Rock,Maybe I'm missing something but I don't see the paradox.In the model (no dividends) the American option will never be exercised early, this is why it has the same value as the stock because its value is S-PV(X) = S.It will always be equal to the value of the stock, so is completely equivalent to owning the stock. If you want to realise the option value you just sell it for the current stock price (ie you would never pay out the exercise price).You on the other hand always have a MTM position of zero. You own a non dividend paying stock (so you never receive any cash) and you are short a position which has the same MV as your long position.In the model world there is no value to controlling the company or the assets. You are long the company's assets (in effect) but you are also short them. You have no way of ever extracting any cash from them.I cannot see any problems for the boundary conditions. I also don't see the probability distribution problem. The relation between S and X is irrelevant because every path will have infinitely many S points arbitrarily higher than X.In the real world obviously nobody would actually pay S for the option, they would buy the stock. All the model price is telling you is that you shouldn't sell the option for any less.

I agree with Mark, I also don't see the paradox. The dilemma:1. The perpetual American call option sells for S at all strike prices in accordance with the theory, in which case I can create wealth at no risk or expense, or 2. The perpetual American call option sells for less than S and strike prices matter, in which case we have to reset the boundary conditions for pricing expiring options.is not a problem since the theoretical value of this perp amer call is S according to BSM. All non-dividend shareholder value (voting rights, % market cap ownership, etc.) is not priced in the model, so one cannot complain about the nonsense real-world results implied in 1 (e.g. that I can eventually take over the company by selling these calls all day long at S). And then, of course, 2. presents no problem: upper boundary condition at S.

Martingale, if you are there perhaps you have an opinion on the question Scholar and I are asking?

In the real world obviously nobody would actually pay S for the option, they would buy the stock. All the model price is telling you is that you shouldn't sell the option for any less.If I own a position in the underlying, I will always be willing to sell the option for less than S since it allows me to take cash out of a position on a riskless, tax-deferred basis. If your point is that B-S&M cannot give us a proper price for a non-expiring American call option on a non-dividend paying stock in the "real" world, I agree. The argument that the option would never be exercised makes this condition worse, not better. Even in a B-S&M world, the sale of a non-expiring call option (and investment at the risk-free rate) creates a dividend on a non-dividend paying stock if the position is covered by ownership of the underlying asset. The combination of the short option and the stock will always create "dominance" [Merton 1973] to ownership of the stock alone, so yes, it is rational to sell the option for less in the "real" world.It will always be equal to the value of the stock, so is completely equivalent to owning the stock. Actually, to be equal to the value of the stock, it must be equivalent to owning the stock, which it will never be given voting rights and the possibility of future dividends.If investors prefer more wealth to less, then they will always be willing to sell non-expiring American call options on a non-dividend paying stock against ownership of the stock BECAUSE it provides a means of extracting cash from the position - the MTM is zero only if interest rates are zero.

All non-dividend shareholder value (voting rights, % market cap ownership, etc.) is not priced in the model, so one cannot complain about the nonsense real-world results implied in 1 (e.g. that I can eventually take over the company by selling these calls all day long at S). And then, of course, 2. presents no problem: upper boundary condition at S.Ok, so in the "real" world, for what price does the non-expiring American call option on a non-dividend paying stock sell?If it sells for less than S, then there is a boundary condition problem.If it sells for S, then there is a zero risk/cost takeover problem.You cannot have it both ways in the "real" world since since ONE of the above MUST be true (setting at zero, of course, the probability that C>S).

Rock,In the real world, C should sell for S but nobody should actually buy C for the obvious "non-model" reasons such as voting and control (and also credit and indeed moral exposure to the seller). They should buy S instead. In the model world S and C are the same instrument (they must be they always have the same value).Suppose you are willing to sell for less to generate the cash flow. Then as a buyer my calculation is how long I have to wait for S to be sufficiently in excess of X plus financing costs. At this point I will exercise and you will be out of pocket. This is why it is not rational for you to sell for less than S. Obviously this assumes I cannot sell the option at that time for in excess of the intrinsic value, but you as the option seller cannot rely on this.These arguments apply in the model world as well because the stock price will be arbitrarily in excess of cost plus financing plus exercise price w.p.1.Looking at the mathematical side I suspect this is a situation where the Optimal Stopping Theorem conditions are not satisfied. Once you get into T = infinity you have to be very careful with the mathematics.Mark

a friend of mine sent me a private mail saying that the literature was right and I was wrong ... I haven't thought it true yet but in case anyone-else has an opinion?Scholar, any ideas?

