Deriving The Moments Of An MA(1) Process

Why the Moments of an MA(1) Process Depend on Shared Innovations

A moving-average process is built from current and past random shocks. To calculate its moments, the key question is which shocks appear in each observation. In an MA(1) process, adjacent observations share one innovation; observations separated by at least two time periods share none. This overlap determines the variance and autocovariances.

The model and notation below use a constant process mean and a zero-mean innovation. Keeping these roles distinct matters: the mean describes the level of the process, while the innovation variance describes the variability of the underlying shocks.

Definition: An MA(1) process is written as \(X_t=\mu+\varepsilon_t+\theta\varepsilon_{t-1}\), where \(\mu\) is a constant, \(\theta\) is the MA(1) coefficient, and \(\{\varepsilon_t\}\) is a sequence of innovations. For the derivations here, assume the innovations are independent and identically distributed, with \(E[\varepsilon_t]=0\) and \(\operatorname{Var}(\varepsilon_t)=\sigma_\varepsilon^2<\infty\). In particular, distinct innovations have covariance zero.

The assumptions are stronger than necessary for these moment formulas: zero means, equal finite variances, and zero covariance between distinct innovations suffice. Independence is a convenient, standard assumption and also guarantees that zero covariance holds. The formulas do not require the innovations to be normally distributed.

Definition: For a weakly stationary process, the autocovariance at lag \(h\) is \(\gamma_h=\operatorname{Cov}(X_t,X_{t-h})\), and its variance is \(\gamma_0=\operatorname{Var}(X_t)\). The process mean is \(E[X_t]\). These quantities do not depend on the choice of time \(t\).

Deriving the Mean, Variance, and Autocovariances

Start with the model \(X_t=\mu+\varepsilon_t+\theta\varepsilon_{t-1}\). Since each innovation has expectation zero, the expected values of both innovation terms are zero. Thus \(E[X_t]=\mu\), which is constant over time.

For the variance, subtract the mean to get \(X_t-\mu=\varepsilon_t+\theta\varepsilon_{t-1}\). Expanding the variance gives the variances of the two innovation terms and their covariance. The covariance is zero because the two terms involve distinct innovations.

\begin{aligned} \gamma_0 &=\operatorname{Var}(X_t)\\ &=\operatorname{Var}(\varepsilon_t+\theta\varepsilon_{t-1})\\ &=\operatorname{Var}(\varepsilon_t)+\theta^2\operatorname{Var}(\varepsilon_{t-1}) +2\theta\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-1})\\ &=\sigma_\varepsilon^2+\theta^2\sigma_\varepsilon^2\\ &=(1+\theta^2)\sigma_\varepsilon^2. \end{aligned}

For lag one, compare \(X_t\) with \(X_{t-1}\). After subtracting the mean, these observations are \(\varepsilon_t+\theta\varepsilon_{t-1}\) and \(\varepsilon_{t-1}+\theta\varepsilon_{t-2}\). Their only shared innovation is \(\varepsilon_{t-1}\). Expanding the covariance shows that the shared term contributes \(\theta\sigma_\varepsilon^2\); all terms involving distinct innovations contribute zero.

\begin{aligned} \gamma_1 &=\operatorname{Cov}(X_t,X_{t-1})\\ &=\operatorname{Cov}(\varepsilon_t+\theta\varepsilon_{t-1}, \varepsilon_{t-1}+\theta\varepsilon_{t-2})\\ &=\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-1}) +\theta\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-2})\\ &\quad+\theta\operatorname{Cov}(\varepsilon_{t-1},\varepsilon_{t-1}) +\theta^2\operatorname{Cov}(\varepsilon_{t-1},\varepsilon_{t-2})\\ &=\theta\sigma_\varepsilon^2. \end{aligned}

At lag two, \(X_t\) involves \(\varepsilon_t\) and \(\varepsilon_{t-1}\), while \(X_{t-2}\) involves \(\varepsilon_{t-2}\) and \(\varepsilon_{t-3}\). There is no shared innovation, so every covariance term in the expansion is zero. The same reasoning applies at every larger lag. Autocovariance is symmetric in its two arguments, so the result also holds for negative lags.

