# Why is the Vega positive?

We know that in the Black-Scholes model, the Vega of a European call option is always positive. This can be proved easily, so my question is not really about the result per se.

My problem is that I find this result somehow counterintuitive. This is my argument: if an option is out of the money and the volatility rises, then the probability that the option end in the money also grows, so that the price grows, and I find it convincing that the vega should be positive in this case.

But think now of an option which is in the money (maybe not "too much" in the money). In this case if the volatility grows, doesn't the same argument show that the probability that the option finishes out of the money in this case increases, and consequently the price of the option should actually decrease, not increase.

But it is a fact that the vega is always positive, also for out of the money options. So, where is the fallacy in my intuition?? Thanks for your answers.

• See also this closely related question - quant.stackexchange.com/questions/30190/…. – LocalVolatility Oct 26 '16 at 16:43
• @LocalVolatility thanks, the link is very useful indeed. – RandomGuy Oct 28 '16 at 8:34
• The price of the call option is what you pay to protect yourself against the downside. If there's more chance of the downside occurring, the seller of the option will require you to pay more for protection. – Chan-Ho Suh Dec 10 '16 at 5:49

• Thanks for your answer, I have a question: in the one-period binomial model, how does the volatility comes up? In Shreve's book (part I) there is no mention of volatility for the binomial model. Am I right if I assume that the volatility enters the picture because we can take the upper move factor u related to $\sigma$ via $u=1+\sigma$ and $d=1/(1+\sigma)$? Is it in this way that I should see volatility in the one-step binomial model? Otherwise I don't see how... – RandomGuy Oct 27 '16 at 9:47