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Cointegration: a stationary relationship to test

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Start with the idea

Cointegration concerns a stationary combination of nonstationary levels. High return correlation alone does not establish it. A stationary fitted residual can motivate a convergence hypothesis, but the economics of two executable positions still need testing.

Symbols, units & horizon

xₜ, yₜ: aligned level series
β: estimated hedge coefficient
zₜ: residual spread yₜ−βxₜ
μ̂_z, σ̂_z: spread mean and positive SD estimated on training data
score: centred spread measured in training standard deviations
Stationarity: a modelling hypothesis requiring testing, not an entry command

When and why to use this

Use a residual model for related assets or cash-versus-derivative research. Check hedge stability, both-leg costs and whether a structural event can permanently change the relationship.

Two non-stationary series are cointegrated if an appropriate nonzero linear combination is stationary under the specified process model. For two synthetic price series, a fitted residual $y_{t} - β x_{t}$ is a candidate relationship to test on data excluded from estimation. Stationarity does not bound the next excursion, guarantee a return by a deadline, or establish profitable convergence after costs.

z_{t} = y_{t} - β x_{t}, {score}_{t} = \frac{z_{t} - {\hat{μ}}_{z}}{{\hat{σ}}_{z}}

Algebra and arithmetic

Construct and normalise a residual spread

Fit $y_{t} = a + β x_{t} + z_{t}$ ; rearrange to $z_{t} = y_{t} - a - β x_{t}$ . The displayed simplified relation omits a, which can instead be absorbed into the spread mean.
For OLS with an intercept, $β = \frac{S_{x y}}{S_{x x}}$ . Standardise a fitted spread with $z_{s c o r e} = \frac{(z - \overline{z})}{s_{z}}$ . A levels residual is not automatically a return or a risk-normalised position.
If residual deviations obey AR(1), expected future spread is $μ + ϕ^{h} (z_{t} - μ)$ . This conditional expectation depends on the model remaining valid.

Work it by hand

y=105, x=50, β=2 gives spread 5. With fitted mean 3 and SD 1, the spread z-score is 2. It says nothing by itself about fill prices or net profitability.

Engle–Granger estimates a level relationship, then tests its residual using cointegration-specific critical values. An AR model can provide a conditional decay forecast if its assumptions hold. That forecast is not a holding-period guarantee or proof that convergence exceeds costs. See the official cointegration test documentation for the null, assumptions and critical-value convention.

Why does a pairs trade on a cointegrated spread not depend on the direction of the market?

Because the position is long one asset and short $β$ of the other, the shared random-walk component (the "common trend", including the market) cancels. What remains is the stationary spread, whose expected path is back toward its mean regardless of where the market goes. The fitted common component cancels only under the assumed model. Factor exposures can remain, and market shocks, changing hedge ratios, borrow and liquidity can all cause losses.

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.

def residual_score(y, x, hedge_beta, training_mean, training_sd):
    """Freeze training parameters before evaluating a new observation."""
    if training_sd <= 0:
        raise ValueError("Positive training spread SD required")
    spread = y-hedge_beta*x
    return spread, (spread-training_mean)/training_sd

print(residual_score(103, 50, 2, 1, 2))

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