When Vanilla Option Vega Is Positive Beyond Black–Scholes

Summary

The document asks whether a vanilla European call or put must have positive vega when the underlying stock follows a process other than the geometric Brownian motion assumed by Black–Scholes. The response cautions that vega is difficult to define without specifying how the model or distribution changes. It reframes the issue in terms of increasing uncertainty in the terminal stock price.

Under the proposed comparison, one terminal price is the product of another price and an independent, mean-one random variable. If that multiplier has positive variance, the response says the option on the more uncertain price is worth more, applying Jensen’s inequality. This supports a qualified positive answer when uncertainty is increased in that particular way. It does not establish that every conceivable change in volatility, distribution, or price process raises option value. The result depends on the stated independence and mean conditions and on how “more uncertain” is defined.

Key ideas

• Vega is not process-free unless the change represented by volatility is clearly defined.
• The response compares terminal prices using an independent multiplier with mean one.
• Positive variance in that multiplier implies greater option value under the stated uncertainty comparison.
• The argument is conditional and does not cover every possible change to the underlying process.

Tags

# Is vega of vanilla European call/put option always positive?


# Is vega of vanilla European call/put option always positive?












We know that the vega of vanilla European call and put option is positive under Black-Scholes framework, where stock price flows a geometric Brownian motion.

The question I want to ask is that is vega always positive for vanilla European call/put option, regardless of the process that the underlying stock follows? In other words, is the statement that 'vega is always positive for vanilla European call/put option', assumption (about the distribution of the stock process) free?

## Answer by Mark Joshi (score 3)

https://quant.stackexchange.com/a/35979

well it's hard to define vega in a process-free manner.

In More Mathematical Finance, I do explore the question, however. My solution is to introduce the concept of increasing uncertainty. Thus model $A$ is more uncertain than $B$ at time $T$ if we can write the rvs for the stock price as $$ B_T = X A_T $$ with $X$ a mean one rv independent of $A_T.$ If the variance of $X$ is positive then the option on $B_T$ is worth more than the one on $A_T.$ This is an application of Jensen's inequality.

So the answer to your question in some sense yes.

( see also Merton 1973 )

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