Fifty Dollars in Hand, Fifty-Five on the Screen

A share at 150, a call struck at 100, a year to run and no dividend: cashing out pays 50 while the option is worth 54.97. The floor S minus X e to the minus rT assumes no distribution at all, and it beats immediate exercise by exactly one year of interest on the strike, 4.8771. The article carries the cusp where the gap peaks at 10.4506, the shelf it settles onto far in the money, and the dividend condition that makes early exercise optimal after all.

A share trades at 150 and pays no dividend. You hold an American call struck at 100 with a year to run, and the interest rate is five per cent. Exercise today and you pay 100, receive the share, sell it, and keep 50. Nearly everybody takes the 50.

The option is worth 54.97. Handing it in for 50 destroys about five dollars of value on every contract, and the reason has nothing to do with volatility, distributions, or which model you prefer. It can be proved in one line of arithmetic that mentions no probability at all.

Where the reflex comes from

The pull of immediate exercise is that 50 is certain and the alternative is not. That framing is the mistake. Nobody is asking you to hold a risky position rather than a safe one. The comparison is between exercising and doing anything else with the same contract, and one of those alternatives is to sell it. Selling is a market transaction, so it pays whatever the option is worth, and the question becomes whether the option is worth more than 50.

It is, and the argument does not need to know what the share will do.

A floor with no model in it

Hold the call and also buy a zero-coupon bond paying $X$ at expiry. That costs $C + Xe^{-rT}$ today. At expiry the pair is worth $\max(S_T - X, 0) + X = \max(S_T, X)$ , which is at least $S_T$ . Something worth at least the share at expiry must cost at least the share today, otherwise there is free money in the difference. So

C \;+\; Xe^{-rT} \;\ge\; S \qquad\Longleftrightarrow\qquad C \;\ge\; S - Xe^{-rT}

(1)

An American call is worth at least the European one, since holding it to expiry is one of the things you are allowed to do with it, so the floor applies to both. Now compare the floor with what exercising pays:

\bigl(S - Xe^{-rT}\bigr) - \bigl(S - X\bigr) \;=\; X\bigl(1 - e^{-rT}\bigr) \;>\; 0 \quad \text{whenever } rT > 0

(2)

No distribution was assumed anywhere in that, and none is needed. At the numbers in the question, the floor is $150 - 100e^{-0.05} = 54.8771$ and the gap over immediate exercise is

100\bigl(1 - e^{-0.05}\bigr) \;=\; 4.8771

(3)

That number is one year of interest on the strike. Exercising early hands over 100 today instead of a year from now, and the interest you would have earned on it in the meantime is what you throw away. Nothing else about the trade changes: you own the share either way in the state of the world where you wanted it.

Early exercise is dominated

For a call on an asset paying no income, with a positive interest rate, the continuation value strictly exceeds the exercise value at every spot and every date before expiry. The optimal stopping time is therefore expiry itself, and the American call has the same price as the European call.

This also settles the practical question. If you want out of the position, sell the option and collect 54.97. Exercising it collects 50. The right of early exercise is worth having and worth never using.

The sketch the question actually asks for

Plot the option price against the spot with the intrinsic value drawn underneath and the gap between them becomes a curve of its own.

Fig. 1 — The option curve never touches the payoff it would collect on exercise. The marked segment at 150 is the roughly five dollars that immediate exercise gives away.

That gap is zero far below the strike, where both the option and the payoff are worthless. It rises to a sharp maximum of 10.4506 at the strike itself. Then it falls again, but it does not fall to zero. The corner at the strike is inherited: the option price is a smooth function of the spot, the intrinsic value has a kink of slope one at the strike, and subtracting a kink from something smooth leaves a kink. The left and right derivatives of the gap differ by exactly one there, and by nothing anywhere else.

Fig. 2 — The gap on its own. A cusp at the strike, and a shelf on the right that the curve settles onto from above and never leaves.

Why the gap never closes

Deep in the money the option is a forward in disguise: exercise is certain, so the holder owns the share and owes the strike at expiry, which is worth $S - Xe^{-rT}$ . Subtract the intrinsic value and the limit falls out:

\lim_{S \to \infty}\; \Bigl[\,C(S) - (S - X)\,\Bigr] \;=\; X\bigl(1 - e^{-rT}\bigr) \;=\; 4.8771

(4)

At ten times the strike the gap is still 4.877058, which is the shelf to six figures. So the familiar line that a deeply in-the-money option has no time value left is false. It has exactly the interest on the strike, and equation (4) is the same quantity as equation (2), which is not a coincidence. Both are the value of not having paid the strike yet.

One clarification, because the phrase is easy to over-extend. The gap never dies as a function of the spot. As a function of time it does: at expiry the option is worth its payoff everywhere, and the shelf collapses with $T$ going to zero.

Six hundred chances to exercise, none taken

The dominance argument deserves a numerical audit, because it is the kind of clean statement that hides an error well. A 600-step binomial tree was built and the American decision was taken explicitly at every node, comparing the exercise value against the continuation value rather than assuming which is larger. It chose continuation at every single node. The American and European values came out bit for bit identical at 54.9698679929, converging to the closed form 54.9701. Repeating the tree at spots from 110 up to 400 still produced zero early exercises.

The floor was also tested against the objection that it smells like a lognormal result. It is not one. Pricing the call under five different terminal laws with the risk-neutral mean pinned to the forward, including a two-point law, a uniform and a heavy tail, gives values from 54.97 to 64.94, and all of them exceed 50.

Where early exercise does pay

The no-income hypothesis is load bearing, and the article would be dishonest without saying so. Add a continuous dividend yield $q$ and the floor becomes $Se^{-qT} - Xe^{-rT}$ , which drops below $S - X$ once $q$ is large enough, at which point exercising can genuinely be optimal. With a single discrete dividend $D$ paid at $t_D$ , exercising just before it goes ex is worth considering when $D$ exceeds the interest saved on the strike over the remaining life, roughly $X(1 - e^{-r(T - t_D)})$ . Deep in the money and close to a fat dividend, that inequality flips.

American puts break the symmetry the other way, and with no dividend at all. Exercising a put brings the strike in early rather than sending it out, so interest works for the holder instead of against, and there is a spot below which exercising immediately is strictly best. The right of early exercise on a put is worth using; on a dividend-free call it is worth owning and leaving alone.

Sources and further reading

The distinction the whole question turns on Option style
The portfolio argument behind equation (1) Rational pricing
What equation (3) is measuring Time value of money
The closed form used for the two figures Black-Scholes model
The optimal stopping problem an American option really is Optimal stopping

The floor was checked over 48 combinations of strike, rate and maturity, the shape of the gap on a 7,000-point grid where it is unimodal with its peak at the strike and monotone on either side, and the dominance claim on a 600-step lattice that took the American decision at every node and never once preferred exercise.