MA(2) Moments And Invertibility Worked Examples

Why MA(2) Moments and Invertibility Belong Together

An MA(2) process depends on the current innovation and the two preceding innovations. Its variance and autocovariances can therefore be calculated by identifying which innovations appear in both observations. Invertibility asks a different but related question: whether the innovations can be recovered from the observed process through a stable representation. The coefficient values determine both calculations, but the moment formulas and the root test answer different questions.

The key practical discipline is to use one polynomial convention consistently. Here the coefficient of the current innovation is one, and the MA polynomial is \(1+\theta_1 z+\theta_2 z^2\). Invertibility requires every root of this polynomial to lie strictly outside the unit circle.

Definition: An MA(2) process is written as \(X_t=\mu+\varepsilon_t+\theta_1\varepsilon_{t-1}+\theta_2\varepsilon_{t-2}\), where \(\mu\) is constant and \(\{\varepsilon_t\}\) is white noise with mean zero and variance \(\sigma_\varepsilon^2\). White noise has zero covariance between distinct time points. Define \(\gamma_k=\operatorname{Cov}(X_t,X_{t-k})\) for \(k\geq 0\). The MA polynomial is \(\theta(z)=1+\theta_1z+\theta_2z^2\).

These assumptions are sufficient for the covariance calculations below; the innovations do not have to be normally distributed. In the worked examples, normal distributions are specified to make each example's distribution explicit, but normality is not needed for the formulas. Since the model is a finite linear combination of white-noise terms with constant coefficients, its mean is constant and its autocovariance depends only on the lag.

Moment Formulas From Shared Innovations

The variance is the covariance at lag zero. At lag one, the observations \(X_t\) and \(X_{t-1}\) share two innovations; at lag two, they share one. Innovations at distinct times contribute zero covariance. This overlap gives the formulas directly.

Write the two observations.
Keep the time indices visible, so the shared and unshared innovations can be identified without guessing.

Match innovation indices.
Use \(\operatorname{Var}(\varepsilon_t)=\sigma_\varepsilon^2\) for a matched term. Products involving innovations at distinct times have covariance zero.

Add the matched contributions.
At lag zero, square each coefficient. At positive lags, multiply coefficients only for innovations appearing in both observations.

\begin{aligned} E[X_t]&=\mu,\\ \gamma_0&=\operatorname{Var}(X_t) =\sigma_\varepsilon^2(1+\theta_1^2+\theta_2^2),\\ \gamma_1&=\operatorname{Cov}(X_t,X_{t-1}) =\sigma_\varepsilon^2(\theta_1+\theta_1\theta_2) =\sigma_\varepsilon^2\theta_1(1+\theta_2),\\ \gamma_2&=\operatorname{Cov}(X_t,X_{t-2}) =\sigma_\varepsilon^2\theta_2,\\ \gamma_k&=0 \qquad (k>2). \end{aligned}

The positive-lag autocovariances also satisfy \(\gamma_{-k}=\gamma_k\), as covariance is symmetric. If autocorrelations are needed, they are \(\rho_k=\gamma_k/\gamma_0\), provided \(\gamma_0>0\). The autocorrelation is dimensionless; the autocovariance retains the squared measurement units of \(X_t\).

Why the lag-one formula holds: After subtracting the constant mean, \(X_t=\varepsilon_t+\theta_1\varepsilon_{t-1}+\theta_2\varepsilon_{t-2}\), while \(X_{t-1}=\varepsilon_{t-1}+\theta_1\varepsilon_{t-2}+\theta_2\varepsilon_{t-3}\). The shared terms are \(\varepsilon_{t-1}\) and \(\varepsilon_{t-2}\). Their coefficient products are \(\theta_1\cdot1\) and \(\theta_2\cdot\theta_1\), respectively. Each contributes its product times \(\sigma_\varepsilon^2\), so \(\gamma_1=\sigma_\varepsilon^2(\theta_1+\theta_1\theta_2)\). All other cross-products have zero covariance.

Worked Examples: Calculating Moments and Testing Roots

Worked Example: All Three Nonzero Autocovariances

A hypothetical monthly index follows \(X_t=10+\varepsilon_t+0.5\varepsilon_{t-1}-0.25\varepsilon_{t-2}\), where the innovations are independent and identically distributed as \(N(0,4)\). Calculate the mean, variance, first two autocovariances, and first two autocorrelations. The method is to apply the overlap formulas; the common mistake to avoid is treating the lag-one covariance as just \(\sigma_\varepsilon^2\theta_1\).

