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FRM Exam Part I · Stationary Time Series

ARMA Models: ACF and PACF Identification for FRM Part I

Updated 11 October 2026 · Fact-checked

An ARMA(p,q) model combines p autoregressive lags of the series with q moving average lags of past shocks. To identify a model, read the ACF and PACF: AR(p) has a PACF that cuts off after lag p, MA(q) has an ACF that cuts off after lag q, and ARMA has both decaying gradually.

Understand ARMA Models

A covariance stationary series has a constant mean, a constant variance and autocovariances that depend only on the lag. Two building blocks describe such series. An AR(p) model says today's value depends on its own past values. An MA(q) model says today's value depends on today's shock and past shocks.

An ARMA(p,q) model uses both. Written in lag form, ARMA(1,1) is Yt = c + φYt-1 + εt + θεt-1, where ε is white noise. You use ARMA when a pure AR needs many lags or a pure MA needs many lags. Mixing the two often fits the data with fewer parameters, which is called parsimony.

The ACF is the correlation between Yt and Yt-k. The PACF is the correlation between Yt and Yt-k after removing the effect of the lags in between. Each model leaves a different fingerprint on these two plots.

AR(p): the ACF decays gradually (geometrically for AR(1)), while the PACF cuts off to zero after lag p. MA(q): the ACF cuts off to zero after lag q, while the PACF decays gradually. ARMA(p,q): both the ACF and the PACF decay gradually, so neither gives a clean cutoff.

Stationarity needs the AR part to satisfy |φ| < 1 for ARMA(1,1). The MA part is always stationary. For a unique, usable forecasting model you also want invertibility, which for MA(1) means |θ| < 1.

Key formulas to remember

ARMA(1,1) process
Yt = c + φYt-1 + εt + θεt-1
εt is white noise with mean 0 and variance σ². Stationary if |φ| < 1.
ARMA(p,q) process
Yt = c + φ1Yt-1 + … + φpYt-p + εt + θ1εt-1 + … + θqεt-q
p counts AR lags, q counts MA lags.
ARMA(1,1) mean
μ = c ÷ (1 − φ)
Requires φ ≠ 1. The MA term does not change the mean.
ARMA(1,1) variance
γ0 = σ² × (1 + 2φθ + θ²) ÷ (1 − φ²)
Valid for |φ| < 1.
ARMA(1,1) autocorrelations
ρ1 = (1 + φθ)(φ + θ) ÷ (1 + 2φθ + θ²); ρk = φ × ρk-1 for k ≥ 2
Decay at rate φ starts from lag 1, not lag 0.
AR(1) ACF
ρk = φ^k
Geometric decay; PACF is φ at lag 1 and zero afterwards.
MA(1) ACF
ρ1 = θ ÷ (1 + θ²); ρk = 0 for k ≥ 2
PACF decays gradually.
Identification rule
AR(p): PACF cuts off after p | MA(q): ACF cuts off after q | ARMA: both decay
Cutoff means statistically insignificant, not exactly zero, in a sample.

How to solve ARMA Models questions

Use this routine for any question asking you to identify, describe or compute with an ARMA model.

  1. 1Write down what is given: a plot or table of ACF and PACF, or the model equation with its parameters.
  2. 2If you are given plots, check each one: does it cut off sharply after a lag, or does it die out gradually?
  3. 3Match the pattern: PACF cutoff only means AR, ACF cutoff only means MA, both gradual means ARMA.
  4. 4For AR or MA, read the order from the lag of the cutoff (the last significant spike).
  5. 5If you are given the equation, count the lagged Y terms for p and the lagged ε terms for q.
  6. 6Check stationarity: for the AR part, |φ| < 1 in ARMA(1,1). Check invertibility for the MA part if asked.
  7. 7For calculations, use the formulas: mean c ÷ (1 − φ), the ρ1 formula, then ρk = φρk-1 for k ≥ 2.
  8. 8Sanity check: autocorrelations must lie between −1 and 1, and the variance must be positive.

Quickest way: Two-plot shortcut

When to use it: Use when the question shows ACF and PACF patterns and asks which model fits.

  1. Ask: does the ACF cut off? If yes, and the PACF decays, choose MA with order equal to the cutoff lag.
  2. Ask: does the PACF cut off? If yes, and the ACF decays, choose AR with order equal to the cutoff lag.
  3. If both decay gradually, choose ARMA.
  4. For ARMA(1,1) numerical questions, compute ρ1 first, then multiply by φ for each later lag.

