# Double Call Option

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 $$T_{2}$$?

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.