The Put That Never Expires Has One Number for a Rule

An American put struck at 100 on a stock at 100, with no expiry date at all, is worth 23.21 when the rate is 5% and the volatility 30%. Removing the clock removes the time derivative from the pricing equation, which turns it into an ordinary differential equation solved by powers, and the exercise boundary collapses from a curve into the single level 1000/19 = 52.63. Its European twin, which cannot be exercised early, is worth exactly nothing, so every cent of the value is the right to stop.

A put struck at 100, on a stock trading at 100 that pays no dividend, with the riskless rate at 5% and volatility at 30%. The contract has no expiry date. It can be exercised on any day from now until the end of time, and it never lapses. What is it worth?

The answer is 23.21, and the exercise rule is the part worth carrying home. It is a single number, 52.63, and that number does not move. Not next week, not in thirty years. Every American option you have ever priced had an exercise boundary that was a curve in time, creeping toward the strike as expiry approached. Remove the expiry and the curve has nothing to creep toward, so it flattens into a level.

Taking the clock out of the equation

Start from the equation any tradeable claim on this stock has to satisfy:

\frac{\partial V}{\partial t} + \tfrac12\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0

(1)

Now ask what the contract knows about the date. Nothing. There is no maturity to be near or far from, and the stock is a time-homogeneous process, so today and next year present the holder with identical problems. The value therefore depends on $S$ alone, the time derivative vanishes, and equation (1) collapses to an ordinary differential equation:

\tfrac12\sigma^2 S^2 V'' + rS V' - rV = 0

(2)

Equation (2) has a symmetry that decides everything that follows. Rescale the stock, writing $S \to cS$ , and every term picks up the same factor. An equation of that shape is solved by powers, because a power is the one function whose scaling is a constant multiple of itself.

One root you can see, one you can read off

Substitute $V = S^{\lambda}$ . Each term carries a common $S^{\lambda}$ , which cancels, and what is left is algebra:

\tfrac12\sigma^2 \lambda(\lambda-1) + r\lambda - r = 0 \quad\Longleftrightarrow\quad \tfrac12\sigma^2\lambda^2 + \left(r - \tfrac12\sigma^2\right)\lambda - r = 0

(3)

One root is $\lambda = 1$ , and you can check it without the quadratic formula: substituting gives $\tfrac12\sigma^2 + r - \tfrac12\sigma^2 - r = 0$ . That root is the stock itself, $V = S$ , which does solve the equation and is no use for a put, since a put cannot grow without bound as the stock rises. Discarding it is a boundary condition at infinity, not a convenience.

The other root now comes for free. For $a\lambda^2 + b\lambda + c = 0$ the product of the roots is $c/a$ , here $-r / (\tfrac12\sigma^2)$ , and one of the two is 1, so the other is the whole product:

\lambda = -\frac{2r}{\sigma^2}

(4)

No square root survives. At 5% and 30% that is $-0.10/0.09 = -10/9$ , and the negative sign is the statement that the option decays as the stock rises.

Where to stop

Below some level the holder should sell the stock at the strike and be done. Call that level $S_b$ . Two conditions pin it. Value matching says the curve has to meet the payoff line there, $V(S_b) = X - S_b$ , which fixes the constant in front of the power and gives $V(S) = (X - S_b)(S/S_b)^{\lambda}$ . Smooth pasting says the curve leaves the payoff line tangentially, $V'(S_b) = -1$ . Differentiating and solving:

S_b = \frac{\lambda X}{\lambda - 1} = X\,\frac{2r}{2r + \sigma^2} = 100\cdot\frac{0.10}{0.19} = \frac{1000}{19}

(5)

The free boundary

The level $S_b$ is not given by the contract. It is an unknown of the problem, solved for alongside the value function, which is what makes this a free boundary problem rather than a boundary value problem. Equation (5) contains $r$ , $\sigma$ and $X$ , and no $t$ anywhere.

So the trigger is 52.631579, ten nineteenths of the strike. The value at the trigger is $900/19 = 47.368421$ , and at a stock of 100 the option is worth 23.214679.

Fig. 1 — Value matching puts the curve on the payoff line at 52.63. Smooth pasting makes it leave tangentially, which is the only slope that survives the next section.

