A Stationary Process With Zero ACF but Dependent Squares

Why Zero Autocorrelation Does Not Settle Dependence

Autocorrelation describes linear association between a process and its lagged values. It is possible for every nonzero-lag autocorrelation to vanish even when observations are not independent. A useful example makes this distinction visible: construct each observation from two consecutive independent innovations. Neighboring observations share one innovation, but the shared term does not produce nonzero autocorrelation when the innovations have mean zero.

The dependence becomes apparent when we examine the squares. Squaring removes the signs of the innovations, so the shared innovation contributes to both neighboring squared observations. The construction therefore gives a direct demonstration that an autocorrelation function (ACF) alone does not capture every form of dependence.

Definition: Let $\{\varepsilon_t\}$ be an independent and identically distributed sequence with $E(\varepsilon_t)=0$, $0<\operatorname{Var}(\varepsilon_t)=\sigma_\varepsilon^2<\infty$, and $E(\varepsilon_t^4)=\mu_{4,\varepsilon}<\infty$. Define $$ X_t=\varepsilon_t\varepsilon_{t-1}. $$ If $\operatorname{Var}(\varepsilon_t^2)>0$, then the squared observations $X_t^2$ and $X_{t-1}^2$ are dependent. The process $X_t$ is stationary and has zero autocorrelation at every nonzero lag.

Stationarity and the Autocorrelation Function

Strict stationarity means that the joint distribution of any finite collection of observations is unchanged when all their time indices are shifted by the same amount. Here, a finite block $X_t,\ldots,X_{t+r}$ is formed from a finite block of innovations $\varepsilon_{t-1},\ldots,\varepsilon_{t+r}$, using the same rule at every time. Since the innovations are identically distributed and independent, shifting all the indices leaves the joint distribution of that innovation block unchanged. Applying the same product rule therefore leaves the joint distribution of the $X$'s unchanged. Thus $\{X_t\}$ is strictly stationary.

The finite second moment also lets us verify weak stationarity directly. Its mean is constant: $$ E(X_t)=E(\varepsilon_t)E(\varepsilon_{t-1})=0. $$ The two innovations in $X_t$ are independent, so its variance is $$ \operatorname{Var}(X_t)=E(X_t^2) =E(\varepsilon_t^2)E(\varepsilon_{t-1}^2) =\sigma_\varepsilon^4. $$ This variance does not depend on $t$. To find the autocovariances, use the definition $\gamma(k)=\operatorname{Cov}(X_t,X_{t-k})$. Because the mean is zero, $\gamma(k)=E(X_tX_{t-k})$.

At lag one the two observations share $\varepsilon_{t-1}$: $$ \gamma(1)=E(\varepsilon_t\varepsilon_{t-1}^2\varepsilon_{t-2}) =E(\varepsilon_t)E(\varepsilon_{t-1}^2)E(\varepsilon_{t-2})=0. $$ The factorization follows from independence; the outer factors have mean zero. The same reasoning applies at lag negative one. At any lag with absolute value at least two, $X_t$ and $X_{t-k}$ are functions of disjoint sets of independent innovations. They are independent and have mean zero, so their covariance is also zero. Consequently, the covariance depends only on the lag: it is $\sigma_\varepsilon^4$ at lag zero and zero otherwise. This verifies weak stationarity as well as strict stationarity.

The autocorrelation is $\rho(k)=\gamma(k)/\gamma(0)$. Since $\gamma(0)=\sigma_\varepsilon^4>0$, the resulting ACF is $$ \rho(0)=1,\qquad \rho(k)=0\quad\text{for every }k\ne 0. $$ A zero autocorrelation says that the covariance is zero; it does not, by itself, imply independence.

Worked Example: Zero ACF With Two-Point Innovations

Suppose the innovations are independent, with each $\varepsilon_t$ equal to $-1$ or $1$, each with probability $1/2$. Then $E(\varepsilon_t)=0$ and $\sigma_\varepsilon^2=1$. Form $X_t=\varepsilon_t\varepsilon_{t-1}$.

Mean and variance. Independence gives $E(X_t)=0$, and $X_t^2=1$ for every outcome. Hence $\operatorname{Var}(X_t)=1$.

Lag-one covariance. For adjacent observations, $$ \gamma(1)=E(\varepsilon_t\varepsilon_{t-1}^2\varepsilon_{t-2}) =E(\varepsilon_t)E(\varepsilon_{t-1}^2)E(\varepsilon_{t-2})=0. $$ At lags of absolute value at least two, the observations use disjoint innovations and are independent. Thus every nonzero-lag autocorrelation is zero.

What happens to the squares? Here $X_t^2=1$ for every $t$, so the squared observations are constant and are not dependent in any informative sense. This example is a useful qualification: zero ACF follows from the assumptions, but dependence among squares requires the squared innovations themselves to vary. The normal-innovation example below satisfies that additional condition.

