One Subtraction Prices the Barrier Option Nobody Has a Formula For

An American call that only wakes up at 80 and dies for good at 125 has no closed form, and it cannot be simulated either, because a path runs forward while the exercise decision looks back. Every path that avoids the ceiling either visited the floor or never did, so the contract is one knock-out minus another and both come off a standard tree. The identity is exact to machine precision for European exercise at all seven grids tested, and for American exercise only in the continuous limit: the finite-tree residual falls from 0.532 percent at 45 steps to 0.043 percent at 3,394.

Here is a contract nobody has a formula for. An American call on a stock at 100. The option does not exist yet: it comes to life the first time the stock touches 80. But it is cancelled for good if the stock ever touches 125, whether or not it had woken up. Exercise is allowed on any day once it is alive. Value it.

Every ingredient is awkward at once. Two barriers, so the payoff depends on the whole path. American exercise, so there is an optimisation inside the expectation. And a knock-in condition, which is the one shape most closed forms are written to avoid. Nothing prices it directly, and the answer is to stop trying.

Why simulation is the wrong first move

The instinct is to generate a million paths and average. Barriers are easy to check on a simulated path, so the hard-looking part of the contract is the part simulation handles best. The exercise decision is where it falls apart, and the reason is worth stating precisely: a path runs forward and the decision looks back. At each date the holder compares the payoff available now with the value of continuing, and the value of continuing is not a number you have while walking forward through a single path.

Cheating on that point does not produce a slightly wrong answer. Exercise each path at whichever moment turned out to be best for that path, and you have priced perfect foresight. On the same lattice used everywhere below, that hindsight valuation comes out at 11.2833 against a true American value of 10.4347 for the plain up-and-out, which is 8.1% high, and it is eight times the European value of 1.4151. The bias is not noise and it does not shrink with more paths.

Do not price it, divide it

Look at the set of paths rather than the contract. Every path either touches the ceiling at some point or never does, and the ones that touch it are worthless regardless of anything else. So restrict attention to the paths that never touch 125. Within those, ask one question: did the path visit 80? Either it did or it did not, there is no third possibility, and the two cases do not overlap.

\mathbf{1}_{\{\text{no ceiling}\}} \;=\; \mathbf{1}_{\{\text{no ceiling, no floor}\}} \;+\; \mathbf{1}_{\{\text{no ceiling, floor visited}\}}

(1)

Multiply equation (1) by the payoff and take expectations. Expectation is linear over a disjoint split, so the values add, and rearranging gives the contract we were asked about as a difference of two contracts that are ordinary:

V_{\text{in at }L,\;\text{out at }H} \;=\; V_{\text{out at }H} \;-\; V_{\text{out at }L\text{ or }H}

(2)

Knock-in, knock-out

A knock-out dies if a barrier is touched; a knock-in does not exist until one is. The contract in the question is both at once, with the knock-in below the spot and the knock-out above it. The two terms on the right of equation (2) are a single-barrier knock-out and a double knock-out, and both come off a standard tree with no new machinery.

Nothing above used a distribution, a volatility, or a model. Equation (1) is a statement about sets, and equation (2) is that statement multiplied by a payoff.

Counting instead of arguing

Because equation (1) is combinatorial, it can be checked by counting rather than reasoned about. Enumerate every path of a binomial lattice and classify each one. At length 8, of 256 paths, 182 avoid the ceiling and they split as 108 and 74. At length 12, of 4,096 paths, 2,508 avoid it and split as 972 and 1,536. At length 16, of 65,536 paths, 35,750 avoid it and split as 8,748 and 27,002, with nothing left over and neither class empty.

Fig. 1 — The surviving paths fall into exactly two classes. That is the whole of the trick, and the counts confirm nothing falls between them.

The European case, exact to machine precision

Now the numbers, on a lattice built so both barriers land on nodes. Take the spot at 100, the ceiling at 125 and the floor at 80. Since $\ln(125/100) = \ln(100/80) = \ln 1.25$ , one step size puts both barriers exactly on lattice levels, which removes the largest source of numerical error in any barrier tree.

