Module 1 — Processes, Dependence, and Stationarity

Represent a stochastic process through finite random vectors, Toeplitz covariance matrices, stationarity, and Wold innovations.

Module 1 — Processes, Dependence, and Stationarity

Core question

We observe one sequence x1,,xnx_1,\ldots,x_n. What assumptions let that one path reveal a population mean, variance, and dependence structure?

Learning outcomes

You will be able to:

  • distinguish a stochastic process, a random variable, and one sample path;
  • construct the mean vector and Toeplitz covariance matrix of a stationary finite block;
  • calculate and interpret autocovariance and autocorrelation;
  • separate strict stationarity, weak stationarity, and ergodicity;
  • distinguish white noise from independence and Gaussianity;
  • derive the dependence and shock persistence of a stationary AR(1);
  • state exactly what the Wold representation does and does not promise.

1. Process versus path

A stochastic process is a collection of random variables indexed by time:

{Xt:tZ}.\{X_t:t\in\mathbb Z\}.

The capital letters describe the population mechanism. The data x1,,xnx_1,\ldots,x_n are one realised path. Observing 120 months is not the same as observing 120 independent copies: adjacent values may share shocks.

ObjectExampleQuestion
random variablenext month's demand Xt+1X_{t+1}what values and probabilities are possible?
process{Xt}\{X_t\} over all monthshow does the joint distribution change with time?
sample path102, 108, 105, ...what can this one history reveal?

2. Finite blocks expose the matrix structure

For any selected times t1,,tnt_1,\ldots,t_n, define

X=(Xt1,,Xtn).\mathbf X=(X_{t_1},\ldots,X_{t_n})^\top.

Its first two moments are

E(X)=μ,Var(X)=Γ=[Cov(Xti,Xtj)]i,j=1n.E(\mathbf X)=\boldsymbol\mu,\qquad \operatorname{Var}(\mathbf X)=\Gamma =\left[\operatorname{Cov}(X_{t_i},X_{t_j})\right]_{i,j=1}^n.

For consecutive observations from a weakly stationary process,

μ=μ1n,Γn=[γ(0)γ(1)γ(n1)γ(1)γ(0)γ(n2)γ(n1)γ(n2)γ(0)].\boldsymbol\mu=\mu\mathbf1_n,\qquad \Gamma_n= \begin{bmatrix} \gamma(0)&\gamma(1)&\cdots&\gamma(n-1)\\ \gamma(1)&\gamma(0)&\cdots&\gamma(n-2)\\ \vdots&\vdots&\ddots&\vdots\\ \gamma(n-1)&\gamma(n-2)&\cdots&\gamma(0) \end{bmatrix}.

Γn\Gamma_n is symmetric Toeplitz: each diagonal is constant. It must also be positive semidefinite because

aΓna=Var(aX)0\mathbf a^\top\Gamma_n\mathbf a =\operatorname{Var}(\mathbf a^\top\mathbf X)\ge0

for every a\mathbf a. The bounds ρ(h)1|\rho(h)|\le1 are necessary but not sufficient; all lags must fit together into valid covariance matrices.

R

Construct and verify a stationary covariance block

Idle

Change ϕ\phi to 0.7-0.7: alternating covariances appear, but validity remains. Move it to 0.980.98: the smallest eigenvalue and condition number reveal near-collinearity even before estimation.

3. Dependence is indexed by lag

For a weakly stationary process,

γ(h)=Cov(Xt,Xth),ρ(h)=γ(h)γ(0).\gamma(h)=\operatorname{Cov}(X_t,X_{t-h}), \qquad \rho(h)=\frac{\gamma(h)}{\gamma(0)}.

γ(h)\gamma(h) retains the unit squared; ρ(h)\rho(h) is unit-free and lies between 1-1 and 11. Neither measures nonlinear dependence completely.

Worked sample calculation

Take one short path:

x=(2,4,3,5,4,6),xˉ=4.x=(2,4,3,5,4,6),\qquad \bar x=4.

The deviations are (2,0,1,1,0,2)(-2,0,-1,1,0,2). Using divisor nn,

γ^(0)=4+0+1+1+0+46=1.667,\widehat\gamma(0)=\frac{4+0+1+1+0+4}{6}=1.667,γ^(1)=(2)(0)+(0)(1)+(1)(1)+(1)(0)+(0)(2)6=0.167,\widehat\gamma(1)=\frac{(-2)(0)+(0)(-1)+(-1)(1)+(1)(0)+(0)(2)}{6}=-0.167,

so ρ^(1)=0.10\widehat\rho(1)=-0.10. The sample is far too short for a strong conclusion; the exercise teaches what each ACF bar aggregates.

Some software divides lag-hh covariance by nhn-h instead of nn. Both conventions occur. State the convention when reproducing an exact number; in large samples the practical difference shrinks for fixed hh.

4. Three kinds of stability

ConceptRequirementWhy it matters
strict stationarityevery finite joint distribution is unchanged by a time shiftcomplete distributional stability
weak stationarityconstant finite mean/variance; covariance depends only on lagARMA, linear prediction, spectra
ergodicitytime averages converge to population quantitieslearning from one long path

Strict stationarity implies weak stationarity only when second moments exist. For a Gaussian process, weak stationarity is enough to determine shift-invariant finite-dimensional distributions. Stationarity alone does not guarantee that one path explores the population adequately; that is the role of ergodic conditions.

