The ACF Cutoff Of An MA(1) Process

Why an MA(1) ACF Has a Cutoff

The autocorrelation function (ACF) describes how a time-series value is linearly associated with values at different time lags. For an MA(1) process, the ACF has a distinctive pattern: it can be nonzero at lag one, but it is zero at every lag greater than one. The reason is structural. Each observation contains only two consecutive innovations, so observations separated by at least two time steps have no innovation in common.

This tutorial derives that pattern and shows how to calculate the lag-one correlation. The treatment is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: Under the convention used here, an MA(1) process is $$ Y_t=\mu+\varepsilon_t+\theta\varepsilon_{t-1}, $$ where $\mu$ and $\theta$ are constants and $\{\varepsilon_t\}$ is white noise with $E[\varepsilon_t]=0$, $\operatorname{Var}(\varepsilon_t)=\sigma^2$, and $\operatorname{Cov}(\varepsilon_t,\varepsilon_s)=0$ for $t\ne s$. Its autocovariance at lag $h$ is $\gamma_h=\operatorname{Cov}(Y_t,Y_{t-h})$, and its autocorrelation at lag $h$ is $\rho_h=\gamma_h/\gamma_0$, provided $\gamma_0>0$.

The white-noise assumptions are about the first two moments; independence is not required for the autocovariance derivation. The finite moving-average representation makes the process covariance stationary: its mean and variance are constant, and its covariance depends only on the time separation. Its mean is $E[Y_t]=\mu$. To calculate covariances, it is useful to center the series by writing $Z_t=Y_t-\mu=\varepsilon_t+\theta\varepsilon_{t-1}$.

Some texts define an MA(1) with a minus sign before the lagged innovation. Under that alternative convention, the reported coefficient changes sign. Here, all formulas use the plus-sign definition above.

Deriving the Autocovariance and ACF

The variance at lag zero comes from the two innovation terms in $Z_t$. They are uncorrelated because they occur at different times. At lag one, the two observations share exactly one innovation, $\varepsilon_{t-1}$. At lags of two or more, they share none.

Calculate the variance.
$Z_t=\varepsilon_t+\theta\varepsilon_{t-1}$. The variance of a sum is the sum of the two variances plus twice their covariance. The cross-covariance is zero, so $\gamma_0=\sigma^2+\theta^2\sigma^2=(1+\theta^2)\sigma^2$.

Calculate the lag-one covariance.
$Z_{t-1}=\varepsilon_{t-1}+\theta\varepsilon_{t-2}$. In the covariance between $Z_t$ and $Z_{t-1}$, only the pair $\theta\varepsilon_{t-1}$ and $\varepsilon_{t-1}$ contributes. Thus $\gamma_1=\theta\sigma^2$.

Check all larger lags.
For $h\ge2$, $Z_{t-h}=\varepsilon_{t-h}+\theta\varepsilon_{t-h-1}$. Neither of its innovations occurs in $Z_t$. Every cross-covariance is therefore zero, so $\gamma_h=0$.

Divide by the variance.
The ACF is the autocovariance divided by $\gamma_0$. This gives the lag-one value and the cutoff.

\gamma_h= \begin{cases} (1+\theta^2)\sigma^2, & h=0,\\ \theta\sigma^2, & |h|=1,\\ 0, & |h|\ge2, \end{cases} \qquad \rho_h= \begin{cases} 1, & h=0,\\ \dfrac{\theta}{1+\theta^2}, & |h|=1,\\ 0, & |h|\ge2. \end{cases}

For negative lags, $\gamma_{-h}=\gamma_h$ and $\rho_{-h}=\rho_h$, as holds for a real-valued covariance-stationary process. In particular, the lag-one correlation has the same value at both lag $1$ and lag $-1$. The innovation variance cancels in the ratio, so $\rho_1$ depends on $\theta$ but not on $\sigma^2$. The coefficient’s sign determines the sign of the lag-one correlation.

Why the cutoff holds: The ACF is based on covariance, and white noise at different times has zero covariance. Since $Y_t-\mu$ uses only $\varepsilon_t$ and $\varepsilon_{t-1}$, a value at time $t$ and one at time $t-h$ have no shared innovation when $|h|\ge2$. Their covariance, and therefore their autocorrelation, is zero.