Here is what I showed Reza earlier: Let X_t=ln(S_t), then X_t = (r-sigma^2/2)t+ sigma B_t, X_0=x <ln(H) let us also denote mu=r-sigma^2. X_t=mu* t+ sigma B_t. In case of mu=0, E(tau)=\infty, that's true( a classical martingale excercise) If not, X_min(tau,t) is a martingale, since [X, X]_min(tau,t)= min(tau,t) and E(tau) finite means that using monotone convergence theorem ln(H)=E(X_tau)=E(X_0)=x which is a contradiction. Similarly that we get the case for positive mu where E(tau) is finite. But I think when mu is negative, E(tau) does not exist, since otherwise we would have X_min{tau,t}-mu*min(tau,t) is a martingale, using monotone convergence theorem we get E(X_tau)-mu*E(tau)=x, i.e, ln(H)-mu*E(tau)=x which is a contradiction, since E(tau)>0.....I might be wrong.

Suppose you are willing to sell for less to generate the cash flow. Then as a buyer my calculation is how long I have to wait for S to be sufficiently in excess of X plus financing costs. At this point I will exercise and you will be out of pocket. This is why it is not rational for you to sell for less than S. Obviously this assumes I cannot sell the option at that time for in excess of the intrinsic value, but you as the option seller cannot rely on this.I cannot be "out of pocket" if you have to pay me an exercise price. Since I received originally the price of S (through sale of the call), which I have been able to place in ANY alternative investment (because the short option was completely hedged through ownership of the underlying), then the exercise price represents excess profit above the market price. Stock sellers would not sell the stock, they would sell the option and hold the stock, since it guarantees at least the same return in all states of the world and a higher return in at least one state of the world - Merton's definition of dominance. If one position is dominant, the two positions cannot have the same price.Given that the average holding period for other non-expiring assets (like stocks) is less than 9 months, this would appeal to a large class of investors, since if exchange-traded, the position could always be unwound.However, I still have no answer to my question:What is the price of a non-expiring American call option on a non-dividend paying stock in the real world?If C=S, we have a zero risk/cost takeover paradox (actually, the paradox applies as C approaches S).If C<S, we have a boundary condition paradox.Since options are supposedly priced in the absence of dominance and paradox, the problem remains unresolved (and I admit we may be bumping up against Godel's Incompleteness Theorem here).THANKS!

I don't think so.Real world situation (1). You are willing to sell option for S. There are no buyers because in the real world they prefer to buy the stock.Real world situation (2). You are willing to sell the option for S - epsilon. Eventually the option will be exercised so you receive X then. PV for you is S - epsilon + X exp(-rt). For large enough t this is less than S therefore you are out of pocket - you would have been better off selling S to begin with. This is true for all choices of X and all epsilon > 0.Mark

(and I admit we may be bumping up against Godel's Incompleteness Theorem here).This sentence is false. But I digress... Hmm -- I must be missing something ... you say If C=S, we have a zero risk/cost takeover paradox (actually, the paradox applies as C approaches S). but I thought we agreed this is not so much of a paradox as it is a foible of the theory? BSM certainly has other wrinkles that are not exploitable in the real world (e.g. volatility smiles).So let's assume the fair value is S_t - epsilon, as Mark pointed out, so epsilon represents some premium on voting / ownership rights, priced in equilibrium (not arbitrage) if enough supply and demand exists. (Perhaps the premium should be proportional to S so the call behaves like stock in a stock split ... but let it be additive for now.) Every option trader I've ever known uses proprietary corrections / enhancements like this in the products they trade, reflecting stuff not priced in the model, supply / demand glitches, volatility corrections, etc.So if we agree that C is really worth less than S in the real world then I'm not sure I understand the paradox involving the upper boundary condition for the identical call that expires in finite time ... why not assert a vanishing second derivative, d^2C/dS^2 = 0, at the upper boundary for instance (option value asymptotically linear).