Why the slope has to be exactly −1

Smooth pasting reads like an extra assumption smuggled in to close the algebra. It is really a first-order condition, and there is a way to see that which uses no calculus of variations at all.

Forget optimality and commit in advance to some constant trigger $b$ , promising to sell the first time the stock ever touches it. Under this model the present value of one dollar paid at that first touch is $(S/b)^{\lambda}$ , with the same $\lambda$ as equation (4). So the strategy is worth $(X - b)(S/b)^{\lambda}$ , a smooth function of the single number $b$ . Maximise it. Setting the derivative in $b$ to zero reproduces equation (5) exactly. Tangency is what optimality looks like once you write the value as a function of the choice.

Fig. 2 — Every constant trigger has a value. The formula picks the top of this curve, and a scan of 998 candidate triggers agrees on 52.6.

The picture says something the closed form hides. The top is flat. Moving the trigger five points either way costs about a quarter of a percent, 22.97 against 23.214679, so a trader who sells at 50 has lost almost nothing. Being badly wrong is expensive, though: insisting on 30 leaves 18.37, and waiting for 80 leaves 15.61. Two independent numerical routes land on the same answer. A value iteration on the stationary problem, which never sees the closed form, returns 23.21 at a stock of 100 and puts its exercise region within one grid step of 52.63. A simulation that simply runs the rule sell the first time the stock touches 52.63 recovers the same value to within two percent over 40,000 paths.

The twin that is worth nothing at all

Change one word. Make the option European, so it can only be exercised at expiry, and keep the expiry infinite. That contract is worth zero, and the argument takes one line: its payoff is at most $X$ , and it arrives infinitely far away, so its present value is at most $Xe^{-rT} \to 0$ . No distributional assumption is needed.

The mirror image is just as blunt. A perpetual European call is worth at least $S - Xe^{-rT}$ and at most $S$ , so in the limit it is worth the stock. Which means every cent of the 23.21 is the right to stop. The contract has no intrinsic value at a stock of 100, no time value in the usual sense, and nothing left over once you remove the choice of when to end it.

Where equation (5) stops working

Push the rate to zero and the formula degenerates in a way that is worth understanding rather than patching. As $r \to 0$ , $\lambda \to 0$ and $S_b \to 0$ : there is no trigger, because receiving the strike sooner buys nothing when money is free. The value climbs to the whole strike. At $r = 0.2\%$ the trigger is already down at 4.26 and the option is worth 83.21. The limit is consistent with the process, since with no drift the log of the stock still falls at $-\sigma^2/2$ and the stock goes to zero, so the payoff tends to $X$ .

Push the volatility to zero instead and the trigger climbs to the strike while the value collapses. At $\sigma = 1\%$ the trigger is 99.90 and the option at 100 is worth 3.7 cents. A stock that only grows at $r$ never comes back down, so a put on it is worth what it is worth today and nothing more. In between, the comparative statics run the way intuition wants: 38.46 at 40% volatility, 52.63 at 30%, 71.43 at 20%, so more volatility waits longer. Raising the rate stops earlier, 68.97 at $r = 10\%$ .

The load-bearing assumption is the one that made equation (4) so clean. The root $\lambda = 1$ was available because the stock pays nothing. Add a continuous dividend yield $q$ and the quadratic becomes $\tfrac12\sigma^2\lambda^2 + (r - q - \tfrac12\sigma^2)\lambda - r = 0$ , where substituting $\lambda = 1$ leaves $-q$ rather than zero. The factorisation is gone and you are back to the quadratic formula. Everything structural survives, including the flat trigger and the tangency; only the closed form for $\lambda$ gets uglier. One last quiet assumption: the argument prices the right to sell at $S_b$ , which needs the stock to arrive there continuously. Let it jump and the holder sells below the trigger, and the value in equation (5) becomes an upper bound.

Sources and further reading

The equation stripped of its time derivative in (2) Cauchy-Euler equation
What makes $S_b$ an unknown rather than a datum Free boundary problem
The problem the holder is actually solving Optimal stopping
The one word that separates 23.21 from zero Option style
Where equation (1) comes from Black-Scholes equation

Every number here was checked three ways: exact symbolic algebra on the quadratic and the two boundary conditions, a value iteration on the stationary problem that recovers both the value and the boundary without ever seeing the closed form, and a simulation of the stopping rule itself alongside a scan of 998 rival triggers.