The magnitude of the lag-one correlation cannot exceed one half. For any real $\theta$, $$ 1+\theta^2-2|\theta|=(|\theta|-1)^2\ge0, $$ so $2|\theta|\le1+\theta^2$. Dividing by the positive quantity $2(1+\theta^2)$ gives $\left|\rho_1\right|=\frac{|\theta|}{1+\theta^2}\le\frac12$. Equality occurs when $|\theta|=1$. This is a useful check on a calculation: a proposed MA(1) lag-one ACF with magnitude greater than $0.5$ cannot follow from this model.

Worked ACF Calculations

Worked Example: A Positive Lag-One Correlation

A hypothetical monthly service-volume index follows $Y_t=8+\varepsilon_t+0.5\varepsilon_{t-1}$. Assume the innovations are independent and normally distributed as $N(0,4)$, so the mean is zero and the variance is $4$. Find the process mean, variance, lag-one autocovariance, and ACF through lag three.

Method. Use the plus-sign MA(1) formulas with $\mu=8$, $\theta=0.5$, and $\sigma^2=4$. The distributional assumption is stated for the example; the ACF formulas themselves require only white-noise first and second moments.

The process mean is $E[Y_t]=\mu=8$. Its variance is $$ \gamma_0=(1+\theta^2)\sigma^2=(1+0.5^2)(4)=(1+0.25)(4)=1.25(4)=5. $$

The lag-one autocovariance is $$ \gamma_1=\theta\sigma^2=(0.5)(4)=2. $$ Therefore, $$ \rho_1=\frac{\gamma_1}{\gamma_0}=\frac{2}{5}=0.4. $$ At lag two and lag three there are no shared innovations, so $\gamma_2=\gamma_3=0$ and $\rho_2=\rho_3=0$. The ACF through lag three is $1, 0.4, 0, 0$, beginning at lag zero. No rounding is needed. The result is checked by $|\rho_1|=0.4\le0.5$.

Common mistake avoided: The lag-one correlation is not $\theta$ itself. It is $\theta/(1+\theta^2)$; here the correct value is $0.4$, not $0.5$.

Worked Example: A Negative Lag-One Correlation

A hypothetical quarterly portfolio indicator is modeled as $Y_t=30+\varepsilon_t-2\varepsilon_{t-1}$. The innovations are independent with distribution $N(0,9)$. Calculate the process variance, the lag-one covariance and correlation, and the autocorrelation at lag two.

Method. Match the model to $Y_t=\mu+\varepsilon_t+\theta\varepsilon_{t-1}$: $\mu=30$, $\theta=-2$, and $\sigma^2=9$. The negative sign is part of the coefficient; the innovation variance is $9$.

First, the variance is $$ \gamma_0=(1+\theta^2)\sigma^2=(1+(-2)^2)(9)=(1+4)(9)=45. $$ The lag-one covariance is $$ \gamma_1=\theta\sigma^2=(-2)(9)=-18. $$ Thus the lag-one correlation is $$ \rho_1=\frac{-18}{45}=-0.4. $$ At lag two, no innovation appears in both observations, so $\gamma_2=0$, and since $\gamma_0=45>0$, $\rho_2=0/45=0$. The results are checked by $|-0.4|\le0.5$, as required. No rounding is needed.

Common mistake avoided: Replacing the coefficient by its magnitude would incorrectly make the lag-one correlation positive. The sign of $\theta$ must be retained in $\gamma_1=\theta\sigma^2$.

Worked Example: Using the ACF to Find a Two-Reading Variance

A hypothetical monthly claims-processing measure follows $Y_t=100+\varepsilon_t+1.5\varepsilon_{t-1}$, with independent innovations $N(0,16)$. Find $\gamma_0$, $\gamma_1$, $\rho_1$, $\rho_2$, and $\operatorname{Var}(Y_t+Y_{t-1})$.

Method. Use $\mu=100$, $\theta=1.5$, and $\sigma^2=16$. The intercept affects the mean, not the variance or autocovariance.

