Why White Noise Has Zero-Lag-Only Covariance

Why the Covariance Pattern Matters

A white-noise series has no covariance between observations at different time points. This is more than a label: it determines the entire autocovariance function and, when the variance is positive, the autocorrelation function. The key is to connect the definition of white noise to the covariance at a given lag.

The result is useful when interpreting time-series calculations. A zero covariance at a nonzero lag follows directly from the white-noise conditions. But that conclusion applies to the white-noise series itself; a series constructed from several white-noise values can have a different covariance pattern.

Definition: For this covariance result, a white-noise series $\{\varepsilon_t\}$ has a finite constant mean $\mu_\varepsilon=0$, finite constant variance $\sigma_\varepsilon^2$, and zero covariance between observations at distinct times: $$ \operatorname{Cov}(\varepsilon_t,\varepsilon_s)=0 \quad \text{when } t\ne s. $$ No particular probability distribution is required by these moment conditions. In particular, the covariance result below does not require a Gaussian distribution.

For a series whose covariance depends on the lag, write its autocovariance at lag $h$ as $\gamma(h)=\operatorname{Cov}(X_t,X_{t-h})$. Lag zero compares an observation with itself. A positive lag compares observations $h$ time steps apart; a negative lag compares them in the reverse order. For white noise, any nonzero lag uses two distinct time points.

Deriving the Autocovariance Function

The covariance of two random variables is the expected product of their deviations from their respective means. Since the white-noise mean is zero, the autocovariance at lag $h$ is $\operatorname{E}[\varepsilon_t\varepsilon_{t-h}]$. At lag zero, both factors are the same observation, so this expected product is its variance. At a nonzero lag, the two observations are at distinct times, so the white-noise covariance condition applies.

Write the autocovariance as a covariance.
By definition, $\gamma_\varepsilon(h)=\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-h})$. The mean is zero, so this is also $\operatorname{E}[\varepsilon_t\varepsilon_{t-h}]$.

Evaluate lag zero.
When $h=0$, the two variables are identical. Therefore, their covariance is the variance.

Evaluate a nonzero lag.
When $h\ne0$, $t$ and $t-h$ are distinct time points. The defining zero-covariance condition therefore gives zero.

\begin{aligned} \gamma_\varepsilon(0) &=\operatorname{Cov}(\varepsilon_t,\varepsilon_t) =\operatorname{Var}(\varepsilon_t) =\sigma_\varepsilon^2,\\ \gamma_\varepsilon(h) &=\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-h}) =0,\qquad h\ne0. \end{aligned}

This is the zero-lag-only covariance pattern: the autocovariance is $\sigma_\varepsilon^2$ at lag zero and zero at every nonzero lag. The argument also covers negative lags. If $h<0$, then $t-h$ is still different from $t$, and the defining condition still gives covariance zero.

The autocorrelation is the autocovariance divided by the variance, provided that the variance is positive. Thus, for $\sigma_\varepsilon^2>0$, the autocorrelation is one at lag zero and zero at every nonzero lag. If the variance is zero, this division is undefined, so the autocorrelation function cannot be obtained by this formula.

\rho_\varepsilon(h) =\frac{\gamma_\varepsilon(h)}{\gamma_\varepsilon(0)} = \begin{cases} 1, & h=0,\\ 0, & h\ne0, \end{cases} \qquad \text{when } \sigma_\varepsilon^2>0.

Worked Examples

Worked Example: Finding Covariance and Correlation at a Specified Lag

A hypothetical monthly measurement error series $\{\varepsilon_t\}$ satisfies the white-noise conditions, with mean zero and variance $16$ measurement-units squared. Its distribution is unspecified. Find the autocovariance at lag zero, the covariance between $\varepsilon_8$ and $\varepsilon_5$, and the autocorrelation at lag $3$.

Method. Identify the lag by comparing the time indices, then apply the white-noise autocovariance result. No distributional assumption is needed.

