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An American call on a continuous dividend paying stock must be above its intrinsic value, i.e $c(t)\geq\max(S_t-K,0)$.

Why is there a critical price above which it is optimal to exercise (i.e. we have equality in the inequality above)? This is shown in Figure 5.1 in these notes but it is not really explained.

In other words, how can we rule out the situation where $c(t)>\max(S_t-K,0)$ for all prices $S_t$, so no critical price $S_t^*$ exists and the curves in that figure asymptote but never meet?

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  • $\begingroup$ Don't you mean why is there a critical option price below which you should exercise? Since when the option value is equal to or below the exercise value, that is the time for optimal exercise. If you mean a critical stock price, then that is just whichever stock price would make the exercise value equal to the option value $\endgroup$ – Slade Jul 14 at 14:42
  • $\begingroup$ No, I think the question is stated correctly. As the notes point out on page 238, there is a critical stock price, below which you continue, and above which you exercise. But my main question, is why does this critical price exist? Why can't it be the case that you should hold for all possible stock prices and never early exercise? $\endgroup$ – leacorv Jul 14 at 15:40
  • $\begingroup$ Oh, I see. I haven't ever thought about that question since I just took things from the point of view of continuation value vs exercise value. I think someone will stop by with an answer soon $\endgroup$ – Slade Jul 14 at 19:44
  • $\begingroup$ Note that such time always exist, for example, for $\tau_0 = T$, where $T$ is the maturity, $c(\tau_0) = \max(S_{\tau}-K, 0)$. $\endgroup$ – Gordon Jul 17 at 17:15
  • $\begingroup$ I'm concerned with the existence of a critical price before expiry, not the existence of a time with a critical price. $\endgroup$ – leacorv Jul 18 at 16:35

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