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European call options with strikes 90, 100 and 110 on the same underlying asset and with the same maturity are trading for 22.50, 18.84 and 13.97 respectively. show that the convexity of the call price as a function of the strike is violated, hence leading to an arbitrage opportunity. Describe in detail a trading strategy that makes a riskless profit.

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  • $\begingroup$ Alvin, this is no place to look for exercises sets solutions. $\endgroup$ – SRKX Nov 4 '12 at 12:51
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    $\begingroup$ @Alvin, did you just try to rewrite the question to something totally different? You can't delete this question because it has an accepted answer. Move on. $\endgroup$ – chrisaycock Nov 13 '12 at 19:54
  • $\begingroup$ @chrisaycock he's probably afraid that the TA finds out where the answer comes from... $\endgroup$ – SRKX Nov 13 '12 at 21:17
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The price of payoff is convex if for every $0\le\lambda\le 1$:

$V(\lambda K_1 + (1-\lambda)K_3 )\le \lambda V(K_1) + (1-\lambda)V(K3)$

, where $V(K)$ is the price of an option with strike $K$.

We want that $\lambda K_1 + (1-\lambda)K_3 = K_2$. Solving it for $\lambda$ we get $\lambda=0.5$. Substituting $\lambda$ back to our inequality, we see that the convexity property doesn't hold.

We see that the portfolio of options with strike $K_1$ and $K_3$ is too cheap. Therefore, we can form an arbitrage portfolio by purchasing 0.5 of $K_1$-strike options and 0.5 of $K_3$-strike options. To finance this purchase we sell one option with strike $K_2$. By checking the payouts of our portfolio for stock prices $S$ in intervals $S<K_1,K_1\le S < K_2, K_2 \le S< K_3$ and $S \ge K_3$ we convince ourselves that this is indeed an arbitrage opportunity.

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