Module 2 — ARMA, ARIMA, and Model Identification

Build AR, MA, and ARIMA models from lag polynomials, roots, shock propagation, and finite-sample diagnostics.

Module 2 — ARMA, ARIMA, and Model Identification

Core question

Can a small number of lagged values and past shocks explain the serial dependence that remains after trend and seasonality are handled?

Learning outcomes

You will be able to:

  • write AR, MA, ARMA, and ARIMA models with the lag operator;
  • check AR causality/stationarity and MA invertibility from polynomial roots;
  • construct an AR companion matrix and solve its Yule–Walker equations;
  • derive benchmark ACF and PACF patterns;
  • explain why invertibility is an identification restriction;
  • detect under- and over-differencing;
  • turn plots into candidate models without treating them as proofs.

1. One notation, three model classes

Let BXt=Xt1BX_t=X_{t-1}. Define

ϕ(B)=1ϕ1BϕpBp,θ(B)=1+θ1B++θqBq.\phi(B)=1-\phi_1B-\cdots-\phi_pB^p, \qquad \theta(B)=1+\theta_1B+\cdots+\theta_qB^q.

Then

ModelEquationWhere memory lives
AR(pp)ϕ(B)(Xtμ)=εt\phi(B)(X_t-\mu)=\varepsilon_tpast observed values
MA(qq)Xtμ=θ(B)εtX_t-\mu=\theta(B)\varepsilon_tcurrent and past shocks
ARMA(p,qp,q)ϕ(B)(Xtμ)=θ(B)εt\phi(B)(X_t-\mu)=\theta(B)\varepsilon_tboth
ARIMA(p,d,qp,d,q)ϕ(B)(1B)dXt=θ(B)εt\phi(B)(1-B)^dX_t=\theta(B)\varepsilon_tARMA after differencing

The intercept in software is not always the unconditional mean. For

Xt=c+ϕXt1+εt,X_t=c+\phi X_{t-1}+\varepsilon_t,

the stationary mean is μ=c/(1ϕ)\mu=c/(1-\phi).

2. AR means decaying feedback

For AR(1),

Xt=ϕXt1+εt.X_t=\phi X_{t-1}+\varepsilon_t.

It is causal and stationary when ϕ<1|\phi|<1, with

Xt=εt+ϕεt1+ϕ2εt2+,ρ(h)=ϕh.X_t=\varepsilon_t+\phi\varepsilon_{t-1}+\phi^2\varepsilon_{t-2}+\cdots, \qquad \rho(h)=\phi^{|h|}.

For AR(pp), all roots of ϕ(z)=0\phi(z)=0 must lie outside the unit circle. Equivalently, the companion-matrix eigenvalues lie inside it.

Worked AR(2) root check

Consider

Xt=0.75Xt10.125Xt2+εt.X_t=0.75X_{t-1}-0.125X_{t-2}+\varepsilon_t.

The polynomial is 10.75z+0.125z2=(10.5z)(10.25z)1-0.75z+0.125z^2=(1-0.5z)(1-0.25z). Its roots are 2 and 4, both outside the unit circle, so the process is stationary. Two positive decay rates create a smoother ACF than a single cutoff rule could describe.

3. Companion and Yule–Walker forms

For AR(pp), define

st=(Xt,Xt1,,Xtp+1),st=Fst1+Gεt,\mathbf s_t=(X_t,X_{t-1},\ldots,X_{t-p+1})^\top, \qquad \mathbf s_t=F\mathbf s_{t-1}+G\varepsilon_t,

where the first row of FF is (ϕ1,,ϕp)(\phi_1,\ldots,\phi_p), its first subdiagonal contains ones, and G=(1,0,,0)G=(1,0,\ldots,0)^\top.

Three stability statements are equivalent:

roots of ϕ(z) outside 1eigenvalues of F inside 1Fh0.\text{roots of }\phi(z)\text{ outside }1 \Longleftrightarrow \text{eigenvalues of }F\text{ inside }1 \Longleftrightarrow F^h\to0.

The same coefficients solve the moment equations

Toeplitz{γ(0),,γ(p1)}Γpϕ=(γ(1),,γ(p))γp.\underbrace{\operatorname{Toeplitz} \{\gamma(0),\ldots,\gamma(p-1)\}}_{\Gamma_p} \boldsymbol\phi = \underbrace{(\gamma(1),\ldots,\gamma(p))^\top}_{\boldsymbol\gamma_p}.

R: verify all three representations

R

AR(2): roots, companion eigenvalues, and Yule-Walker solve

Idle

The polynomial roots are reciprocals of the companion eigenvalues. The Toeplitz solve returns the original coefficients because it uses population autocorrelations.

