Twenty Daily Variances Do Not Make a Monthly One, and the 18 Percent That Explains It

Daily, weekly and monthly returns give per-day variance estimates of 1.0000, 1.3225 and 1.4000, and the reflex is to average them into 1.2408, a figure no horizon produced. The variance ratio is a weighted sum of autocorrelations, so a forty percent overshoot at twenty periods measures dependence rather than noise, and the coefficient that reproduces it is 0.17554. With twenty years of daily data that ratio sits 4.6 standard errors above one and with five years only 2.3, which is why the number means nothing without the sample size attached.

You have a price series and you want a volatility for it. Estimating from daily returns gives one number. Estimating from weekly returns gives a bigger one. Estimating from monthly returns gives a bigger one still, and the monthly variance lands forty percent above twenty daily variances. Nothing is wrong with the arithmetic. Now go price an option.

The temptation is to treat the disagreement as measurement noise and split the difference. It is not noise. It is a measurement, and what it measures is that the returns are not independent of each other.

Why the three numbers should have agreed

Write $r_1, r_2, \dots$ for successive one-period returns and suppose only that they are stationary with common variance $\sigma^2$ . The variance of a $q$ -period return is the variance of their sum, and a variance of a sum is never just a sum of variances unless the cross terms vanish:

\operatorname{Var}\!\Bigl(\sum_{i=1}^{q} r_i\Bigr) = q\,\sigma^2 + 2\sum_{k=1}^{q-1} (q-k)\,\gamma_k, \qquad \gamma_k = \operatorname{Cov}(r_i, r_{i+k})

(1)

A random walk sets every $\gamma_k$ to zero, which leaves $q\sigma^2$ . That is where the square root of time comes from, and it is why twenty daily variances are supposed to make exactly one monthly variance. Divide (1) by $q\sigma^2$ and the whole comparison collapses into one dimensionless number.

Variance ratio

The variance of a $q$ -period return divided by $q$ times the variance of a one-period return, $\mathrm{VR}(q) = \operatorname{Var}(r_1 + \cdots + r_q)/(q\sigma^2)$ . It equals one at every horizon under a random walk, and reading (1) gives it in terms of the autocorrelations $\rho_k = \gamma_k/\sigma^2$ alone.

\mathrm{VR}(q) = 1 + 2\sum_{k=1}^{q-1}\Bigl(1 - \frac{k}{q}\Bigr)\rho_k

(2)

Two periods is the case worth memorising, because the weight collapses: $\mathrm{VR}(2) = 1 + \rho_1$ exactly. The ratio is not a proxy for serial correlation, it is a weighted sum of it.

The average of three incompatible numbers

Quote all three estimates in units of the daily variance and they read 1.0000, 1.3225 and 1.4000. The reflex is to average them, get 1.2408, and price off that. Look at what the average is. It is not the daily figure, not the weekly figure and not the monthly figure. It misses the nearest of the three by 0.0817, and because it sits strictly inside the range there is no horizon at all whose returns have that variance.

Fig. 1 — The three estimates and their mean. The dashed line is the number the reflex prices off, and it belongs to no horizon.

Averaging is the right move when three instruments measure one quantity with independent error. Here they measure three different quantities. The spread between them carries the information, so an average throws away exactly the thing the data was trying to say.

What a forty percent overshoot measures

Read (2) backwards. If a single autoregressive coefficient generated the dependence, so that $\rho_k = \rho^k$ , then one number fixes the whole term structure of variance ratios. Solving $\mathrm{VR}(20) = 1.40$ for that coefficient gives

\rho = 0.17554 \quad \Longrightarrow \quad \mathrm{VR}(2) = 1.1755,\; \mathrm{VR}(5) = 1.3225,\; \mathrm{VR}(10) = 1.3742

(3)

About eighteen percent of yesterday's move survives into today. Notice that the weekly ratio in (3) is not an extra assumption, it is a prediction, and it comes out at the 1.3225 that was measured. One coefficient reproduces the whole pattern.

Fig. 2 — Total variance against horizon. The dashed line is the walk; the measured points curve away from it and finish forty percent high.

A closed form deserves a check that does not use it. Simulating four million steps of a series with that coefficient and block-summing the returns returns a lag-one correlation of 0.1753 and variance ratios of 1.175, 1.322 and 1.400 at two, five and twenty periods, against a closed form of 1.176, 1.323 and 1.400.

How much of forty percent is sampling error

The exercise hands you 1.40 as a fact, and a real desk has to ask how many standard errors it is worth. Under independence the non-overlapping estimator of $\mathrm{VR}(q)$ has an asymptotic standard error that a simulation confirms to two digits:

\mathrm{s.e.}\bigl(\widehat{\mathrm{VR}}(q)\bigr) \approx \sqrt{\frac{2(q-1)}{n}}, \qquad q = 20

(4)

With twenty years of daily data, $n = 5000$ , that is 0.087 against a measured 0.0871, so a ratio of 1.40 sits about 4.6 standard errors above one and the dependence is real. Cut the sample to five years and the standard error triples to 0.17, which puts the same 1.40 at 2.3 standard errors, where a careful person says the walk is doubtful rather than dead. The number 1.40 means nothing without the sample size attached to it.

What the overshoot does not prove

Given stationarity, a ratio away from one is exactly equivalent to a non-zero weighted sum of autocovariances, and that equivalence runs in both directions. A ratio of one at every horizon forces every autocovariance to zero, and any non-zero weighted sum shows up as a ratio away from one. So the deduction that the moves are not independent needs no model at all.

The eighteen percent is a different kind of statement. It is one calibration of the fact, and any other dependence structure with the same weighted sum in (2) would produce the same 1.40. The specific coefficient is a summary, not an identification.

Stationarity is doing real work too. Variance that changes over time can push the ratio away from one with no serial correlation anywhere, which is why the honest conclusion is that the walk is out rather than that the returns are autoregressive. What does survive is the direction: the same simulation with a coefficient of $-0.20$ returns a twenty-period ratio of 0.678, below one, and an independent series returns 0.999 and 1.001 at five and twenty periods. The sign of the departure tells you which way the memory runs.

So what do you price the option off

Not the average. The option formula is less fragile than it looks here, because what it needs is that the accumulated variance over the life of the contract match the variance you priced with. It does not need the increments to be independent for that one purpose. Estimate the swing over the horizon you are actually pricing, and a one-month option gets the 1.40 figure rather than the daily one.

The part that genuinely breaks is the hedge. The derivation of the formula from replication needs the underlying to be a diffusion, and a series with memory is not one. So a month-horizon volatility gives you a defensible price for a European payoff while the delta that comes with it no longer replicates, and anything path dependent is outside the model entirely. Keep the shape of the formula, keep the horizon honest, and stop claiming the hedging argument.

Sources and further reading

The hypothesis that (1) tests — Random walk hypothesis
The quantity that (2) is built out of — Autocorrelation
The one-coefficient model used in (3) — Autoregressive model
The alternative explanation the last section refuses to rule out — Heteroscedasticity
The formula whose assumptions are at stake — Black-Scholes model

The closed form in (2) was checked against the trace of the full autocovariance matrix at four horizons and agreed to nine decimals, the coefficient in (3) was recovered from a four-million-step simulation, and the standard error in (4) was measured over thousands of independent samples rather than taken from the asymptotics.