Delta-Gamma Hedge Derivation

Article Quant Q&A · Author: Dark

Summary

The document derives a local delta-gamma hedge for a target option using a second option and the underlying asset. It writes the hedge as a combination of the underlying and the second option, then matches the target option's gamma by setting the second option's weight to the ratio of their gammas. Matching delta determines the underlying position after accounting for the second option's delta.

The apparent extra term in the question arises from differentiating hedge weights as though they vary with the underlying. The answer's derivation treats the weights as fixed when computing the portfolio's instantaneous delta and gamma; the hedge is established at the current point and may be rebalanced as conditions change. Thus, the coefficient formulas match the local sensitivities under that convention, rather than asserting that a portfolio with continuously changing weights has the same total derivative. The exchange does not discuss transaction costs, rebalancing frequency, or risks beyond these local Greeks.

Key ideas

A second option can supply the gamma needed to match a target option locally.
The underlying position is chosen to make the combined portfolio delta match the target.
The hedge weights are treated as fixed when calculating instantaneous portfolio Greeks.
A hedge based on local Greeks may need rebalancing as market conditions change.

Tags

Options Derivatives pricing Risk management

Full text

# Delta-Gamma Neutral portfolio, derivation issue


# Delta-Gamma Neutral portfolio, derivation issue












Let $C$ be an option on an underlying $S$. I want to construct a portfolio $V$ using another asset $C_0$ such that the delta and the gamma of $V$ is the same as the delta/gamma of $C$, in order to hedge the option.

Let : $\gamma = \frac{\frac{\partial^{2}C}{\partial S^{2}}}{\frac{\partial^{2}C_0} {\partial S^{2}}} = \frac{\Gamma_C}{\Gamma_{C_0}}$

$\delta = \frac{\partial C}{\partial S} - \frac{\partial C_0}{\partial S}\gamma$

Apparently, if $V = \gamma C_0 + \delta S$, then $\Delta_V = \Delta_C$ and $\Gamma_V = \Gamma_C$

However, when I try to derive the delta of $V$, I get :

$\Delta_V = \frac{\partial V}{\partial S} = \Delta_C + \frac{\partial \gamma}{\partial S} (C_0 - S\frac{\partial C_0}{\partial S})$

So the second term in the sum must be equal to 0, but I don't see why ? Maybe it isn't and we just choose $C_0$ such that $\frac{\partial \gamma}{\partial S}$ is small ?

Thanks for your help.

## Answer by Quantifeye (score 1)

https://quant.stackexchange.com/a/16716

let $\frac{\partial C}{\partial S}=\delta_c$

let $\frac{\partial^2 C}{\partial S^2}=\Gamma_c$

let $\frac{\partial C_0}{\partial S}=\delta_0$

let $\frac{\partial^2 C_0}{\partial S^2}=\Gamma_0$

we want

$\frac{\partial V}{\partial S}=\frac{\partial C}{\partial S}=\delta_c$

and

$\frac{\partial^2 V}{\partial S^2}=\frac{\partial^2 C}{\partial S^2}=\Gamma_c$

let

$V=aS+bC_0$

then

$\delta_c=\frac{\partial}{\partial S}\left( aS+bC_0 \right)=a+b \delta_0$

and

$\Gamma_c = \frac{\partial}{\partial S}\left( a+b \delta_0 \right)=b\frac{\partial}{\partial S} \left( \delta_0 \right) = b \Gamma_0$

therefore

$b=\frac{\Gamma_c}{\Gamma_0} \quad (=\gamma)$

and

$\delta_c=a+ \frac{\Gamma_c}{\Gamma_0} \delta_0$

showing that

$a= \delta_c- \frac{\Gamma_c}{\Gamma_0} \delta_0$

or

$a= \frac{\partial C}{\partial S}- \frac{\Gamma_c}{\Gamma_0} \frac{\partial C_0}{\partial S} \quad (=\delta)$

Hope this helps

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Deriving a Local Delta-Gamma Hedge with an Option and the Underlying

Summary

Key ideas

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