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?
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2 Answers
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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.
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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).
