Proof on Greeks for a delta neutral portfolio

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Suppose that $ \Delta S $ is the price change of an underlying asset during a small time interval $\Delta t$ and $ \Delta \pi$ is the corresponding price change in the portfolio. Show that $ \Delta \pi = \Theta \Delta t + \frac{1}{2} \Gamma \Delta S^2$ for a delta neutral portfolio, where $\Theta$ is the theta of the portfolio, and where terms of higher order that $\Delta t$ are ignored.


For a delta nuutral portfolio, it is required that

$\frac{\Delta \pi }{\Delta S} =0 $

For theta :

$\Theta = \frac{\Delta \pi}{\Delta t}$

For Gamma :

$\Gamma = \frac{\Delta \pi ^2}{\Delta ^2 S}$

I am confused whether to write $\Delta$ or use $\partial$ instead. And I don't know how to get the equation required.

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Let the portfolio be $\pi=\pi(S,t)$. So we have the Taylor expansion $$ \Delta\pi\approx\underbrace{\frac{\partial\pi}{\partial S}}_{\text{Delta}\,\Delta_\pi}\times\Delta S+\underbrace{\frac{\partial\pi}{\partial t}}_{\text{Theta}\,\Theta_\pi}\times\Delta t+\frac{1}{2}\underbrace{\frac{\partial^2\pi}{\partial S^2}}_{\text{Gamma}\,\Gamma_\pi}\times(\Delta S)^2+\underbrace{\frac{1}{2}\frac{\partial^2\pi}{\partial t^2}\times(\Delta t)^2+\cdots}_{\text{ignored}} $$ and for a delta neutral portfolio $\Delta_\pi=0$, ignoring terms of higher order that $\Delta t$, we have $$\boxed{ \Delta\pi\approx\Theta_\pi\times\Delta t+\frac{1}{2}\Gamma_\pi\times(\Delta S)^2} $$