Autocovariances Of MA(q) Processes

Why MA(q) Autocovariances Have a Finite Cutoff

An MA(q) process is built from a finite number of current and past innovations. At nearby times, two observations can share some of those innovations, creating covariance. Once the observations are separated by more than q time periods, they share none, so their covariance is zero. The key calculation is therefore to identify exactly which innovation terms overlap.

The autocovariance function records the covariance between observations separated by a given lag. Its values describe the scale and dependence pattern of the process; they also provide the basis for the autocorrelation function, which rescales each autocovariance by the variance.

Definition: Let \(\{\varepsilon_t\}\) be a white-noise sequence with \(E[\varepsilon_t]=0\), \(\operatorname{Var}(\varepsilon_t)=\sigma^2\), and \(\operatorname{Cov}(\varepsilon_t,\varepsilon_s)=0\) for \(t\ne s\). A finite MA(q) process is \[ X_t=\mu+\sum_{j=0}^{q}\theta_j\varepsilon_{t-j}, \qquad \theta_0=1, \] where \(q\) and the coefficients are fixed. Its autocovariance at lag \(h\) is \(\gamma(h)=\operatorname{Cov}(X_t,X_{t-h})\). The formula below requires only the stated zero means and covariance properties; independence of the innovations is not required.

Deriving the General Formula

Write each observation as a sum of its innovation terms. An innovation contributes to the covariance only when it occurs in both sums: distinct white-noise terms have covariance zero. For \(h\ge 0\), the term \(\varepsilon_{t-j}\) in \(X_t\) matches the term \(\varepsilon_{t-h-k}\) in \(X_{t-h}\) when \(t-j=t-h-k\), or \(j=h+k\). Thus \(j\) ranges from \(h\) through \(q\), and the matching coefficient product is \(\theta_j\theta_{j-h}\).

\begin{aligned} \gamma(h) &=\operatorname{Cov}\left(\sum_{j=0}^{q}\theta_j\varepsilon_{t-j}, \sum_{k=0}^{q}\theta_k\varepsilon_{t-h-k}\right)\\ &=\sum_{j=0}^{q}\sum_{k=0}^{q}\theta_j\theta_k \operatorname{Cov}(\varepsilon_{t-j},\varepsilon_{t-h-k})\\ &=\sigma^2\sum_{k=0}^{q-h}\theta_{h+k}\theta_k =\sigma^2\sum_{j=0}^{q-h}\theta_j\theta_{j+h}, \qquad 0\le h\le q. \end{aligned}

The mean \(\mu\) disappears from the covariance because subtracting the mean leaves only the innovation sums. At lag zero, every innovation term is matched with itself, giving the variance. At a lag greater than q, no terms match. Covariance is symmetric, so \(\gamma(-h)=\gamma(h)\).

Theorem: For the MA(q) process above, \[ \gamma(0)=\sigma^2\sum_{j=0}^{q}\theta_j^2,\qquad \gamma(h)=\sigma^2\sum_{j=0}^{q-h}\theta_j\theta_{j+h}\quad (1\le h\le q), \] and \(\gamma(h)=0\) for \(|h|>q\). The autocorrelation is \(\rho(h)=\gamma(h)/\gamma(0)\), provided \(\gamma(0)>0\).

The cutoff follows directly from the innovation indices: if \(h>q\), the intervals of innovation times used by \(X_t\) and \(X_{t-h}\) do not overlap. The formula does not say that every autocovariance from lag one through q must be nonzero. Coefficients can make a sum cancel, or all products at a lag can be zero.

Worked Example: Computing Every Autocovariance of an MA(2)

Suppose \(X_t=12+\varepsilon_t+0.5\varepsilon_{t-1}-0.25\varepsilon_{t-2}\), where the innovations are white noise with variance \(\sigma^2=4\). Find \(\gamma(0)\), \(\gamma(1)\), \(\gamma(2)\), and the values beyond lag two.

Method. Read the coefficients as \(\theta_0=1\), \(\theta_1=0.5\), and \(\theta_2=-0.25\), then apply the overlap formula at each lag. The constant mean 12 does not enter the covariance.

