FRM Exam Part I · Stationary Time Series

Moving Average (MA) Models: MA(1), MA(q) and Invertibility

Q: What is the difference between AR and MA models?

An AR model regresses the series on its own past values, so shocks persist and the ACF decays gradually while the PACF cuts off. An MA model uses current and past shocks, so shocks last only q periods and the ACF cuts off after lag q while the PACF decays. AR needs a stationarity condition. MA needs an invertibility condition.

Q: What is the MA(1) autocovariance formula?

For y(t) = μ + ε(t) + θε(t−1), the lag-0 autocovariance is σ²(1 + θ²), the lag-1 autocovariance is θσ², and all higher lags are zero. The lag-1 autocorrelation is θ ÷ (1 + θ²).

Q: What is invertibility in an MA process?

A process is invertible if it can be written as an AR(∞) in which current shocks are expressed by current and past observations. For MA(1) this needs |θ| < 1. For MA(q), all roots of the lag polynomial must be outside the unit circle. It makes the model uniquely identified, since θ and 1/θ give the same ACF.

Q: Why does the autocorrelation of an MA(q) cut off after lag q?

Two observations more than q periods apart share no common shocks. Shocks are uncorrelated over time, so their covariance is exactly zero. Observations within q periods share at least one shock and are correlated.

Updated 11 October 2026 · Fact-checked

A moving average model writes today's value as a constant plus the current shock and a weighted sum of past shocks. For MA(1), y(t) = μ + ε(t) + θε(t−1). Variance is σ²(1 + θ²), lag-1 autocorrelation is θ ÷ (1 + θ²), and autocorrelation is zero beyond lag q. It is invertible if |θ| < 1.

Understand Moving Average (MA) Models

A moving average (MA) model builds a series from white noise shocks. Each shock ε is uncorrelated over time, has mean zero and constant variance σ². The series y(t) is the mean plus a weighted sum of the current shock and the last q shocks. Past values of y itself do not appear on the right side.

The MA(1) model is y(t) = μ + ε(t) + θε(t−1). A shock hits the series today and again tomorrow, with weight θ. After that it is gone. That is why the autocorrelation function (ACF) cuts off: y(t) and y(t−1) share the shock ε(t−1), so they are correlated. y(t) and y(t−2) share no shock, so their correlation is exactly zero. An MA(q) process has non-zero autocorrelation up to lag q and zero after lag q.

Compare this with an AR model. An AR(1) series depends on its own past value, so a shock echoes forever and the ACF decays gradually. An MA(q) shock has a finite memory of q periods and the ACF cuts off. The reverse pattern holds for the partial autocorrelation function (PACF): for MA it decays gradually, for AR it cuts off.

Invertibility means an MA process can be rewritten as an AR(∞) process, so the shocks can be recovered from current and past observations of y. For MA(1) this needs |θ| < 1. For MA(q), all roots of the lag polynomial 1 + θ1·z + ... + θq·z^q must lie outside the unit circle. Invertibility matters because two different θ values can give the same autocorrelations, for example θ = 0.5 and θ = 2 both give ρ1 = 0.4. We pick the invertible one, θ = 0.5.

An MA process is always covariance stationary when q is finite, because the weights are finite. Stationarity is never the issue for MA. Invertibility is.

Key formulas to remember

MA(1) model: y(t) = μ + ε(t) + θ·ε(t−1), ε ~ white noise (0, σ²); Mean is μ at every date. Past shocks enter, not past values of y.
MA(1) variance: γ0 = σ²(1 + θ²); Always larger than σ² unless θ = 0.
MA(1) autocovariance: γ1 = θσ²; γk = 0 for k ≥ 2; Only lag 1 is non-zero.
MA(1) autocorrelation: ρ1 = θ ÷ (1 + θ²); ρk = 0 for k ≥ 2; The largest possible |ρ1| is 0.5, reached at θ = 1 or −1. Sign of ρ1 equals sign of θ.
MA(q) model: y(t) = μ + ε(t) + θ1·ε(t−1) + ... + θq·ε(t−q); Mean is μ. ACF is zero for all lags above q.
MA(q) variance: γ0 = σ²(1 + θ1² + θ2² + ... + θq²); Sum of squared weights, with weight 1 on the current shock.
MA(q) autocovariance: γk = σ²(θk + θ1·θ(k+1) + ... + θ(q−k)·θq) for 1 ≤ k ≤ q; γk = 0 for k > q; Add the products of weights that line up on the same shock, with θ0 = 1. For MA(2): γ1 = σ²(θ1 + θ1θ2), γ2 = σ²θ2.
Invertibility condition (MA(1)): |θ| < 1; Then ε(t) = y(t) − μ − θ(y(t−1) − μ) + θ²(y(t−2) − μ) − ... , an AR(∞) form.
MA(1) forecasts: 1-step: μ + θ·ε(t); h-step for h ≥ 2: μ; The forecast equals μ for all horizons h > q. For MA(1) that means h ≥ 2. The 1-step forecast (h = q) is not the mean. One-step error variance is σ².

