For a American call option on a stock with continuous dividend yield, show that there exists a critical price, that is a price $S^*_t$ such that if the stock price is above this at time $t$, then it is optimal to early exercise.

Is anyone aware of a proof of this fact in continuous time, ideally in a model-free way (i.e. not needing to assume the Black-Scholes model, that is not assuming the Black-Scholes PDE)?

The usual explanation is this: using the Black-Scholes equation, if a European call is very deep in the money, then its value is $c_t \sim e^{-qT}S_t -e^{-rT}K$ which will be below its intrinsic value $S-K$, but the holder of an American call would never let the call value fall below its intrinsic value, so they will early exercise. But why? Why won't they let the call fall below its intrinsic value?

And secondly, this assumes the Black-Scholes model. What if the Black-Scholes model doesn't hold. Can purely no arbitrage arguments be used to show that there exists a critical price?

A citation would be sufficient.

  • $\begingroup$ Ask yourself why would you let it fall below intrinsic value? Hint: if you exercise now, you get intrinsic value. I don't think this requires Black Scholes Merton, you just need a price for the call option (which is frequently a market quoted price). $\endgroup$
    – AKdemy
    Commented Jul 11, 2023 at 8:32
  • $\begingroup$ @AKdemy Maybe if you wait, you get more than intrinsic value + time value of money? If you don't have BSM, how can you conclude that when the price is large enough it falls below the intrinsic value? According to what model? $\endgroup$
    – user67149
    Commented Jul 11, 2023 at 9:48
  • $\begingroup$ Intrinsic value of a call is S-K (there is no model). $\endgroup$
    – AKdemy
    Commented Jul 11, 2023 at 10:40
  • $\begingroup$ @AKdemy Yes, but that doesn't answer the question. What rules out: call is about to fall below its intrinsic value, but instead of exercising now, the stock price may rise so much that waiting until maturity will lead to a higher payoff? $\endgroup$
    – user67149
    Commented Jul 11, 2023 at 11:07
  • $\begingroup$ "I am waiting to exercise because I know the stock price will go up tomorrow" does not seem to be compatible with market efficiency. You don't know if the stock price will go up or not, all you know is the current state. $\endgroup$
    – nbbo2
    Commented Dec 13, 2023 at 13:55

1 Answer 1


Sketch proof: what do you get from early exercise : the PV of future dividends. (A). What do you give up from early exercise : the benefit of potentially not exercising if S were to fall below K in the future (B) and you have to pay the strike price K earlier, which costs $Ke^{-r(T-t)}$. (C)

Now as S -> infinity , A tends to infinity assuming constant dividend yield. However B-> 0 , and C is not dependent on S.

Hence above critical value of S we must have A>B+C. Hence optimal to exercise early at that point.


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