Beginner here learning about black scholes looking for a high level/intuitive explanation. So I've learned that the physical/"true" probabilities of S_t (or whatever underlying asset) do not matter for the option values, and that one should use risk-neutral probabilities to ensure no arbitrage. At the same time, it seems one of the assumptions of black scholes is an explicit physical distribution for stock price (from assuming its process follows a GBM). Why does it need to assume a physical property of the underlying if only the risk-neutral probabilities matter?


You are correct, in that you need not (explicitly) specify real world dynamics to calculate option prices. Indeed in many rates derivatives models, you simply assume a unique risk neutral measure exists (completeness), specify the dynamics under the risk neutral measure (risk neutral probabilities) and price your options.

At the same time, it is important to note that any model cannot give a price without explicitly/implicitly saying (implying) something about the real world physical process:

Recall that ultimately every model is giving you a cost of (real world) dynamic hedging. Since you hedge dynamically (reactively), you are obviously exposed to the real world dynamics of the underlying. You therefore cannot have a model that does not have a comment (implicitly or explicitly) on the true dynamics of the underlying.

Ultimately, risk neutral probabilities are a mathematical convenience to find out the cost of this real world hedging strategy. So when you say only risk neutral probabilities matter, you are actually saying that 'only the cost of this real world dynamic hedging strategy matters'. You cannot completely separate the two.

Black Scholes assumes a physical specification because as soon as that is done, completeness guarantees that the cost of this hedging strategy is determined. Risk neutral probabilities are just a mathematical utility to calculate what this cost is.

  • $\begingroup$ That was a beautiful answer. $\endgroup$
    – mark leeds
    Jul 19 at 2:27
  • $\begingroup$ @markleeds thank you, much appreciated, $\endgroup$
    – Arshdeep
    Jul 19 at 2:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.