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Hello, I am trying to find the Expected Shortfall corresponding to a 95% confidence level on a lognormal distribution. Given x~LogNorm solve for Alpha. E[ x | x > LogNormInv( Alpha%) ] = LogNormInv(95%) Any idea?Many thanks

Are you sure your equations are correct?Should'nt it be ES = E[x | x > LNORMINV(0.95) ] This expectation, the partial expectation of a lognormal variate, is available in closed form.

Hello acastaldo,Do you know of a clear textbook or other reference which walks through the derivation of this partial expectation?Thanks.

teddydavis asks for a clear derivation of the partial expectation of a lognormal r.v. Here is mine:we want to findMake a change of variablesone more change of variableQEI

Expected Shortfall is usually defined as:Also, is typically defined through the percentile .Putting it all together:

Hi acastaldo,I wanted to ask if g(k) gives final partial expectation or we should multiply it with probability to get the expectation.Thanks

I believe fizik's remark is correct: the integral g(k) needs to be divided by the probability of occurrence of an extreme event to give ES.