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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.

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    $\begingroup$ See also this closely related question - quant.stackexchange.com/questions/30190/…. $\endgroup$ – LocalVolatility Oct 26 '16 at 16:43
  • $\begingroup$ @LocalVolatility thanks, the link is very useful indeed. $\endgroup$ – RandomGuy Oct 28 '16 at 8:34
  • $\begingroup$ 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. $\endgroup$ – Chan-Ho Suh Dec 10 '16 at 5:49
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The payout from a call option is non-linear with limited downside and unlimited upside. So while volatility can increase or decrease the value of the underlying, the payoff is greater on an up-move than the loss on a down-move, hence the positive vega.

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  • $\begingroup$ so basically the problem with my "intuition" is that although it is true that the increasing volatility for a deep-in-the money call option increases the probability that the option finishes out of the money, it also increases the payoff of the option whenever it finishes in the money, so this latter effect counterbalances the first and in the end the price reflects them both, and hence it raises, correct? $\endgroup$ – RandomGuy Oct 28 '16 at 9:41
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    $\begingroup$ @RandomGuy. Yes that is correct. $\endgroup$ – RRG Oct 29 '16 at 1:03
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If volatility increases, then not only probability to be in the money increases but also the money you get when you are in the money will increase too.

If you want to convince yourself, try it on the one-step binomial model.

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  • $\begingroup$ 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... $\endgroup$ – RandomGuy Oct 27 '16 at 9:47

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