In convex arbitrage, we say that if the convexity of call(put) price as a function of the strike is violated, we can have arbitrage strategy. For instance, $$ C_{K_2}\geq \lambda C_{K_1}+(1-\lambda) C_{K_3} $$ where $\lambda=\frac{K_3-K_2}{K_3-K_1}$, $C_{K_i}$ is the call price with strike $K_i$ at present and $K_1<K_2<K_3$.

We can get arbitrage by selling $C_{K_2}$ and buying $\lambda C_{K_1}+(1-\lambda)C_{K_3}$.

My question is: on the opposite, if the convexity is perfectly satisfied: $$ C_{K_2}\leq \lambda C_{K_1}+(1-\lambda) C_{K_3}. $$ Why can't we get arbitrage by selling $\lambda C_{K_1}+(1-\lambda)C_{K_3}$ and buying $C_{K_2}$? If so, there also exists arbitrage strategy under convex situation.

  • $\begingroup$ Hint: look at the Profit-and-loss profile of your two examples $\endgroup$ Aug 4, 2021 at 5:16
  • $\begingroup$ 2 bananas are more expensive than one banana. Can we create arbitrage by selling 2 bananas and buying 1 banana? $\endgroup$
    – Arshdeep
    Aug 4, 2021 at 9:30

1 Answer 1


See the graph below. Let's define the PNL as the position's payoff at expiry plus accrued initial investment, i.e. collected / paid option premia.

Assuming $K_1=95,K_2=100,K_3=105$ (i.e. $\lambda=0.5$), the orange payoff diagram below belongs to a setup where $C_2<\lambda C_1 + (1-\lambda) C_3$: You paid some net fee initially, and you obtain a position that will either end in the money or out of the money $\Rightarrow$ no-arbitrage opportunity. If you simply FLIP this trading strategy, as you have suggested, you will see the same result: You then obtain some money now, but in the future you will either win or loose. With an arbitrage-opportunity (grey graph), you will pay nil (or even less) for a (probabilistically) strictly positive payoff in the future.

enter image description here

NB: Do note that there exists, of course, a lower bound on $C_2$ : $ C_2\geq e^{-r(T-t)}\left(S_t-K_2\right)^+$


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