At lag zero, square each coefficient and add:

\gamma(0)=4\left(1^2+0.5^2+(-0.25)^2\right) =4(1+0.25+0.0625)=5.25.

At lag one, include both adjacent coefficient products:

\gamma(1)=4\left(\theta_0\theta_1+\theta_1\theta_2\right) =4\left(1(0.5)+0.5(-0.25)\right) =4(0.5-0.125)=1.5.

At lag two, only one pair overlaps:

\gamma(2)=4(\theta_0\theta_2)=4(1)(-0.25)=-1.

The process has order two, so \(\gamma(h)=0\) for \(h>2\); for negative lags, symmetry gives \(\gamma(-h)=\gamma(h)\). As a check, the variance is positive, the negative lag-two covariance agrees with the negative coefficient \(\theta_2\), and no innovation is shared at lag three or larger.

Common mistake avoided: using only \(4\theta_1=2\) for \(\gamma(1)\) omits the overlap between the lag-one and lag-two innovation terms. The complete sum is \(4(\theta_0\theta_1+\theta_1\theta_2)=1.5\).

Working from an Unknown Coefficient

An observed or specified autocovariance can also give an equation for an unknown MA coefficient. The equation must include every product at that lag, not just the first product in the sum. After solving for the coefficient, substitute it into the other lag formulas; do not reuse an autocovariance value calculated under a different coefficient.

Worked Example: Solving for a Coefficient from the Lag-One Autocovariance

Let \(X_t=\varepsilon_t+0.4\varepsilon_{t-1}+a\varepsilon_{t-2}+\varepsilon_{t-3}\), with white-noise innovations of mean zero and variance \(\sigma^2=2\). If \(\gamma(1)=0.7\), find \(a\), then calculate \(\gamma(0)\) and \(\gamma(2)\). Round final values to four decimal places.

Method. The coefficient sequence is \((1,0.4,a,1)\). At lag one, the products are \(1(0.4)\), \(0.4a\), and \(a(1)\). Therefore,

$$ 0.7=2(0.4+0.4a+a)=2(0.4+1.4a). $$

Divide both sides by two and solve:

0.35=0.4+1.4a,\qquad a=\frac{0.35-0.4}{1.4}=-0.0357142857\ldots

Thus \(a=-0.0357\) to four decimal places. For the variance, square all four coefficients:

\begin{aligned} \gamma(0) &=2\left(1^2+0.4^2+a^2+1^2\right)\\ &=2\left(1+0.16+0.0012755102\ldots+1\right)\\ &=4.3225510204\ldots=4.3226. \end{aligned}

At lag two, the matching products are \(\theta_0\theta_2=a\) and \(\theta_1\theta_3=0.4(1)\):

\gamma(2)=2(a+0.4) =2(0.3642857143\ldots) =0.7285714286\ldots=0.7286.

The calculation checks against the given lag-one value: \(2(0.4+1.4a)=2(0.4-0.05)=0.7\). Since this is an MA(3), its autocovariances are zero beyond lag three. The common mistake is to report a positive coefficient from an incorrect rearrangement; the equation gives a negative \(a\), and the variance and lag-two covariance above use that negative value.

Overlap Sums, Signs, and the Cutoff

Autocovariances can be positive, negative, or zero. A negative coefficient does not automatically make every autocovariance negative: each lag is a sum of products, and different products can have different signs. A zero at a lag can result from cancellation even when the process has nonzero coefficients.

Worked Example: A Sparse MA(3) and a Corrected Overlap Sum

Suppose \(X_t=\varepsilon_t+0\varepsilon_{t-1}+0.5\varepsilon_{t-2}-0.2\varepsilon_{t-3}\), where the innovations are white noise with variance \(\sigma^2=3\). Find the autocovariances from lag zero through lag three and state the cutoff. Round final values to four decimal places.

Method. Use \((\theta_0,\theta_1,\theta_2,\theta_3)=(1,0,0.5,-0.2)\). First, the variance is

\gamma(0)=3\left(1^2+0^2+0.5^2+(-0.2)^2\right) =3(1+0+0.25+0.04)=3.87.

