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here is an example I am studying to prepare for a quant interview:

Nine months call options with strikes 20 and 25 on a non-dividend-paying underlying asset with spot price 22 are trading for 5.5 and 1, respectively. Can you find an arbitrage?

I already know the answer: we consider one unit of the asset as a third option, namely we have $c_1 = 22$, $c_2 = 5.5$ and $c_3 = 1$ with $k_1 = 0$, $k_2 = 20$ and $k_3 = 25$. Where $c$ is the option price and $k$ in the strike price.

At this point $$k_2 = \frac{1}{5}k_1 + \frac{4}{5}k_3$$ at the same time $$\frac{1}{5}c_1 + \frac{4}{5}c_3 = 5.2 < 5.5 = c_2$$

therefore the convexity is violated and there is an arbitrage. The arbitrage strategy is "buy 100 units of asset for 2200, buy 400 calls ($k_3$) for 400 and sell 500 calls ($k_2$) for 2750." This way the profit is 150.

Here is my question:

If the convexity wasn't violated and let's say $c_2 = 5$, couldn't we just flip the strategy like the following?

"buy 500 calls ($k_2$) for 2500, sell 100 units of asset for 2200, sell 400 calls ($k_3$) for 400."

This way we would make a profit of 100.

where am I going wrong?

Andrea
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If you do that what happens if at maturity the stock price is $S_T = k_2 = 20$? Then your portoflio's value is $\Pi_T = -100 (k_2 - k_1) + 0 + 0 = -2000$.

An arbitrage opportunity means that you have a positive profit with probability $1$ or in other words for all scenario.

Gábor Pálovics
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  • Follow up: I just came across a case where the c(k) is convex but we still have the opportunity for an arbitrage. So I assume that an arbitrage can in principle exist both in the concave and convex case and we have to check "by hand" every time? – Andrea Oct 14 '21 at 12:44
  • @Andrea, Could you share that example? – Gábor Pálovics Oct 14 '21 at 18:49