A Dollar at the Barrier, Seventy-Five Cents Today

A share sits at 75, the rate is zero, and a perpetual claim pays one dollar the first time the price ever touches 100. It is worth exactly 75 cents, and no volatility number is needed to say so. The reflex answer of a dollar assumes the barrier is always reached, which a price with a floor at zero never promises: a quarter of the paths fade away without paying anything.

A share trades at 75. The interest rate is zero. Somebody offers you a piece of paper that pays exactly one dollar the first time the share price ever touches 100, with no expiry date attached. What is the paper worth?

Seventy-five cents. You are not told the volatility and you do not need it, which is the part of the answer worth remembering. The price is $S/B$ , the ratio of where the share is now to where the barrier sits, and nothing about how violently the share moves enters the calculation.

Why one dollar feels right

The reflex answer is a dollar, and the argument behind it is nearly good. A share price wanders. The paper never expires. Given unlimited time, a wandering price will visit 100 sooner or later, and with rates at zero there is no cost to waiting. So the paper is a certain dollar with the timing left open, which is a dollar.

Only one step in that chain is false, and it is the one that sounds like common sense. Unlimited time does not guarantee arrival, because a share price is not free to wander in both directions equally. It has a floor at zero and it can approach that floor forever without a lower bound to bounce off. The intuition that a random walk visits every level belongs to walks that are unbounded below, and a price is not one of those.

The trade that prices it without any probability

Own one share. Write one hundred of these papers. Now follow the position to the end.

If the share ever prints 100, sell it at that moment for 100 and pay the hundred paper holders their dollar each. The two cash flows are equal and the position closes at zero. If the share never prints 100, you owe nothing forever, and the share itself, being a positive price with no drift under the pricing measure, drifts down towards zero without a floor to stop it. Either way the position ends worth nothing.

Fig. 1 — One share against a hundred papers. The position is worth nothing at the end on every path, so it must cost nothing at the start.

A position that is worth zero on every path has to cost zero today, or somebody is being handed money. Writing that down prices the paper with no distribution, no integral and no volatility:

75 - 100\,V = 0 \quad\Longrightarrow\quad V = \frac{75}{100} = \frac{3}{4}

(1)

Notice which sentence carries the weight. It is the one about never touching, where the share fades to zero and the writer keeps a worthless asset against a debt that never comes due. That is where the perpetuity and the floor at zero both do their work, and it is the sentence to say out loud if somebody asks you to defend the trade.

The same number as a first-passage probability

With rates at zero the price of a claim paying one dollar on an event is the risk-neutral probability of that event. So 0.75 should also be the chance the share ever reaches 100, and it is. The cleanest derivation puts a second barrier in first, then removes it.

The model, pinned once

Under the pricing measure the share follows a driftless geometric Brownian motion, so it is a martingale bounded below by zero, and $S_t \to 0$ almost surely. The paper is perpetual, the rate is zero and the share pays no dividend. Write $\tau$ for the first time the price touches the barrier $B = 100$ .

Suppose the position is closed out either at 100 or at some lower level $A < 75$ . Both are reached in finite time with certainty, so optional stopping applies to the martingale without any integrability worry, and the starting price is the average of the two exit prices:

75 = 100\,p + A\,(1-p) \quad\Longrightarrow\quad p = \frac{75 - A}{100 - A}

(2)

Now let $A$ fall towards zero, which is the honest way to say and no lower barrier at all. Writing $A = 75e^{-L}$ and letting $L \to \infty$ gives the limit in closed form:

\lim_{L \to \infty} \frac{1 - e^{-L}}{\tfrac{4}{3} - e^{-L}} = \frac{3}{4}, \qquad \mathbb{P}(\tau < \infty) = \frac{S_0}{B} = 0.75

(3)

Fig. 2 — Removing the lower barrier is the left-hand end of this curve. It tops out at 3/4, and the reflex answer of 1 is not on the graph.

So a quarter of the paths never pay anything. The paper is not a delayed dollar, it is a dollar that arrives three times in four.