How to solve Moving Average (MA) Models questions

Use this routine for any MA question, whether it asks for moments, autocorrelations, identification or invertibility.

1Write the model in standard form y(t) = μ + ε(t) + θ1·ε(t−1) + ... and list μ, every θ and σ². If the shock variance is given as a standard deviation, square it.
2Note the order q. This is the number of lagged shocks. Everything above lag q is zero for autocovariance and autocorrelation.
3Compute the mean. It is μ, the constant. The shock terms have zero expected value.
4Compute the variance: σ² times (1 + sum of squared θ).
5For each lag k from 1 to q, add the products of weights that share the same shock (θ0 = 1) and multiply by σ². This gives γk.
6Divide each γk by γ0 to get ρk. State that ρk = 0 for k > q.
7If asked about invertibility, check |θ| < 1 for MA(1), or that all roots of the lag polynomial lie outside the unit circle for MA(q).
8For forecasts, use the known shock estimates for steps up to q and the mean beyond. Check that your final answer matches the question: covariance, correlation or variance.

Quickest way: Shortcut for MA(1) and MA(2) questions

When to use it: Use when the question gives θ values and asks for variance, covariance or autocorrelation, or when it asks you to identify the process from a correlogram.

MA(1): compute ρ1 = θ ÷ (1 + θ²) directly. You do not need σ² for any autocorrelation.
Sanity check: |ρ1| must be at most 0.5. If you get more, you have a mistake.
MA(2): numerator for ρ1 is θ1(1 + θ2); numerator for ρ2 is θ2; the denominator for both is 1 + θ1² + θ2².
You need σ² only for γ values. Autocorrelations do not depend on it.
From a correlogram: sharp cutoff after lag q with slow decay in the PACF points to MA(q). Slow decay in the ACF with a PACF cutoff points to AR.
Invertibility for MA(1): just test |θ| < 1.

Common mistakes in Moving Average (MA) Models

Writing the MA(1) variance as σ²(1 + θ) or σ² + θ².
Students forget that each shock term has its own variance and the weight is squared.
Fix: Variance of a sum of uncorrelated terms is the sum of weight² × σ². The current shock has weight 1, so γ0 = σ²(1 + θ²).
Using ρ1 = θ for an MA(1).
Confusion with AR(1), where ρ1 = φ.
Fix: For MA(1), ρ1 = θ ÷ (1 + θ²). Check that |ρ1| ≤ 0.5.
Claiming the ACF of an MA(q) is non-zero beyond lag q or that it decays gradually.
Mixing up the ACF and PACF patterns of AR and MA.
Fix: MA: ACF cuts off after lag q, PACF decays. AR: ACF decays, PACF cuts off after lag p.
Saying an MA process is nonstationary when |θ| ≥ 1.
Confusing the invertibility condition with the stationarity condition for AR models.
Fix: A finite-order MA is always covariance stationary. |θ| < 1 is needed only for invertibility.
Forgetting the cross terms when computing MA(2) γ1.
Students use only θ1 and miss that y(t) and y(t−1) share two shocks, ε(t−1) and ε(t−2). ε(t−1) has weight θ1 in y(t) and weight 1 in y(t−1). ε(t−2) has weight θ2 in y(t) and weight θ1 in y(t−1).
Fix: Line up the two expressions and multiply the weights on each shared shock: γ1 = σ²(θ1 × 1 + θ2 × θ1) = σ²(θ1 + θ1θ2). Check by listing which shocks appear in both.
Forecasting an MA(1) two steps ahead using θ·ε(t).
Students carry the one-step formula forward.
Fix: For h > q the forecast is the mean μ. Only the first q steps use known shocks.

