The Hedge That Lands on the Right Answer and Still Swings Nine Dollars

You are short one call struck at 100 on a share that starts at 100, and you collected $11.9235$ for it. You propose to hedge with one line of code: hold one share whenever the price is above 100, hold nothing whenever it is below. Check the position at expiry and it is exactly right. If the call is exercised you own the share to deliver; if it expires worthless you own nothing and owe nothing.

So the rule reproduces the obligation on every path, without exception. It is still not a hedge, and the reason is not transaction costs.

The part of the rule that is true

Start by conceding the strong half, because refusing to is how people talk past this problem. The rule's terminal position is the indicator that the share finished above the strike, which is precisely the exercise indicator. Path by path, on every one of the 40 000 paths I ran, the final holding matched what the call owed. A rule that matches the terminal payoff of the thing you are hedging deserves an argument, not a dismissal.

The argument is that a hedge is not a promise about the last day. It is a promise about the whole journey, and the journey is where this rule spends money.

Every round trip loses, by construction

Follow the rule through one crossing. The share is below 100, you hold nothing, the price rises past 100 and you buy. You did not buy at 100. You bought at the first monitored price above it, at $100 + \varepsilon$ for some $\varepsilon > 0$. Later the share crosses back down and you sell, again not at 100 but at $100 - \delta$. The round trip returns $-(\varepsilon + \delta)$, and both terms are positive by the definition of the trigger.

This is worth stating carefully because it is a structural fact, not a statistical tendency. There is no market condition, no volatility, no drift under which a completed round trip of this rule makes money. Buying high and selling low is what the trigger instructs.

Why the average profit is not evidence

The tempting punchline is that those round trips hand back the premium. They do, on average, and the measurement is uncannily clean: 11.90 against a premium of 11.9235. It is also completely uninformative, and this is the trap I want to spend a paragraph on.

The control settles it. A genuine delta hedge run on the same paths hands back 11.91, the same number to two decimal places. If handing back the premium condemned a strategy, it would condemn the correct one too. Any argument built on the average is an argument that cannot distinguish the rule from its replacement, which means it is not an argument about the rule.

The number that does discriminate

What separates a hedge from a gamble is dispersion, so measure that instead. Write the residual as premium plus trading gain minus payoff, a quantity that should be near zero for a working hedge. At daily monitoring its standard deviation is $9.07$ on a premium of $11.92$. Then refine the monitoring, which is the move that fixes almost every discretisation problem in this business:

Nothing happens. Now the same three refinements for a real delta hedge on the same paths:

which halves each time the interval is quartered, the familiar $1/\sqrt{m}$ convergence of discrete replication, and is heading to zero. Equation (2) is what a hedge looks like. Equation (1) is what a rule that happens to land on the right answer looks like.

Where the nine dollars actually lives

The rule's failure has an exact closed form, and it is the reason refinement cannot help. Tanaka's formula extends the chain rule to the kinked function $x \mapsto (x - X)^{+}$:

where $L_T^{X}$ is the local time of the share at the strike, a non-negative random variable that measures how much time the path spends arbitrarily close to $X$. The middle term is exactly the continuous-monitoring limit of our rule's trading gain. So with $S_0 = X$, rearranging (3) gives

Three consequences fall straight out. Taking expectations recovers $\tfrac12\mathbb{E}[L_T^{X}] = C$, so the mean is zero for the reason the definition box gave and not by coincidence. The variance is the variance of half the local time, which does not mention the monitoring interval anywhere, which is why (1) is flat. And since $L_T^{X} \ge 0$, the residual is bounded above by $C$and unbounded below: the best case is that the share leaves the strike and never comes back, and there is no corresponding cap on the loss. A distribution with mean zero, a hard ceiling at 11.9235 and a long left tail is a short-volatility position wearing a hedge's clothing.

The fee, separately, diverges

Costs are a second and independent objection, and they behave differently from the whipsaw. Position changes measured over a year come out at 4.37 at weekly monitoring, 9.88 at daily, 28.34 hourly and 69.68 at five-minute intervals. Weekly to daily is a fivefold refinement and the count grows by 2.26, against the $\sqrt5 = 2.24$ the square-root law predicts:

so quartering the interval doubles the trades. A Brownian path that touches a level crosses it infinitely often in any interval containing the touch, so the count has no finite limit. The whipsaw loss converges to something wrong; the fee converges to nothing at all. Put any fixed cost per trade on the rule and it becomes arbitrarily expensive as the monitoring tightens, while a delta hedge pays a cost that grows like $\sqrt{m}$ against an error that falls like $1/\sqrt m$, which is a tradeoff with an optimum.

When the rule is nevertheless the right choice

Nothing above says never. If the option is deep out of the money and stays there, the local time at the strike is zero, equation (4) says the residual is the whole premium, and the rule was free. If the underlying is illiquid enough that a delta hedge's continual small trades are impossible, a coarse rule with a wide band around the strike may dominate an unachievable one. And the version of this rule with two triggers, buying at $X + b$ and selling at $X - b$, deliberately trades a bigger loss per round trip against fewer of them, which is the honest engineering version of the same idea.

What none of those cases restore is convergence. The band-widened rule has a residual that shrinks with $b$ only down to the local-time floor. Sitting at half the local time of the share at the strike is not an implementation detail to be tuned away. It is what the rule is.

Sources and further reading

• The extension of the chain rule used in (3) — Tanaka's formula
• The quantity $L_T^{X}$ in (4) — Local time
• The hedge that equation (2) belongs to — Greeks (finance)
• The martingale property behind the definition box — Risk-neutral measure
• The premium in (4) — Black-Scholes model

The numbers in (1) and (2) come from 40 000 paths per monitoring frequency and were independently reproduced while writing this, at 9.06 to 9.11 for the rule and with the control halving on the same schedule. The terminal position was checked against the exercise indicator on every path. The bound in (4) shows up in the simulation as a residual whose maximum is exactly 11.9235 and never a fraction more, which is the cleanest confirmation of the local-time decomposition I know how to run.