Delta-Gamma Neutral portfolio, derivation issue

Question

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 1

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