The lag-zero autocovariance equals the variance: $$ \gamma_\varepsilon(0)=16. $$

The observations $\varepsilon_8$ and $\varepsilon_5$ are three time steps apart, since $8-5=3$. Their covariance is therefore $$ \operatorname{Cov}(\varepsilon_8,\varepsilon_5) =\gamma_\varepsilon(3) =0. $$

The variance is positive, so the autocorrelation at lag $3$ is defined. Dividing by the lag-zero autocovariance gives $$ \rho_\varepsilon(3)=\frac{\gamma_\varepsilon(3)}{\gamma_\varepsilon(0)} =\frac{0}{16}=0. $$

Check. The zero-lag value is the stated variance, and the two observations in the requested covariance have distinct time indices. The common mistake avoided here is treating “three steps apart” as a covariance of $16$; $16$ applies only when an observation is paired with itself.

Worked Example: Scaling White Noise and Adding a Constant

Let $\{\varepsilon_t\}$ be white noise with mean zero, variance $9$, and unspecified distribution. Define $X_t=11+2\varepsilon_t$. Find the mean and variance of $X_t$, then find its autocovariance and autocorrelation at lag $2$.

Method. Use the effect of a constant shift and a constant multiplier on means and covariances. The added constant changes the mean but disappears from centered deviations; the multiplier of $2$ multiplies variance by $2^2$.

First calculate the mean: $$ \operatorname{E}[X_t] =\operatorname{E}[11+2\varepsilon_t] =11+2\operatorname{E}[\varepsilon_t] =11+2(0) =11. $$

Next calculate the variance. A constant has no effect on variance, and multiplying a random variable by $2$ multiplies its variance by $4$: $$ \operatorname{Var}(X_t) =\operatorname{Var}(11+2\varepsilon_t) =2^2\operatorname{Var}(\varepsilon_t) =4(9) =36. $$

For lag $2$, use the covariance directly: $$ \begin{aligned} \gamma_X(2) &=\operatorname{Cov}(X_t,X_{t-2})\\ &=\operatorname{Cov}(11+2\varepsilon_t,11+2\varepsilon_{t-2})\\ &=4\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-2})\\ &=4(0)=0. \end{aligned} $$ The two error terms are at distinct times, which is why their covariance is zero. Since $\gamma_X(0)=36$, the lag-2 autocorrelation is $$ \rho_X(2)=\frac{\gamma_X(2)}{\gamma_X(0)}=\frac{0}{36}=0. $$

Check. The variance is positive and the requested lag is nonzero, so a zero autocorrelation is consistent with the covariance pattern. The common mistake avoided here is adding $11$ to the variance or multiplying the variance by $2$ instead of $2^2$.

Worked Example: Overlapping Terms Can Create Nonzero Covariance

Suppose $\{\varepsilon_t\}$ is white noise with mean zero, variance $9$, and unspecified distribution. Define $U_t=\varepsilon_t+\varepsilon_{t-1}$. Calculate the variance of $U_t$, its autocovariance at lags $1$ and $2$, and its autocorrelation at lag $1$. Is $U_t$ itself white noise under the stated covariance conditions?

Method. Expand each covariance using bilinearity. Terms from distinct times have covariance zero, but a term shared by both expressions contributes its variance.

At a single time, the two component errors are from distinct times: $$ \begin{aligned} \operatorname{Var}(U_t) &=\operatorname{Var}(\varepsilon_t+\varepsilon_{t-1})\\ &=\operatorname{Var}(\varepsilon_t)+\operatorname{Var}(\varepsilon_{t-1}) +2\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-1})\\ &=9+9+2(0)=18. \end{aligned} $$ Thus, $\gamma_U(0)=18$.

At lag $1$, $U_{t-1}=\varepsilon_{t-1}+\varepsilon_{t-2}$. Expanding all four covariance terms gives $$ \begin{aligned} \gamma_U(1) &=\operatorname{Cov}(\varepsilon_t+\varepsilon_{t-1}, \varepsilon_{t-1}+\varepsilon_{t-2})\\ &=\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-1}) +\operatorname{Cov}(\varepsilon_t,\varepsilon_{t-2})\\ &\quad+\operatorname{Cov}(\varepsilon_{t-1},\varepsilon_{t-1}) +\operatorname{Cov}(\varepsilon_{t-1},\varepsilon_{t-2})\\ &=0+0+9+0=9. \end{aligned} $$ The third term is $9$, not zero, because it pairs the same error at time $t-1$ with itself.

