Proton — ACF Pattern Recognition

Proton: ACF Pattern Recognition

Goal: Given an ACF description or plot characteristics, identify the model. This is the reverse of the atom/molecule pipeline (which goes model → ACF).

Format

Each item gives you an ACF signature. You identify the model and justify.

Problem 1

ACF: $\rho(0) = 1$, $\rho(1) = 0.4$, $\rho(2) = 0$, $\rho(h) = 0$ for $h \geq 2$.

Answer 1

MA(1). ACF cuts off after lag 1. From $\rho(1) = \theta/(1+\theta^2) = 0.4$, solve: $\theta^2 - 2.5\theta + 1 = 0$ → $\theta = 2$ or $\theta = 0.5$. Both are valid MA(1) models (they share the same ACF — this is the non-uniqueness/invertibility issue).

Atoms used: MA(1) ACVF and ACF, MA(q) ACF Cutoff Property

Problem 2

ACF: $\rho(1) = 0.6$, $\rho(2) = 0.36$, $\rho(3) = 0.216$, $\rho(4) = 0.1296$, … The pattern is $\rho(h) = 0.6^h$.

Answer 2

AR(1) with $\phi = 0.6$. The ACF decays geometrically as $\phi^h$ with no cutoff — the defining signature of causal AR(1). Monotone positive decay confirms $\phi > 0$.

$\gamma(0) = \sigma^2/(1 - 0.36)$.

Atoms used: AR(1) — Causal Case, Autocorrelation Function (ACF)

Problem 3

ACF: $\rho(1) = -0.7$, $\rho(2) = 0.49$, $\rho(3) = -0.343$, … Alternating sign, absolute values decay as $0.7^h$.

Answer 3

AR(1) with $\phi = -0.7$. ACF $= \phi^h = (-0.7)^h$ — alternating sign with exponential decay in absolute value (damped oscillation). This is the negative-$\phi$ signature.

Atoms used: AR(1) — Causal Case

Problem 4

ACF: $\rho(1) = 0.35$, $\rho(2) = -0.22$, $\rho(h) = 0$ for $h \geq 3$.

Answer 4

MA(2). ACF cuts off after lag 2. Two nonzero lags followed by exact zeros is the MA(2) signature. The specific values relate to $\theta_1, \theta_2$ via:

$\rho(1) = \frac{\theta_1 + \theta_1\theta_2}{1+\theta_1^2+\theta_2^2}$, $\rho(2) = \frac{\theta_2}{1+\theta_1^2+\theta_2^2}$.

Atoms used: MA(2) ACVF and ACF, MA(q) ACF Cutoff Property

Problem 5

ACF: Very slow decay. $\rho(1) = 0.95$, $\rho(2) = 0.90$, $\rho(10) = 0.55$, $\rho(20) = 0.12$. Near-linear decay.

Answer 5

Non-stationary (likely random walk or unit root). Slow linear decay of the sample ACF is the visual hallmark of $|\phi| = 1$. The series needs differencing before model fitting.

This is NOT an AR(1) with $\phi = 0.95$ — true AR(1) with $\phi = 0.95$ decays as $0.95^h$ (exponential, reaching 0.36 at $h = 20$), whereas this shows near-linear decay.

Atoms used: AR(1) — Random Walk, Trend Elimination — Differencing

Problem 6

ACF: $\rho(1) = 0.7$, $\rho(2) = 0.3$, $\rho(3) = 0.15$, $\rho(4) = 0.08$, … Tails off but doesn’t cut off, and the decay pattern is not a clean $\phi^h$.

Answer 6

ARMA or AR(p) with $p \geq 2$. The ACF tails off (ruling out pure MA), but the decay is not a simple geometric $\phi^h$ (ruling out AR(1)). Could be AR(2), AR(3), or ARMA(p,q). Need PACF to distinguish: if PACF cuts off at lag $p$ → AR($p$); if PACF also tails off → ARMA.

Atoms used: ARMA(p,q) Model, AR(p) Model, Autocorrelation Function (ACF)

(Note: PACF is a future topic — for now, recognize this as “not pure MA, not AR(1)”)

Summary Decision Tree

ACF cuts off at lag q?
  ├── Yes → MA(q)
  └── No (tails off) →
        ACF = φ^h exactly?
        ├── Yes → AR(1) with that φ
        └── No →
              Very slow/linear decay?
              ├── Yes → Non-stationary (difference first)
              └── No → AR(p≥2) or ARMA (need PACF)

Links

• STA457