Why Option Gamma Appears in Delta-Hedging Price Changes

Summary

The document explains why an option’s price change includes a gamma term in addition to theta and delta. It frames the question around a call option under constant volatility and asks whether gamma corrects for the curvature of price with respect to the underlying.

The answer derives the term from Itô’s lemma: for a diffusion driven by Brownian motion, the underlying’s quadratic variation is proportional to elapsed time. As a result, the squared price increment contributes at the same order as the time increment and cannot be dropped in the expansion. Higher-order terms can be neglected at this scale. A symmetric random-walk step of size proportional to the square root of time offers an intuition, since its square is proportional to time. This is a concise conceptual explanation; it does not discuss hedging errors, discrete rebalancing, or assumptions beyond the Brownian setting described.

Key ideas

• Itô’s lemma produces the gamma contribution when option value is modeled as a function of time and spot.
• The underlying’s quadratic variation is of the same order as elapsed time in a Brownian model.
• A price increment’s squared term matters even when the increment itself is small.
• Higher-order Taylor terms can be neglected relative to the time and quadratic-variation terms.

Tags

# Change in call price Value as time goes by


# Change in call price Value as time goes by












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?

## Answer by Antoine Conze (score 3, accepted)

https://quant.stackexchange.com/a/38983

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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