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When Risk-Neutral Valuation Supports an Arbitrage-Free Market

Article Quant Q&A · Author: Richard
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Summary

The document asks whether risk-neutral valuation implies the absence of arbitrage, using a Black–Scholes setting with a bank account, a stock, and a call option priced by an expectation. One response argues that if all discounted asset prices are martingales under a pricing measure, self-financing portfolios are martingales as well. A zero-cost strategy then has zero expected value, which rules out arbitrage under the stated reasoning. Since the pricing measure is equivalent to the real-world measure, the response says the absence of arbitrage carries over.

Another response cautions that assigning an arbitrage-free price to one claim does not establish that the whole market is arbitrage-free. This is an important scope distinction: the argument depends on market-wide martingale assumptions, not merely on one option’s price formula. The replies do not work through the expectation or prove the full result under specific technical conditions, so the exchange serves as a conceptual outline rather than a complete proof.

Key ideas

  • A market-wide martingale condition for discounted prices can rule out arbitrage in self-financing strategies under the response’s assumptions.
  • Equivalence of the pricing and real-world measures is used to transfer the no-arbitrage conclusion between measures.
  • An arbitrage-free price for one claim alone does not establish that every traded asset is arbitrage-free.
  • The question’s option expectation is not evaluated in the provided answers.
  • The exchange outlines the argument but does not state all technical conditions for a formal proof.

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# risk-neutral valuation implies no arbitrage?


# risk-neutral valuation implies no arbitrage?












It is known that in an arbitrage-free continuous time market, the price of every asset is evaluated as the corresponding price in the replicating strategy using risk-neutral valuation.

I want to know is the converse true? There is actually a question that asks to show that for a Black-Scholes model with a bank account ($dB_t = B_t r dt$) and a stock satisfying $dS_t = r S_t dt + \sigma S_t dW_t$, then there is no arbitrage if the time-t price of a call with maturity $T$ and strike $K$ is \begin{equation} X_t= S_t \mathbb{E} \big\{ \big( e^{-\frac{(T-t)\sigma^2}{2} + \sqrt{T-t} \sigma Z} - \frac{K e^{-r(T-t)}}{S_t} \big)^{+} \big\}. \end{equation} Any ideas of how to show this? I don't know how to evaluate this expectation, as it involves two random variables and I don't know the joint density of $S_t$ and $Z$.

## Answer by Mark Joshi (score 4)

https://quant.stackexchange.com/a/15689

I do this question to death in Concepts and ...

If (discounted price of) everything is a martingale then every trading strategy is a martingale. Therefore any self-financing portfolio of initial value zero and has expectation zero. Therefore there are no arbitrages (since these have positive expectation and initial value zero).

So there is no arbitrage in the martingale measure.

However, the pricing measure and the real-world measure are equivalent so they have the same set of arbitrages. So there is no arbitrage in the real-world measure either.

So if we set the price to be its expectation price, we get a martingale in the pricing measure and the price is arbitrage free.

## Answer by emcor (score 0)

https://quant.stackexchange.com/a/15687

The fact that one claim has an arbitrage-free price, does not imply that the entire market (for all claims) is arbitrage-free. E.g. $C_T=0$ is always arbitrage-free.

## Answer by Keith A. Lewis (score 0)

https://quant.stackexchange.com/a/15690

You need to replace Z by Brownian motion at time t. Also, the expectation should be conditional expectation with respect to the sigma-algebra at time t. See http://kalx.net/fms/fms.html for a more complete explanation.

## Answer by user151781 (score -3)

https://quant.stackexchange.com/a/15688

put call parity implies no arbitrage , along with a expected value of stock price of $pe^{rt}$ . working out the integrals yields this outcome

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