Why At-the-Money Option Vega Changes Little with Volatility

Summary

The discussion explains why an at-the-money option’s vega is nearly linear in volatility, leaving it with relatively little vomma, or sensitivity of vega to volatility. The main argument uses a small-volatility Taylor expansion of the at-the-money option value: its leading term is proportional to volatility, while the next stated contribution is of higher order. This provides an approximation for why vega changes little as volatility moves near that regime.

A second explanation contrasts at-the-money and out-of-the-money options. As volatility rises, an out-of-the-money option can behave more like an at-the-money option and gain vega, while very far out-of-the-money options remain far from the money. That intuition helps explain variation in volga across strikes. The exchange answers are concise; the expansion is local to low volatility, and neither response supplies a full derivation or broad numerical evidence.

Key ideas

• At the money, the option value has a leading term that is linear in volatility.
• The Taylor expansion suggests that vega is nearly constant near low volatility.
• Vomma measures how vega changes as volatility changes.
• Out-of-the-money options can gain vega as rising volatility brings them closer to at the money.

Tags

# Answer by Mark Joshi (score 2)


# Why is the vega of an at the money option so insensitive to movements in volatility? I.e, why do ATM options have such little Vomma?












I've been trying to understand why at the money options have very little vomma. I was reading and came across a graph that showed vega as volatility changes and I couldn't grasp how the relationships work. Why is the vega of an ATM option just constant with respect to volatility?

## Answer by Mark Joshi (score 2)

https://quant.stackexchange.com/a/16369

well you are really asking why is the ATM value so linear in $\sigma. $ If you take a Taylor about $\sigma =0$ when ATM you get the well-known expression

$$ \frac{1}{\sqrt{2\pi}} \sigma \sqrt{T} S_t +{\cal O}(\sigma^3 T^{3/2}) $$ which gives the approximate linearity.

For details of the derivation see for example my book Concepts etc.

## Answer by user15229 (score 1)

https://quant.stackexchange.com/a/16406

Perhaps the better question is why is it that OTM options do have volga? ATM options have the most vega. As volatility rises, OTM options look more like ATM options, so you would expect them to increase in vega as well. (And really really far OTM options are still gonna look like really really far OTM options, giving rise to the bimodal shape of a volga by strike graph).

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