4. MA means finite shock memory

For MA(1),

Xt=εt+θεt1,X_t=\varepsilon_t+\theta\varepsilon_{t-1},ρ(1)=θ1+θ2,ρ(h)=0(h>1).\rho(1)=\frac{\theta}{1+\theta^2}, \qquad \rho(h)=0\quad (|h|>1).

With θ=0.5\theta=0.5, ρ(1)=0.4\rho(1)=0.4. A shock affects two observations, so the theoretical ACF cuts off after lag 1.

Why invertibility is needed

θ=2\theta=2 and θ=1/2\theta=1/2 generate the same autocorrelation because

21+22=1/21+(1/2)2=0.4.\frac{2}{1+2^2}=\frac{1/2}{1+(1/2)^2}=0.4.

Invertibility selects the representation whose MA roots lie outside the unit circle—here θ<1|\theta|<1 under this sign convention—so innovations are recoverable from current and past observations and parameters are identified.

5. ACF and PACF are signatures, not verdicts

Population patternAR(pp)MA(qq)ARMA(p,qp,q)
ACFtails offcuts off after qqtails off
PACFcuts off after pptails offtails off

“Cuts off” is a population result. A sample ACF has noise of roughly order 1/n1/\sqrt n under simple white-noise conditions, and inspecting many lags creates multiple-comparison ambiguity. Use ACF/PACF to propose a small candidate set, then estimate, diagnose, and backtest it.

6. Differencing changes the model

First and seasonal differences are

ΔXt=(1B)Xt,ΔsXt=(1Bs)Xt.\Delta X_t=(1-B)X_t, \qquad \Delta_sX_t=(1-B^s)X_t.
SymptomCandidate operationBoundary
stochastic level driftΔXt\Delta X_tdo not use merely because a deterministic trend exists
stochastic annual drift in monthly dataΔ12Xt\Delta_{12}X_tmonth indicators may suit stable seasonal means better
variance proportional to levellogXt\log X_trequires positive data and changes interpretation to relative change

Over-differencing stationary white noise gives ΔXt=εtεt1\Delta X_t=\varepsilon_t-\varepsilon_{t-1}, an MA(1) boundary case with ρ(1)=0.5\rho(1)=-0.5. A strong negative lag-1 ACF after transformation is a warning.

7. See the signatures, then disturb them

Edit phi, theta, and n. Small samples show why clean textbook cutoffs often look ambiguous.

R

AR(1) and MA(1): sample ACF/PACF versus theory

Idle

8. Identification workflow

  1. Define target, frequency, origin, and horizon.
  2. Plot levels; mark trend, seasonality, breaks, and missing values.
  3. Choose transformations for stated reasons.
  4. Inspect ACF/PACF and propose at most a few parsimonious orders.
  5. Check estimated roots and near-cancellation between AR and MA factors.
  6. Audit residual dependence and parameter stability.
  7. Compare against naive and seasonal-naive forecasts on rolling origins.
  8. Prefer the smallest model whose remaining errors are not predictably useful.
An ARMA model with nearly identical AR and MA factors can mimic a simpler process while producing unstable estimates. A high-order fit is not evidence of rich dynamics when factors almost cancel.

Practice

  1. For Xt=0.6Xt1+εtX_t=0.6X_{t-1}+\varepsilon_t, calculate ρ(1)\rho(1) through ρ(4)\rho(4).
  2. For MA(1) with θ=0.5\theta=-0.5, calculate ρ(1)\rho(1) and describe the visual pattern.
  3. Factor 10.9z+0.2z21-0.9z+0.2z^2 and check stationarity.
  4. Explain why an MA coefficient outside the invertible region is not necessarily an impossible data-generating process but is a non-unique parameterisation.
  5. Write the companion matrix for the AR(2) in the worked root check and verify that its eigenvalues are 0.50.5 and 0.250.25.
Answers
  1. 0.6, 0.36, 0.216, 0.1296.
  2. 0.5/1.25=0.4-0.5/1.25=-0.4; adjacent observations tend to alternate around the mean.
  3. (10.5z)(10.4z)(1-0.5z)(1-0.4z) has roots 2 and 2.5, so it is stationary.
  4. Its autocovariance can be reproduced by an invertible representation after rescaling the innovation variance; invertibility chooses the unique recoverable-shock representation.
  5. F=[0.750.12510]F=\begin{bmatrix}0.75&-0.125\\1&0\end{bmatrix} has characteristic polynomial λ20.75λ+0.125\lambda^2-0.75\lambda+0.125 and eigenvalues 0.5,0.250.5,0.25.

Takeaway

ARMA is a disciplined approximation to stable linear dependence. The next module asks what that approximation implies for future values and uncertainty: Prediction.

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