Pricing an Option in Your Head, and the 0.4 Nobody Explains

A three-month at-the-money call on a stock at 100 with 40% volatility is worth about eight dollars, and you can get there in two multiplications. The constant four tenths turns out to be the height of the normal bell at its peak, and the whole error of the mental rule is one rounding plus one cubic term. Scaling volatility linearly with time instead of with its square root gives ten dollars, which is 25.5% too high.

A stock trades at 100. Someone asks what a three-month call struck at 100 is worth, says volatility is 40% a year and rates are zero, and then starts counting to ten. There is no way to evaluate the closed form in your head, and nobody is asking you to. What they want is the number 8, and they want to hear the route.

The route has two ingredients. One is a scaling law that most people get backwards under pressure. The other is a constant, four tenths, that looks arbitrary until you see where it comes from. Both are worth carrying around, because together they price any at-the-money option to better than half a percent.

The ten seconds, spent

Three months is a quarter of a year. The square root of a quarter is a half, so 40% a year becomes 20% over the quarter. On a stock at 100 that is a twenty dollar move, one standard deviation of where the price ends up. The call is worth four tenths of that move, so eight dollars.

C_{\mathrm{ATM}} \;\approx\; 0.4\,\sigma S\sqrt{T}

(1)

Black-Scholes with these inputs gives 7.965567. The mental answer of 8 is high by 0.432%, which is three and a half cents on an eight dollar option. Ten seconds is generous.

Why the square root, and what happens if you forget it

The mistake this question is built to catch is scaling volatility with time instead of with its square root. A quarter of the year, a quarter of the swing: 40% times a quarter is 10%, so ten dollars. The arithmetic is exact and the answer is 25.5% too high.

Variance, not standard deviation, is what adds up over independent increments. If the log price accumulates independent shocks of variance $\sigma^2$ per year, then over a horizon $T$ the total variance is $\sigma^2 T$ and the standard deviation is its square root.

\operatorname{Var}\!\big[\ln S_T\big] = \sigma^2 T \quad\Longrightarrow\quad \operatorname{sd}\!\big[\ln S_T\big] = \sigma\sqrt{T}

(2)

Ten dollars is not a slow route to eight. It is the answer to a question about a process whose increments reinforce each other perfectly, which no traded stock does. The two laws agree at one year, where $\sqrt{T} = T = 1$ , and they disagree everywhere else, with the gap widest at short horizons. That is unfortunate, because short horizons are where the quoting happens.

Fig. 1 — Both laws pass through zero and through 40% at one year. Between them they part company, and the three-month reading is 20% against 10%.

Where the four tenths comes from

Start from the formula and let it collapse. Writing $v = \sigma\sqrt{T}$ for the total volatility to expiry, the Black-Scholes call with zero rates and strike equal to spot has

d_1 = \frac{\ln(S/K) + \tfrac12\sigma^2 T}{\sigma\sqrt T} = \frac{v}{2}, \qquad d_2 = d_1 - v = -\frac{v}{2}

(3)

The two arguments are equal and opposite. With $K = S$ and no discounting, the price collapses onto a single term:

C = S\big[\Phi(d_1) - \Phi(d_2)\big] = S\big[2\Phi(v/2) - 1\big] = S\,\operatorname{erf}\!\left(\frac{v}{2\sqrt2}\right)

(4)

Now expand. Near zero the normal distribution function is almost a straight line, with slope equal to the density at the origin, and that density is $1/\sqrt{2\pi}$ . So $\Phi(x) \approx \tfrac12 + x/\sqrt{2\pi}$ , the two halves cancel, and equation (4) becomes

C \;\approx\; \frac{S\,v}{\sqrt{2\pi}} \;=\; 0.39894\,\sigma S\sqrt{T}

(5)

There is the constant. It is not fitted to this example and it does not depend on the inputs. It is the height of the standard normal bell at its peak, rounded up by a fraction of a percent so that it can be multiplied in your head. Anyone who tells you the coefficient is a half has confused it with the value of $\Phi(0)$ .

How wrong the estimate is, and in which direction

Two separate approximations sit between equation (1) and the truth, and they both push the same way. The first is the rounding of 0.39894 up to 0.4, which costs a flat 0.265% no matter what the inputs are. The second is the truncation. Taking one more term of the error function, $\operatorname{erf}(x) = \tfrac{2}{\sqrt\pi}\big(x - x^3/3 + \dots\big)$ , gives a cubic correction:

C = \frac{S\,v}{\sqrt{2\pi}}\left(1 - \frac{v^2}{24} + O(v^4)\right)

(6)

The correction is negative, so the linear rule always sits above the true price. At $v = 0.2$ it is $0.04/24 = 1/600$ , or 0.167%. Add the 0.265% from the rounding and you get 0.432%, which is exactly the discrepancy between 8 and 7.965567. The error is fully accounted for, which means you can say out loud that your estimate is high and roughly by how much.

