If there is no arbitrage there is no dominant trading strategy, but there may be arbitrage opportunities even if there are no dominant trading strategies.

Could you explain this statement and bring an example?


2 Answers 2


Please clarify rigorously what you mean by each term. It is not true that no dominance is a consequence of no arbitrage. Think of the put-call parity:

$C-P=S-K$, assuming $r=0$ since it's inconsequential.

If there is no short selling then we can have:

$C-P \geq S-K$ without arbitrage but No Dominance would not hold.

If you think very deeply about this, and I am assuming conventional meanings since no definition is given, then no arbitrage can be the consequence of a single trader taking huge positions and removing arb opportunities from the market. But ND is something that will occur only if the asset prices correspond to an equilibrium process in some economy.

Conventionally, $ND \implies NA$, but don't confuse this for risk-neutral valuation. For that to be the true value of an asset, we need $ND$ in some economy.


If there are no arbitrage opportunities there is no dominant hedge or long position.

Why would there be an arbitrage opportunity if everything was priced correctly?

There may be arbitrage opportunities even if there are no dominant hedges or long positions.

Put-call parity shows arbitrage opportunities of badly priced options regardless of long position mispricing.

Using put-call parity, only, to try and explain this is a bit limiting, and I would like to warn you about people who offer answers only relating to put-call parity.

  • 1
    $\begingroup$ ND $\implies$ NA is well known. NA $\implies$ ND, doesnt seem to be true. Can you rigorously define both and then give me an example. Please, be a Bourbakist about this, I think that level of mathematical rigor will clarify everything $\endgroup$
    – Drew
    Dec 23, 2014 at 15:28
  • $\begingroup$ Let's say this was a market that functioned well. No foreign stock market, and no OTC market. No Arbitrage implies no contango and no backwardization. No contango and no backwardization doesn't imply no arbitrage. Theortically NA => ND could happen, because a functional market requires three market participants: hedgers, speculators, and arbitragers at all time. $\endgroup$
    – JTHouseCat
    Dec 24, 2014 at 1:50
  • $\begingroup$ once again please provide a concrete example. A proof works by providing a concrete example. For example, no backwardation implies: $$F_{t,T} \leq S_te^{r (T-t)}$$ and Contango implies: $$F_{t,T} \geq S_te^{r (T-t)}$$ Which means Contango + Backwardation $\implies F_{t,T} = S_te^{r (T-t)}$, $\implies$ NA. Once again, rigorous mathematics would help you push aside meaningless ideas easily. $\endgroup$
    – Drew
    Dec 24, 2014 at 2:26
  • $\begingroup$ Your above proof is something I like better than the put-call parity proof. NA ===> ND. NA could imply zero or no market activity or a non-functional market, which could imply no dominant position. $\endgroup$
    – JTHouseCat
    Dec 24, 2014 at 3:25
  • $\begingroup$ Put-Call parity was not a proof but an intuition building example. I proved ND $\implies$ NA. But this is a small case of that. NA doesnt imply ND. For example, if no short selling was allowed: then you could have the backwardation inequality without aribtrage but not without no dominance. And there can be cases where they are the same thing, but ND is a more stringent condition. What you are saying is a useless degenerate case. $\endgroup$
    – Drew
    Dec 24, 2014 at 3:26

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