A Ticket Worth Fifty-Three Dollars, Priced Without a Forecast

A ticket paying a hundred dollars if a share finishes above its strike is squeezed between two ordinary call spreads at every width, so its value is pinned by prices already quoted with no distribution assumed anywhere. The limit is minus the derivative of the call price in the strike, which equals e to the minus rT times N(d2) because two density terms cancel exactly at every strike. Here that is 53.2325, against the 62.35 a real-world drift would give.

A ticket pays a hundred dollars if a share finishes above a hundred in a year, and nothing otherwise. The share sits at a hundred today, the interest rate is five per cent and the volatility is twenty. What is the ticket worth?

The usual first answer reasons that shares drift upward, puts the drift at ten per cent, works out a 65.5 per cent chance of finishing above the strike, discounts it, and arrives at 62.35. Call it sixty-two. The ticket is worth 53.23, so that is nine dollars of overpayment, and the cleanest argument for the right number never mentions a probability at all.

Two ordinary options, squeezed

Buy a call struck at 100, sell a call struck at 105, and hold twenty of each. The payoff of that package is nothing below 100, a straight ramp from nothing to a hundred dollars between 100 and 105, then a hundred dollars flat. Compare it with the ticket point by point: the two agree below 100 and above 105, and in between the ramp pays strictly less. So the package is worth less than the ticket, and its price is a floor under the ticket's price.

Now slide the same package down: buy twenty calls struck at 95 and sell twenty struck at 100. The ramp now finishes at the strike instead of starting there, so it pays a hundred dollars over the whole interval where the ticket pays nothing, and matches it everywhere else. That package is worth more than the ticket, so its price is a ceiling.

\frac{H}{w}\bigl[C(X) - C(X{+}w)\bigr] \;\le\; V \;\le\; \frac{H}{w}\bigl[C(X{-}w) - C(X)\bigr]

(1)

Fig. 1 — Two ordinary call spreads, one above the step and one below. Both are made of options whose prices are already quoted.

Every quantity in equation (1) is a price you can read off a screen. No distribution was assumed, no volatility was used, and nothing was said about where the share is going. The two spreads just described, at $w = 5$ , already bracket the ticket between 48.58 and 57.92. Narrowing $w$ tightens it further: at a width of ten it runs 44.10 to 62.49, at two it is 51.36 to 55.11, at 0.2 it is 53.045 to 53.420, and at 0.02 it is 53.214 to 53.251. The trap's 62.35 scrapes inside the widest of those by a tenth of a dollar and is dead at every tighter one.

Fig. 2 — The bracket tightens monotonically onto 53.2325. The trap answer survives exactly one width of it.

Static replication

A claim is statically replicated when a fixed portfolio of traded instruments, bought once and never adjusted, matches or dominates its payoff in every state. The price of the portfolio then bounds the price of the claim by arbitrage alone, for any terminal distribution whatsoever. The bounds in equation (1) are of that kind, which is why they hold before any model is chosen.

The limit is a slope in the strike

Both sides of equation (1) are difference quotients of the call price in the strike, and they squeeze together as $w \to 0$ . Whenever the call price is differentiable in the strike, the common limit is

V \;=\; -H\,\frac{\partial C}{\partial X}

(2)

which is worth pausing on, because it says the value of the ticket is a property of the shape of the observed price surface rather than of anyone's view. Options at neighbouring strikes are quoted; the slope between them is the ticket. If you can see the strike ladder, you have already seen the answer.

The step where two terms vanish

Now put a model in, only to evaluate equation (2). The Black-Scholes call is $C = S\,N(d_1) - Xe^{-rT}N(d_2)$ , and both $d_1$ and $d_2$ depend on the strike, so differentiating produces three terms rather than one. Two of them cancel, and the identity that kills them is this:

S\,\varphi(d_1) \;=\; Xe^{-rT}\varphi(d_2) \qquad \text{at every strike}

(3)

It comes out of the exponents. Since $d_1 = d_2 + \sigma\sqrt{T}$ , a short expansion gives $d_1^2 - d_2^2 = 2\log(S/X) + 2rT$ , so the ratio of the two densities is exactly $(S/X)e^{rT}$ , which is what equation (3) says. The same cancellation is what makes the Black-Scholes delta come out as $N(d_1)$ with no density term attached, so it is a step worth recognising rather than rediscovering. What survives is one term:

-\frac{\partial C}{\partial X} \;=\; e^{-rT}N(d_2) \qquad\Longrightarrow\qquad V \;=\; H\,e^{-rT}N(d_2)

(4)

So the closed form is not a separate result. It is the evaluation of a bound that was already true.

