Why is the gamma for an at the money option less when volatility increases. Intuitively ,I thought that increasing volatility means more uncertainty,hence the option price will be more sensitive to the underlying price. Hence more volTility in my mind would mean a higher gamma for ATM option.

But the reverse is true. What am I missing?


Think of moving volatility in the other direction.

As volatility approaches zero, any call strike strictly smaller than the ATM strike, $K<K_{ATM}$, will have zero probability of ending in the money, and the corresponding option value will be zero. An infinitesimally small change in stock price will not move $K$ past $K_{ATM}$, so the option value remains zero nearby. Thus, the sensitivity is zero.

Similarly for $K>K_{ATM}$, all options end up in the money, so the $\Gamma$ is also zero (though for the ITM options $\Delta=1$ rather than 0, ignoring interest and dividend rates).

Only strikes very close to $ATM$ have any likelihood of changing between $\Delta=0$ and $\Delta=1$.

Now, note that that $\Gamma = \frac{\partial}{\partial S}\Delta$ for any volatility. Furthermore, for a call $\Delta(S) \rightarrow 1$ as $ S \rightarrow \infty$.


$$ \int_0^\infty \Gamma(S) dS = 1 $$

That is to say, the area under the gamma curve is always 1.

In high-volatility cases, the $\Gamma$ is "spread out" over a wide range of $S$, so it never gets very big yet adds up to 1. When volatility is low, the $\Gamma$ is all concentrated near $K_{ATM}$ so it has to get very big.

We conclude that as volatility increases, $\Gamma$ decreases near $K_{ATM}$ and increases for other strikes.

(This is a more formal version of SolitonK's answer, which I have upvoted)


Victor123, let's start from $\Delta$. This is the expected change in the price of an option if the underlying asset moves by a currency unit, say 1 USD. For the case of a call option, the Delta varies between 0 and 1. Everything else been equal, the Delta of OTM calls will approach to 0 as the price moves out of the target barrier. Conversely for the case where ITM, Delta will approach to 1. It is thus safe to assume that the delta of an ATM option (i.e $S=K$) equals around 0.5. Relaxing any language formalisms, we can say that at the ATM point, the Stock has around 50% chance of going up or down.

The $\Gamma$ of an option, shows the Delta's sensitivity to changes in the price of the underlying -or how volatile the option is relative to the underlying asset movements. Following the above, given that the expiration doesn't change, ATM options exhibit maximum Gamma value which washes out as we move away from $K$.


Increase in volatility increases the price of options which in turn increases the IV. Increase in IV is equivalent to pushing out the expiry date with the old IV. You know ATM Gamma increases as you get closed to expiry. Assume your DTE increases now .. Gamma has to decrease :)


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