The Share Fell and the Hedge Says Buy: How the Clock Beats a 1.2 Percent Slip

Short a call struck at 100 with the share at 113.40 and two months left, the hedge holds 0.90028 shares; a month later at 112 it wants 0.94591, so you buy 4.56 per hundred. The reflex to sell is not a blunder, because freezing the clock and letting the same fall happen alone really does take the hedge to 0.87726, but the time effect is 2.7 times larger and points the other way. Holding the hedge constant traces the curve S(T) = X exp(d1 sigma root T minus half sigma squared T), which puts the break-even fall at 3.51 percent and, at expiry, at the strike itself.

Ten months ago you sold a call struck at 100. The share now trades at 113.40, two months of life are left, and your hedge holds 90 shares for every 100 options. Over the following month the share drifts gently down to 112, a slip of 1.2 percent. Do you buy stock or sell it?

Sell is the reflex, and unlike most reflexes in this business it is not stupid. It is half of a real calculation. The other half is bigger, points the other way, and the answer is buy.

Which way the hedge points

Sign conventions decide this problem before any arithmetic, so fix them first. You are short the call. A short call loses money when the share rises, so the hedge is long shares, and the number of shares you hold is the option's sensitivity to the share price. When that sensitivity rises, the hedge is short of stock and you buy. Reverse the position and every instruction below reverses with it.

The hedge ratio

For a European call with strike $X$ , zero rates and no dividend, the hedge holds $\Delta = \Phi(d_1)$ shares per option, where $\Phi$ is the standard normal distribution function and $d_1 = \big[\ln(S/X) + \tfrac12\sigma^2 T\big]/\big(\sigma\sqrt T\big)$ with $T$ the time left. Everything here uses $\sigma = 25\%$ a year.

At $S = 113.40$ and $T = 2/12$ that gives $d_1 = 1.2831$ and $\Delta = 0.90028$ , the ninety shares you started with. A month later at $S = 112$ and $T = 1/12$ it gives $d_1 = 1.6064$ and

\Delta = 0.94591, \qquad \text{so you buy } 4.56 \text{ shares per hundred.}

(1)

Both forces, measured separately

The change in $\Delta$ over the month comes from two moves happening at once, and each can be isolated by freezing the other. Freeze the clock at two months and let the share fall alone:

\Delta(112,\ 2/12) = 0.87726, \qquad \text{a fall of } 0.02301.

(2)

So the reflex is describing something real. A share that drops does reduce the hedge, by curvature, and the size of that reduction is roughly $\Gamma \cdot \Delta S = 0.01513 \times 1.40 = 0.0212$ . Now freeze the share at 113.40 and let a month pass alone:

\Delta(113.40,\ 1/12) = 0.96234, \qquad \text{a rise of } 0.06206.

(3)

The clock is worth 2.7 times the price move here, and in the same direction as buying. Add the two one at a time and you get $+0.03905$ ; the true change is $+0.04563$ , the missing $+0.00658$ being the interaction that a one-at-a-time decomposition always leaves behind over a step this large. A month is far too long a step for first-order Greeks, which is worth remembering before quoting gamma and charm at each other across a desk.

Fig. 1 — The clock alone lifts the hedge from 0.9003 to 0.9623. The price alone would have pulled it to 0.8773. The realised move is the dashed segment.

Why the clock is on the side of buying

No formula is needed for the direction. A call that will finish in the money is, at expiry, worth exactly $S_T - X$ , and the sensitivity of that to $S$ is one. So for any share strictly above the strike the hedge has to arrive at one whole share per option, and since $\Delta < 1$ always, it arrives from below. Time can only push it up.

In symbols, $\ln(S/X) > 0$ is a fixed positive number while $\sigma\sqrt T \to 0$ , so

d_1 = \frac{\ln(S/X)}{\sigma\sqrt T} + \frac{\sigma\sqrt T}{2} \;\xrightarrow[T \to 0^+]{}\; +\infty, \qquad \Delta \to 1.