With European exercise at the finest grid tested, 3,394 steps:

1.35183 \;-\; 1.14213 \;=\; 0.20970

(3)

and 0.20970 is what the out-in contract prices at when computed independently. The agreement is not approximate. Across all seven grids from 45 to 3,394 steps the discrepancy runs between $5.55 \times 10^{-17}$ and $2.16 \times 10^{-15}$ , which is floating-point noise. That is what you should expect: equation (2) is linearity over a disjoint split, so the lattice does not have to be fine for it to hold, only internally consistent.

The American case, where the identity needs a caveat

American exercise is a genuinely different situation, and the identity is not exact on a finite tree. At the same finest grid:

10.98491 \;-\; 10.58643 \;=\; 0.39848 \quad\text{against}\quad 0.39865

(4)

A residual of 0.00017, or 0.043%. The reason there is anything at all is that an American value is a supremum over stopping times, and a supremum is not additive. Splitting the paths into two groups and optimising inside each group is not, in general, the same as optimising once across all of them.

So why does the subtraction work as well as it does? Because of a feature of this particular contract. An up-and-out call on a stock paying no dividend has no interior early-exercise region: away from the barrier, holding always beats exercising, since there is no carry to lose by waiting. The only advantage early exercise ever offers is capturing $H - K$ at the instant the ceiling is touched, and that right lives on the same paths the partition already splits. In the continuous limit the two sides of equation (2) therefore agree exactly.

The measurement confirms that reading. Refine the grid and the residual falls monotonically: 0.532, 0.313, 0.199, 0.139, 0.091, 0.061 and 0.043 percent across seven grids from 45 to 3,394 steps. It is positive at every grid, so the subtraction underprices slightly, and it dies as the discretisation dies. The correct statement of the result is therefore in two parts: equation (2) is exact for European exercise at any grid, and exact for American exercise in the continuous limit, with a small positive error on any finite tree.

Fig. 2 — The American gap is a discretisation artefact rather than a violation. Seven grids, monotone decay, and always the same sign.

Where the subtraction stops being safe

The load-bearing assumption in the American argument was the absence of an interior exercise region. Add a dividend yield and there is one, because it becomes worth exercising early to collect the dividend at spots nowhere near the ceiling. The supremum objection then has nothing to answer it, and equation (2) becomes an approximation with no guaranteed sign. Nothing above licenses using it on a dividend-paying underlying.

Two failures are numerical rather than mathematical, and they are the ones that actually bite in practice. If $\ln(H/S)$ and $\ln(S/L)$ are not integer multiples of the step, the tree quietly relocates each barrier to the nearest node, and the resulting error is far larger than the 0.043% measured above. A residual computed on a misaligned lattice tells you about the lattice, not about equation (2). Separately, a knock-out that pays a rebate on being hit has to carry that rebate consistently in both terms on the right, since the difference of two contracts is only the third contract if the two agree on everything except the barrier condition.

The same warning applies to monitoring. Continuous monitoring in one leg and daily monitoring in the other breaks equation (1) outright, because the events being split are no longer the same events. Once those conditions hold, though, the trick generalises further than the example suggests. It is a partition and nothing more, so it applies to any payoff and to any arrangement of barriers whose events nest the way these two do. What it cannot do is make anything easier when the two contracts on the right are themselves hard to price.

Sources and further reading

The family of contracts equation (2) moves between Barrier option
The one property that makes equation (1) into equation (2) Indicator function
Why an American value is a supremum and not an integral Optimal stopping
The machine both right-hand terms run on Lattice model
The reflex this article opens by rejecting, and what it takes to repair it Monte Carlo methods for option pricing

The partition was verified by enumerating every lattice path at three lengths rather than by argument, the European identity at seven grid sizes with barriers pinned to nodes, and the American residual measured at those same seven grids so its decay is a measurement rather than a hope. A reviewer briefed only to refute the identity attacked the American case and was right to, which is why the result above is stated in two parts.