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advanced41 min studyLesson 869 in Mathematics

Unit roots, cointegration and trend

Distinguish stochastic accumulation from a removable deterministic trend, and state the nonstandard inference and common-trend structure precisely.

Before you begin
On this page

The question for today

What you’ll understand

  • Explain why a unit-root level series lacks a stationary variance.
  • State a no-intercept Dickey–Fuller limit with its assumptions and Brownian functional.
  • Identify a stationary cointegrating combination without inferring causality.

Two series can drift together because they share an accumulating random component, because both follow deterministic time trends, or merely by chance in a short record. These explanations imply different transformations and different inference. A unit root makes shocks persist in the level and changes the limiting distribution of autoregressive estimates. Cointegration describes a stationary combination of integrated levels, not a generic synonym for high correlation. Deterministic detrending and stochastic differencing must therefore be chosen from an explicit model.

Accumulated shocks are not a fixed time trend

Consider y_t=rho y at t−1 plus epsilon_t, starting at zero. Under the unit-root null rho=1, let epsilon be iid with mean zero and positive finite variance sigma-squared. Then y_t is the cumulative sum of shocks and its variance is t times sigma-squared, so the level is not weakly stationary. Its first difference equals the stationary innovation sequence. This is the elementary integrated-of-order-one example; integration order refers to the stochastic differencing needed to obtain stationarity.

A deterministic trend plus stationary noise has a different structure. Removing its specified trend leaves stationary noise, while first differencing also differences that noise and can introduce negative serial covariance. A stable coefficient close to one is another possibility: it has a stationary variance but can resemble a random walk over a short window. Breaks in level or trend can complicate the visual distinction further. A plot alone cannot reliably decide which source of persistence generated the data.

ρ=1,y0=0:yt=∑s=1tεs,Var⁡(yt)=tσ2,Δyt=εt;ρ^=∑t=1nyt−1yt∑t=1nyt−12\rho=1,\quad y_0=0:\qquad y_t=\sum_{s=1}^t\varepsilon_s,\quad\operatorname{Var}(y_t)=t\sigma^2,\quad\Delta y_t=\varepsilon_t;\qquad\widehat\rho=\frac{\sum_{t=1}^n y_{t-1}y_t}{\sum_{t=1}^n y_{t-1}^2}
The estimator shown is the no-intercept least-squares autoregressive coefficient. The stated null model has neither drift nor a deterministic trend.↔ Scroll wide equations horizontally.

A unit root changes the reference distribution

For that iid null model, a functional central limit theorem converts scaled partial sums into standard Brownian motion on the unit interval. The no-intercept estimator converges at rate n rather than the usual square-root rate. Its limit is the ratio of the stochastic integral of W against itself to the integral of W-squared. The numerator equals one half of W at one squared minus one, by Itô's formula. It is not an ordinary centered normal numerator independent of its denominator.

The corresponding usual regression t statistic has the same numerator divided by the square root of the Brownian squared integral, not a standard normal limit. Including an intercept or deterministic trend changes these distributions and their critical values. Augmented Dickey–Fuller regressions add lagged differences to address serially correlated disturbances under further assumptions; they do not justify using ordinary t tables. Failure to reject a unit root can also reflect limited power against persistent stationary alternatives, rather than proving the null model true.

n(ρ^−1) ⇒ ∫01W(r) dW(r)∫01W(r)2 dr,tn ⇒ ∫01W(r) dW(r)∫01W(r)2 dr,∫01W dW=W(1)2−12n(\widehat\rho-1)\ \Rightarrow\ \frac{\int_0^1W(r)\,dW(r)}{\int_0^1W(r)^2\,dr},\qquad t_n\ \Rightarrow\ \frac{\int_0^1W(r)\,dW(r)}{\sqrt{\int_0^1W(r)^2\,dr}},\qquad\int_0^1W\,dW=\frac{W(1)^2-1}{2}
These limits use iid mean-zero finite-positive-variance innovations, y-zero=0, and a regression without intercept or trend. t-n uses the ordinary no-intercept residual-variance standard error; adding deterministic regressors changes the limiting functionals.↔ Scroll wide equations horizontally.