Here \(\mu=10\), \(\theta_1=0.5\), \(\theta_2=-0.25\), and \(\sigma_\varepsilon^2=4\). Thus \(E[X_t]=10\).

For the variance, square the two MA coefficients and add the squared coefficient of the current innovation:

\gamma_0=4\left(1+0.5^2+(-0.25)^2\right) =4(1+0.25+0.0625)=4(1.3125)=5.25.

At lag one, both shared innovations must be counted:

\gamma_1=4\left(0.5+(0.5)(-0.25)\right) =4(0.5-0.125)=4(0.375)=1.5.

At lag two, there is one shared innovation:

\gamma_2=4(-0.25)=-1.

Divide each autocovariance by \(\gamma_0=5.25\):

\rho_1=\frac{1.5}{5.25}=\frac{2}{7}\approx0.2857, \qquad \rho_2=\frac{-1}{5.25}=-\frac{4}{21}\approx-0.1905.

The mean is 10, the variance is 5.25, and the autocovariances are \(\gamma_1=1.5\) and \(\gamma_2=-1\). The autocorrelations are approximately 0.2857 and -0.1905, rounded to four decimal places. As a check, \(\gamma_1\) is positive because the positive \(\theta_1\) contribution outweighs the negative shared-term contribution \(\theta_1\theta_2=-0.125\).

Worked Example: A Quadratic With Two Real Roots

A hypothetical quarterly process has \(X_t=\varepsilon_t-\frac{5}{6}\varepsilon_{t-1}+\frac{1}{6}\varepsilon_{t-2}\), with independent innovations distributed as \(N(0,9)\). Find its variance and first two autocovariances, and test whether the process is invertible. The method is to calculate the moments from the coefficients and then solve \(\theta(z)=0\). The common mistake to avoid is checking only one of the two roots.

The parameters are \(\mu=0\), \(\theta_1=-5/6\), \(\theta_2=1/6\), and \(\sigma_\varepsilon^2=9\). The variance is

\gamma_0=9\left(1+\frac{25}{36}+\frac{1}{36}\right) =9\left(\frac{62}{36}\right)=\frac{31}{2}=15.5.

The first two autocovariances are

\begin{aligned} \gamma_1 &=9\left(-\frac{5}{6}+\left(-\frac{5}{6}\right)\left(\frac{1}{6}\right)\right) =9\left(-\frac{30}{36}-\frac{5}{36}\right) =-\frac{35}{4}=-8.75,\\ \gamma_2&=9\left(\frac{1}{6}\right)=\frac{3}{2}=1.5. \end{aligned}

The MA polynomial is \(\theta(z)=1-\frac{5}{6}z+\frac{1}{6}z^2\). Set it equal to zero and multiply by 6:

z^2-5z+6=0 \quad\Longrightarrow\quad (z-2)(z-3)=0.

The roots are \(z=2\) and \(z=3\), and both have absolute value greater than 1. Therefore, the process is invertible. The moment calculations and root test are consistent: the moments use the coefficients and innovation variance, while the invertibility decision depends on the locations of all polynomial roots.

Worked Example: One Root Inside the Unit Circle

A hypothetical monthly signal is modeled as \(X_t=\varepsilon_t-2.5\varepsilon_{t-1}+\varepsilon_{t-2}\), where the innovations are independent and identically distributed as \(N(0,1)\). Calculate its first three autocovariances and decide whether it is invertible. The method is to solve the full characteristic polynomial. The common mistake to avoid is declaring invertibility because one root is outside the unit circle.

Here \(\mu=0\), \(\theta_1=-2.5\), \(\theta_2=1\), and \(\sigma_\varepsilon^2=1\). The moments are

\begin{aligned} \gamma_0&=1\left(1+(-2.5)^2+1^2\right)=1+6.25+1=8.25,\\ \gamma_1&=1\left(-2.5+(-2.5)(1)\right)=-5,\\ \gamma_2&=1(1)=1. \end{aligned}

The polynomial is \(\theta(z)=1-2.5z+z^2\). Solving the quadratic gives

z^2-2.5z+1=0 \quad\Longrightarrow\quad z=\frac{2.5\pm\sqrt{2.5^2-4}}{2} =\frac{2.5\pm1.5}{2},

so the roots are \(z=2\) and \(z=0.5\). Since \(|0.5|<1\), not every root is outside the unit circle, and the process is not invertible. A wrong conclusion would be “invertible because 2 is greater than 1.” The correction is to check every root: a single root inside or on the unit circle fails the strict condition.

