The Share Falls Ten Dollars and the Hedge Says Buy

You own a call option, so you have the right to buy a share at a fixed price later. You are hedged, which means you hold enough stock short that a small move in the share leaves your total position flat. The share now falls by ten dollars. To restore the hedge, do you buy stock or sell it?

Most people sell. The share went down, so cutting exposure feels like the disciplined move. The correct answer is to buy, and the reason has nothing to do with a view on the share. It is forced by the shape of the option's value.

The hedge is a slope

Three properties of $c$settle the question without naming a pricing model. It is increasing, since a call is worth more when the share is worth more. Its slope lies strictly between zero and one, since the option cannot gain more than the share it is written on. And it is convex, bending upward everywhere, because the option's payoff is convex and taking expectations preserves that.

Convexity is the one doing the work. A convex function has an increasing derivative, so

and a smaller short is reached by buying stock back. That is the whole argument. Notice what it never mentions: volatility, time to expiry, interest rates, or any distribution at all. You cannot escape the conclusion by choosing different numbers, because no numbers were used.

The same thing inside a pricing model

In the standard model for a European call with no dividends, the hedge ratio is

with $N$ the standard normal distribution function. Two derivatives make (1) formal. First, $N' = \varphi$, the normal density, which is positive everywhere, so $N$ is strictly increasing and $0 < h < 1$ for every choice of parameters. Second,

so $d_1$ rises with the share and the composition $N(d_1)$ rises with it. Chaining the two gives the curvature explicitly,

which is the second derivative of the call price and the model's version of convexity. It is strictly positive for any positive volatility and any time left, so there is no corner of the parameter space where the direction of the trade flips.

Sixty short becomes forty short

Numbers make the size concrete, and they can be made exact rather than illustrative. Take zero interest, one year to expiry, a volatility of $20.7937\%$ and a strike of $96.9416$. Then (2) returns

to six decimal places, which is why those are the numbers I use. A position of one hundred calls is hedged with sixty shares short at a price of $100$, and with forty short at $90$. Going from sixty short to forty short means buying twenty shares. The share fell ten dollars and the hedge instruction is to buy into it.

Why buying the drop is not a loss

This is where the puzzle stops being about a sign and starts being about what a hedged option position actually earns. Under the same parameters the call is worth $9.7736$ at $100$ and $4.7560$ at $90$, so the option lost $5.0176$. The short of sixty shares gained $0.60 \times 10 = 6.00$. Net:

per share of option. The hedged position made money on a fall of ten dollars, and by symmetry of the curvature it would have made money on a rise as well. That is what convexity buys, and the second-order estimate confirms where the figure comes from:

close to the exact $0.9824$, with the small gap being the third and higher terms.

Run this repeatedly and the rebalancing rule reads: buy when the share falls, sell when it rises, forever. A hedged long option is a mechanical buy-low-sell-high machine, which sounds like free money and is not, because the position also loses value every day that nothing happens. The premium you paid is the price of the machine.

The side you are on decides everything

Sell the call instead of buying it and every sign reverses. A short call is hedged by holding stock long, the required holding shrinks when the share falls, and shrinking a long means selling. So the seller sells into the drop and buys into the rally, which is why the same market move that pays the buyer costs the seller. Any statement of this problem has to fix which side you are on before any number means anything.

Where the picture stops being enough

Continuous dividends at rate $q$ change the hedge to $e^{-qT} N(d_1)$, which is still increasing in $S$. The direction survives, the size does not, and a candidate who says "buy, and the amount shrinks by the dividend discount" has said something more useful than the bare answer.

The straight-line reading of the hedge breaks when the curvature is large relative to the move. A call with a day to run, struck right at the current price, has enormous $(4)$, and a ten-dollar move on a hundred-dollar share takes the hedge from about a half to nearly zero. In that regime the tangent in Fig. 1 is useless over the whole move and rebalancing has to be continuous or nearly so. The exhibit above sits deliberately far from that case, with a full year to expiry.

And deep in the wings the answer is technically the same and practically nothing. If the share is at $20$ against a strike of $97$, both hedges round to zero shares and the required rebalance is a fraction of a share. Across two hundred thousand randomly generated markets, not one ever called for a bigger short after a fall, and about two percent of them called for a change too small for a double-precision float to see.

Sources and further reading

• The quantity in (2) and its derivative in (4) — Greeks (finance)
• The pricing model those formulas come from — Black-Scholes model
• The property that makes (1) model-free — Convex function
• The practice equation (6) describes — Delta neutral

Every claim above was checked before publication: the monotonicity symbolically, the two hedges in (5) both from the closed form and independently from a central finite difference of the option price which never mentions formula (2), and the direction across two hundred thousand random markets spanning share prices from eleven to four hundred, volatilities from two to two hundred percent and maturities from one day to five years. Not one produced a sale.