# 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?

• Vega is defined as the sensitivity of the option price to the constant diffusion coefficient under a geometric Brownian motion process. So if you consider different underlying dynamics then there isn't necessarily a natural analogue of "vega" anymore. You could e.g. be interested in the sensitivity w.r.t. to the initial variance of a stochastic vol. model but there is no general answer to this. Could you please clarify your question? – LocalVolatility Sep 10 '17 at 19:46
• @LocalVolatility That's a valid way of thinking about things, but isn't the way I'm used to seeing options traders think about vega. Nobody 'believes' black-scholes, of course, but everybody still has an implied volatility surface fit to the market and vega is the sensitivity of an option's price to when the surface moves up or down (i.e. vol starts trading higher/lower). So in this view all options have positive vega (which is something of a tautology... the option's price goes up when their implied volatility goes up)... atm options and long-dated have the most. – spaceisdarkgreen Sep 10 '17 at 23:17

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.