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Say we are looking at European Call options in a toy environment with zero deterministic interest rates, a stock paying no dividends, no repo rates etc...

Let $C(T,K)$ be the price of a call with expiry $T$ and strike $K$.

If for $T1 < T2$, $C(T1,K) > C(T2,K)$ then this is calendar arbitrage.

Please explain how should one exploit this arbitrage opportunity.

Thank you.

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    $\begingroup$ Buy the T2 call and sell the T1 call. $\endgroup$ – barrycarter Oct 30 '14 at 16:27
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The answer by @HenriK is certainly correct. However, for justification, technique such as the Jensen inequality is needed. For example, since $x^+$ is a convex function, assuming zero interest and zero divdiend, \begin{align*} E\big((S_{T_{2}}-K)^+ \mid \mathcal{F}_{T_1} \big) &\ge \big(E(S_{T_{2}} \mid \mathcal{F}_{T_1})-K\big)^+\\ &=(S_{T_1}-K)^+. \end{align*} That is, $C(T_2) - (S_{T_1}-K)^+\ge 0$. Then, \begin{align*} C(T_2) - (S_{T_1}-K)^+ + x > 0. \end{align*}

Alternatively, if we short the option with maturity $T_1$ and long the option with with maturity $T_2$, then we have the initial profit $x= C(T_1)-C(T_2) > 0$.

At time $T_1$, if $S_{T_1}\le K$, the shorted option expires worthless, and then we have the total profit $(S_{T_2}-K)^++x$.

On the other hand, if $S_{T_1} > K$, the option is exercised, then, we short sell the stock (i.e., borrow and sell) and receive $K$. At time $T_2$, if $S_{T_2} > K$, we buy the stock by paying the amount $K$ that we had received at $T_1$, and return the stock that we had short sold at $T_1$. The net profit for our trading strategy is $x$, that is, the initial profit. On the other hand, if $S_{T_2} < K$, we buy the stock by paying $S_{T_2}$ and return the stock that we short sold at $T_1$. Note that, we had received $K$ at time $T_1$, our net profit is then $K-S_{T_2} + x>x>0$.

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You should always think: I buy the one which is to cheap and sell the one that is too expensive and figure it out.

The figuring out in this case is noting that:

  • $C\geq 0$ since it will never cost you money
  • The option is strictly better than $S-K$ so has a higher price.

Now to your strategy: You buy $C(T_2)$ (the cheap) and sell $C(T_1)$ (the expensive), call the difference $x>0$. At $T_1$ your position is

$C(T_2) - \max\{S_{T_1}-K,0\} + x$

The first term we have just argued is non-negative, the second is strictly positive. Arbitrage :)

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  • $\begingroup$ Hello! Appreciate your answer. Can you please elaborate how does one exploit if a call is priced below S-K? $\endgroup$ – Alexander Chertov Jan 8 '15 at 10:58
  • $\begingroup$ @AlexanderChertov Sorry, but you should really be able to see that for yourself. $\endgroup$ – Henrik Jan 8 '15 at 14:51
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you need to check implied volatility... Calendars must be opened when IV is low, hopping that will increase, that way you will capture the volatility increase.

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  • $\begingroup$ If you're hoping it's not really an arbitrage or do I misunderstand you? $\endgroup$ – Bob Jansen Jan 6 '15 at 7:16
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    $\begingroup$ This answer doesn't make much sense... $\endgroup$ – SmallChess Oct 10 '15 at 14:06

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