Where the volatility went

There is a third route, and it explains the absence of $\sigma$ more directly than the other two. A perpetual claim has no clock, so its value depends on the price alone and the time derivative drops out of the pricing equation. With $r = 0$ what remains is

\tfrac{1}{2}\sigma^2 S^2 \, V''(S) = 0 \quad\Longrightarrow\quad V'' = 0 \quad\Longrightarrow\quad V(S) = aS + b

(4)

The two boundary conditions are $V(0) = 0$ , because a share stuck at zero never reaches 100, and $V(B) = 1$ , because the paper pays at once. They force $a = 1/100$ and $b = 0$ whatever $\sigma$ happens to be. Volatility divides out of the equation before the boundary conditions are even applied.

The discrete version says the same thing with no calculus. A symmetric walk on $\{0, 1, \dots, 100\}$ reaches 100 before 0 with probability $h(s)$ , and $h$ is harmonic:

h(s) = \tfrac{1}{2}\big(h(s-1) + h(s+1)\big), \quad h(0) = 0, \; h(100) = 1 \;\Longrightarrow\; h(s) = \frac{s}{100}

(5)

That is the gambler’s ruin problem, and the step size of the walk plays the role of the volatility. A simulation of that walk at three different step sizes gives 0.7496, 0.7496 and 0.7438, a spread of under six thousandths across a factor of twenty-five in step size. Three agreeing runs at three volatilities is the empirical half of the claim.

The intuition to keep is that volatility governs when, not whether. A wild share reaches 100 fast when it reaches it at all, and dies fast when it does not, and with rates at zero the arrival time is worth nothing to you.

Where the formula stops working

Push a nonzero rate through the same argument and something surprising survives. With $r > 0$ the discounted price $e^{-rt}S_t$ is the martingale, and applying optional stopping to it gives

S_0 = B \cdot \mathbb{E}\big[e^{-r\tau}\mathbf{1}_{\{\tau < \infty\}}\big] \quad\Longrightarrow\quad V = \frac{S_0}{B}

(6)

The price is still 0.75. The probability, however, is not. Under the pricing measure the log price now drifts at $r - \sigma^2/2$ , and the standard first-passage result gives

\mathbb{P}(\tau < \infty) = \Big(\frac{S_0}{B}\Big)^{1 - 2r/\sigma^2} \;\; \text{if } r < \tfrac{\sigma^2}{2}, \qquad 1 \;\; \text{otherwise}

(7)

At $r = 0$ that collapses back to $S_0/B$ and the two numbers coincide, which is why the problem can be posed with rates at zero and answered twice over with one figure. Raise the rate and they separate: at $r = 5\%$ and $\sigma = 40\%$ the barrier is reached on about ninety percent of paths, yet the paper is still worth 0.75, because the extra arrivals are exactly cancelled by the discounting of a payment that comes later.

Four other departures matter. A dividend breaks it downward, since the hedger now collects cash along the way and needs less than 75 of paper income to fund the position, so the paper is worth less than $S/B$ . A finite maturity breaks it downward too, and brings volatility straight back in, because reaching 100 before a deadline is very much a question about how fast the price moves. A barrier below the current price is reached with probability one, since the share fades to zero, so the paper is worth a full dollar and the general answer is $\min(S/B,\, 1)$ . And monitoring only at fixed times, rather than continuously, can only lower the value, because an excursion above the barrier between two observations goes unrecorded.

The last one is easy to underestimate. A discretely monitored simulation of this exact problem comes in around 0.73 rather than 0.75, and the gap is not numerical noise. It is the price of the peeks you did not take.

Sources and further reading

The theorem doing the work in equations (2) and (6) — Optional stopping theorem
The discrete skeleton behind equation (5) — Gambler's ruin
Why a driftless price fades to zero instead of visiting every level — Geometric Brownian motion
The family this paper belongs to — Barrier option
First passage of a Brownian motion with drift, which is equation (7) — Hitting time

The 0.75 was confirmed three independent ways before publication: as the cost of the static hedge, as the solution of the perpetual pricing equation with its two boundary conditions, and as a first-passage probability from optional stopping, with a seeded simulation of the symmetric walk agreeing to within six thousandths across three step sizes.