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The change in a call portfolio ($f$), derived from Itô's Lemma, is: \begin{align*} \left( \frac{\partial f}{\partial t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2 f}{\partial S^2}\right)\mathrm{d}t &=r\left( f-rS\frac{\partial f}{\partial S}\right)\mathrm{d} t, \\ \implies\frac{\partial f}{\partial t}+rS\frac{\partial f}{\partial S}+\frac{1}{2}\sigma^2S^2\frac{\partial^2 f}{\partial S^2} -rf&=0 \end{align*}

where $\frac{\partial f}{\partial t}$ denotes theta, $\frac{\partial f}{\partial S}$ denotes delta and $\frac{\partial^2 f}{\partial S^2}$ denotes gamma.

So gamma's PnL would be $\frac{1}{2}\Gamma \sigma^2 \mathrm{d}S^2$, where $\mathrm{d}S^2$ is the underlying price's change.

But why is gamma's PnL in reallity $\frac{1}{2}\Gamma \mathrm{d}S^2$, and not the previous formula? Why shouldn't volatility be included gamma's PnL?

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$$ \frac{1}{2} \frac{\partial^2 f}{\partial S^2} dS^2 \approx \frac{1}{2} \sigma^2 S^2\frac{\partial^2 f}{\partial S^2} dt$$

(for small $dt$, ignoring $(dt)^2$ terms )

$\sigma$ is embedded in $dS = \mu S dt + \sigma S dW$ and $$ dS^2 = \mu^2 S^2 dt^2 + 2\mu \sigma S^2 dt dW + \sigma^2 S^2 dt \approx \sigma^2 S^2 dt$$

You picked up $1/2\Gamma \sigma^2$ from the PDE and for some (unknown) reason you multiplied it by $dS^2$. You can only multiply it by $S^2$ as in the PDE (to get PnL per unit of time) or by $S^2 dt$ like in the SDE (to get dollar PnL).

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  • $\begingroup$ Thanks for answering, but I don't get why, even for small $dt$, the volatility variable dissapears. Could you elaborate a little bit about it, please? $\endgroup$
    – DUM03
    Jun 30, 2020 at 0:07
  • $\begingroup$ @DUM03 Just added an explanation. $\endgroup$
    – ir7
    Jun 30, 2020 at 0:12
  • $\begingroup$ Tks, things are becoming clearly. Why we have $≈ σ^2S^2 dt$, instead of $≈ σ^2S^2 dW^2$? It would be because $dW = e√dt $, that turns into $e^2*dt$, where "$e$" follows a Wiener process? If it is correct, what should we do with $e^2$? Thanks $\endgroup$
    – DUM03
    Jun 30, 2020 at 0:47
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    $\begingroup$ See this link for a formal explanation: math.stackexchange.com/questions/81865/wiener-process-db2-dt $\endgroup$
    – ir7
    Jun 30, 2020 at 0:53
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    $\begingroup$ @DUM03 Great. Don't forget to accept the answer so I get the point bounty! :) $\endgroup$
    – ir7
    Jun 30, 2020 at 1:03

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