(4)

Read equation (4) twice, because it also says what happens elsewhere. Below the strike the logarithm is negative, $d_1$ runs to minus infinity and the hedge collapses: $\Delta(95, 1/252) = 0.00058$ , so a share four days from expiry five percent below the strike is hedged with nothing. Exactly at the strike the logarithm vanishes, $d_1$ reduces to $\sigma\sqrt T/2 \to 0$ , and the hedge settles at a half. The clock has three destinations, and which one it drives toward is decided entirely by which side of the strike the share is on.

The fall that would have cancelled the month

"Gentle" is not a number, so put one on it. Since $\Phi$ is strictly increasing, the hedge is unchanged exactly when $d_1$ is unchanged, and holding $d_1 = d_1^{*}$ fixed can be solved for the share price:

S(T) = X \exp\!\left(d_1^{*}\,\sigma\sqrt T - \tfrac12\sigma^2 T\right).

(5)

With $d_1^{*} = 1.2831$ that curve passes through 113.40 at two months, and at one month it sits at

S(1/12) = 109.4172, \qquad \text{a fall of } 3.51\%.

(6)

Anything gentler than 3.51 percent and you buy; anything sharper and you sell. The realised 1.2 percent is not close to the boundary, which matters: a sweep of 20 000 slips between zero and three percent raised the hedge in every single case, so the answer is not a knife edge.

Equation (5) also shows how hopeless the reflex becomes near expiry. As $T \to 0$ the exponent vanishes and $S(T) \to X$ . To hold a hedge of 0.90 constant, the share would have to walk all the way down to the strike by the last day. Four trading days out, the level that keeps the hedge flat is already 102.03. Any share that stays comfortably in the money is going to see its hedge march to one whole share, and the trade that does that is buying.

Fig. 2 — Equation (5). Land above the curve and the hedge has grown, so you buy. The curve ends at the strike, which is why staying in the money means buying eventually.

What the answer does and does not depend on

Barely on the volatility, which surprised me. The starting hedge moves a lot with $\sigma$ , from 0.9814 at 15 percent to 0.7637 at 50 percent, and yet the break-even fall stays inside 3.20 to 3.58 percent over that entire range, drifting down slowly as the volatility rises. So a 1.2 percent slip lands on buy no matter what you assume about the swing, and a candidate who does not know the volatility can still answer the question.

Not on the normal distribution either. A 3000-step binomial lattice, which contains no $\Phi$ anywhere, returns 90.0289 shares at two months, 94.5943 at one, a break-even fall of 3.513 percent and 87.7268 at frozen time. Those are the same four numbers to the precision the lattice can offer.

It does depend on being genuinely in the money. Put the share at 100.50 with two months left and the picture inverts, because $\Gamma$ is concentrated at the strike and the price effect becomes the dominant one. It also depends on there being no dividend: a continuous yield $q$ multiplies the hedge by $e^{-qT}$ , which rises toward one as expiry approaches and so reinforces the conclusion rather than threatening it. And it depends on the option surviving: an American call on a dividend-paying share can be exercised early, at which point the hedge jumps to one whole share and stops being a question.

Sources and further reading

The sensitivities used above, including gamma and the derivative with respect to time — Greeks (finance)
The formula behind $d_1$ — Black-Scholes model
The lattice used as an independent check — Binomial options pricing model
The distribution function that makes (4) a limit statement — Normal distribution

Every figure above was checked twice before publication. The closed form was written with $\Phi(z) = \tfrac12\big(1 + \operatorname{erf}(z/\sqrt2)\big)$ so the symbolic channel never touches a numerical integral, and the limit in (4) was taken formally from the positive side. The same four hedge ratios were then reproduced by a binomial lattice, the break-even fall by bisection on that lattice, and the direction of the trade by a 20 000-draw sweep over gentle slips, which produced no sale.