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 FORUMS > General Forum < refresh >
 Topic Title: delta gamma VAR Created On Wed May 21, 08 01:31 AM Topic View: Branch View Threaded (All Messages) Threaded (Single Messages) Linear

tw
Senior Member

Posts: 577
Joined: May 2002

Wed May 21, 08 01:31 AM

Hi

Does anyone know a simple approximation for delta-gamma VAR for a single asset?

I feel it should be something like
VAR(delta-gamma)=VAR(delta-only) + (gamma^2/(4*delta))*volatility*NORMINV(1-0.95)

Cheers.

daveangel
Senior Member

Posts: 16967
Joined: Oct 2003

Wed May 21, 08 08:15 PM

i feel that this is not correct.

make h be the change in the asset price for the required confidence level

h = vol * S * f

where f is the confidence factor ie number of sds that are equivalent to your confidence

then

VaR = delta * h + 1/2 * gamma * h^2

-------------------------
Knowledge comes, but wisdom lingers.

tw
Senior Member

Posts: 577
Joined: May 2002

Tue May 27, 08 06:42 PM

Hi

So if you have zero delta position, positive gamma gives larger VAR?

I was a bit puzzled how to come up with a rule of thumb where one can adjust the delta in the standard
formula to take into account gamma. Or indeed if that's possible.

I came across the paper "Delta -gamma four ways" :it seemed to suggest an approximation of the form

VAR=mu_1+sqrt(mu_2)*za

With mu_1 and mu_2 the moments of the Johnson transformed variable and za a normal of the appropriate size.
for the confidence limit.

It gives formulas for the moments as
mu_1=1/2*gamma and
mu_2=delta^2+gamma^2/2

(with a priced scaled definition of gamma/delta, it looks like).

Hence
VAR=gamma/2+sqrt(delta^2+gamma^2/2)*za

Expanding for gamma<<delta

VAR(delta-gamma)=VAR(delta)+[za*gamma^2/4*delta+gamma/2]

Make any sense?

Quote

Originally posted by: daveangel
i feel that this is not correct.

make h be the change in the asset price for the required confidence level

h = vol * S * f

where f is the confidence factor ie number of sds that are equivalent to your confidence

then

VaR = delta * h + 1/2 * gamma * h^2

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 Delta-Gamma 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