At lag $2$, $U_{t-2}=\varepsilon_{t-2}+\varepsilon_{t-3}$. Each error in $U_t$ is at a distinct time from each error in $U_{t-2}$, so every term in the expansion is zero: $$ \gamma_U(2)=0+0+0+0=0. $$ The lag-1 autocorrelation is $$ \rho_U(1)=\frac{\gamma_U(1)}{\gamma_U(0)} =\frac{9}{18}=0.5. $$

Conclusion and check. $U_t$ is not white noise under the stated covariance conditions because its lag-1 covariance is $9$, not zero. The calculation avoids the common mistake of assuming that any construction from white-noise terms must itself have zero covariance at nonzero lags. Here the shared term $\varepsilon_{t-1}$ creates the nonzero covariance.

What the Result Does—and Does Not—Say

For the white-noise series itself, the full covariance pattern is determined by its defining conditions. The following table summarizes the result and a useful contrast.

Quantity	At lag zero	At a nonzero lag
White-noise autocovariance	$\sigma_\varepsilon^2$	0
White-noise autocorrelation, if $\sigma_\varepsilon^2>0$	1	0
Autocovariance after forming a sum of overlapping terms	Depends on the construction	Can be nonzero

The last row is important: the zero-covariance condition applies to pairs of distinct white-noise observations, not automatically to every pair of quantities built from those observations. In the third example, $U_t$ and $U_{t-1}$ both contain $\varepsilon_{t-1}$, so those two quantities have a shared random component.

Also, zero covariance is a statement about a second moment. By itself, it does not establish that two observations are independent. The derivation here needs only the stated means, variances, and covariances; it makes no stronger distributional claim.

Key takeaway: White noise has $\gamma_\varepsilon(0)=\sigma_\varepsilon^2$ and $\gamma_\varepsilon(h)=0$ for every $h\ne0$. If the variance is positive, its autocorrelation is one at lag zero and zero at every nonzero lag. Check which random variables are actually being compared before applying this result.

Check Your Understanding

For each problem, use only the information stated about the white-noise series and show the relevant lag or covariance calculation.

A white-noise error series has mean zero, variance $25$, and unspecified distribution. Find $\gamma(0)$, $\gamma(3)$, and $\rho(-3)$. (Answer: $\gamma(0)=25,\ \gamma(3)=0,\ \rho(-3)=0$.)
Let $X_t=5+2\varepsilon_t$, where $\{\varepsilon_t\}$ is white noise with mean zero and variance $9$. Find $\operatorname{E}[X_t]$, $\gamma_X(0)$, and $\gamma_X(1)$. (Answer: $\operatorname{E}[X_t]=5,\ \gamma_X(0)=36,\ \gamma_X(1)=0$.)
Let $\{\varepsilon_t\}$ be white noise with variance $4$, and define $R_t=\varepsilon_t+\varepsilon_{t-2}$. Find $\gamma_R(0)$, $\gamma_R(2)$, and $\rho_R(2)$. (Answer: $\gamma_R(0)=8,\ \gamma_R(2)=4,\ \rho_R(2)=0.5$.)
Let $\{\varepsilon_t\}$ be white noise with variance $5$, and define $V_t=\varepsilon_t-\varepsilon_{t-1}$. Find $\gamma_V(0)$ and $\gamma_V(1)$. Is $V_t$ white noise under the stated covariance conditions? (Answer: $\gamma_V(0)=10,\ \gamma_V(1)=-5$; no, because its lag-1 covariance is nonzero.)
Let $\{\varepsilon_t\}$ be white noise with variance $7$, and define $W_t=\varepsilon_t+\varepsilon_{t-1}$. Find $\gamma_W(1)$ and state which term in the covariance expansion is nonzero. (Answer: $\gamma_W(1)=7$; the nonzero term is $\operatorname{Cov}(\varepsilon_{t-1},\varepsilon_{t-1})=7$.)

Quantity	At lag zero	At a nonzero lag
White-noise autocovariance	\(\sigma_\varepsilon^2\)	0
White-noise autocorrelation, if \(\sigma_\varepsilon^2>0\)	1	0
Autocovariance after forming a sum of overlapping terms	Depends on the construction	Can be nonzero

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