The title, and might I add, that this question is in relation to the Black-Scholes model and why the concepts are important for option pricing in general.


2 Answers 2


A market model is arbitrage-free if and only if it has a risk-neutral probability measure. This is the fundamental theorem of asset pricing.

That is, in a securities model, the two concepts are one and the same. You can think of the risk-neutral probabilities as those that give the arbitrage free prices of derivatives.

Suppose the interest rate, $r$, is constant. Then, for example, the time $T$ expected value of a share, $S_t$, where the expectation is taken using risk-neutral probabilities, is $S_t e^{r(T-t)}$, which you might recognise as the no-arbitrage forward price.

In a similar way, the no-arbitrage value of a call option at time 0 is the discounted expectation (using risk-neutral probabilities) of the call payoff, i.e., $$C = e^{-rT} \mathbb{E_Q}[\max(S_T - K,0)].$$

This method can be used to obtain the no-arbitrage price of derivatives in general. You'd want this price to ensure that you do not expose yourself to any losses from being arbitraged.


Absence of arbitrage is in general considered equivalent to the existence of a risk neutral probability measure.

The measure is unique iif the market is complete (meaning any payoff can be replicated with a self financing hedging portfolio).

While it is easy to prove that the existence of a risk neutral measure implies absence of arbitrage, the reverse is more complicated to show. In the discrete case it makes use of the Hahn Banach theorem, and in the continuous case it requires quite a bit of technical conditions, so when working on a particular model one simply builds the risk neutral measure to show its existence, usually using the Girsanov theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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