Why persistence reduces effective information

For a stationary mean-zero series,

Var(Xˉn)=1n[γ(0)+2h=1n1(1hn)γ(h)].\operatorname{Var}(\bar X_n) =\frac{1}{n}\left[\gamma(0)+2\sum_{h=1}^{n-1}\left(1-\frac{h}{n}\right)\gamma(h)\right].

Positive autocovariances make the sample mean noisier than the iid formula γ(0)/n\gamma(0)/n. One hundred highly persistent months need not contain one hundred months' worth of independent information.

5. White noise is not one assumption

LabelRequired claim
white noisezero mean, constant variance, zero autocovariance at non-zero lags
independent white noisewhite noise observations are also independent
Gaussian white noiseindependent observations follow N(0,σ2)N(0,\sigma^2)

Uncorrelated does not imply independent outside special families such as jointly Gaussian variables. A squared-noise series can be predictable in variance even when the level has zero autocorrelation.

6. The AR(1) as a complete dependence example

Let

Xt=ϕXt1+εt,εtWN(0,σε2).X_t=\phi X_{t-1}+\varepsilon_t, \qquad \varepsilon_t\sim WN(0,\sigma_\varepsilon^2).

If ϕ<1|\phi|<1, repeated substitution gives the causal representation

Xt=j=0ϕjεtj.X_t=\sum_{j=0}^{\infty}\phi^j\varepsilon_{t-j}.

Therefore

Var(Xt)=σε21ϕ2,ρ(h)=ϕh.\operatorname{Var}(X_t)=\frac{\sigma_\varepsilon^2}{1-\phi^2}, \qquad \rho(h)=\phi^{|h|}.

With ϕ=0.7\phi=0.7 and σε2=4\sigma_\varepsilon^2=4:

  • long-run variance is 4/(10.49)=7.844/(1-0.49)=7.84;
  • lag-3 correlation is 0.73=0.3430.7^3=0.343;
  • a unit shock has about half its effect after log(0.5)/log(0.7)=1.94\log(0.5)/\log(0.7)=1.94 periods.

If ϕ<0\phi<0, signs alternate; if ϕ=1\phi=1, shocks do not decay and the process becomes a random walk.

7. Linear processes and Wold's result

A causal linear process has the form

Xt=μ+j=0ψjεtj,X_t=\mu+\sum_{j=0}^{\infty}\psi_j\varepsilon_{t-j},

with coefficients that decay sufficiently for the sum to exist. Its autocovariance is

γ(h)=σε2j=0ψjψj+h.\gamma(h)=\sigma_\varepsilon^2\sum_{j=0}^{\infty}\psi_j\psi_{j+|h|}.

Wold's decomposition says that every covariance-stationary, purely non-deterministic process can be represented as an infinite moving average of uncorrelated innovations. It does not say that a short finite ARMA model is true, that innovations are independent, or that the coefficients are easy to estimate.

8. Diagnostic distinctions

ObservationSupported interpretationUnsupported leap
ACF decays slowlystrong linear persistence or unremoved low-frequency structure“the process has a unit root”
residual ACF is near zerolittle remaining linear serial correlation at inspected lags“residuals are iid Gaussian”
sample mean stabilisesevidence consistent with mean ergodicityproof of stationarity
variance changes after an interventionpossible break or volatility shiftautomatic need for first differencing

Practice

  1. For Xt=0.5Xt1+εtX_t=-0.5X_{t-1}+\varepsilon_t with innovation variance 3, calculate Var(Xt)\operatorname{Var}(X_t) and ρ(1),ρ(2),ρ(3)\rho(1),\rho(2),\rho(3).
  2. Construct two dependent variables with zero correlation; explain why an ACF cannot detect their dependence.
  3. For the six-point path above, recompute γ^(1)\widehat\gamma(1) using divisor n1n-1.
  4. Explain why stationarity and ergodicity answer different questions.
  5. Construct Γ3\Gamma_3 for an AR(1) with ϕ=0.5\phi=0.5 and innovation variance 0.750.75, then find Var(X1X2)\operatorname{Var}(X_1-X_2) as a quadratic form.
Answers
  1. Variance =3/(10.25)=4=3/(1-0.25)=4; correlations are 0.5,0.25,0.125-0.5,0.25,-0.125.
  2. If ZZ is symmetric around zero, ZZ and Z2Z^2 are dependent but can have zero covariance because E(Z3)=0E(Z^3)=0.
  3. 1/5=0.2-1/5=-0.2.
  4. Stationarity concerns invariance across calendar shifts; ergodicity concerns whether one long path recovers population quantities.
  5. γ(0)=1\gamma(0)=1, so Γ3=[10.50.250.510.50.250.51]\Gamma_3=\begin{bmatrix}1&0.5&0.25\\0.5&1&0.5\\0.25&0.5&1\end{bmatrix} and (1,1,0)Γ3(1,1,0)=1(1,-1,0)\Gamma_3(1,-1,0)^\top=1.

Takeaway

Time-series inference begins by stating what is random, what is observed once, and which form of stability connects one to the other. Continue to ARMA and ARIMA to encode that stable dependence with lag polynomials.

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