\gamma_h= \begin{cases} (1+\theta^2)\sigma_\varepsilon^2, & h=0,\\ \theta\sigma_\varepsilon^2, & |h|=1,\\ 0, & |h|\ge 2. \end{cases}

Theorem: Under the stated innovation assumptions, an MA(1) process has mean \(\mu\), variance \((1+\theta^2)\sigma_\varepsilon^2\), lag-one autocovariance \(\theta\sigma_\varepsilon^2\), and zero autocovariance at every lag of absolute value at least two. These are moment results; they do not require a normal distribution or an invertibility condition.

A short justification for the zero covariances makes the general pattern clear. If \(|h|\ge2\), the innovation indices in \(X_t-\mu\), namely \(t\) and \(t-1\), do not overlap the indices in \(X_{t-h}-\mu\), namely \(t-h\) and \(t-h-1\). Each product in the covariance expansion therefore involves distinct innovations and has expectation zero. This argument relies on the zero covariance of distinct innovations; without that assumption, the conclusion need not hold.

Worked Moment Calculations

Worked Example: Calculating Moments from the Model Parameters

A hypothetical portfolio’s monthly index is modeled as \(X_t=12+\varepsilon_t+0.4\varepsilon_{t-1}\). Assume the innovations are independent with mean zero and variance \(25\). Find the process mean, variance, lag-one autocovariance, and lag-one correlation. Also give the lag-two autocovariance.

Method and assumptions. This is an MA(1) with \(\mu=12\), \(\theta=0.4\), and \(\sigma_\varepsilon^2=25\). Use the derived formulas; no distributional assumption beyond the stated innovation moments is needed.

Mean. The process mean is \(E[X_t]=\mu=12\).

Variance. Substitute the coefficient and innovation variance: \(\gamma_0=(1+\theta^2)\sigma_\varepsilon^2=(1+0.4^2)(25)=(1+0.16)(25)=1.16(25)=29\).

Lag-one autocovariance. \(\gamma_1=\theta\sigma_\varepsilon^2=(0.4)(25)=10\).

Lag-one correlation. Correlation is covariance divided by the common variance: \(\gamma_1/\gamma_0=10/29=0.344827\ldots\), or \(0.3448\) to four decimal places.

Lag two. Since \(|2|\ge2\), \(\gamma_2=0\). The result follows because \(X_t\) and \(X_{t-2}\) share no innovation.

Check and common mistake avoided. The variance \(29\) exceeds the innovation variance \(25\), as expected from the additional independent term \(0.4\varepsilon_{t-1}\). The lag-one covariance is positive because \(\theta\) is positive. Using \(\theta^2\sigma_\varepsilon^2\) for the lag-one covariance would incorrectly give \(4\); the correct coefficient is \(\theta\), not \(\theta^2\).

Worked Example: A Negative MA(1) Coefficient

A hypothetical weekly claims measure follows \(X_t=100+\varepsilon_t-0.5\varepsilon_{t-1}\), where the independent innovations have mean zero and standard deviation \(8\). Calculate the variance, lag-one autocovariance, lag-one correlation, and lag-three autocovariance.

Method and assumptions. Here \(\mu=100\), \(\theta=-0.5\), and the innovation variance is \(\sigma_\varepsilon^2=8^2=64\). The sign of \(\theta\) must be retained in the autocovariance calculation.

Variance. \(\gamma_0=(1+\theta^2)\sigma_\varepsilon^2=(1+(-0.5)^2)(64)=(1+0.25)(64)=1.25(64)=80\).

Lag-one autocovariance. \(\gamma_1=\theta\sigma_\varepsilon^2=(-0.5)(64)=-32\).

Lag-one correlation. \(\gamma_1/\gamma_0=-32/80=-0.4\).

Lag three. Since \(|3|\ge2\), \(\gamma_3=0\).

Check and common mistake avoided. The negative lag-one covariance agrees with the negative coefficient: the shared innovation enters the two neighboring observations with opposite signs. Squaring the coefficient in the covariance formula would give a positive answer and lose this sign information. The coefficient is squared in the variance formula, but not in the lag-one covariance formula.

Worked Example: Finding the Innovation Variance from the Process Variance

A hypothetical monthly expense index has the model \(X_t=3+\varepsilon_t-0.25\varepsilon_{t-1}\). Its process variance is known to be \(34\). Assuming independent, zero-mean innovations with a common finite variance, find the innovation variance, lag-one autocovariance, and lag-two autocovariance.

