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In various papers and discussions in here I have seen that in delta hedging setup people compute the Change in value/Price of Call option by: $$ dC_t = \Theta_t dt + \Delta_t dS + \frac{1}{2} \Gamma_t dS^2 $$ assuming constant volatility. As tome goes by (and spot stays the same) the option is loosing value so the the first part make sense (theta is negative in BS model).

The second part also make sense: when spot change by $dS$ (relative small changes) the option value changes by $\Delta$.

My question: Why do we include the last part? Is that because we want to make up for the for non-linearity of delta? If that is the case; why only include the second derivative and why multiplying by 1/2?

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    $\begingroup$ I heard a funny anecdote about the 1/2 recently: the quants at ING where so annoyed by having to explain the 1/2 to management everytime that they just decided to redefine their $\hat{\Gamma}=\frac{1}{2}\Gamma$. Problem solved. Except it created a new problem that you now have to explain to the new quants at ING where the 1/2 went to. $\endgroup$ – Raskolnikov Mar 25 '18 at 14:12
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From the mathematical standpoint it is a consequence of Ito's Lemma. The intuition is that $dS^2$ (or rather $<dS,dS>$ - the quadratic variation of $S$) is of the same order of magnitude as $dt$ so that when you do a Taylor expansion you cannot neglect it, but you can neglect higher order terms. In the Brownian case if $dS = \alpha dt + \sigma dW$ then $<dS,dS> = \sigma^2 dt$. Again you can gain some intuition by noting that a random walk $\pm \sqrt{dt}$ with probabilities $1/2$ and $1/2$ is an approximation of the Brownian motion, and that $(\pm \sqrt{dt})^2 = dt$.

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    $\begingroup$ Yes. One rule of approximation is "you must not drop any terms which are of the same order as terms which are being retained" and $\frac{1}{2}dS^2$ cannot be dropped for this reason. (Don't be misled by the exponent of 2, this is not a small term. Stochastic calculus works differently). $\endgroup$ – Alex C Mar 25 '18 at 12:34

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