Count the Outcomes Before You Count on a Hedge

A share at 100 that jumps to either 80 or 130 gives a call an exact price of 12, from two equations in two unknowns and no probability at all. Let the jump size be random, so 110 is also reachable, and that same hedge pays 18 where the option pays 10 while no other portfolio does better. The arbitrage-free prices then fill the whole interval from 20/3 to 12, and the obstruction turns out to be the kink in the payoff rather than the number of states.

A share sits at 100. In one period it jumps, and the jump has a known size: the share ends at 80 or at 130, nothing in between. Interest is zero. Price a call struck at 100.

The answer is exactly 12, and getting there uses no probability at all. Now change one thing. Let the jump size be random, so the share can also end at 110. The same question now has no answer. Not a hard answer, not an approximate answer. No-arbitrage pricing stops delivering a number and delivers an interval instead, from 6.67 to 12, and the top of it is 1.8 times the bottom.

The reflex, and why it is nearly right

The instinct is that a jump is just a large move, and delta hedging copes with large moves. The instinct is almost sound. It works when the number of things that can happen matches the number of instruments you can trade, and it fails the instant it does not. That count is the whole subject.

Two outcomes, two instruments, one linear solve

Hold $\Delta$ shares and $B$ in cash. With no interest that portfolio is worth $\Delta S_T + B$ at the end, and matching it to the call in both states is two equations in two unknowns:

\begin{aligned} 80\,\Delta + B &= 0 \\ 130\,\Delta + B &= 30 \end{aligned} \qquad\Longrightarrow\qquad \Delta = \frac{30}{50} = \frac{3}{5}, \quad B = -48

(1)

That portfolio costs $\tfrac35 \cdot 100 - 48 = 12$ to set up, and it pays the option’s payoff in every state that can occur, so 12 is the price. If the option traded anywhere else you could buy the cheap one, sell the expensive one, and collect the difference with no exposure left over. The matrix of the system is square with independent rows, which is the entire reason a unique price exists.

Attainable, and complete

A claim is attainable when some portfolio of traded assets reproduces its payoff in every state. A market is complete when every claim is attainable, which happens exactly when the traded payoffs span the space of possible payoffs. Two states and two instruments span it. Three states and two instruments do not.

A fixed jump size is a binomial step wearing unfamiliar clothes, which is why the answer feels so familiar. The size of the jump never entered as a probability, only as the two endpoints of the span.

One extra outcome and the hedge is off by 8

Keep the same portfolio and evaluate it at 110. It pays $\tfrac35\cdot 110 - 48 = 18$ , where the option pays 10. It misses by 8. Refitting does not help: all three ways of running a straight line through two of the three outcomes miss the one left out, which is three failures out of three.

The reason is structural rather than numerical. A portfolio of the share and cash is an affine function of the terminal share price, and the call payoff is not affine on any set that straddles the strike, because it bends there. Write the three outcomes as $\ell < m < h$ with the strike $K$ between $\ell$ and $h$ . The line through the outer two overshoots the middle one by

(K - \ell)\,\frac{h - m}{h - \ell} \;=\; \frac{20 \cdot 20}{50} \;=\; 8

(2)

which is strictly positive whenever the strike is straddled. Equation (2) also says how the failure shrinks: push the middle outcome toward either end and the miss goes to zero, which is the statement that a nearly two-state model is a nearly complete one.

Fig. 1 — The hedge is a straight line and the payoff bends. Two points can be matched exactly; a third point between them cannot.

What no-arbitrage still pins down

No hedge means no single price. It does not mean no information. A price is arbitrage-free exactly when it is a discounted expectation under some measure that prices the share correctly, and with no interest that condition is one linear equation:

80\,q_\ell + 110\,q_m + 130\,q_h = 100, \qquad q_\ell + q_m + q_h = 1, \qquad q \ge 0

(3)

Three unknowns and two equations leave a one-parameter family. Write $t = q_m$ , solve, and you get $q_\ell = \tfrac35 - \tfrac25 t$ and $q_h = \tfrac25 - \tfrac35 t$ , which stay non-negative for $t \in [0, \tfrac23]$ . The option’s value along that segment is

10\,t + 30\left(\tfrac25 - \tfrac35 t\right) = 12 - 8t, \qquad t \in \left[0, \tfrac23\right]

(4)

Linear in $t$ , so its range is the interval between the two endpoint values: 12 at $t = 0$ and $20/3$ at $t = \tfrac23$ . Both endpoints come from two-point measures. The first, $(\tfrac35, 0, \tfrac25)$ , puts no weight on the middle state and returns the hedgeable answer of 12. The second, $(\tfrac13, \tfrac23, 0)$ , ignores the top state and returns 6.67. Both genuinely price the share at 100, which is easy to check by hand. The same interval was computed twice, by enumerating those extreme points and by scanning 200,002 measures on a fine grid.

