Volatilities Add Like Arrows, Not Like Numbers

Two stocks, each swinging 20% a year, and they move half together. Price a European call on their product, so the payoff is $\max(S_1 S_2 - X, 0)$. The pricing turns out to be Black-Scholes with the product in place of the spot, which means the only genuinely new number the formula needs is one volatility. What is it?

It is $20\sqrt3 = 34.641\%$, and the two natural wrong answers are 40% and 28.284%. Both are off by more than a tenth of the option’s value, and they miss in opposite directions.

The reflex is the answer to a different question

Forty percent. Two twenties, add them. It is worth noticing that this is not a random guess: it is the correct answer under a specific assumption, and finding out which one is more useful than being told it is wrong. Solve $\sigma_1^2 + \sigma_2^2 + 2\rho\sigma_1\sigma_2 = 0.16$ for the correlation and you get exactly 1. So whoever adds the volatilities has silently assumed that the two stocks move in perfect lockstep, and the question said they move half together.

Why the product is tractable at all

Products of random variables are usually unpleasant. This one is not, for one reason:

A sum of two jointly normal variables is normal, so the product of two correlated lognormals is itself lognormal. That single fact is what lets the ordinary formula run unmodified on $P = S_1 S_2$, and it also decides the volatility, because the variance of the log is now the variance of a sum:

Equation (2) is the law of cosines

Written with $\rho = \cos\theta$, equation (2) is $c^2 = a^2 + b^2 + 2ab\cos\theta$, which is the law of cosines for the third side of a parallelogram. So draw each volatility as an arrow of that length, set the angle between them so its cosine is the correlation, join them head to tail, and the composite volatility is the length of the arrow across.

The picture is not a mnemonic. Covariance is an inner product on centred random variables, standard deviation is the norm that inner product induces, and correlation really is the cosine of the angle between two vectors in that space. To keep that honest rather than decorative, the formula was checked against literal plane vectors at correlations of −0.9, −0.5, 0, 0.25, 0.5, 0.75 and 1, adding them componentwise and measuring the length. The two agree to nine decimals at every one.

At $\rho = \tfrac12 = \cos 60^{\circ}$, two 20% arrows sit at 60 degrees and the arrow across has length $20\sqrt3 = 34.641\%$.

The whole range, and where the reflex is right

Sweeping the correlation shows how much room there is. At $\rho = 1$ the arrows point the same way and the lengths simply add, giving 40%, which is the one point on the axis where the reflex is correct. At $\rho = 0$ they are perpendicular and Pythagoras gives $\sqrt{0.08} = 28.284\%$. At $\rho = -0.9$ they nearly oppose and the composite drops to 8.944%. And at $\rho = -1$ with equal volatilities the arrows cancel exactly, which says something startling and true: the product of two perfectly anti-correlated lognormals with matching volatilities is deterministic.

The dividend that comes out negative

The composite volatility is not the only thing that changes when the product takes the spot price’s place, and the second change looks at first like a mistake. Black-Scholes on an asset paying a continuous dividend yield needs that yield, and for the product it is

With no dividends on either stock and a zero rate, equation (3) gives $q = -\rho\sigma_1\sigma_2 = -2\%$. A negative dividend yield, on an asset paying nothing.

It is the drift correction rather than an error. What makes the discounted product a martingale is not the same as what makes each stock one, because the product’s risk-neutral mean grows faster than either factor:

At a zero rate and $\rho\sigma_1\sigma_2 = 2\%$, the expected product grows at 2% a year even though neither stock has any expected growth at all. The covariance between the two logs pushes the mean of the product above the product of the means, and equation (3) subtracts that back out so the formula sees a properly drifting asset. Over 2,000,000 simulated paths the mean product came out at 102.0164 against the 102.0201 equation (4) predicts.

What each wrong volatility costs

Put a strike of 100 on a product that starts at 100, one year, both stocks at 20%, correlation a half, and a zero rate. The closed form gives 14.9228, and a 2,000,000-path simulation of the two correlated stocks gives 14.9198 with a standard error of 0.018421, which is agreement inside four standard errors.

Now feed it the two wrong volatilities. Adding them, at 40%, prices the call at 17.0419, which is 14.2% too high. Ignoring the correlation, at 28.284%, prices it at 12.3982, which is 16.9% too low. Neither error is a rounding difference, and because they point opposite ways there is no version of “close enough” that covers both.

Where the arrows stop being the maths

The trick is specific to products, and the reason is equation (1). A sum of two lognormals is not lognormal, so none of this transfers to a basket option on $S_1 + S_2$, which has no closed form of this shape at all. Anyone who takes “volatilities add like vectors” as a general rule will reach for it on a basket and get a number with nothing behind it.

Ratios do work, with one sign flipped. Since $\ln(S_1/S_2) = \ln S_1 - \ln S_2$, the composite volatility becomes $\sqrt{\sigma_1^2 + \sigma_2^2 - 2\rho\sigma_1\sigma_2}$, which is the other diagonal of the same parallelogram. At these parameters that is exactly 20%, so the ratio of the two stocks swings as much as either stock alone. That is the volatility sitting inside the standard formula for an option to exchange one asset for another.

Two assumptions hold the geometry up. The arrows have fixed lengths and a fixed angle, so a volatility that moves is not a length and a correlation that moves is not an angle. Once either is stochastic, equation (2) is at best an average of something, and the picture stops being exact.

A last confusion worth heading off: this is not a quanto. A quanto converts a payoff denominated in another unit at a rate agreed in advance, which puts the correlation into the drift and leaves the payoff’s own volatility alone. Here the payoff is already in the product’s own units and the correlation lands in the volatility instead. Both setups involve two assets and one correlation, which is exactly why they get swapped. The cheapest guard against all of it is the triangle inequality: $|\sigma_1 - \sigma_2| \le \sigma' \le \sigma_1 + \sigma_2$, with equality only at $\rho = -1$ and $\rho = +1$. A composite volatility outside that band is a bug, and you can check it in your head.

Sources and further reading

• The identity equation (2) turns out to be Law of cosines
• Why the cross term is there in the first place Covariance
• The family the product stays inside, and the sum does not Log-normal distribution
• Where the difference diagonal shows up as a published formula Margrabe’s formula
• The neighbouring contract this one is not Quanto

The composite volatility was measured from the standard deviation of the simulated log product, by a path generator that never computes equation (2), and separately reproduced as plane geometry at seven correlations. The negative dividend was checked against the simulated mean product rather than argued for, and the two wrong volatilities were priced so their errors are measurements.