Equation (6) also gives a sharper rule for anyone willing to spend two more seconds, provided you keep the exact constant rather than the rounded one. The refined estimate $\tfrac{\sigma S\sqrt{T}}{\sqrt{2\pi}}\,\big(1 - \sigma^2 T/24\big)$ returns 7.96555 against a true 7.96557, agreeing to five figures. That is a curiosity rather than a working tool, but it does confirm that nothing else is going on: the whole error of the mental rule is one rounding plus one cubic term.

Fig. 2 — The rule overstates the price at every horizon, and the overstatement is cubic. It stays under 1% while total volatility stays under about 0.42.

What happens when rates are not zero

The derivation above used $r = 0$ twice, once to drop the discount factor and once to make the strike coincide with the forward. Only the second use matters, and it is repairable.

At the money, forward

A call is at the money forward when its strike equals the forward price $F = S e^{rT}$ rather than the spot. With $r = 0$ the two coincide, which is why the distinction never appears in the quick version.

Price a forward-struck call with the Black-76 form, $C = e^{-rT}[F\Phi(d_1) - K\Phi(d_2)]$ . Setting $K = F$ makes $d_1 = v/2$ and $d_2 = -v/2$ again, so

C = e^{-rT} F\big[2\Phi(v/2) - 1\big] = S\big[2\Phi(v/2) - 1\big]

(7)

because $e^{-rT}F = S$ by definition of the forward. Equation (1) therefore prices a call struck at the forward for any interest rate at all, with the same accuracy it has at zero rates. The rate cancels, which is a pleasant surprise the ten-second version never gets to mention.

Struck at the spot, though, the story changes. Take the same inputs with $r = 5\%$ : the true price of the spot-struck call rises to 8.5526 while the rule still says 8, so it is now 6.5% low rather than 0.4% high. The direction of the error flips with a rate of five percent, which is worth knowing before you quote confidently. Put-call parity fills in the rest: at the forward strike the call and the put are worth the same, so the single number 8 prices both.

Where it stops working

Everything in equation (1) is controlled by the single quantity $v = \sigma\sqrt T$ , the total volatility to expiry. Neither $\sigma$ nor $T$ matters on its own. The relative error passes 1% at $v \approx 0.42$ , 2% at $v \approx 0.64$ , and 5% at $v \approx 1.06$ . A 40% name at three months has $v = 0.2$ and is comfortably inside all of that. The same name at four years has $v = 0.8$ , where the rule is 2.9% high. A 100% volatility name at one year has $v = 1$ and the rule is 4.5% high, which is no longer a rounding error.

Push further and the rule fails on principle rather than on precision. A call can never be worth more than the stock, because owning the stock dominates the payoff. But $0.4\,\sigma S\sqrt{T}$ grows without bound in $v$ , so at $v = 2.5$ it returns the entire spot price and past that it returns more. The true price, meanwhile, climbs towards $S$ and stops there, exactly as equation (4) says it must, since the error function is bounded by one. The linear approximation knows nothing about that ceiling.

The last restriction is the strike. Equation (5) is a first-order expansion around the forward, and it says nothing whatsoever about options away from it. Twenty percent out of the money, the price depends on the shape of the distribution rather than on its width, and the pleasant cancellation of $\Phi$ against $1 - \Phi$ is gone.

There is one last thing worth noticing about equation (5). If you price an at-the-money call on a stock that moves by absolute rather than proportional amounts, with normal instead of lognormal terminal prices and absolute volatility $\sigma S$ , the answer is $\sigma S \sqrt{T/2\pi}$ , which is equation (5) exactly. So the trader’s rule is not a crude fit to Black-Scholes. It is the exact price under the older arithmetic Brownian model Bachelier wrote down in 1900, and the two models agree to first order in total volatility. That agreement is the reason a constant from 1900 still prices a 2026 option book to half a percent.

Sources and further reading

The closed form and its assumptions — Black-Scholes model
Why volatility carries a square root of time — Volatility (finance)
The constant $1/\sqrt{2\pi}$ and the function behind equation (4) — Error function
The forward-strike version of the formula — Black model
The 1900 model that equation (5) prices exactly — Bachelier model
Why the same number prices the put — Put-call parity

Every price quoted here was computed three ways that share no machinery: the closed form through the error function, a Simpson quadrature of the payoff against the lognormal density with no option formula anywhere in it, and sixteen million antithetic Monte Carlo paths. All three agree on 7.9656.