A Down-and-Out Call Climbs Faster, and the Up-and-Out Climbs Backwards

Three calls struck at 100: one plain, one that dies at 90, one that dies at 120. The knock-outs cost less, and their slopes can be ranked from the two ends of the picture instead of by differentiating a barrier formula. That argument only bounds an average slope, so the article also carries the exact pointwise gap, the strike times a normal tail at the reflected share price divided by the barrier, which comes to 0.138146 and turns 0.539828 into 0.677974.

Three calls, all struck at 100, all running a year, on a share sitting at 100 with a twenty percent volatility and a zero rate. The first is plain. The second dies worthless if the share ever touches 90. The third dies if the share ever touches 120. Rank them by how much each one gains when the share goes up a dollar.

The answer is that the down-and-out is the steepest, the plain call is in the middle, and the up-and-out is at the bottom, and not merely at the bottom: it goes the wrong way. Its value falls when the share rises. What makes the question worth asking is that you are not meant to get there by differentiating anything.

Three calls and one question

Prices first, so the rest of the article has something to check itself against. The plain call is worth 7.965567 and gains 0.539828 per dollar. The down-and-out barriered at 90 is worth 6.467368 and gains 0.677974. The up-and-out barriered at 120 is worth 1.104953 and gains about $-0.0086$ , so the share rising by a dollar costs its holder just under a cent.

Knock-out

A knock-out option is a plain option with a rule attached: if the share ever trades through a stated level before expiry, the contract is cancelled and pays nothing. Because that rule only ever deletes payoffs, a knock-out is worth no more than the plain option at every share price, and strictly less wherever the barrier can still be reached.

The answer from the two ends

Take the down barrier at 90 and look at the two ends of the picture. Just above the barrier the knock-out is nearly certain to be cancelled, so it is worth nearly nothing while the plain call is worth something. Far above the barrier the chance of ever falling to 90 is negligible, so the two contracts are effectively the same and their prices meet.

A curve that starts below another one and finishes level with it has to close the gap in between, which means its average slope over that stretch is larger. That is the whole argument, it takes one sentence to say out loud, and it gets you the correct ordering without touching a barrier formula.

Fig. 1 — Starts lower at the barrier, finishes level out to the right. The tangents at 100 make the average-slope statement visible at a point.

What the endpoints argument does not prove

It bounds an average. Nothing about a mean value theorem tells you the inequality holds at every share price, and in general two curves can meet after one of them crosses steeply in a narrow region and stays flat elsewhere. The endpoints story is the right answer arrived at by an argument that stops one step short of the claim people quote from it, and it is worth saying so before someone points it out for you.

The pointwise statement is true here, and proving it costs one line of algebra rather than a page of differentiation.

The reflection that gives the exact gap

At a zero rate, with the barrier at or below the strike, the down-and-out call has a closed form built out of the plain call evaluated at a reflected share price:

C_{\text{do}}(S) = C(S) - \frac{S}{B}\,C\!\left(\frac{B^2}{S}\right), \qquad S > B

(1)

The boundary condition is immediate: at $S = B$ the reflected argument $B^2/S$ is $B$ itself and the factor $S/B$ is one, so the subtracted term equals the plain call exactly and the price is zero. That is the contract, and no fitting was needed to get it.

Now differentiate the subtracted term, writing $u = B^2/S$ so that $du/dS = -B^2/S^2$ :

\frac{d}{dS}\left[\frac{S}{B}C(u)\right] = \frac{C(u)}{B} - \frac{B}{S}\,\Delta(u)

(2)

At a zero rate the plain call satisfies $C(u) = u\,\Delta(u) - X\Phi(d_2(u))$ , and since $u/B = B/S$ , the two terms carrying $\Delta(u)$ cancel when (2) is subtracted from the plain call's slope. What survives is a single positive quantity:

\Delta_{\text{do}}(S) - \Delta(S) = \frac{X\,\Phi\big(d_2(B^2/S)\big)}{B} > 0

(3)

The normal distribution function is strictly positive everywhere, so the inequality is strict at every share price above the barrier. At our numbers the reflected share price is $90^2/100 = 81$ , the argument is $d_2(81) = -1.153605$ , and the gap is $100 \times 0.124331 / 90 = 0.138146$ . Adding it to 0.539828 gives 0.677974, which is the measured slope of the down-and-out to the last digit.

