Quick Value of an At-the-Money Call: 0.4 Sigma S Root Tau
A European call on a non-dividend-paying stock is at the money, $S = K$. Interest rates are low and the time to maturity $\tau$ is short, so that both $r\tau$ and $\sigma^2\tau$ are small.
Give a quick approximation for the call price $c$ in terms of $S$, $\sigma$ and $\tau$, and explain where it comes from. Illustrate with $S = K = 100$, $\sigma = 20\%$ per year and $\tau = 3$ months.
Hints
- Set $r \approx 0$ and $S = K$ in Black-Scholes: $c = S\,[N(d_1) - N(d_2)]$ with $d_1 = \sigma\sqrt{\tau}/2$ and $d_2 = -\sigma\sqrt{\tau}/2$.
- Both $d_1$ and $d_2$ are close to zero, so $N(d_1) - N(d_2) \approx N'(0)\,(d_1 - d_2)$ by a first-order Taylor expansion of $N$ around $0$.
- $N'(0) = 1/\sqrt{2\pi} \approx 0.4$ and $d_1 - d_2 = \sigma\sqrt{\tau}$, so $c \approx 0.4\,\sigma S\sqrt{\tau}$.
Worked Solution
How to Think About It: At the money and with negligible rates, Black-Scholes reduces to $S$ times a difference of two normal CDFs evaluated at $\pm\sigma\sqrt{\tau}/2$. Because those arguments are tiny for short maturities, the difference of CDFs is just the density at zero times the gap between the arguments. That single Taylor step gives a formula you can do in your head.
Quick Estimate: With $S = 100$, $\sigma = 0.2$, $\tau = 0.25$: $\sigma\sqrt{\tau} = 0.1$, so $c \approx 0.4 \times 0.1 \times 100 = 4$. The exact Black-Scholes value with $r = 0$ is $3.99$; with $r = 2\%$ it is $4.23$, so the approximation is within a few percent when rates are low.
Formal Solution:
*Step 1 -- Specialize Black-Scholes.* With $S = K$ and $r \approx 0$, $$c = S\,N(d_1) - Ke^{-r\tau}N(d_2) \approx S\,[N(d_1) - N(d_2)], \qquad d_1 = \frac{(r + \sigma^2/2)\tau}{\sigma\sqrt{\tau}} \approx \frac{\sigma\sqrt{\tau}}{2}, \quad d_2 = d_1 - \sigma\sqrt{\tau} \approx -\frac{\sigma\sqrt{\tau}}{2}.$$
*Step 2 -- Linearize the normal CDF.* For small $x$, $N(x) \approx N(0) + N'(0)\,x = \tfrac12 + x/\sqrt{2\pi}$. Hence $$N(d_1) - N(d_2) \approx \frac{d_1 - d_2}{\sqrt{2\pi}} = \frac{\sigma\sqrt{\tau}}{\sqrt{2\pi}}.$$ (The second-order terms cancel because $d_1$ and $d_2$ are symmetric about zero; the error is third order in $\sigma\sqrt{\tau}$.)
*Step 3 -- Assemble.* Since $1/\sqrt{2\pi} = 0.3989 \approx 0.4$, $$c \approx \frac{\sigma S\sqrt{\tau}}{\sqrt{2\pi}} \approx 0.4\,\sigma S\sqrt{\tau}.$$ Equivalently $c/S \approx 0.4\,\sigma\sqrt{\tau}$: an at-the-money call costs about $40\%$ of one standard deviation of the stock's move over the option's life. By put-call parity with $r \approx 0$, the at-the-money put has the same value, so the straddle costs about $0.8\,\sigma S\sqrt{\tau}$.
*Step 4 -- Example.* $S = 100$, $\sigma = 0.20$, $\tau = 0.25$: $c \approx 0.4 \times 0.20 \times 100 \times 0.5 = 4.0$ versus the exact $3.99$ (with $r = 0$).
Answer: $c \approx 0.4\,\sigma S\sqrt{\tau}$ (more precisely $\sigma S\sqrt{\tau}/\sqrt{2\pi}$), because $c = S[N(d_1) - N(d_2)]$ with $d_1 \approx -d_2 \approx \sigma\sqrt{\tau}/2$ and $N(d_1) - N(d_2) \approx (d_1 - d_2)/\sqrt{2\pi}$. For $S = 100$, $\sigma = 20\%$, $\tau = 0.25$ this gives about $4$.
Intuition
At the money the call is a bet on the size of the move, not its direction, so its value is proportional to the expected absolute move: a normal variable with standard deviation $\sigma S\sqrt{\tau}$ has $E|X| = \sqrt{2/\pi}\,\sigma S\sqrt{\tau}$, and half of that, the positive part, is $\sigma S\sqrt{\tau}/\sqrt{2\pi} \approx 0.4\,\sigma S\sqrt{\tau}$. Traders use this constantly: an ATM straddle costs about $0.8\,\sigma S\sqrt{\tau}$, and reading implied volatility off a quoted ATM price is just inverting this formula. The square-root-of-time scaling is why a one-month option costs roughly half a four-month option, not a quarter.