eredhuin
Senior Member
Posts: 263
Joined: Jul 2002

Wed May 28, 08 04:49 PM


Let's start by admitting VaR isn't perfect. My view is that it's probably not a good idea to use Gamma in VaR. I say this having seen disasters with books showing a bazillion dollars of scenario Gamma PnL because the option is ATM and close to expiry.
tw, one way to look at VaR is that it's just the taylor series using a shock of a certain confidence interval. For Delta VaR you don't even need to care which shock (up or down) hurts you. Just take the absolute value and you're done.
Once you start including Gamma you need to worry about signage of your shock. You have two choices (up, down) then (here I borrow DaveAngel's nomenclature, which I like)
PnL_0 = 0 // base case PnL_1a = Delta * h + 0.5 * Gamma * h ^2 PnL_1b = Delta * h + 0.5 * Gamma * (h)^2
If you're short gamma you need to look at abs(min(0,(PnL_1b  PnL_0),(PnL_1a  PnL_0))) to know what VaR is.
But this raises another question. Once you get positive gamma into the VaR calculation, you're looking for a minimum inside a confidence interval of width [h,h]. It need not be at +/h (which does occur for negative gamma). Consider where you are short delta and long gamma. By differentiating the quadratic form of TSE PnL you find a slope of zero (i.e. minimum) at hstar=Delta/Gamma. So to be totally pedantic the DeltaGamma VaR would be something like
if (Gamma > 0 and hstar > h and hstar < h) then // special case, positive 2nd deriv parabola  find the worst PnL change DGVaR = abs(Delta * hstar + 0.5 * hstar^2 * Gamma) = 0.5 * Delta ^2 / Gamma else // worst PnL will be either h or +h and the min(0) part is there for the case where delta is zero and gamma is positive DGVaR= abs(min(0,(PnL_1b  PnL_0),(PnL_1a  PnL_0))) end
Note that as I have said I would be careful using Gamma in VaR esp. if shocks are large and your options are atm and close to expiry.
Edited: Wed May 28, 08 at 04:51 PM by eredhuin