Fig. 2 — The arbitrage-free prices form a segment, not a point. Its endpoints are the two measures that ignore one outcome each.

The obstruction is the kink, not the count

It is tempting to summarise all of this as three states beating two instruments. That summary is wrong in a way that matters. A payoff that is affine across the reachable states is reproduced by one share plus cash however many states there are, which was checked over 200 random state sets with no failures. Move the strike below 80 and the call is affine on all three outcomes, hedgeable again, and priced by no arbitrage again.

So the real condition is that the payoff bends somewhere inside the reachable range. Across 1000 randomly drawn three-state models, the option is unhedgeable whenever the strike sits strictly between the outer two outcomes, and deleting the middle state makes every one of them hedgeable again.

What is done instead, and what the substitution costs

A real jump-diffusion is worse than three states. The diffusion alone supplies a continuum of outcomes, so the interval above is where the problem starts. The standard route out is to import an assumption from outside no-arbitrage: jump risk is diversifiable, and therefore earns no risk premium. Grant that and the rest is bookkeeping, because conditional on the number of jumps the terminal price is lognormal, so the value is a Poisson-weighted sum of ordinary Black-Scholes prices:

c = \sum_{n \ge 0} \frac{e^{-\lambda T}(\lambda T)^n}{n!}\; C_{\mathrm{BS}}\!\left(S, K, r_n, \sigma_n, T\right), \qquad \sigma_n^2 = \sigma^2 + \frac{n v^2}{T}, \qquad r_n = r - \lambda\kappa + \frac{n\!\left(\mu + \tfrac12 v^2\right)}{T}

(5)

with $\kappa = e^{\mu + v^2/2} - 1$ the mean proportional jump. The series behaves: its first forty weights sum to one to within $10^{-12}$ , and a simulation of the same jump-diffusion agrees with the sum. What it is not is a no-arbitrage price. Change the assumption about the jump premium and equation (5) returns a different number, with nothing in the market able to contradict either. That is worth saying out loud, because the formula looks exactly like the ones that are forced.

Where the clean result stops working

The two-state result is fragile in one specific direction. Add any diffusion at all to a single fixed jump size and the outcomes become a continuum, so the hedge fails again for exactly the reason it failed at three states. The headline is a statement about a limiting case, not a technique to reach for.

Incompleteness is also a statement about your instrument set rather than about the asset. One more traded option on the same share, with the same maturity and a different strike, supplies a third equation. Three outcomes and three instruments make the system square again and the price unique again, at which point the extra option’s market price is doing the work the model could not.

One technical edge deserves a mention, because it is the sort of thing that gets an argument rejected. The two endpoint measures each assign probability zero to a state that can genuinely occur, so they are not equivalent to the physical measure. If you insist on equivalence, as the fundamental theorem does, the set of prices is the open interval and 6.67 and 12 belong to its closure. Nothing above depends on which convention you take, but the endpoints are attained only under the looser one.

Sources and further reading

The property equation (1) quietly relies on Complete market
Why equation (3) is the right definition of an arbitrage-free price Fundamental theorem of asset pricing
The interval, and what is done with it in practice Incomplete markets
The process equation (5) prices Jump diffusion
The model equation (1) turns out to be Binomial options pricing model
R. C. Merton, Option pricing when underlying stock returns are discontinuous, Journal of Financial Economics 3 (1976), where equation (5) and its diversifiability assumption come from

The arithmetic here is exact rational arithmetic throughout, with no floating point anywhere in the hedge or the interval. The interval was found twice, once from the extreme points and once by a grid scan, and the general claim about the kink was tested on 1000 random three-state models and 200 random state sets with a linear payoff.