Why Vega Can Be Misleading for Options with Sign-Changing Gamma

Summary

The document examines why vega may give a misleading picture of volatility risk for options whose gamma changes sign, using a binary option as its example. Under Black–Scholes, it differentiates the discounted probability-based binary price with respect to volatility. The resulting vega contains a factor that can become zero at a particular relationship among spot, strike, volatility, and time to expiry.

At that point, the instantaneous first-order sensitivity to volatility is zero even though the option may be highly exposed to volatility around that region. The explanation emphasizes that this is a local measure: small changes in spot or volatility can move the option into a state with substantial vega. The example illustrates a limitation of relying on vega alone for instruments with more complicated risk profiles. It does not provide a broader risk-management framework or quantify the effects for barrier options, which are mentioned in the question but not analyzed in the answer.

Key ideas

• For a Black–Scholes binary option, vega can equal zero at a specific spot, strike, volatility, and maturity relationship.
• A zero instantaneous vega does not imply that volatility exposure remains small after market inputs change.
• Sign-changing gamma can make a single local sensitivity an incomplete description of option risk.
• The worked explanation covers a binary option and does not develop a corresponding barrier-option analysis.

Tags

# Why is Vega meaningful only for options which have single-signed gammas


# Why is Vega meaningful only for options which have single-signed gammas












I have been reading Wilmott Frequently Asked Question book and this was mentioned that Vega is not useful when measuring risk for options that have gammas changing signs such as Digital option or Barrier option. In particular, even though Vega is 0 when spot is around strike level, it is where the option is most sensitive to volatility. Unfortunately, it was not explained in details. Can anyone please elaborate on this ?

## Answer by Brian B (score 2, accepted)

https://quant.stackexchange.com/a/21222

In the Black-Scholes model the price of a binary option is

$$ B = e^{-r(T-t)}N(d_2) $$

with

$$ d_2 = \frac{\log(\frac{S}{K})-\frac12 \sigma^2 (T-t)}{\sigma\sqrt{T-t}} $$

Differentiation with respect to $\sigma$ gives our our volatility risk, or vega

$$ \frac{\partial B}{\partial\sigma} = e^{-r(T-t)} N^\prime(d_2)\frac{d_2+\sigma\sqrt{T-t}}{\sigma} $$

Therefore, if we happen to have

$$ d_2 = -\sigma\sqrt{T-t} $$

or equivalently

$$ S = Ke^{-\frac12 \sigma^2(T-t)} $$

Then the apparent risk is zero. Of course the instant any of these parameters, especially $\sigma$ or $S$, changes you will find yourself with considerable volatility risk.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.