Delta of an At-the-Money Call: Above or Below One Half?
Consider a European call on a non-dividend-paying stock that is exactly at the money: $S = K$. Under Black-Scholes-Merton with $r > 0$,
(a) Estimate the delta of the call. Is it above or below $0.5$, and why?
(b) How does the delta of an at-the-money call change with the time to maturity $\tau$? What does it converge to as $\tau \to 0$?
Hints
- Delta of a European call on a non-dividend stock is $N(d_1)$. Write down $d_1$ when $S = K$; the $\ln(S/K)$ term vanishes.
- With $S = K$, $d_1 = (r + \sigma^2/2)\sqrt{\tau}/\sigma = (r/\sigma + \sigma/2)\sqrt{\tau}$, which is strictly positive whenever $r \ge 0$.
- $d_1$ grows like $\sqrt{\tau}$, so the delta rises with maturity; as $\tau \to 0$, $d_1 \to 0$ and $N(d_1) \to N(0) = 1/2$.
Worked Solution
How to Think About It: For a non-dividend stock, $\Delta_{\text{call}} = N(d_1)$. At the money the messy $\ln(S/K)$ term disappears and $d_1$ collapses to a single expression in $r$, $\sigma$ and $\tau$. Everything about the answer, including the maturity dependence, is read off from the sign and size of that expression.
Quick Estimate: Take $r = 5\%$, $\sigma = 20\%$, $\tau = 1$ year: $d_1 = (0.05/0.2 + 0.1)\cdot 1 = 0.35$ and $N(0.35) \approx 0.64$. For a 3-month option, $d_1 = 0.35 \times 0.5 = 0.175$ and $\Delta \approx 0.57$. For a 1-day option, $d_1 \approx 0.35 \times 0.052 = 0.018$ and $\Delta \approx 0.507$. So the delta is a bit above one half and moves toward one half as expiry approaches.
Formal Solution:
Part (a): Size of the at-the-money delta
*Step 1 -- Specialize $d_1$ to $S = K$.* $$d_1 = \frac{\ln(S/K) + (r + \sigma^2/2)\tau}{\sigma\sqrt{\tau}} = \frac{(r + \sigma^2/2)\tau}{\sigma\sqrt{\tau}} = \left(\frac{r}{\sigma} + \frac{\sigma}{2}\right)\sqrt{\tau}.$$
*Step 2 -- Sign.* Both $r/\sigma$ and $\sigma/2$ are positive, so $d_1 > 0$ and $\Delta = N(d_1) > N(0) = 0.5$. The delta of an at-the-money call is slightly above one half. (For typical parameters $d_1$ is small, so $\Delta \approx 0.5 + d_1/\sqrt{2\pi} \approx 0.5 + 0.4\,d_1$.)
Part (b): Dependence on maturity
*Step 3 -- Monotonicity.* $d_1 = (r/\sigma + \sigma/2)\sqrt{\tau}$ is increasing in $\tau$, and $N$ is increasing, so the at-the-money delta rises with time to maturity: a longer-dated at-the-money call has a higher delta.
*Step 4 -- Limit.* As $\tau \to 0$, $d_1 \to 0$ and $\Delta \to N(0) = 1/2$. Intuitively, right before expiry an at-the-money call is a coin flip between finishing worth $S_T - K > 0$ (delta 1) and expiring worthless (delta 0), and the average is $1/2$.
*Step 5 -- Caveat.* With dividends, $\Delta = e^{-y\tau}N(d_1)$ and $d_1 = ((r - y)/\sigma + \sigma/2)\sqrt{\tau}$, so a large enough yield can push the at-the-money delta below one half.
Answer:
(a) $\Delta = N\!\left((r/\sigma + \sigma/2)\sqrt{\tau}\right)$, slightly above $0.5$ because $d_1 > 0$ (for example $0.64$ with $r = 5\%$, $\sigma = 20\%$, $\tau = 1$).
(b) The delta increases with $\tau$ (as $\sqrt{\tau}$ inside $N$) and tends to $N(0) = 0.5$ as $\tau \to 0$.
Intuition
An at-the-money call is not a fifty-fifty bet in the risk-neutral world: the stock has positive drift $r$ and the lognormal distribution is skewed, so $d_1$ is positive and the delta sits a bit above one half. The gap is $(r/\sigma + \sigma/2)\sqrt{\tau}$ in standard-normal units, so it widens with maturity and shrinks to zero at expiry, where the option becomes a coin flip. Traders use "ATM delta is a little over 50" as a daily sanity check, and the $\sqrt{\tau}$ scaling is why short-dated ATM options have deltas glued to $0.5$ while long-dated ones (LEAPS) can have deltas of $0.6$ to $0.7$.