Risk Modelling and Survival Analysis · Core concepts of time series models

Moving Average (MA) Models and Invertibility Condition

Q: What is the largest possible lag-1 autocorrelation of an MA(1)?

It is 0.5 in absolute value. This occurs at θ = 1 or θ = −1, since θ ÷ (1 + θ²) peaks there. If a sample shows a lag-1 value well above 0.5, an MA(1) is not a good fit.

Updated 11 October 2026 · Fact-checked

An MA(q) process writes X_t as a constant plus a weighted sum of the current and last q white noise terms. Its autocorrelation is zero beyond lag q. It is invertible if all roots of the characteristic polynomial θ(z) = 0 lie outside the unit circle, so it can be written as an AR(∞).

Understand Moving Average (MA) Models and Invertibility

A moving average process of order q, written MA(q), builds today's value from today's random shock and the last q shocks. The shocks are white noise: uncorrelated, mean zero, constant variance σ². The model is X_t = μ + e_t + θ₁e_{t-1} + ... + θ_q e_{t-q}.

Because X_t is a finite sum of white noise terms, it is always weakly stationary. You never need a stationarity condition for an MA process. This is the main contrast with AR models, which are built from past values of X and do need a condition.

The key shape is the autocorrelation function (ACF). Two values X_t and X_{t+k} share a noise term only if k ≤ q. So the ACF is non-zero up to lag q and exactly zero after that. This cut-off is how you identify the order q from data. For an AR process the ACF decays gradually, and the partial ACF cuts off instead.

The second idea is invertibility. Different θ values can give the same ACF. For MA(1), θ and 1/θ give identical autocorrelations. To get a unique model, we choose the invertible one. Invertible means you can write e_t as a convergent sum of present and past X values, so the MA(q) becomes an AR(∞). It matters for forecasting, because you estimate the shocks from observed data.

The condition is on the polynomial θ(B) = 1 + θ₁B + ... + θ_q B^q, where B is the backshift operator (B X_t = X_{t-1}). The process is invertible if all roots of θ(z) = 0 lie outside the unit circle, meaning each has modulus greater than 1.

Key rules to remember

MA(q) definition: X_t = μ + e_t + θ₁e_{t-1} + ... + θ_q e_{t-q}, with e_t white noise with variance σ²; Always weakly stationary. Mean is μ.
Variance of MA(q): γ₀ = σ²(1 + θ₁² + ... + θ_q²); Shocks are uncorrelated, so squared coefficients add.
Autocovariance of MA(q): γ_k = σ²(θ_k + θ₁θ_{k+1} + ... + θ_{q-k}θ_q) for 1 ≤ k ≤ q, with θ₀ = 1; γ_k = 0 for k > q; Sum the products of coefficients that line up on the same shock.
MA(1) ACF: ρ₁ = θ ÷ (1 + θ²); ρ_k = 0 for k ≥ 2; |ρ₁| ≤ 0.5 always. The maximum is at θ = ±1.
MA(2) ACF: ρ₁ = θ₁(1 + θ₂) ÷ (1 + θ₁² + θ₂²); ρ₂ = θ₂ ÷ (1 + θ₁² + θ₂²); ρ_k = 0 for k ≥ 3; Check lag 1 numerator carefully: θ₁ + θ₁θ₂.
Invertibility condition: All roots of 1 + θ₁z + ... + θ_q z^q = 0 satisfy |z| > 1; Equivalent to all roots outside the unit circle.
MA(1) invertibility: |θ| < 1; Root of 1 + θz = 0 is z = −1/θ.
MA(2) invertibility region: θ₂ + θ₁ > −1, θ₂ − θ₁ > −1, |θ₂| < 1; This is the triangle region for the parameters. Using the roots directly also works.

How to solve Moving Average (MA) Models and Invertibility questions

Use this method for ACF and invertibility questions on MA models. State the model first so the marker can see your notation.

1Write the model in the form X_t = μ + e_t + θ₁e_{t-1} + ... and note the sign convention given in the question. Some texts use minus signs.
2Identify q, the highest lag with a non-zero coefficient. This is the order.
3Compute γ₀ = σ²(1 + Σθᵢ²). Then compute γ_k for k = 1 to q by summing products of coefficients that match the same shock.
4Divide each γ_k by γ₀ to get ρ_k. State that ρ_k = 0 for k > q.
5For invertibility, write the polynomial θ(z) = 1 + θ₁z + ... + θ_q z^q and set it to zero.
6Find the roots. For MA(1) solve directly. For MA(2) use the quadratic formula or the parameter region.
7Check every root has modulus greater than 1. If so, say the process is invertible. If not, state it is not invertible.
8If asked, give the AR(∞) form or the invertible alternative with the same ACF.

Quickest way: Fast check for MA(1) and MA(2)

When to use it: Use in multiple-choice questions or when time is short and you only need the ACF or an invertibility verdict.

For MA(1), ρ₁ = θ ÷ (1 + θ²) and invertible if |θ| < 1. No roots needed.
For MA(2), compute the denominator 1 + θ₁² + θ₂² once. Then ρ₁ = (θ₁ + θ₁θ₂) ÷ denominator and ρ₂ = θ₂ ÷ denominator.
For MA(2) invertibility, test the three inequalities: θ₂ + θ₁ > −1, θ₂ − θ₁ > −1, |θ₂| < 1. All must hold.
Sanity check: any MA(1) with |ρ₁| above 0.5 is impossible.
If the ACF is zero beyond lag q, the model is MA(q). Do not look for a decay.