Why Neighboring Squares Are Dependent

The squared process has the form $X_t^2=\varepsilon_t^2\varepsilon_{t-1}^2$. Consecutive squared observations both contain $\varepsilon_{t-1}^2$: $$ X_t^2=\varepsilon_t^2\varepsilon_{t-1}^2,\qquad X_{t-1}^2=\varepsilon_{t-1}^2\varepsilon_{t-2}^2. $$ This shared random factor can create dependence even though $\operatorname{Cov}(X_t,X_{t-1})=0$. The covariance calculation makes the condition precise.

\begin{aligned} E(X_t^2)&=\sigma_\varepsilon^4,\\ E(X_t^2X_{t-1}^2) &=E(\varepsilon_t^2\varepsilon_{t-1}^4\varepsilon_{t-2}^2)\\ &=\sigma_\varepsilon^2\mu_{4,\varepsilon}\sigma_\varepsilon^2 =\sigma_\varepsilon^4\mu_{4,\varepsilon},\\ \operatorname{Cov}(X_t^2,X_{t-1}^2) &=E(X_t^2X_{t-1}^2)-E(X_t^2)E(X_{t-1}^2)\\ &=\sigma_\varepsilon^4(\mu_{4,\varepsilon}-\sigma_\varepsilon^4) =\sigma_\varepsilon^4\operatorname{Var}(\varepsilon_t^2). \end{aligned}

The fourth-moment assumption ensures these quantities are finite. If $\operatorname{Var}(\varepsilon_t^2)>0$, the covariance is positive. Since independent variables with finite second moments have covariance zero, this positive covariance proves that $X_t^2$ and $X_{t-1}^2$ are dependent. At lags of absolute value at least two, the squares depend on disjoint innovations and are independent.

Worked Example: Standard Normal Innovations

Let the innovations be independent with $\varepsilon_t\sim N(0,1)$. Then $\sigma_\varepsilon^2=1$ and $\mu_{4,\varepsilon}=3$. The product process has $\operatorname{Var}(X_t)=1$, and its ACF is zero at every nonzero lag.

For the squares, $E(X_t^2)=1$. The covariance at lag one is $$ \operatorname{Cov}(X_t^2,X_{t-1}^2) =1^2(3-1^2)=2. $$ Thus neighboring squares are dependent. To express the size of this relationship as a correlation, first calculate $E(X_t^4)=E(\varepsilon_t^4)E(\varepsilon_{t-1}^4)=3\cdot3=9$. Therefore $\operatorname{Var}(X_t^2)=9-1^2=8$, and the lag-one correlation of the squared process is $$ \operatorname{Corr}(X_t^2,X_{t-1}^2)=\frac{2}{\sqrt{8\cdot8}}=0.25. $$ The original observations have zero lag-one correlation, while their squares have correlation $0.25$. These are different statements about different variables.

Worked Example: A Non-Normal Innovation Distribution

Suppose each innovation equals $0$ with probability $1/2$, and equals either $\sqrt{2}$ or $-\sqrt{2}$ with probability $1/4$ each. The distribution is symmetric, so its mean is zero. Its second and fourth moments are $$ E(\varepsilon_t^2)=\tfrac12(0^2)+\tfrac14(\sqrt2)^2+\tfrac14(-\sqrt2)^2=1, $$ $$ E(\varepsilon_t^4)=\tfrac12(0^4)+\tfrac14(\sqrt2)^4+\tfrac14(-\sqrt2)^4=2. $$

Thus $\sigma_\varepsilon^2=1$, $\mu_{4,\varepsilon}=2$, and $\operatorname{Var}(\varepsilon_t^2)=2-1^2=1$. The product process has variance $1$, with zero autocorrelation at every nonzero lag. But its neighboring squares have covariance $$ \operatorname{Cov}(X_t^2,X_{t-1}^2)=1^2(2-1^2)=1. $$ This positive covariance establishes dependence between adjacent squares, without requiring normal innovations. At lag two the squared observations use disjoint innovations, so their covariance is zero and they are independent.

Interpretation and a Common Mistake

This construction matters because inspecting only the ACF of $X_t$ can miss structure in the data. The observations may have no linear correlation across time, yet a transformation such as squaring can reveal that neighboring values are statistically dependent. Here, a large magnitude of $X_t$ tends to contain information about the magnitude of $X_{t-1}$, because both are affected by the same innovation.

Common mistake: “All nonzero-lag autocorrelations are zero, so the observations are independent.” That conclusion does not follow. Zero autocorrelation rules out linear correlation at those lags, not all forms of dependence. In this construction, the covariance between neighboring squares is positive whenever $\operatorname{Var}(\varepsilon_t^2)>0$, which directly contradicts independence. Conversely, the two-point example shows why the qualification matters: if $\varepsilon_t^2$ is constant, the squared process is constant and this particular dependence does not appear.

The key distinction is between a statement about covariances and a statement about joint distributions. Independence implies zero covariance when the relevant moments exist, but zero covariance alone does not imply independence. Checking transformed observations, such as squares, can expose dependence that the original ACF does not show.

Key takeaway: For $X_t=\varepsilon_t\varepsilon_{t-1}$ with independent, identically distributed, mean-zero innovations of finite fourth moment, the process is stationary and has zero ACF at all nonzero lags. If the innovation squares vary, neighboring squared observations have positive covariance $\sigma_\varepsilon^4\operatorname{Var}(\varepsilon_t^2)$, so zero ACF does not mean independence.

Check Your Understanding

Answer each question using the assumptions and calculations developed above.

Which innovation terms appear in $X_t$ and $X_{t-2}$, and why does that imply independence?
Why does $\gamma(1)$ equal zero even though $X_t$ and $X_{t-1}$ share an innovation?
State the condition on $\varepsilon_t^2$ that makes the covariance between neighboring squares positive.
For standard normal innovations, what is the covariance between $X_t^2$ and $X_{t-1}^2$, and what does it imply?
Why does zero autocorrelation at every nonzero lag not establish that the process is independent?

A Stationary Process With Zero ACF but Dependent Squares

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