43 Percent of the Variance Survives One Reversion Time, and the Model Does Not

Give a pulled-back log price the same 20 percent instantaneous swing as a free-wandering one and its horizon variance stops being sigma squared times T: at one reversion time only 0.432332 of it survives, the volatility that prices a one-year call is 13.1504 percent, and the call falls from 7.9656 to 5.2425. The same pull makes consecutive returns fight each other, with a first-order autocorrelation of exactly minus half of one minus phi, and that is the independence the pricing model rests on. The formula still returns the right European price and has lost the hedging argument that justified it.

Two shares are built from the same shocks and wobble by the same 20 percent a year. One wanders freely. The other is pulled back towards a level whenever it strays. Which one carries the more expensive one-year call?

The pulled one is cheaper, and by a third rather than by a rounding. That is only half the answer. The other half is that the pull also destroys the assumption the pricing formula was derived from, so the candidate who says cheaper and stops has answered the easy question.

Same wobble, different accumulation

An option cares about the spread of the terminal price, not about the size of a typical daily move. Those two are linked by the way variance accumulates, and the square root of time rule is a statement about accumulation rather than about volatility. Model the log price with a pull towards a level and the accumulation changes.

Ornstein-Uhlenbeck log price

$dx = -\kappa(x - m)\,dt + \sigma\,dW$ , with $x = \log S$ . The parameter $\sigma$ is the same instantaneous swing a free-wandering price would have, and $1/\kappa$ is the time the pull needs to undo a displacement. Setting $\kappa = 0$ recovers the free walk.

Solving that equation gives $x(T)$ as a Gaussian whose variance is not $\sigma^2 T$ :

\operatorname{Var}\bigl(x(T)\bigr) = \sigma^2\,\frac{1 - e^{-2\kappa T}}{2\kappa} \qquad\text{against}\qquad \sigma^2 T \ \ \text{for the free walk}

(1)

Divide one by the other and everything collapses onto a single dimensionless argument $u = \kappa T$ , the number of pull-back times in the horizon:

\mathrm{VR}(u) = \frac{1 - e^{-2u}}{2u} = 1 - u + \tfrac{2}{3}u^2 - \cdots

(2)

The expansion says the discount starts linearly, so a horizon a tenth of a reversion time loses about ten percent of its variance. The function is strictly below one for every $u > 0$ and strictly decreasing, so a pull can only shrink the horizon variance. It never grows it.

Fig. 1 — Equation (2). One reversion time in the horizon costs more than half the variance, and the dashed line is the 1 − u opening of the expansion.

One reversion time, and 34 percent off

Take a horizon of exactly one reversion time, $\kappa T = 1$ , with a measured swing of 20 percent, a spot of 100, a strike of 100 and rates at zero. Then

\mathrm{VR}(1) = \frac{1 - e^{-2}}{2} = 0.432332, \qquad \sigma_{\text{eff}} = \sqrt{0.432332}\times 20\% = 0.657520 \times 20\% = 13.1504\%

(3)

Price the two calls at their own volatilities and the free-wandering one is 7.9656 while the pulled one is 5.2425, a ratio of 0.6581 and a discount of 34.19 percent. The Gaussian quadrature of the terminal law agrees with both to eight decimals, and a crude Euler simulation of the path itself, which never touches the closed form in (1), lands inside three standard errors of the price.

The direction of that answer is parameter-free. A call's vega is $S\sqrt{T}\,\varphi(d_1)$ , strictly positive, so value is strictly increasing in volatility and any reduction in the effective volatility is a strictly cheaper option. Only the size of the discount depends on the numbers.

Fig. 2 — The same instantaneous swing in every row. Only the accumulation differs, and the price falls monotonically with the pull.

The half the reflex never reaches

Sample the log price at spacing $h$ . The solution of the process is exactly a stationary first-order autoregression with coefficient $\phi = e^{-\kappa h}$ , and the returns are its increments. Compute the first-order autocorrelation of those increments and the algebra is clean:

\rho_1 = \frac{-V(1-\phi)^2}{2V(1-\phi)} = -\frac{1-\phi}{2} \;<\; 0 \quad \text{for every } \kappa > 0

(4)

Every pull, however gentle, makes consecutive returns fight each other. At a reversion speed of one per year the predicted figure is $-0.039978$ on monthly returns, $-0.009524$ weekly and $-0.001980$ daily, and a 400,000-step stationary path measures $-0.0391$ , $-0.0087$ and $-0.0011$ . A free walk would give exactly zero at every spacing, and zero is what the pricing model assumes.

So the two halves of the answer come from the same fact seen twice. Moves that undo each other stop piling up, which is the discount in (3), and moves that undo each other are correlated, which is the assumption in (4).

The formula still fits the number and still does not apply

Here is the subtlety worth having, because it is where people who have understood everything above still overshoot. The terminal law of a pulled log price is Gaussian, so the terminal price is lognormal, and a European call on a lognormal terminal price is given exactly by the Black-Scholes formula at the volatility that reproduces the terminal variance. The 5.2425 above is not an approximation of the right answer. It is the right answer, produced by the formula whose assumptions have just been broken.

What has actually been lost is the hedging argument. Over the life of the contract the pulled path accumulates quadratic variation $\sigma^2 T = 0.04$ , while a walk priced at $\sigma_{\text{eff}}$ would accumulate $0.0173$ . A hedger who trades the delta that comes with the 13.15 percent number is rebalancing against a process with more than twice the local variance he is being paid for, so he does not break even and the replication that justifies the price is gone. Anything whose payoff depends on the path rather than on the terminal value, a barrier or an average, is not given by the formula at any volatility at all.

Where cheaper stops being the answer

The discount is a long-horizon statement. As $\kappa T \to 0$ the ratio in (2) tends to one, so a one-day option on a pulled share and a one-day option on a wandering share have essentially the same price. Nothing in this argument says anything about short-dated options, and the sweep bears that out: the surviving share of variance is 0.990 at $\kappa T = 0.01$ and 0.0625 at $\kappa T = 8$ .

The comparison also has a fixed input that is easy to lose. Both shares were given the same instantaneous $\sigma$ , which is what makes cheaper the correct answer. If instead you measure the realised annual volatility of the pulled series and put that number into the formula, you get 13.15 percent straight out of the data and price the European option correctly with no reversion model at all. The mistake being punished is not using the formula, it is annualising a daily volatility by the square root of time when the series has memory.

Finally, the sign of $\kappa$ is doing more work than it looks. Equation (2) is defined and decreasing for negative arguments too, so a share with momentum rather than reversion has $\mathrm{VR} > 1$ and dearer options: at $\kappa T = -0.25$ the ratio is 1.2974. Reversion and momentum are the same formula read on opposite sides of zero, and only one of them makes options cheap.

Sources and further reading

The process in the definition box — Ornstein-Uhlenbeck process
The statistic in (2), under its econometric name — Mean reversion (finance)
The sampled process behind (4) — Autoregressive model
The formula that fits the number and loses its derivation — Black-Scholes model
The quantity the hedger is actually exposed to — Quadratic variation

The terminal variance in (1) was confirmed against a seeded Euler simulation that never uses the closed form, both prices were reproduced by Gaussian quadrature of the terminal law, the autocorrelations in (4) were measured on a 400,000-step stationary path at three sampling frequencies, and the monotonicity of (2) was checked across eight values of the pull.