Why Convex Call Prices Do Not Create Arbitrage by Themselves
Summary
The note addresses why a call spread built from two outer strikes does not create arbitrage simply because its price exceeds the price of an intermediate-strike call. The intermediate call is cheaper than the weighted combination, so buying it and selling the combination produces an initial credit, but the terminal payoff can be negative in some price regions. The premium difference alone does not guarantee a nonnegative payoff at expiry.
The answer contrasts this position with a genuine convexity violation, where selling the overpriced middle-strike call and buying the weighted outer calls can generate a nonnegative payoff for no net cost. It illustrates the distinction with a payoff diagram for three nearby strikes. The note also points out that the middle call has a lower bound tied to the discounted intrinsic value. The example is explanatory rather than a complete treatment of transaction costs, early exercise, or market frictions.
Key ideas
- A call price curve that satisfies convexity does not make the reverse spread an arbitrage.
- Selling the weighted outer calls and buying the middle call gives an initial credit but may create future losses.
- A convexity violation can allow a nonnegative payoff for no initial cost.
- The middle-strike call is also bounded below by discounted intrinsic value.
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Full text
# Question in convex arbitrage
# Question in convex arbitrage
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.
## Answer by Kermittfrog (score 3, accepted)
https://quant.stackexchange.com/a/66294
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.
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)^+$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.