The number, three ways

At the money the logarithm in $d_2$ vanishes and the expression collapses:

d_2 \;=\; \frac{\bigl(r - \tfrac{1}{2}\sigma^{2}\bigr)T}{\sigma\sqrt{T}} \;=\; \frac{0.05 - 0.02}{0.20} \;=\; 0.15

(5)

with $N(0.15) = 0.559618$ , so the ticket is worth $100\,e^{-0.05} \times 0.559618 = 53.2325$ . A 2,001-step binomial lattice, which contains no normal distribution anywhere in it, gives 53.2333. Four hundred thousand risk-neutral paths give 53.1899. A central difference of the call price in the strike matches $e^{-rT}N(d_2)$ to ten decimal places. And a share-or-nothing ticket, which pays $S_T$ rather than a hundred dollars in the same states and is worth $100\,N(d_1) = 63.6831$ , minus this ticket reproduces the ordinary call to ten decimals at 10.4505835722, which cross-checks the whole construction against a formula nobody disputes.

What that number is not

$N(d_2) = 0.5596$ is not a forecast. It is the chance of finishing above the strike under the measure that makes the discounted share price a martingale, which is a device for pricing and not a belief about the world. Swap in a real-world drift $\mu$ and the whole difference is a shift in the argument:

d_2(\mu) - d_2(r) \;=\; \frac{(\mu - r)\sqrt{T}}{\sigma} \;=\; \frac{0.05}{0.20} \;=\; 0.25

(6)

which turns 0.5596 into 0.6554 and 53.23 into 62.35, an overpayment of 17.1 per cent. The bracket in equation (1) is what makes that indefensible rather than merely debatable. Paying 62.35 is inconsistent with the quoted prices of the calls either side of the strike, so you cannot hold a private view about the drift and simultaneously trade at the market's option prices. Whoever sells you the ticket at 62.35 buys the spread and locks in the difference.

The other reflex, that an at-the-money digital must be worth about half its payoff, is also wrong, and interestingly it is wrong in both directions depending on the volatility. Here $d_2 = 0.15$ and the ticket is worth 53.23, above half. Raise volatility to forty per cent and $d_2 = (0.05 - 0.08)/0.40 = -0.075$ , so the ticket is worth 44.72. The crossover sits near $\sigma = 25.8\%$ , where the discount factor and the drift term exactly offset. More volatility makes this ticket cheaper, which is a sign that a step payoff does not obey the intuitions built on convex ones.

Where the argument stops

Equation (1) needs neighbouring strikes to exist. On a real strike ladder the spacing is fixed, so the bracket is only as tight as the grid, and the finite-width form matters for that reason: the argument survives a discrete market, it just does not pin the price to the cent.

The price converges as $w \to 0$ and the hedge does not. The replicating package holds $H/w$ of each option, so the position size diverges as the spread tightens. That is why digitals are quoted and risk-managed as narrow spreads in practice rather than as limits, and why a digital sitting on its strike close to expiry is one of the least pleasant positions in a book: the thing you are short behaves like an enormous spread with a vanishing gap.

Equation (2) also assumes the call price is differentiable in the strike, which is a statement about the terminal law having no atom at the strike. Put an atom there, a pinned closing auction for instance, and the left and right slopes differ. The bracket still holds; the limit does not exist, and the price genuinely depends on a convention about what happens exactly at the boundary.

Sources and further reading

The contract itself Binary option
The arbitrage principle equation (1) rests on Rational pricing
The package being bought and sold in it Bull spread
Where equation (4) is evaluated Black-Scholes model
What the probability in equation (6) is and is not Risk-neutral measure
What differentiating twice in the strike would have given instead Probability density function

Equations (3) and (4) were proved symbolically rather than checked numerically, so the cancellation is an identity in the strike and not a coincidence at one point. The bracket was evaluated at five widths and tightens monotonically onto 53.2325 at every one, and the value was reproduced by a lattice and by simulation, neither of which uses the closed form.