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Autocorrelation: momentum, mean reversion, or coin flips

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Start with the idea

Autocorrelation asks whether past deviations line up with later deviations. The sign can suggest continuation or reversal at a chosen lag, but the same series can have different signs at different horizons.

Symbols, units & horizon

xₜ: observation at time t
x̄: full-sample mean
n: observation count
k: lag in periods
ρ_k: sample autocorrelation at lag k
φ: coefficient of zero-mean AR(1)
εₜ: unpredictable innovation
|φ|: absolute decay coefficient
ln: natural logarithm
h or half-life: number of periods for expected shock magnitude to halve

When and why to use this

Use ACF as a diagnostic for forecastable dependence and for invalid IID standard errors. Use estimated half-life to compare a reversion thesis with holding costs and execution latency.

The autocorrelation at lag $k$ is the correlation of a series with itself $k$ steps ago:

ρ_{k} = \frac{\sum_{t = k + 1}^{n} (x_{t} - \overline{x}) (x_{t - k} - \overline{x})}{\sum_{t = 1}^{n} (x_{t} - \overline{x})^{2}}

Algebra and arithmetic

Compute sample autocorrelation

Centre the series using its sample mean. Multiply each deviation by the deviation k observations earlier and sum over overlapping pairs.
Divide by the full squared-deviation sum. This is the biased ACF convention shown here; other normalisations differ in small samples. The approximate white-noise band scales as $\frac{1.96}{\sqrt{n}}$ .

Work it by hand

For x=(1,2,1,2), mean 1.5 and denominator 1. Lag-one products sum to −.75; estimated ρ₁=−.75.

ρ₁ of returns	behaviour	the trade
> 0	momentum — moves continue	buy strength, sell weakness
< 0	mean reversion — moves reverse	fade extremes
≈ 0	random walk	no timing edge at this horizon

Under the null of no autocorrelation, sample $ρ_{k}$ is approximately normal with standard deviation $\frac{1}{\sqrt{n}}$ . Bars outside $\pm \frac{1.96}{\sqrt{n}}$ are "significant". With 250 daily returns the band is $\pm 0.12$ ; a lag-1 autocorrelation of 0.05 in daily equity returns — which is about what exists — is invisible at that sample size and only becomes tradeable with thousands of observations or many assets.

ARIMA: the grammar of linear forecasting

AR(p) says today depends on the last $p$ values; MA(q) says today depends on the last $q$ shocks; I(d) says difference $d$ times first to make the series stationary. The AR(1) case is the workhorse:

x_{t} = ϕ x_{t - 1} + ε_{t}, half-life of a shock = \frac{\ln 0.5}{\ln | ϕ |}

Algebra and arithmetic

Solve an autoregressive shock half-life

Ignoring future mean-zero innovations, a deviation d becomes $ϕ^{h} d$ after h periods. Its magnitude halves when $| ϕ |^{h} = \frac{1}{2}$ .
Take logs: $h \ln | ϕ | = \ln (\frac{1}{2})$ , then divide. Require 0<|φ|<1. A negative φ alternates direction; φ=0 removes the conditional deviation in one step.

Work it by hand

φ=.8 gives h=ln(.5)/ln(.8)=3.106 periods. φ=.5 gives exactly one period.

$ϕ$ close to 1 means shocks persist (near random walk). $ϕ$ close to 0 means fast reversion. The half-life is the natural holding period for a mean-reversion trade on that series — it converts a statistical parameter into a decision.

A spread follows AR(1) with

ϕ = 0.9

on daily data. Half-life in days?

$\frac{\ln 0.5}{\ln 0.9} = \frac{- 0.693}{- 0.105} = 6.6$ . Expect to hold about a week for half the deviation to close; two weeks for three-quarters.

Python implementation

Self-contained teaching example. Python 3.10+; dependencies and input conventions are shown in the code and notation. Run in your own Python environment.

from statistics import mean
from math import log

def acf(values, lag):
    if not 0 <= lag < len(values):
        raise ValueError("Lag outside sample")
    centered = [x-mean(values) for x in values]
    denominator = sum(x*x for x in centered)
    if denominator == 0:
        raise ValueError("ACF undefined for constant series")
    return sum(centered[t]*centered[t-lag] for t in range(lag,len(values)))/denominator

def shock_half_life(phi):
    if not 0 < abs(phi) < 1:
        raise ValueError("Require 0 < |phi| < 1")
    return log(.5)/log(abs(phi))

print(shock_half_life(.8))

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