An At-the-Money Call's Delta Is Never Exactly a Half

At the money the log term in d1 vanishes and what remains is strictly positive for every non-negative rate and every volatility, so the delta always beats 0.5 and is 0.6554 at twenty percent. A square rather than a derivative gives the sharp floor: at six percent over a year the delta can never fall below 0.6355. The article also kills the sentence that sounds like a restatement of the answer, since the chance of finishing in the money falls to 0.4801 at forty percent volatility.

A one-year European call is struck exactly at today's share price. The rate is six percent, there are no dividends. The share goes up a dollar. Does the call gain more or less than fifty cents?

More. Always more, for every non-negative rate and every positive volatility, and at twenty percent volatility the gain is 65.5 cents rather than 50. The reflex that says a half is answering a question about a coin, and the object in front of it is not a coin.

What the fifty cents would have to be

Delta

The delta of an option is the derivative of its value with respect to the share price. It is the number of shares that replicates the option locally, so it is also the hedge ratio a desk holds against it. For a European call under Black-Scholes it is exactly $\Phi(d_1)$ , with $\Phi$ the standard normal distribution function.

Writing $S$ for the share, $X$ for the strike and $\sigma$ for the volatility:

d_1 = \frac{\ln(S/X) + \left(r + \tfrac12\sigma^2\right)T}{\sigma\sqrt{T}}, \qquad \Delta = \Phi(d_1)

(1)

A delta of exactly one half would need $d_1 = 0$ , because $\Phi$ is strictly increasing and $\Phi(0) = 1/2$ . So the question is not about probabilities at all. It is about whether one expression can vanish.

At the money the logarithm disappears

At the money $S = X$ , so $\ln(S/X) = \ln 1 = 0$ and the numerator loses its first term entirely:

d_1 = \frac{\left(r + \tfrac12\sigma^2\right)\sqrt{T}}{\sigma} > 0

(2)

Every factor left is positive. The rate is non-negative, the half-variance term is strictly positive, the volatility and the maturity are positive. There is no configuration of inputs that makes (2) vanish, so the delta is never one half, and it is always on the high side. The result does not even need a positive rate: set $r = 0$ and the surviving term is $\sigma\sqrt{T}/2$ , still strictly positive.

At the stated inputs, $T = 1$ and $r = 0.06$ with $\sigma = 0.20$ , that gives $d_1 = 0.30 + 0.10 = 0.40$ and:

\Delta = \Phi(0.40) = 0.6554217

(3)

Fig. 1 — The delta is this curve read at d₁. The half at the centre is what you get only if d₁ is exactly zero, and at the money it never is.

That is fifteen cents away from a half, which is far too much to be a rounding matter. It is also money: a desk that hedges a thousand of these calls at 0.50 rather than at 0.655 is carrying the exposure of a hundred and fifty-five loose shares.

How low the delta can go

Equation (2) says the delta beats a half. It does not say by how little, and someone could reasonably ask whether a well-chosen volatility drives $d_1$ arbitrarily close to zero. It cannot. Read (2) at $T = 1$ as a function of $\sigma$ alone and it is $r/\sigma + \sigma/2$ , a sum of a falling term and a rising one. Its minimum comes from an identity rather than a derivative:

\frac{r}{\sigma} + \frac{\sigma}{2} - \sqrt{2r} = \left(\sqrt{\tfrac{\sigma}{2}} - \sqrt{\tfrac{r}{\sigma}}\right)^{2} \;\ge\; 0

(4)

A real square is never negative, so $d_1 \ge \sqrt{2r}$ with equality exactly when $\sigma = \sqrt{2r}$ . This is the arithmetic-geometric mean inequality wearing different clothes, and proving the bound by exhibiting a square is more convincing than differentiating, because there is no second-order condition left to check.

At six percent that floor is $\sqrt{0.12} = 0.34641$ , reached at a volatility of 34.64 percent, and the delta there is $\Phi(0.34641) = 0.6354828$ . So over every volatility whatsoever, a one-year at-the-money call at this rate has a delta of at least 63.5 cents. It cannot get near a half from any direction.

Fig. 2 — Sweeping the volatility. The curve has a floor, marked by the dashed line, and the floor is nowhere near zero.

The right answer with the wrong reason

There is a sentence that sounds like a restatement of the result and is false: that the call finishes above the strike more often than not. It gets attached to this problem constantly, and it deserves to be killed on sight.

The chance of finishing in the money is not $\Phi(d_1)$ but $\Phi(d_2)$ , and at the money:

d_2 = d_1 - \sigma\sqrt{T} = \frac{\left(r - \tfrac12\sigma^2\right)\sqrt{T}}{\sigma}

(5)

That numerator changes sign. As soon as $\sigma^2 > 2r$ it is negative and the call is more likely than not to expire worthless. At six percent and forty percent volatility $d_2 = -0.05$ and the risk-neutral chance of finishing above the strike is 0.4801, below a half, at the same time as the delta is 0.6368. Both figures are correct together. The delta is not a probability, and the two quantities separate by exactly $\sigma\sqrt{T}$ .

The reason they differ is that the delta weights outcomes by the share price as well as by their likelihood. Under the measure in which $\Phi(d_1)$ is a probability, the share itself is the numeraire rather than the money, and the mean of a lognormal sits above its median. The half-variance term in (2) is that gap, written down.

Two conventions and one exception

At the money has a second reading: strike at the forward rather than at the spot. That kills the rate term, since $\ln(S/X) = -rT$ exactly cancels it, and leaves:

d_1 = \tfrac12\sigma\sqrt{T}

(6)

Still strictly positive. At twenty percent volatility that is $d_1 = 0.10$ and a delta of 0.5398, closer to a half but on the same side of it. The conclusion survives the convention, which is worth saying out loud before anyone argues about which reading was intended.

Dividends are the real exception. With a continuous yield $q$ the delta becomes $e^{-qT}\Phi(d_1)$ , and that leading factor is less than one. A large enough yield does push the at-the-money delta below a half, and it does so for a reason worth knowing: holding the share pays you something the option does not, so the replicating position needs fewer shares. The question above states no dividends, which is exactly why the answer is unconditional.

Sources and further reading

The formula that equation (1) comes from — Black-Scholes model
Delta among the other sensitivities — Greeks (finance)
The function being evaluated at $d_1$ — Normal distribution
The inequality behind equation (4) — AM-GM inequality

Every figure here was checked three ways: symbolically, by a twelve-thousand-point sweep over volatility that located the worst case at 0.3465 against the exact $\sqrt{2r}$ of 0.346410, and by a finite-difference Monte Carlo over four hundred thousand common random numbers that measured the hedge ratio without ever evaluating $\Phi(d_1)$ . That last channel returned 0.65453 at twenty percent volatility and 0.63552 at thirty-five percent.