A double call option allows the holder to either exercise at time $T_{1}$ or time $T_{2}$, where $T_{2}$>$T_{1}$. With corresponding strike prices $K_{1}$ and $K_{2}$, it can be shown that it is never optimal to exercise at $T_{1}$ if $K_{1}e^{-rT_{1}}>K_{2}e^{-rT_{2}}$. This is shown by the idea that, at $T_{1}$ you have two options (if $S_{T_{1}}>K_{1}$):

  1. Exercise and put $S_{T_{1}}>K_{1}$ in the bank
  2. Sell the stock short, put $S_{T_{1}}$ in the bank and wait until $T_{2}$ to exercise the call

I understand how this proves the required inequality, but I don't understand why these are the two options - why are the options not just to either exercise at $T_{1}$ or exercise at $T_{2}$?


1 Answer 1


1st option: Your payoff in $T_2$ is $$(S_{T_1} - K_1)e^{r (T_2 - T_1)}.$$ 2nd option: Your payoff in $T_2$ is $$S_{T_1}e^{r (T_2 - T_1)} + \max (S_{T_2} - K_2,0) - S_{T_2} \geq S_{T_1}e^{r (T_2 - T_1)} + S_{T_2} - K_2 - S_{T_2}$$ and the RHS is equal to $$S_{T_1}e^{r (T_2 - T_1)} - K_2.$$ The term $S_{T_1}e^{r (T_2 - T_1)}$ is in the payoff of the 1st option and in the payoff which is dominated by the payoff of the 2nd option. As $$K_1 e^{-r T_1} > K_2 e^{-r T_2}$$ implies $$K_2 < K_1 e^{r (T_2 - T_1)},$$ the 2nd option, i.e. not executing in $T_1$, is always better.

Your suggestion for the 2nd option is to just exercise in $T_2$. Your payoff in $T_2$ would then be $$\max (S_{T_2} - K_2,0).$$ Compare this to the payoff in $T_2$ of the 1st option. As $S_{T_1}$ and $S_{T_2}$ are random and therefore unknown, you can't tell which of the two options is better. The short selling in the 2nd option is a construction to introduce $S_{T_1}$ in the payoff of that option and get rid of $S_{T_2}$ in the dominated payoff.


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