Common mistakes in ARMA Models

  • Swapping the cutoff rules for AR and MA.

    Both models have one plot that cuts off and one that decays, so the pairs are easy to mix up.

    Fix: Remember: MA cuts off in the ACF (its memory is finite in shocks). AR cuts off in the PACF.

  • Claiming ARMA has a cutoff in the ACF or PACF at lag p or q.

    Students carry over the pure AR and MA rules.

    Fix: In ARMA both functions decay gradually, so neither identifies p and q cleanly.

  • Using ρk = φ^k for ARMA(1,1).

    It is the AR(1) formula and looks similar.

    Fix: For ARMA(1,1), compute ρ1 with the full formula, then ρk = φρk-1 for k ≥ 2, so ρ2 = φρ1.

  • Forgetting the stationarity condition on φ.

    Candidates focus on the shape of the autocorrelations.

    Fix: Check |φ| < 1 before using any variance or autocorrelation formula.

  • Treating sample ACF spikes as exactly zero beyond the cutoff.

    Textbook plots are clean, but real data is noisy.

    Fix: Judge cutoffs by whether spikes lie inside the significance bands, usually about ±2 ÷ √T.

Worked examples

Example 1

An analyst fits ARMA(1,1): Yt = 0.5Yt-1 + εt + 0.3εt-1. Find ρ1 and ρ2.

Show the solution
  1. φ = 0.5 and θ = 0.3. Stationary because |0.5| < 1.
  2. Numerator of ρ1: (1 + φθ)(φ + θ) = (1 + 0.15)(0.8) = 1.15 × 0.8 = 0.92.
  3. Denominator: 1 + 2φθ + θ² = 1 + 0.30 + 0.09 = 1.39.
  4. ρ1 = 0.92 ÷ 1.39 = 0.6619.
  5. ρ2 = φ × ρ1 = 0.5 × 0.6619 = 0.3309.

Answer: ρ1 ≈ 0.662 and ρ2 ≈ 0.331.

Example 2

A sample ACF decays slowly and smoothly from lag 1 to lag 10. The sample PACF shows significant spikes at lags 1 and 2 only, then lies inside the bands. Which model is suggested, and which of these would you reject: AR(1), AR(2), MA(2), ARMA(1,1)?

Show the solution
  1. The ACF decays gradually, so the process is not a pure MA (an MA would show an ACF cutoff).
  2. The PACF cuts off after lag 2, which is the AR fingerprint with p = 2.
  3. So AR(2) fits both plots.
  4. MA(2) is rejected because its ACF would cut off after lag 2.
  5. AR(1) is rejected because the PACF has a significant spike at lag 2.
  6. ARMA(1,1) is less consistent because it would give a gradually decaying PACF, not a sharp cutoff.

Answer: AR(2) is suggested. MA(2) is clearly rejected, and AR(1) and ARMA(1,1) fit worse.

Exam tips

  • Questions often give plots or a verbal description. Decide first which function cuts off, then pick the model.
  • If both ACF and PACF decay, the answer is ARMA, even if the question does not say so directly.
  • For ARMA(1,1) calculations, do ρ1 carefully with the full formula and then apply ρk = φρk-1.
  • Eliminate options using the cutoff lag: it equals p for AR in the PACF and q for MA in the ACF.
  • Remember that ARMA is chosen for parsimony, so a mixed model with few parameters is preferred over a long pure AR or MA.

Practice questions from Stationary Time Series

ARMA Models in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

ARMA Models: frequently asked questions

What is the difference between AR, MA and ARMA models?

An AR model regresses the series on its own past values. An MA model expresses it as a weighted sum of current and past shocks. An ARMA model includes both parts, which often fits with fewer parameters.

How do I identify AR, MA or ARMA using ACF and PACF?

If the PACF cuts off after lag p and the ACF decays, it is AR(p). If the ACF cuts off after lag q and the PACF decays, it is MA(q). If both decay gradually, it is ARMA.

What does the ACF of an ARMA(1,1) look like?

It decays gradually. The first autocorrelation ρ1 depends on both φ and θ. From lag 2 onward, each autocorrelation equals φ times the one before it.

Why not always use a high-order AR or MA instead of ARMA?

More parameters raise estimation error and the risk of overfitting. An ARMA with few terms can capture the same pattern more efficiently.