1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks. But I don't understand why $\frac{\partial a}{\partial S} = 0$ and $\frac{\partial b}{\partial S} =0$. $\endgroup$ – Dark Feb 24 '15 at 15:32
  • 1
    $\begingroup$ Because they are constants. You start off by saying lets find these two constants, so that over some very small increment of time the replicating portfolio will behave in the same way as the instrument being replicated. After that increment of time has passed, you have to go and adjust those values again. So you can actually index all of the elements above according to time. $\Gamma_c$, $\Gamma_0$, $\delta_0$, and $\delta_c$ are all values that you observe now... or the theoretical values that they should have now... $\endgroup$ – Quantifeye Feb 24 '15 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.