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I have the following problem. Let C(K) be the market price of a Option Call with respect to the strike K. Let $C(100) = \frac{C(110)+C(90)}{2}$, then show that there exists an arbitrage opportunity.

I could solve this if I replace the K (i.e. strike) with S (i.e. the initial spot price of the asset) using the options delta, but i have no idea how to solve this.

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    $\begingroup$ Hint: what is the payout of longing one call at 90, another call at 110 and at the same time shorting two calls at 100? $\endgroup$
    – achille hui
    Commented Mar 27, 2014 at 10:27
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    $\begingroup$ Thanks, was able to solve it... I will post an answer once i 'm allowed to. $\endgroup$
    – Ace7k3
    Commented Mar 27, 2014 at 12:18

1 Answer 1

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With the hint by achille i was able to do this myself.

Set up the following portfolio:

Sell (i.e. go short): 2x the strike=100 stock

Buy (i.e. go long): 1x the strike=110 stock and 1x the strike=90 stock

Then at time $t=0$ we have a value of the portfolio of: $V = -2 C(100) + 1 C(110) + 1 C(90) = -2 \frac{C(110)-C(90)}{2} + 1 C(110) + 1 C(90) = 0$

To calculate the value at time $t=T$ we consider the following cases:

Case $S_T < 90$: $\quad V = -2 \cdot 0 + 1 \cdot 0 + 1 \cdot 0 = 0 \ge 0$

Case $90 \le S_T < 100$: $\quad V = -2 \cdot 0 + 1 \cdot 0 + S_T - 90 \ge 0$

Case $100 \le S_T < 110$: $\quad V = -2 \cdot (S_T - 100) + 1 \cdot 0 + S_T - 90 = -S_T + 110 > 0$

Case $110 \le S_T$: $\quad V = -2 \cdot (S_T - 100) + 1 \cdot (S_T-110) + S_T - 90 = 0 \ge 0$

Therefore in any case the payoff at expiration date is greater equal zero. And in some cases even stricly greater zero! Together with the fact that it did not cost anything to set this portfolio up at $t=0$ this is an arbitrage opportunity.

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