Forty-Two Dollars in Six Months, and the Quarter Million That Is Not There

Gold trades at 400. The interest rate is zero. Its price wanders with a standard deviation of 60 dollars over a year, quoted in dollars rather than in percent, so the model is an arithmetic random walk. A ticket pays one million dollars if gold is above 430 in six months. What is the ticket worth?

Two things have to be right, and the second one is where most published solutions go wrong, including the one this problem is usually copied from.

Six months is not half the wander

The reflex is to halve the annual figure and work with 30 dollars. That treats uncertainty as something that accumulates at a steady rate, and it does not. For a process with independent increments the variances add, so over a period $T$ the variance is $\sigma^2 T$ and the standard deviation is the square root of that.

Not 30. The difference is 41 percent, and it goes the same way at every horizon: three months carries 30 dollars, one month carries $60/\sqrt{12} = 17.3$, two years carries 84.9. Cutting the horizon in four is what halves the spread.

You can rebuild equation (1) from the bottom if the identity feels like a slogan. Split six months into 126 trading days. Each day carries variance $3600 \times (0.5/126) = 14.286$, the 126 of them sum to 1800, and $\sqrt{1800} = 42.4264$. The same construction with 2, 6 or 26 sub-periods gives the same answer to nine decimals, which is what "variance is additive" means in practice.

From the spread to the price

With the rate at zero there is no drift to add and no discount factor to apply, so the ticket is worth its notional times the probability of finishing above 430. Standardise the threshold:

That the answer comes out as $1/\sqrt2$ exactly is an accident of the numbers chosen, and a pleasant one, because it means the whole problem reduces to one tail of the standard normal at a memorable point.

About 240,000 dollars. The halved-volatility route puts 430 exactly one standard deviation up, giving $1 - \Phi(1) = 0.15866$ and a price of 158,655, which is 34 percent light.

The quarter of a million is wrong, and it is instructive

This exercise circulates with the answer "about 250,000". It is not 250,000, and the way that figure is reached is worth pulling apart, because the same move produces wrong answers in problems that have nothing to do with gold.

The published chain rounds the $z$ score of 0.7071 up to 0.75. That step is defensible in a room with no table to hand. Then it reads the upper tail at 0.75 as 0.25, and that step is a slip: $\Phi(0.75) = 0.773373$, so the tail there is 0.226627. Somewhere between the two lines, 0.75 stopped being an argument to $\Phi$ and became a probability.

Notice the direction. Rounding $z$ from 0.7071 up to 0.75 moves the threshold further out, so the tail must get smaller, from 0.2398 down to 0.2266. The quoted 0.25 is larger. Two roundings compounding in the same direction would have been forgivable; an approximation that moves the answer the wrong way is a sign that a symbol has changed meaning mid-calculation, and it is worth stopping for.

The honest verbal answer in a room is "a bit under a quarter of a million". The number is 239,750, and a Monte Carlo of 200,000 paths lands at 0.2388, which excludes both 0.1587 and 0.25.

The general contract

Nothing above depended on the particular numbers. For an arithmetic random walk started at $S$ with annual volatility $\sigma$ in price units, a rate $r$, and a digital paying $Q$ above a threshold $K$ at time $T$:

At $r = 0$ this collapses to equation (3). Two features of equation (4) are worth reading off directly. The value depends on $\sigma$ and $T$ only through the product $\sigma\sqrt T$, so a contract with double the volatility and a quarter of the life is worth exactly the same. And at $K = S$ with $r = 0$ the value is $Q/2$ whatever the volatility, since a symmetric walk finishes above its start half the time.

Where the model stops describing anything

An arithmetic walk allows the price to go negative, which for gold is nonsense. The question is whether it matters here, and it does not: a fall from 400 to zero over six months is $400/42.43 = 9.4$ standard deviations, a probability around $3 \times 10^{-21}$. At longer horizons or higher volatility that defence stops working, and a multiplicative model becomes necessary rather than optional. Under a lognormal walk with the same six-month spread, the tail past 430 is slightly thinner, because the density has been skewed to the right.

The bigger practical caveat is not about the density at all. Equation (4) prices a discontinuous payoff, and a discontinuity is hard to hedge. As expiry approaches with the price sitting near 430, the sensitivity of equation (4) to a one-dollar move grows without bound, so a position that looks like a modest bet on a probability becomes a very large bet on a very small price move. The number is exact. The instruction it gives you about what to do is not.

Sources and further reading

• Why variances add over disjoint intervals — Wiener process
• The scaling in equation (1) — Volatility (finance)
• The function that is worth 0.773373 at 0.75 — Normal distribution
• The exact form used in equation (3) — Error function

The value 239,750 was confirmed three ways before publication: by an exact error-function evaluation, by rebuilding the six-month spread out of 2, 6, 26 and 126 independent sub-periods, and by a seeded simulation of 200,000 paths. No step in the verification code ever divides 60 by 2, so the halved-volatility answer had no route in.