hello all, Assume the underlying stock price S(t) follows the usual Geometric brownian motion with NO DIVIDEND. It is well-known that the time t price of a perpetual American call on this stock with a fixed strike is simply equal to S(t). Moreover, the associated optimal stopping rule is that it should never be excercised, i.e. . I well understand the derivation of the results above. But now I have a definitional question. According to many references,the discounted value of the payoff when we use the infinite stopping time is by definition equal to the limsup of the discounted payoff at infinity, i.e.For the perpetual American call without dividend, we know that the optimal stopping time is the infinite stopping time. Using the definition above, I find that the limsup mentioned above is equal to 0 (maybe I am wrong). So the value of the perpetual call based on the infinite stopping time (i.e. taking the expectation of the payoff under the infinite stopping time ) is 0, which is not the same as the well-known result : S(t). I am wondering what is my mistake here or if or not I misunderstand the definition given in those references.Thanks a lot, - TW813

I think in your formula, the expectation operator after the limsup is missing. Then I would expect that this limsup equals S(t). I wonder how you derive 0 for this limit

It is not S(t) discountet with an infinite time (which gives 0), but the limit of exp(-rt) max(S(t)-K,0) when t tends to infinity. there was a confusion between t and tau in your notation. As t tends to infinity, so will the stock price (assuming GBM). Do these perpetual options really exist? there will be a time when S(t)>>K and delta ~1 so that the writer will be totally hedged no matter how high the stock rises. I don't understand why the call wouldn't be exercised because you would keep a call which is worth a lot to increase your profit but you wouldn't be able to invest with this money.After some time the call will stay in the money forever, so exercising and buying the stock instead would produce the same payoff eventually.

The equation i wrote down which involves the limsup thing is not calculation, but a definition. In fact, I do not think perpetual American call exists in reality. But they are useful in that they provide an upper bound for their finite-lived counterpart. In reality, of cousre if the price is high enough, every rational individual (in the realistic sense, not the theorectical sense) will exercise the option.

How about this explanation:Let the current time be 0, then the value of an American perpetual call is S(0) [not S(t) for t>0]. From your formula, in the large t limit exp(-rt) S(t) = S(0) and exp(-rt) K = 0, what is the problem?