The variance of one reading is $$ \gamma_0=(1+1.5^2)(16)=(1+2.25)(16)=3.25(16)=52. $$ The lag-one covariance and correlation are $$ \gamma_1=(1.5)(16)=24, \qquad \rho_1=\frac{24}{52}=\frac{6}{13}=0.4615 $$ to four decimal places. Since the ACF cuts off after lag one, $\gamma_2=0$ and $\rho_2=0$.

For two consecutive readings, use $\operatorname{Var}(A+B)=\operatorname{Var}(A)+\operatorname{Var}(B)+2\operatorname{Cov}(A,B)$. Both readings have variance $52$, and their covariance is $\gamma_1=24$, so $$ \operatorname{Var}(Y_t+Y_{t-1})=52+52+2(24)=104+48=152. $$ The mean $100$ does not enter this variance calculation. The results are checked by the correlation bound, since $0.4615<0.5$. Only $\rho_1$ is rounded, to four decimal places.

Common mistake avoided: Treating the two consecutive readings as uncorrelated would give $104$ and omit the $2\gamma_1=48$ contribution. The cutoff applies starting at lag two, not lag one.

Interpreting the Cutoff Carefully

An ACF cutoff is a statement about theoretical autocorrelation, not a guarantee that a sample correlogram will show exact zeros after lag one. A sample ACF is calculated from a finite set of observed values and will generally fluctuate around the theoretical ACF. In practice, the theoretical cutoff is a pattern to compare with the sample pattern, not a requirement that every estimated value after lag one equal exactly zero.

There is also an important distinction between being uncorrelated and being independent. The MA(1) derivation establishes zero covariance at lags of two or more. It does not, by itself, establish independence of observations at those lags. If the innovations are independent and Gaussian, the observations form a jointly Gaussian process, and zero covariance implies independence for pairs of observations. Without a Gaussian assumption, zero correlation alone is not enough to conclude independence.

The most common exam traps are to confuse the coefficient $\theta$ with $\rho_1$, to forget that a negative coefficient gives a negative lag-one correlation, or to place the cutoff at lag one rather than after lag one. Keep the indexing clear: lag zero has correlation $1$; lag one can be nonzero; every lag of two or more has correlation zero. A nonzero mean does not change this pattern.

Key takeaway: For $Y_t=\mu+\varepsilon_t+\theta\varepsilon_{t-1}$, the lag-one autocorrelation is $\theta/(1+\theta^2)$, and the ACF is zero at every lag with absolute value at least two. The cutoff follows because observations two or more periods apart share no innovation.

Check Your Understanding

Use the plus-sign MA(1) convention in each question. Assume the stated innovations are white noise; where a distribution is given, it describes those innovations.

A hypothetical index is $Y_t=5+\varepsilon_t+0.25\varepsilon_{t-1}$, with independent $\varepsilon_t\sim N(0,4)$. Find $\gamma_0$, $\gamma_1$, and $\rho_1$. (Answer: $\gamma_0=4.25,\ \gamma_1=1,\ \rho_1=0.2353$.)
A hypothetical process has $Y_t=12+\varepsilon_t-\varepsilon_{t-1}$, with independent $\varepsilon_t\sim N(0,5)$. Find its variance and lag-one correlation, including the sign. (Answer: $\gamma_0=10,\ \rho_1=-0.5000$.)
For $Y_t=40+\varepsilon_t+2\varepsilon_{t-1}$, suppose the innovations are independent $N(0,3)$. Calculate $\gamma_0$, $\gamma_1$, $\rho_1$, and $\rho_2$. (Answer: $\gamma_0=15,\ \gamma_1=6,\ \rho_1=0.4000,\ \rho_2=0$.)
A hypothetical monthly measure is $Y_t=12+\varepsilon_t+0.5\varepsilon_{t-1}$, with independent innovations $N(0,8)$. Find $\operatorname{Cov}(Y_t,Y_{t-1})$, $\operatorname{Var}(Y_t)$, and $\rho_1$. (Answer: covariance $=4$, variance $=10$, and $\rho_1=0.4000$.)
An analyst claims that any two MA(1) observations at least two periods apart must be independent because their ACF is zero. Is that conclusion justified by the white-noise assumptions alone? (Answer: No. Those assumptions imply zero covariance at those lags, not independence; independence follows for such pairs under additional conditions such as jointly Gaussian observations.)

What You'll Learn