A Double Knock-Out Is Worth Less Than Half the Two One-Sided Options

The sharp statement is stronger than the usual one: on every path, missing the ceiling plus missing the floor counts the paths that miss both exactly twice, so the pair is twice the double plus the value of the one-sided survivors. That is a polynomial identity in indicators, so it holds under every pricing measure with no volatility anywhere, and one half is the tight bound. In a worked instance the pair is 10.317 against a double of 1.494, and the fastest refutation of the trap is that the pair exceeds the plain call with no barriers at all.

A call on a share dies if the price ever touches 80, and it also dies if the price ever touches 125. Is that contract worth the same as owning two calls, one that dies only at 80 and one that dies only at 125?

No, and the honest statement is a good deal stronger than “worth less”. The double knock-out is worth strictly less than half the pair, one half is the tight bound, and none of that depends on a volatility, a distribution or an infinite series. It follows from one line of algebra about indicators.

Three contracts that all look alike

The three knock-outs

All three pay $\max(S_T - X, 0)$ at expiry unless killed first, with barriers monitored continuously and knock-out on touch. The down-and-out dies if the price ever reaches the lower level $L$ , the up-and-out dies if it ever reaches the upper level $U$ , and the double knock-out dies at either.

Write $a$ for the indicator that the path touches $U$ at some point and $b$ for the indicator that it touches $L$ . Then the up-and-out survives on $1-a$ , the down-and-out on $1-b$ , and the double on $(1-a)(1-b)$ . Everything below is bookkeeping on those three expressions.

Why adding them feels right

The pull of the wrong answer comes from the contract description rather than from any calculation. Each single-barrier option carries one barrier, the double carries two, and barriers feel like the sort of thing that accumulates. Two hazards, so two prices, added.

There is a one-line refutation before any algebra. In the instance priced below the pair is worth 10.317 while the plain call with no barriers at all is worth 8.883. A contract that can be killed two different ways cannot be worth more than the same contract that cannot be killed at all, so the pair is not the price of anything with two barriers on it.

One line of indicators settles it

For any $a, b \in \{0,1\}$ , expand and collect:

(1-a) + (1-b) \;=\; 2(1-a)(1-b) \;+\; \underbrace{(1-a)b + a(1-b)}_{\text{touched exactly one}}

(1)

Both sides equal $2-a-b$ , so this is an identity rather than an approximation. Read it as a counting statement: the pair counts every path that avoids both barriers twice, once in each book, and counts the paths that touch exactly one barrier once.

Now multiply through by the payoff, which is non-negative, and take expectations under whatever pricing measure you like. Writing $d$ for the double knock-out and $s$ for the value carried by paths that touch exactly one barrier and still finish in the money:

\text{up-and-out} + \text{down-and-out} \;=\; 2d + s

(2)

Nothing in this step used lognormal dynamics, a constant volatility, or even continuity of the price. It used the payoff being non-negative and the two survival events being what they are. Equation (2) holds for a put, for a digital, for any contract that never pays a negative amount.

Fig. 1 — The solid path is dead in the double knock-out and dead in the down-and-out, but alive and paying 12 in the up-and-out. The dashed path survives in all three, which is exactly why the pair counts it twice.

Less than half, and half is tight

Divide (2) by the pair and write $\varepsilon = s/d$ :

\frac{d}{\text{pair}} \;=\; \frac{1}{2 + \varepsilon} \;=\; \frac{1}{2} - \frac{\varepsilon}{2(2+\varepsilon)}

(3)

Strictly below one half whenever $\varepsilon > 0$ , meaning whenever some path can touch one barrier and still finish in the money. And one half is not a loose bound: as the barriers move far away the one-sided value becomes negligible relative to the double, $\varepsilon \to 0$ , and the ratio climbs to one half without ever reaching it.

Fig. 2 — Quadrupling the log distance to each barrier gets the ratio to 0.4998. The bound in equation (3) is tight and it is never attained.

