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As of time $t$, the price of an American call option struck at $K$ and expiring at $T $ is: $$ V_t = \text{sup}_{\tau} E^\mathbb{Q} \left [ e^{-r(\tau-t)} (S_\tau - K)^+ \vert \mathcal {F}_t \right] $$ where $\tau$ figures a family of stopping times with values in $[t,T]$. Now setting $K=0$ and letting $T \rightarrow \infty$ we have: $$ V_t = \text{sup}_{...


In your question, you reduced your derivative to just a stock. No doubt, it is unlikely to have a derivative with zero strike price and never ending expiry. But let's assume it exists. Keeping aside ownership advantage of owning of a stock [as mentioned by OP], from here, it is appear to be a dilemma between buying a stock or call. Theoretically, ...


Here is a good paper which can help you.


I think the chain of logic should be as follows: We have put value >= intrinsic. Therefore either put value > intrinsic or put value= intrinsic. If put value > intrinsic, then it is not optimal to exercise. If put value = intrinsic , it may be optimal to exercise.. Hence there is no contradiction.


So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value. Yes indeed However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever ...

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