Cointegration removes a common stochastic component

A vector of integrated-of-order-one series is cointegrated when a nonzero linear combination of its levels is stationary. The coefficients cancel a common stochastic trend, leaving a stable relation despite nonstationary individual levels. More than one independent combination can exist in a larger system. An estimated relation requires inference designed for integrated regressors; treating fitted residuals as if the coefficients had been specified in advance gives the wrong reference distribution for a residual stationarity test.

Cointegration does not establish which variable causes another, nor does a large sample correlation establish cointegration. Independent random walks can generate apparently impressive regressions whose usual stationary-regression t statistics are invalid. Their sample fit remains random; it is incorrect to assert that every such regression has R-squared converging to one. Economic or physical structure, deterministic terms and possible structural changes matter when interpreting a candidate long-run relation and deciding whether it remains useful for forecasting.

Work it through

Cancel one shared random walk

Let S_t be a random walk starting at zero with iid unit-variance centered increments. Set Y-one,t=S_t and Y-two,t=2S_t+v_t, where v is independent iid centered noise of variance nine, independent of S. Find the common-trend-free combinations.

  1. The first level has variance t. The second has variance 4t+9, so neither has time-invariant level variance. Their first differences are stationary, making each integrated of order one.

  2. The combination cY-one+dY-two equals (c+2d)S_t+d v_t. Its variance is (c+2d)²t+9d², so stationarity requires c+2d=0 for a nonzero combination.

  3. Choose (c,d)=(−2,1). The resulting level difference is v_t, an iid stationary process of variance nine. All nonzero multiples of this vector describe the same one-dimensional cointegrating space.

The result

Y-two−2Y-one is stationary even though both levels accumulate random shocks. This algebra identifies a common trend and does not assign a causal direction.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

Separate homogeneous persistence from random accumulation

Pause and reset, then set both controls to eta = 0 and initial position one. Predict the recurrence remains at one. For a fresh comparison, pause and reset, then set eta = 0.05 and initial position one; the multiplier is 0.95, giving the first two iterates 0.95 and 0.9025. This deterministic quadratic-descent recurrence illustrates a unit versus a near-unit homogeneous multiplier. It adds no random innovations, so its constant path is not a random walk and its display cannot calculate a unit-root test.

Separate homogeneous persistence from random accumulationGradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Current readings: Position x0: 2; Objective x²/2: 2; Iteration behavior: Stable step. All parameters are adjustable with the labeled controls below.−4−202402.174.356.528.7Parameter xf(x) = x²/2n = 0
ObjectiveDiscrete iterate
Position x0
2
Objective x²/2
2
Iteration behavior
Stable step
n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA coefficient describes persistence only in conjunction with the forcing and initial-state assumptions.

Pause & reason

Check your understanding

Which statement holds for the no-intercept unit-root model and limit specified here?

Your turn

Try it yourself

Let Y_t=2+3t+U_t with iid centered U of variance four. Compare removing the known trend with first differencing: give their residual variance and the first-difference lag-one covariance. Is the original stochastic component a random walk?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Write the difference as three plus two neighboring noise terms with opposite signs.

Compare with the solution

Removing 2+3t leaves U_t, stationary white noise of variance four. Differencing gives three plus U_t−U at t−1, with variance eight and lag-one covariance minus four; higher nonzero-lag covariances vanish. The stochastic component of the original series is stationary, not an accumulated random walk. Differencing removes the deterministic slope but unnecessarily creates an MA(1) noise structure with coefficient minus one, illustrating why trend stationarity and a unit-root model call for different explanations.

Keep the idea

Take it with you

  • A unit root accumulates shocks and destroys stationary level variance in the stated random-walk model.
  • Unit-root statistics have deterministic-term-specific nonnormal reference distributions.
  • Cointegration cancels a shared stochastic trend and does not by itself establish causality.

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