There is a companion identity worth having, since it explains the refutation from the previous section exactly. Rewrite $2-a-b$ as $1 + (1-a)(1-b) - ab$ and take expectations again:

\text{pair} \;=\; \text{plain call} \;+\; d \;-\; v_{\text{both}}

(4)

where $v_{\text{both}}$ is the value carried by paths that touch both barriers and still finish in the money. So the pair exceeds the plain call exactly when the double knock-out is worth more than that small residual, which is almost always. In the instance below $v_{\text{both}} = 0.060$ , and $8.883 + 1.494 - 0.060 = 10.317$ to the cent.

Numbers on a lattice whose barriers are nodes

Share at 100, strike 100, barriers at 80 and 125, half a year to run, zero rate, volatility 31.56 percent. The two barriers are placed symmetrically in the logarithm, since $\ln(125/100) = -\ln(80/100) = \ln(1.25)$ , so a single lattice spacing puts both of them exactly on nodes. The four prices:

\begin{array}{lc} \text{plain call} & 8.883 \\ \text{down-and-out} & 8.695 \\ \text{up-and-out} & 1.622 \\ \hline \text{the pair} & 10.317 \\ \text{double knock-out} & 1.494 \end{array}

(5)

The pair is 6.9 times the double, so “less than half” is satisfied with a great deal of room. Anyone quoting the sum would be asking about seven times the fair price. The reason the gap is this wide rather than merely under a half is that the up-and-out is nearly worthless on its own, 1.622 against a plain call of 8.883, because a call needs the share to rise and the upper barrier is only 25 percent up. Most of the down-and-out's 8.695, some 7.20 of it, sits on paths that go on to touch 125, and those paths are worth nothing to the double.

The identity was also checked path by path rather than only in expectation. On a 16-step walk whose barriers land exactly four steps from the start, all 65,536 paths were enumerated: 22,288 avoid both barriers, 42,940 touch exactly one, and 308 touch both. Of the ones touching exactly one, 19,652 still finish in the money, which is what makes $\varepsilon > 0$ a fact about this corridor rather than an assumption.

What monitoring and the strike do to all of this

Two conditions sit underneath equation (5), and both change the numbers rather than the inequality.

Monitoring first. Everything above assumes the barrier is watched continuously. Watch it once a day instead and some excursions happen and reverse between observations, so paths that should have been killed survive, and a knock-out priced under discrete monitoring comes out too high. The bias is worst for the double knock-out, since there are two levels whose excursions can be missed, which is why the figures here come from lattices whose barriers are exact nodes rather than from a simulation with a fixed observation schedule. A daily-monitored contract is a genuinely different and more expensive contract, not an approximation to this one.

The strike next. The double knock-out is worth something only when $U > X$ . If the ceiling sits at or below the strike, then finishing in the money forces the path to have crossed the ceiling, so every surviving path is a dead path and both the up-and-out and the double are worth exactly zero. Equation (1) still holds; it just becomes a statement about zeros. The interesting regime is a ceiling above the strike, which is where a one-sided survivor can exist and where $\varepsilon$ is strictly positive.

One last note on what equation (2) is not. It is not a hedging recipe. Knowing that the pair equals twice the double plus $s$ does not let you build the double out of the two one-sided options, because $s$ is itself an exotic payoff. What the identity gives is a bound available before any model is chosen, which is the kind of thing worth carrying around: any quote at or above half the pair is wrong, whatever the volatility turns out to be.

Sources and further reading

Wikipedia: Barrier option, for the contract definitions and the standard classification.
Wikipedia: Inclusion–exclusion principle, which equation (1) is the two-set case of, and Binomial options pricing model for the lattice the prices come from.
Robert Merton, “Theory of Rational Option Pricing”, Bell Journal of Economics and Management Science4 (1973), 141–183, where the first closed-form down-and-out call appears.

No Monte Carlo appears anywhere in the verification, deliberately, since a fixed observation schedule biases exactly the quantity under test. The lattice priced all four contracts at 16, 36, 100, 400 and 1,600 time steps with the barriers exact at every refinement, and the double moved by under two cents between the last two, with separate runs at 3,600 and 6,400 steps giving 1.4931 and 1.4939. The weighted form of equation (2) was confirmed in exact rational arithmetic on the full enumeration, with zero exceptions among the 65,536 paths.