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I am supposed to calculate the value at risk and expected shortfall of an asset with revenue given by a density function: $f(x)=0.5\exp{(-|x-0.05|)}$.

My workings:

If I understand it correctly, than $VaR_\alpha=F^{-1}(\alpha)$. Thus I need to compute the distribution function:

For $x_1<0.05$ $$\int^{x_1}_{-\infty}\frac12 e^xe^{-0.05}dx=\frac12e^{-0.05}e^{x_1}$$

For $x_2\geq 0.05$ $$ \frac12+\int^{x_2}_{0.05}f(x)dx=1-\int^{\infty}_{x_2}\frac12e^{-x}e^{0.05}dx= 1-\frac12e^{0.05}e^{-x_2} $$

Thus, for $y<\frac12$, $VaR_\alpha=\log{2\alpha e^{0.05}}$ and for $y\geq\frac12$, $VaR_\alpha=\log{\frac{e^{0.05}}{2(1-\alpha)}}$

Now, again, if I understand it correctly (and this time I really am not sure) $$CVaR_\alpha=\frac{1}{1-\alpha}\int^{\infty}_{VaR_\alpha} xf(x)dx$$

This is where I'd like to stop and check with Math.SE.

My questions:

  1. Is this correct, so far?
  2. Is the CVaR formula correct?
  3. Is there any good text about CVaR on the internet? I've been unable to find one that'd treat the subject more mathematically. Especially an example including a density function would be very helpful.

Thank you for help.

Dahn
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    Your VaR formulas are correct. Generally you won't need the case $x>0.05$ as VaR is usually a left-tail percentile. CVar also looks correct - it the the expected loss conditoned on the loss exceeding VaR. – RRL Jun 03 '14 at 14:51
  • Reference http://www.ise.ufl.edu/uryasev/files/2011/11/cvar2_jbf.pdf – RRL Jun 03 '14 at 14:55
  • @RRL Thank you. Are you sure the bounds in the CVaR should not then be reversed, since I am dealing with a profit, rather than a loss density function? – Dahn Jun 03 '14 at 18:22
  • Yes that is correct the upper limit should be VaR – RRL Jun 03 '14 at 18:41

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