Worked examples

Example 1

An MA(1) process is y(t) = 2 + ε(t) + 0.6·ε(t−1), where ε is white noise with variance σ² = 4. Find the mean, variance, lag-1 autocovariance, lag-1 autocorrelation and lag-2 autocorrelation. Is the process invertible?

Show the solution

Mean = μ = 2.
Variance γ0 = σ²(1 + θ²) = 4 × (1 + 0.36) = 4 × 1.36 = 5.44.
Autocovariance γ1 = θσ² = 0.6 × 4 = 2.4.
Autocorrelation ρ1 = γ1 ÷ γ0 = 2.4 ÷ 5.44 = 0.4412. Check: θ ÷ (1 + θ²) = 0.6 ÷ 1.36 = 0.4412.
Lag 2 is beyond q = 1, so ρ2 = 0.
Invertibility: |θ| = 0.6 < 1, so the process is invertible.

Answer: Mean 2; variance 5.44; γ1 = 2.4; ρ1 ≈ 0.441; ρ2 = 0; invertible.

Example 2

An MA(2) process is y(t) = ε(t) + 0.5·ε(t−1) − 0.4·ε(t−2), with σ² = 1. Find γ0, γ1, γ2, ρ1, ρ2 and ρ3.

Show the solution

Variance γ0 = σ²(1 + θ1² + θ2²) = 1 + 0.25 + 0.16 = 1.41.
γ1 = σ²(θ1 + θ1·θ2) = 0.5 + 0.5 × (−0.4) = 0.5 − 0.2 = 0.3.
γ2 = σ²·θ2 = −0.4.
ρ1 = 0.3 ÷ 1.41 = 0.2128.
ρ2 = −0.4 ÷ 1.41 = −0.2837.
q = 2, so γ3 = 0 and ρ3 = 0.

Answer: γ0 = 1.41; γ1 = 0.3; γ2 = −0.4; ρ1 ≈ 0.213; ρ2 ≈ −0.284; ρ3 = 0.

Exam tips

Memorize ρ1 = θ ÷ (1 + θ²) for MA(1). Many questions need only this and the cutoff after lag 1.
Check your answer with the bound |ρ1| ≤ 0.5. It catches most arithmetic and formula slips.
Read whether the question gives σ² or σ. Square the standard deviation before using the formulas.
For identification questions, look at where the ACF drops to zero. A cutoff at lag q means MA(q). A slow decay means AR or ARMA.
Remember that MA stationarity is automatic. Invertibility is the condition the exam tests, and it is what lets you rewrite the model as AR(∞).

Practice questions from Stationary Time Series

Moving Average (MA) Models in other exams

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

Actuarial Core Principles (IAI)Moving Average (MA) Models and Invertibility

Moving Average (MA) Models: frequently asked questions

What is the difference between AR and MA models?

An AR model regresses the series on its own past values, so shocks persist and the ACF decays gradually while the PACF cuts off. An MA model uses current and past shocks, so shocks last only q periods and the ACF cuts off after lag q while the PACF decays. AR needs a stationarity condition. MA needs an invertibility condition.

What is the MA(1) autocovariance formula?

For y(t) = μ + ε(t) + θε(t−1), the lag-0 autocovariance is σ²(1 + θ²), the lag-1 autocovariance is θσ², and all higher lags are zero. The lag-1 autocorrelation is θ ÷ (1 + θ²).

What is invertibility in an MA process?

A process is invertible if it can be written as an AR(∞) in which current shocks are expressed by current and past observations. For MA(1) this needs |θ| < 1. For MA(q), all roots of the lag polynomial must be outside the unit circle. It makes the model uniquely identified, since θ and 1/θ give the same ACF.

Why does the autocorrelation of an MA(q) cut off after lag q?

Two observations more than q periods apart share no common shocks. Shocks are uncorrelated over time, so their covariance is exactly zero. Observations within q periods share at least one shock and are correlated.