This is true for all choices of X and all epsilon > 0.There are infinite examples where this is false.1. Any company that goes out of business - in the real world, bankruptcy risk is positive. (Enron?)2. Any company that is taken over in an all cash bid - in the real world, every company is subject to takeover risk if it underperforms.3. Any company (A) which underperforms its competitors (B). Investors holding stock in A can sell covered expirationless calls and invest the proceeds in B. No matter how large t is, the return to the combination of A and B will always surpass A alone.For large enough t this is less than S therefore you are out of pocket - you would have been better off selling S to begin with.3. You are assuming that EVERY S outperforms EVERY alternative investment (even an investment in another S) - an impossibility by definition. If the underlying stock is held and the expirationless option is sold against it, the cash is free to use for ANY alternative investment - even an investment in another company. Therefore, you cannot say that if t is large you are out of pocket, the compounded return (or alternative investment) may be larger than your surrender value and you are most definitely NOT better off selling S to begin with as long as X > 0.4. Under your logic, the following is true:X = 0, C = SX = 5, C = S...X = 100, C = S...X = 1000, C = SThis is not sustainable in a competitive market.THANKS!

but I thought we agreed this is not so much of a paradox as it is a foible of the theory? BSM certainly has other wrinkles that are not exploitable in the real world (e.g. volatility smiles).1. Why is it not exploitable? Either the expirationless call sells for less than S or it does not. My question is simply: which one? If the call sells for less than S, that's certainly exploitable, because buyers would rather buy the call rather than the non-dividend paying stock. If the option sells for S, that's exploitable, because stock holders would rather sell the option than sell the stock. In either case, either buyers or sellers have a preference (especially once you include taxes). This doesn't even address the preferences that companies would have in selling non-expiring options on their treasury stock.2. If the expirationless option sells for less than S, how can we evaluate an otherwise identical expiring option over the range zero to S? Wouldn't this require an assumption that the expiring option can have a greater value than a non-expiring option? A change in the max value of an expiring option should impact its price just as a change in the price of the underlying does.3. I don't want to get diverted here, but does Godel not apply because BSM is not self-referencing or because its not complete?why not assert a vanishing second derivative, d^2C/dS^2 = 0, at the upper boundary for instance (option value asymptotically linear).4. Again, I don't want to get diverted, but I can't see how this helps. Isn't it just a restatement of the original proposition?THANKS!

Reza,I don't have any new ideas. I thought my calculation was accurate, and did not see a loophole there. Martingale, I think it would be best to see that E[tau] = infty for r - sig^2/2 < 0 (if it is true !) by some direct method as e.g. a divergence of an integral.

Rock,I think we are beginning to go round in circles.For bankruptcy you would have been better off selling the stock. For a cash takeover this is you selling the shares which you can't do. Investment in something else applies if you sell the stock.I think the underlying fallacy is that you are assuming that by controlling the shares you personally can extract some economic value from the company. Given that you can't sell the shares without having a risky position and we are assuming no dividends, this is not the case. eg if you borrowed the company yacht this is a dividend equal to the PV of the rental of it. Once you enter into the call you can extract no further cash from ownership of the stock.What my calculation proved was that for any choice of C less than S there is no stochastic dominance problem, not that under no circumstances would you not be better off. For you to be better off however the buyer of the call has to act irrationally ie exercise "early". For you to be worse off all that has to happen is for the buyer to not exercise until your PV is less than S.For C=S obviously no buyer in their right mind would prefer C to S, but it still does not give you an advantage because the only way to profit is to take money out of the company aka a dividend.I'm afraid there really is no paradox, no stochastic dominance problem and no implications for boundary values. Any real transactions would be priced on an expectations pricing basis which affects none of these.Mark