Applying the Invertibility Test Correctly

For the convention \(X_t-\mu=\theta(B)\varepsilon_t\), where \(B\) denotes the backshift operation, the MA polynomial is \(\theta(z)=1+\theta_1z+\theta_2z^2\). An MA(2) process is invertible when every solution of \(\theta(z)=0\) has modulus strictly greater than 1. A root on the unit circle does not pass the test.

The condition uses roots of the polynomial as written, not roots of a polynomial formed by reversing the coefficients. This distinction matters because solving \(1+\theta_1z+\theta_2z^2=0\) yields roots whose locations are tested against the unit circle. Do not take reciprocals unless deliberately changing to a different polynomial convention, and do not treat a root of modulus exactly 1 as acceptable.

The reason the root locations matter is that inversion expands \(1/\theta(B)\) as a power series in the backshift operator. For example, a factor \(1-B/r\) has reciprocal \(1+(B/r)+(B/r)^2+\cdots\), which converges when \(|1/r|<1\), equivalently \(|r|>1\). For a quadratic polynomial, factoring it into its root-based factors applies this requirement to both roots. Thus the strict outside-unit-circle condition gives a convergent representation in current and past observations.

Common Mistakes and Exam Checks

Dropping the second lag-one contribution. The correct formula is \(\gamma_1=\sigma_\varepsilon^2(\theta_1+\theta_1\theta_2)\), not just \(\sigma_\varepsilon^2\theta_1\). Write both adjacent observations and match their shared innovation indices.
Squaring the lag-two coefficient in the variance formula. The variance includes \(\theta_2^2\), but \(\gamma_2=\sigma_\varepsilon^2\theta_2\). Variance and lag-two covariance are different calculations.
Forgetting the innovation variance. The coefficient-only expressions must be multiplied by \(\sigma_\varepsilon^2\). If the innovation variance changes, every autocovariance scales by that variance.
Using a non-strict root condition. Invertibility requires \(|z|>1\) for every root. A root with \(|z|=1\) is not sufficient.
Checking only one root. Both roots of the quadratic must satisfy the condition, including complex roots, for which the relevant quantity is their modulus.

Key takeaway: For an MA(2), calculate moments by matching innovations shared across observations. Test invertibility separately by solving \(1+\theta_1z+\theta_2z^2=0\) and verifying that every root has modulus strictly greater than 1.

Check Your Understanding

For each problem, calculate the requested moments or root test using the stated MA polynomial convention.

A process has \(X_t=3+\varepsilon_t+0.4\varepsilon_{t-1}+0.2\varepsilon_{t-2}\), with independent innovations distributed as \(N(0,5)\). Find \(\gamma_0,\gamma_1,\gamma_2\), and state \(\gamma_3\). (Answer: \(\gamma_0=6,\ \gamma_1=2.4,\ \gamma_2=1,\ \gamma_3=0\).)
An MA(2) has \(\theta_1=0,\theta_2=-0.36\) and innovation distribution \(N(0,2)\). Find \(\gamma_0,\gamma_1,\gamma_2\), solve its MA polynomial, and determine invertibility. (Answer: \(\gamma_0=2.2592,\ \gamma_1=0,\ \gamma_2=-0.72\); roots are approximately \(\pm1.6667\), so it is invertible.)
For \(\theta(z)=1-1.2z+0.32z^2\), solve for both roots and decide whether the MA(2) is invertible. (Answer: roots \(1.25\) and \(2.5\); invertible.)
A process has \(\theta_1=-1.5,\theta_2=0.5\) and innovation distribution \(N(0,4)\). Calculate \(\gamma_0,\gamma_1,\gamma_2\), then test invertibility. (Answer: \(\gamma_0=14,\ \gamma_1=-9,\ \gamma_2=2\); roots are \(1\) and \(2\), so it is not invertible.)
A process has mean 6, \(\theta_1=0.25,\theta_2=0.5\), and independent innovations distributed as \(N(0,4)\). Calculate the first three autocovariances and determine invertibility from the polynomial roots. (Answer: \(\gamma_0=5.25,\ \gamma_1=1.5,\ \gamma_2=2\); the roots are \(-0.25\pm i\sqrt{31}/4\), each with modulus \(\sqrt{2}>1\), so it is invertible.)

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