A Ten Dollar Swing Prices a Four Dollar Call

An at-the-money call with a zero interest rate is worth one over the root of two pi, which is 0.39894, times the absolute swing of the terminal price, so a 10 dollar standard deviation prices it at 3.9894 and the closest of the offered 1, 5 and 10 is 5. The two-step estimate is exact under a symmetric terminal law, where the option finishes above the strike exactly half the time and the average gain when it does is 7.979. Calibrate a lognormal to the same 10 dollar swing and those two factors become 0.4801 and 8.285, whose product is still 3.98, which is why the qualifier about half cannot be cut.

A stock trades at 100. The interest rate is zero. Over the coming year the standard deviation of its price at expiry is 10 dollars. What is a one-year call struck at 100 worth: closer to 1 dollar, 5, or 10?

The value is 3.99, so the closest of the three is 5. The number to remember is the ratio: an at-the-money call with no interest rate is four tenths of the absolute swing, whatever the swing happens to be.

Why ten is the answer people give

Ten is the number printed in the problem, and under pressure the standard deviation gets read as a price. It is not one. The standard deviation describes how far the stock is likely to travel in either direction; the call only collects the upward half, and only the part of it above the strike. Quoting 10 is more than six dollars wrong on a four dollar option, which is the kind of error that ends an interview rather than costing a mark.

There is a second route that produces the right multiple-choice letter for the wrong reason: roughly half the time the option finishes in the money, and the average gain is about 10, so half of 10 is 5. That lands on the offered answer and gets the second factor wrong. The average gain when the option pays is not 10. It is 8.

Two steps, both of which can be checked

Take the terminal price to be centred on 100 and symmetric, and write $X$ for the terminal price minus the strike, so $X$ is centred at zero with standard deviation $\sigma = 10$ . With a zero rate there is no discounting and no drift, so the call is worth the expected value of what it pays:

C = \mathbb{E}\big[X^{+}\big] = \mathbb{P}(X > 0)\cdot \mathbb{E}\big[X \mid X > 0\big]

(1)

The first factor is a half, by symmetry. The second is the mean of a half-normal, which is $\sigma\sqrt{2/\pi} = 7.9788$ , the eight dollars from a moment ago. Multiplying:

C = \frac{1}{2}\cdot \sigma\sqrt{\frac{2}{\pi}} = \frac{\sigma}{\sqrt{2\pi}} = 0.39894\,\sigma

(2)

The decomposition in (1) is not an approximation here. It is an identity, and the two factors multiply to exactly $\sigma/\sqrt{2\pi}$ . At $\sigma = 10$ that is 3.9894 dollars. The direct route confirms it in one line, integrating the payoff against the density:

\int_0^{\infty} \frac{x}{\sigma\sqrt{2\pi}}\,e^{-x^2/2\sigma^2}\,dx = \frac{1}{\sigma\sqrt{2\pi}}\left[-\sigma^2 e^{-x^2/2\sigma^2}\right]_0^{\infty} = \frac{\sigma}{\sqrt{2\pi}}

(3)

Fig. 1 — The two factors in (1), drawn. Half the probability sits to the right of the strike, and the centre of mass of that half is 7.98 dollars away from it.

Half-normal mean

If $X$ is normal with mean 0 and standard deviation $\sigma$ , then $|X|$ has mean $\sigma\sqrt{2/\pi} \approx 0.798\,\sigma$ . Since the distribution is symmetric, that is also the mean of $X$ conditional on $X > 0$ , and the expected positive part is half of it. The constant $1/\sqrt{2\pi} = 0.39894$ is the same number that sits in front of the normal density, which is why it is worth memorising once.

The rule scales, and survives the model

Equation (2) is linear in $\sigma$ , so nothing about 10 was special. A 20 dollar swing gives 7.98, a 40 dollar swing gives 15.96, and the ratio to the swing is 0.39894 in every case. That linearity is the reason the rule is usable at all: you do not need the option, only the swing.

The symmetric terminal law used above is not the market convention, so the honest test is whether the answer moves when the model does. Calibrate a lognormal terminal price to the same 10 dollar standard deviation. The log volatility is $\sqrt{\ln(1 + (10/100)^2)} = \sqrt{\ln 1.01} = 0.099751$ , and the zero-rate at-the-money value has a compact closed form:

C = S\left(2N\!\left(\tfrac{\sigma_{\log}}{2}\right) - 1\right) = 100\left(2N(0.049876) - 1\right) = 3.9779

(4)

Just over one cent from the symmetric answer. Both were also reproduced by four-million-draw simulations with fixed seeds, agreeing with their closed forms inside three standard errors, and the simulated terminal standard deviation of the lognormal came back at 10.00 as designed.

Fig. 2 — The four-tenths rule under both models. The agreement is within a cent at a 10 dollar swing and drifts to 68 cents at 40.

The two averages are not the same two averages

Something interesting happens when you look inside (1) under lognormality. The probability of finishing above the strike is no longer a half. With a zero rate the expected terminal price is $S$ , so the median is $Se^{-\sigma_{\log}^2/2}$ , which sits below the strike. The probability of finishing in the money is $N(-\sigma_{\log}/2) = 0.4801$ .

The other factor moves in the opposite direction. The average gain when the option pays rises from 7.98 to 8.285, because a lognormal has a longer right tail. Their product is 3.98 either way, which is why the qualifier in about half the time is load-bearing rather than decorative. Deleting the word about turns a true sentence into a false one while leaving the final number untouched, and that is the sort of thing an interviewer listens for.

Where the four-tenths rule stops working

The constant belongs to the money, not to options in general. Move the strike and the value falls away fast. For a strike $k$ dollars above the centre, the symmetric-law value is:

\mathbb{E}\big[(X - k)^{+}\big] = \sigma\,\varphi\!\left(\frac{k}{\sigma}\right) - k\left(1 - N\!\left(\frac{k}{\sigma}\right)\right)

(5)

At $k = 0$ the second term vanishes and (5) collapses to (2). At $k = 5$ with the same 10 dollar swing it gives 1.978, so half a standard deviation of moneyness has halved the option. Quoting 0.4 times the swing for a strike that is not at the money is a bigger error than the mistake this problem is designed to catch.

The two models also stop agreeing once the swing is large relative to the spot. At a 40 dollar swing on a 100 dollar stock the gap is 68 cents rather than one cent, and the symmetric law has started to misbehave in its own right: it assigns probability $N(-2.5) = 0.62\%$ to a negative stock price. At a 10 dollar swing that same event is ten standard deviations out and costs nothing to ignore.

Two smaller conditions round it off. The zero rate matters, because with a positive rate the relevant centre is the forward $Se^{rT}$ rather than the spot, and a call struck at the spot is then in the money against the forward. And the swing in the rule is the standard deviation of the terminal price in dollars, not an annualised percentage volatility; at $S = 100$ over one year the two look interchangeable, which is exactly why it is worth writing down which one you meant.

Sources and further reading

The distribution behind the second factor in (1) — Half-normal distribution
Where the constant 0.39894 comes from — Normal distribution
The model behind the arithmetic in (2), with prices moving in dollars rather than percentages — Bachelier model
The lognormal cross-check in (4) — Black-Scholes model
The general shape of (5) — Truncated normal distribution

Both models were priced twice, once in closed form and once by a four-million-draw seeded simulation, at swings of 5, 10, 20 and 40 dollars. The probability of finishing in the money and the average gain given that it does were measured separately under each law, because they differ: exactly one half and 7.979 under symmetry, 0.4801 and 8.285 under lognormality, with the product landing on 3.98 in both cases.