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The market is arbitrage-free iff there exists an equivalent martingale measure for the discounted price process of the stock.

So in a world with a finite amount of possible outcomes $\Omega$ that follow the probability distribution $P$, where we can change $P$ by an equivalent probability measure that is a martingale, there are no arbitrage opportunities.

Now this martingale measure is supposed to model a 'fair game/price'. Could someone please shed some light on why a 'fair price' implies there is no arbitrage? What is a 'fair price'?

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    $\begingroup$ A fair price is neither too high nor too low in the sense that the expectation of where it is tomorrow is the same as the price today. To be sure, the price will go up or down, but with no "bias" in either direction. Which is exactly what a "martingale" is. $\endgroup$ – noob2 Jan 18 '17 at 22:37
  • $\begingroup$ @noob2 That's a nice description of a fair price! Thanks. However, I just don't see how a fair price doesn't allow arbitrage. If 2 co-integrated stocks are "fairly priced", they would never exhibit any kind of mean reversing behaviour on which arbitrage exists? $\endgroup$ – xrdty Jan 18 '17 at 22:51
  • $\begingroup$ The keyword "in expectation" is important here. You are talking about a single realisation of the process as far as I can understand. $\endgroup$ – Quantuple Jan 19 '17 at 8:09
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    $\begingroup$ Two co-integrated stocks allow a statistical arbitrage, as long as they are not perfectly co-integrated (i.e. their price differs by a constant). They do not offer an arbitrage, as convergence is not guaranteed. The fact that co-integrated stocks offer a statistical arbitrage should make you skeptical about the existence of co-integrated stocks. $\endgroup$ – Chris Taylor Jan 19 '17 at 8:28

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