At lag one, the three adjacent products are \(1(0)\), \(0(0.5)\), and \(0.5(-0.2)\):

\gamma(1)=3(0+0-0.1)=-0.3.

At lag two, there are two overlap products, \(1(0.5)\) and \(0(-0.2)\):

\gamma(2)=3\left(1(0.5)+0(-0.2)\right)=3(0.5)=1.5.

At lag three, only the first and last coefficient pair overlaps:

\gamma(3)=3\left(1(-0.2)\right)=-0.6.

Therefore the values, to four decimal places, are \(3.8700\), \(-0.3000\), \(1.5000\), and \(-0.6000\). For \(h>3\), \(\gamma(h)=0\); negative lags repeat the corresponding positive-lag values. The check is the overlap count: lag one has three candidate pairs, lag two has two, and lag three has one. In particular, the lag-two value is \(1.5\), not \(1.2\): the displayed lag-two sum contains exactly the two products shown.

Common Errors and Practical Checks

The most reliable way to calculate an MA(q) autocovariance is to write the coefficient sequence with its indices and list the products for the lag in question. This prevents a missing endpoint term or an extra product from changing the answer. The following checks are useful:

At lag zero, square every coefficient. The variance is \(\sigma^2\sum_{j=0}^q\theta_j^2\), so it cannot be negative.
At a positive lag, pair coefficients separated by that lag. For lag \(h\), include \(\theta_j\theta_{j+h}\) for each \(j=0,\ldots,q-h\), and no other terms.
Apply the innovation variance once. Multiply the sum of products by \(\sigma^2\); do not multiply individual terms by it again.
Use the correct cutoff. An MA(q) has zero autocovariance when \(|h|>q\), although some autocovariances at smaller lags may also be zero.
Use symmetry for negative lags. The autocovariance function satisfies \(\gamma(-h)=\gamma(h)\); it is not generally an odd function.

These calculations are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus. In an exam problem, a clear coefficient list and a separate overlap sum for each requested lag make both the reasoning and any arithmetic errors easier to check.

Check Your Understanding

For each problem, identify the overlap terms before calculating. Answers are given at the end of each question.

A hypothetical quarterly reserve index follows \(X_t=5+\varepsilon_t-0.4\varepsilon_{t-1}\), where the innovations are white noise with variance \(2\). Find \(\gamma(0)\), \(\gamma(1)\), and \(\gamma(2)\). (Answer: \(\gamma(0)=2.3200\), \(\gamma(1)=-0.8000\), and \(\gamma(2)=0\).)
A hypothetical monthly service-cost series has coefficient sequence \((1,0.2,0.6)\) and innovation variance \(1.5\). Calculate \(\gamma(0)\), \(\gamma(1)\), and \(\gamma(2)\), and state the cutoff. (Answer: \(\gamma(0)=2.1000\), \(\gamma(1)=0.4800\), \(\gamma(2)=0.9000\), and \(\gamma(h)=0\) for \(h>2\).)
A hypothetical quarterly claims index has coefficient sequence \((1,0,-0.5,0.25)\) and innovation variance \(4\). Find the autocovariances from lag zero through lag three, including their signs. (Answer: \(\gamma(0)=5.2500\), \(\gamma(1)=-0.5000\), \(\gamma(2)=-2.0000\), and \(\gamma(3)=1.0000\).)
For a hypothetical MA(2), the coefficient sequence is \((1,a,0.5)\), the innovation variance is \(2\), and \(\gamma(1)=1.2\). Solve for \(a\), then find \(\gamma(0)\) and \(\gamma(2)\). (Answer: \(a=0.4000\), \(\gamma(0)=2.8200\), and \(\gamma(2)=1.0000\).)
A hypothetical MA(3) has coefficient sequence \((1,0,0.5,-0.2)\) and innovation variance \(3\). A calculation reports \(\gamma(2)=1.2\). Identify the error and give \(\gamma(1)\), \(\gamma(2)\), and \(\gamma(3)\). (Answer: the lag-two overlap is \(1(0.5)+0(-0.2)=0.5\), so \(\gamma(2)=1.5000\), not \(1.2000\); \(\gamma(1)=-0.3000\) and \(\gamma(3)=-0.6000\).)

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