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I tried a lot of different things to check for arbitrage on the following calls but didn't succeed.

Let's suppose we have a stock that is currently valued at 40. The interest rate is 0.05 and the strike price of the european calls is 35.
Call 1: price 6.75, run time 6 months
Call 2: price 7.93, run time 9 months
Call 3: price 7.76, run time 12 months

I know that a lower price of Call 3 with longer duration in comparison to call 2 indicates arbitrage, but I don't know how to prove it. I hope someone can assist with an arbitrage strategy.

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  • $\begingroup$ Short sell call 2 and buy call 3 from the proceeds and put the (positive) remainder in your bank account. In 9 months time, since call 3 will always be worth more than its extrinsic value (which is the value of call 2 at expiration) you will either still have money left if call 2 is exercised at expiration, and if call 2 is not exercised at the very least you'll have a non-negative amount at expiration of call 3. Plus you still have the positive amount in your bank account from the original short sale of 2 and purchase of 3. Arbitrage. $\endgroup$
    – Frido
    Jun 8, 2023 at 7:51

1 Answer 1

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What I wrote in my comment is one way to show the arbitrage. For your own benefit it might be good if you try to understand also the following more mathematical reasoning.

Let $T = 12$ months, $t = 9$ months and $0$ is today. Then $$ C_3 = e^{-rT} E_0 (S_T - K)_+ $$ and $$ C_2 = e^{-rt} E_0 (S_t - K)_+ $$

By conditioning and using Jensen's inequality \begin{align} C_3 &= e^{-rT} E_0 (S_T - K)_+ \\ &= e^{-rT} E_0 E_t (S_T - K)_+ \\ &\geq e^{-rT} E_0 (E_t S_T - K)_+ \\ &= e^{-rT} E_0 (S_t e^{r(T-t)} - K)_+ \\ &= e^{-rt}e^{-r(T-t)}E_0 (S_t e^{r(T-t)} - K)_+ \\ &= e^{-rt}E_0 (S_t - Ke^{-r(T-t)})_+ \\ &\geq e^{-rt}E_0 (S_t - K)_+ \\ &= C_2 \end{align} So $C_3$ must always be more expensive than $C_2$ if risk-free rate is positive.

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  • $\begingroup$ Can you also show what happens when interest rates are different (which they essentially always are) and the stock pays dividends? $\endgroup$
    – AKdemy
    Jun 8, 2023 at 9:03
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    $\begingroup$ @AKdemy Yes, the inequalities will be slightly different. But I think that will only complicate matters for the OP so I'll leave as is for now. $\endgroup$
    – Frido
    Jun 8, 2023 at 9:37

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