Method and assumptions. The model gives \(\mu=3\) and \(\theta=-0.25\). Rearrange the MA(1) variance formula to solve for the innovation variance.

Innovation variance. Since \(34=(1+(-0.25)^2)\sigma_\varepsilon^2=(1+0.0625)\sigma_\varepsilon^2=1.0625\sigma_\varepsilon^2\), we obtain \(\sigma_\varepsilon^2=34/1.0625=34/(17/16)=34(16/17)=32\).

Lag-one autocovariance. \(\gamma_1=\theta\sigma_\varepsilon^2=(-0.25)(32)=-8\).

Lag-two autocovariance. \(\gamma_2=0\), because observations two periods apart do not share an innovation.

Check and common mistake avoided. Substitution gives \(\gamma_0=(1+0.0625)(32)=1.0625(32)=34\), matching the stated process variance. Do not set the process variance equal to the innovation variance: the lagged innovation contributes an additional variance of \(\theta^2\sigma_\varepsilon^2=0.0625(32)=2\).

Common Errors and Why the Distinctions Matter

These moment formulas are useful because they connect the model’s parameters to observable features of a time series. They also provide direct checks on a proposed calculation. A variance must be nonnegative; the sign of the lag-one covariance matches the sign of \(\theta\) when the innovation variance is positive; and all autocovariances at lags of two or more must be zero under the stated model and assumptions.

Several errors arise from treating the coefficient, its square, and the innovation variance as interchangeable. They play different roles:

Using the wrong variance. \(\sigma_\varepsilon^2\) is the variance of an innovation, while \(\gamma_0=(1+\theta^2)\sigma_\varepsilon^2\) is the variance of the process.
Squaring the lag-one coefficient. The lag-one covariance is \(\theta\sigma_\varepsilon^2\), not \(\theta^2\sigma_\varepsilon^2\). Its sign carries information about whether the shared innovation enters the two observations with the same or opposite signs.
Assuming lag-one covariance is always positive. It is negative when \(\theta<0\), zero when \(\theta=0\), and positive when \(\theta>0\), provided \(\sigma_\varepsilon^2>0\).
Misreading the lag. A lag of one compares neighboring observations and produces one shared innovation. A lag of two or more produces no shared innovations and hence zero covariance.
Assuming normality is needed. The moment derivations use only the innovation means, variances, and zero covariances. Normality is not needed for these results.

Key takeaway: In an MA(1), each observation contains two consecutive innovations. Its variance includes both innovation variances, while adjacent observations share exactly one innovation. That overlap gives \(\gamma_1=\theta\sigma_\varepsilon^2\); no overlap at larger lags gives zero autocovariance.

Check Your Understanding

For each problem, use the MA(1) moment formulas and keep the sign of the coefficient when calculating covariance.

A hypothetical index has \(X_t=4+\varepsilon_t+0.3\varepsilon_{t-1}\), with independent innovations of mean zero and variance \(10\). Find its mean, variance, and lag-one autocovariance. (Answer: mean \(4\), variance \(10.9\), lag-one autocovariance \(3\).)
A hypothetical weekly count is modeled by \(X_t=20+\varepsilon_t-0.8\varepsilon_{t-1}\), where the innovation standard deviation is \(3\). Find the process variance and lag-one autocovariance. (Answer: variance \(14.76\), lag-one autocovariance \(-7.2\).)
An MA(1) process has \(\theta=0.5\) and process variance \(45\). Find the innovation variance and lag-one autocovariance. (Answer: innovation variance \(36\), lag-one autocovariance \(18\).)
A hypothetical monthly measure has \(X_t=7+\varepsilon_t+0.2\varepsilon_{t-1}\), with innovation variance \(25\). Find its variance, \(\operatorname{Cov}(X_t,X_{t-1})\), and \(\operatorname{Cov}(X_t,X_{t-3})\). (Answer: variance \(26\), lag-one covariance \(5\), lag-three covariance \(0\).)
An MA(1) process has innovation variance \(30\), mean \(20\), and lag-one autocovariance \(-12\). Find \(\theta\), the process variance, and the lag-three autocovariance. (Answer: \(\theta=-0.4\), process variance \(34.8\), lag-three autocovariance \(0\).)

What You'll Learn