Common mistakes in Moving Average (MA) Models and Invertibility

Saying an MA process needs a stationarity condition on θ.
Students carry over the AR rule that roots must lie outside the unit circle for stationarity.
Fix: An MA process is always stationary. The root condition on an MA polynomial is for invertibility, not stationarity.
Forgetting the 1 in the variance formula.
The coefficient of e_t is 1 and is easy to leave out of the sum.
Fix: Write γ₀ = σ²(1 + θ₁² + ... + θ_q²) every time, with θ₀ = 1.
Writing the MA(2) lag-1 autocovariance as θ₁ only.
Students miss the product θ₁θ₂ from the shared term e_{t-1}... e_{t-2}.
Fix: Line up X_t and X_{t-1}. Shared shocks give θ₁ + θ₁θ₂ times σ². Write the two lined-up rows if unsure.
Claiming the ACF of an MA(q) decays slowly.
Mixing up the MA and AR patterns.
Fix: MA: ACF cuts off after lag q. AR: ACF decays and PACF cuts off after lag p.
Checking the roots of θ₁ + z instead of 1 + θ₁z, or concluding |θ| > 1 is invertible.
Confusing the polynomial in B with its reciprocal.
Fix: Use 1 + θ₁z = 0, giving z = −1/θ₁. Need |z| > 1, so |θ₁| < 1.
Ignoring the sign convention in the question.
Some sources write X_t = e_t − θe_{t-1}, which flips the sign of every coefficient.
Fix: Rewrite the model in the plus form first, then apply formulas.

Worked examples

Example 1

The process X_t = e_t + 0.6e_{t-1} − 0.2e_{t-2} has e_t white noise with variance σ². (a) Find ρ₁ and ρ₂ and ρ₃. (b) Is the process invertible?

Show the solution

Here θ₁ = 0.6 and θ₂ = −0.2, so q = 2.
Denominator: 1 + 0.36 + 0.04 = 1.40. So γ₀ = 1.40σ².
γ₁ = σ²(θ₁ + θ₁θ₂) = σ²(0.6 − 0.12) = 0.48σ².
ρ₁ = 0.48 ÷ 1.40 = 0.342857, about 0.343.
γ₂ = σ²θ₂ = −0.2σ². So ρ₂ = −0.2 ÷ 1.40 = −0.142857, about −0.143.
ρ₃ = 0 because 3 > q.
Invertibility: θ₂ + θ₁ = −0.2 + 0.6 = 0.4 > −1. θ₂ − θ₁ = −0.2 − 0.6 = −0.8 > −1. |θ₂| = 0.2 < 1.
All three conditions hold, so the process is invertible.

Answer: ρ₁ ≈ 0.343, ρ₂ ≈ −0.143, ρ₃ = 0. The process is invertible.

Example 2

An MA(1) process is X_t = e_t + 2e_{t-1}. (a) Is it invertible? (b) Find ρ₁. (c) Find an invertible MA(1) with the same ACF.

Show the solution

(a) Solve 1 + 2z = 0. Then z = −0.5. Its modulus is 0.5, which is less than 1. So the process is not invertible. Equivalently |θ| = 2 > 1.
(b) ρ₁ = θ ÷ (1 + θ²) = 2 ÷ (1 + 4) = 0.4.
(c) Replace θ by 1/θ = 0.5. Then ρ₁ = 0.5 ÷ (1 + 0.25) = 0.5 ÷ 1.25 = 0.4. This matches.
The invertible model is X_t = e*_t + 0.5e*_{t-1}, where e*_t is white noise with a different variance. The variance of X_t is equal in both forms: σ²(1 + 4) = 5σ² must equal σ*²(1 + 0.25) = 1.25σ*², so σ*² = 4σ².

Answer: Not invertible. ρ₁ = 0.4. The invertible equivalent has θ = 0.5 with noise variance 4σ².

Exam tips

Always state that an MA(q) is stationary for any θ. Examiners often award a mark for this line.
In written answers, show the shared-shock working for each γ_k. Method marks are given even if arithmetic slips.
Use the ACF cut-off to justify the order when asked to identify a model from a sample ACF, and compare against the roughly ±2/√n bounds.
For invertibility, give the polynomial, the roots and the modulus comparison. A bare yes or no scores little.
In the computer-based paper, check whether R's arima uses the plus sign for MA coefficients and then read the fitted values accordingly.

Practice questions from Core concepts of time series models

Moving Average (MA) Models and Invertibility in other exams

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

FRM Part IMoving Average (MA) Models

Moving Average (MA) Models and Invertibility: frequently asked questions

What is the difference between an AR and an MA process?

An AR process regresses X_t on its own past values. An MA process uses current and past white noise terms. The ACF of an AR decays gradually while the ACF of an MA(q) cuts off after lag q. An AR needs a stationarity condition and an MA needs an invertibility condition.

Why does invertibility matter?

Without it, more than one MA model gives the same ACF, so the model is not unique. An invertible model lets you express the shocks in terms of observed data, which you need for forecasting and estimation.

How do I check invertibility of an MA(2) model?

Form 1 + θ₁z + θ₂z² = 0 and find the roots. Both must have modulus above 1. Alternatively check that θ₂ + θ₁ > −1, θ₂ − θ₁ > −1 and |θ₂| < 1.

What is the largest possible lag-1 autocorrelation of an MA(1)?

It is 0.5 in absolute value. This occurs at θ = 1 or θ = −1, since θ ÷ (1 + θ²) peaks there. If a sample shows a lag-1 value well above 0.5, an MA(1) is not a good fit.