to vary with the deductible u. This is different from what I have in mind when I allow the threshold to move.
Finally thanks a lot for your explanation and the website. I will continue enjoying your book!
Joy
After checking the website, I finally am able to get a consistent result. The implicit assumption in the website is that the cutoff threshhold for using the GPD is zero. And we calculate an excess loss when loss is greater than deductible. The expected tail loss now in 1.3.69 is consistent with Hull's expected shortfall. We only need to set his cuoff threshhold to be 0.
And if the imlicit threshhold is set to be zero, certainly the mean excess loss (greater than an arbitrary deductible u) is going to vary with the deductible u. This is different from what I have in mind when I allow the threshold to move.
Finally thanks a lot for your explanation and the website. I will continue enjoying your book!
Joy
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