In an interview, I was once told that I should not consider probability when talking about option greeks since from a mathematical point of view greeks have nothing to do with probability. That is of course true, from a mathematical point of view, but the way greeks are taught always takes into account the probability reasoning.

For example, in this website Volatility it says this:

A rise in the implied volatility of a call will decrease the delta for an in-the-money option, because it has a greater chance of going out-of-the-money, whereas for an out-of-the-money option, a higher implied volatility will increase the delta, since it will have a greater probability of finishing in the money.

Again, it is taking into account probability to explain vega. I understand that because I was also taught greeks through probability reasoning. Therefore, how can I understand greeks functioning without taking into account the probability reasoning? How would you explain the relationship vega vs delta without considering "chances of ending in the money"?

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    $\begingroup$ Honestly I think it's wrong to say that Greeks have nothing to do with probabilities. The value of an option is the integral of the payoff over the probability weighted possibilities. Many Greeks come about by changing the probabilities of those various possible scenarios. 9f someone thinks about Greeks purely from the formulae, then great, but it's wrong to say that they have nothing to do with probabilities. $\endgroup$
    – will
    Commented Oct 21, 2021 at 23:34
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    $\begingroup$ How can we say they’re unrelated to probabilities when the Gaussian CDF or PDF is present in almost all of them? (Sincere question) $\endgroup$
    – CasusBelli
    Commented Oct 22, 2021 at 13:07
  • $\begingroup$ To add to the brilliant comments and answers, the otm call logic given in the question's link is not really accurate either as shown here and here. At the end of the day, even looking at mathematical concepts is less objective as one would think, especially if it's concerning applied mathematics. $\endgroup$
    – AKdemy
    Commented Oct 29, 2022 at 21:43

1 Answer 1


There is a really technical issue here. Both Ito's and Stratonovich's methods presume that all parameters are known. It is necessary for the math to work out correctly. If you drop that assumption and rework the rules of math, you get very different models because you have added parameter uncertainty as well.

There is no probability in the Greeks because there is no parameter uncertainty possible. If you were to calculate Apple's option price and various derivatives, it must come to a fixed number based on the parameters you know must be true and that you have not estimated. Of course, that is the rub. You cannot put estimators into Black-Scholes or similar models. That would produce a different result and a different equation for that matter. Of course, that is what everybody does, but without the adjustments.

The probability/certainty issue is a core element of the problems with these models.

Also, "chances of ending in the money" is a multi-faceted issue. If the parameters are known, then there is still the uncertainty created by possible realizations of the price path. If the parameters are not known, then every possible parameterization (of which there are an infinite number) has an infinite number of paths.

To understand the potential differences, consider the differences between a prediction interval and a confidence interval. If you choose a fixed point as your parameter estimate, then you will have too little uncertainty. Your distribution of possible realizations will be too narrow. You have only accounted for the uncertainty in the price process, not the uncertainty in the sample you used to create the estimate.


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