The mechanism is worth stating in words. A rising share does two good things for the holder of a down-and-out call at once: it moves the option further into the money, and it moves the share away from the level that would cancel the contract. The plain call only gets the first of those. Equation (3) is the price of the second.

The up-and-out half needs a different argument

The reflection trick does not transfer to the up barrier, and the fix is a decomposition instead. A knock-in and a knock-out with the same barrier add up to the plain option, since exactly one of them survives on every path, so their slopes add too:

\Delta = \Delta_{\text{ui}} + \Delta_{\text{uo}}

(4)

The up-and-in payoff is nondecreasing in the starting share price: scale the start up and every path scales up with it, which can only make the upper barrier easier to reach and the terminal payoff bigger. So $\Delta_{\text{ui}} \ge 0$ , and (4) forces $\Delta_{\text{uo}} \le \Delta$ . That is the second half of the ordering, and it needed no formula at all.

The same argument refuses to work on the down side, which explains why (1) was necessary there. A down-and-in call is not monotone in the starting price: raising the share makes the eventual payoff bigger but makes the knock-in less likely, and the two effects fight.

The up-and-out slope is also not merely smaller, it is negative. As the share climbs towards 120 the contract approaches certain cancellation, so its value has to fall to zero there. Somewhere below the barrier the slope must therefore turn, and at these parameters it has already turned twenty dollars early: at a share price of 100 the up-and-out is worth 1.104953 and losing about 0.86 cents per dollar of rally.

Fig. 2 — The ordering, with the zero line drawn in. The third bar is on the wrong side of it, which no amount of averaging would have suggested.

Where the ordering and the formula stop working

Equation (1) is a zero-rate special case. In general the reflected term carries a power, $(S/B)^{1-2r/\sigma^2}$ , which is one exactly when $r = 0$ . The boundary condition survives for any rate, since that power is one at $S = B$ , but the clean gap in (3) does not: differentiating the power adds a term and the answer stops being a single normal distribution function.

Equation (1) also needs the barrier at or below the strike. Put the down barrier above the strike and the option can be knocked out while it is in the money, which is a different contract and a different image.

Two modelling assumptions matter more than they look. The barrier here is monitored continuously; a barrier checked once a day is harder to breach, so the knock-out is worth strictly more, and the gap grows with the monitoring interval. That is also why none of these numbers was checked with a Monte Carlo: a simulated path only sees the barrier at its sampling times, so it misses crossings and prices knock-outs too high. The three channels used instead were the image solution, the general barrier formulas, and a trinomial lattice with the barrier placed exactly on a node. The two closed forms agree to nine decimals and the lattice to 0.008 percent on the down-and-out.

Finally, the argument in the second section rests on a knock-out being worth less than the plain call everywhere. Attach a rebate paid on cancellation and that stops being true near the barrier, so the ordering has to be re-derived rather than assumed.

Sources and further reading

The contracts and their standard closed forms — Barrier option
The slope being ranked here — Greeks (finance)
Why a reflected solution kills the value on a boundary — Method of images
The lattice used as the third pricing channel — Trinomial tree
The equation that (1) solves — Black-Scholes equation

The pointwise gap in (3) was measured against the closed form at 1,500 share prices, and the ordering itself was confirmed at 3,000 down-barrier and 2,400 up-barrier placements with no violations, plus 5,400 points spread over nine barrier levels for the up-and-out half. The average-slope argument is the one to say in an interview; equation (3) is the one to have ready when the interviewer asks whether it holds everywhere.