# Question in convex arbitrage [closed]

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.

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.

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

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