Long-Run Variance of the Sample Mean

Why the Variance of a Time-Series Mean Is Different

For independent observations with common variance, averaging $n$ observations divides that variance by $n$. In a time series, however, neighboring observations may move together or in opposite directions. The sample mean can therefore contain less or more information than the same number of independent observations. Its uncertainty depends not only on the variance at an individual date, but also on covariance across dates.

The quantity that collects this dependence is the long-run variance. It determines the leading, large-sample variance of the sample mean and its asymptotic standard error. Positive serial correlation usually increases uncertainty; negative serial correlation can reduce it.

Definition: Let $\{X_t\}$ be a covariance-stationary process with mean $\mu$ and finite variance. Define its lag-$h$ autocovariance by $\gamma(h)=\operatorname{Cov}(X_t,X_{t-h})$, for integer $h$. Covariance stationarity makes this depend on the lag, not on $t$, and $\gamma(-h)=\gamma(h)$. For a sample of $n$ consecutive observations, the sample mean is $\bar X_n=n^{-1}\sum_{t=1}^n X_t$.

From the Exact Variance to the Long-Run Variance

The exact variance of $\bar X_n$ follows by expanding the variance of a sum. There are $n$ diagonal terms, each equal to $\gamma(0)$. For each lag $h$ from 1 to $n-1$, there are $n-h$ pairs of observations separated by that lag, in each of the two possible orders. Thus

\begin{aligned} \operatorname{Var}(\bar X_n) &=\frac{1}{n^2}\left[n\gamma(0)+2\sum_{h=1}^{n-1}(n-h)\gamma(h)\right]\\ &=\frac{1}{n}\left[\gamma(0)+2\sum_{h=1}^{n-1}\left(1-\frac{h}{n}\right)\gamma(h)\right]. \end{aligned}

The factor $1-h/n$ is a finite-sample weight: at a fixed lag, fewer pairs fit inside the sample as the lag grows. If autocovariances are absolutely summable, meaning $\sum_{h=1}^{\infty}|\gamma(h)|<\infty$, these weights approach 1 at each fixed lag and the distant-lag tail can be controlled by the summable sequence. Consequently, $n\operatorname{Var}(\bar X_n)$ converges to the sum of all autocovariances.

Definition: If the autocovariances are absolutely summable, the long-run variance is $$ \Omega=\sum_{h=-\infty}^{\infty}\gamma(h) =\gamma(0)+2\sum_{h=1}^{\infty}\gamma(h). $$ It follows that $\operatorname{Var}(\bar X_n)$ is approximately $\Omega/n$ for large $n$, and the corresponding asymptotic standard error is $\sqrt{\Omega/n}$.

In terms of the autocorrelation function $\rho(h)=\gamma(h)/\gamma(0)$, when $\gamma(0)>0$ and the sum exists,

\Omega=\gamma(0)\left[1+2\sum_{h=1}^{\infty}\rho(h)\right].

This expression separates the marginal variance $\gamma(0)$, measured in squared units of $X_t$, from the adjustment for serial dependence. The long-run variance has the same squared units. It is nonnegative: it is the limit of $n\operatorname{Var}(\bar X_n)$, and each variance is nonnegative. In special cases the sum can be zero because positive and negative autocovariances cancel; then the usual nondegenerate normal approximation with positive variance does not apply.

Worked Example: A Geometric Autocorrelation Pattern

Suppose a hypothetical stationary process, measured in index points, has $\gamma(0)=4$ index-points squared and $\rho(h)=0.5^h$ for $h\geq 1$. The autocovariances are absolutely summable because the autocorrelation sequence is geometric with ratio less than 1. Find the long-run variance and compare a finite-sample variance with its large-sample approximation.

Since $\gamma(h)=\gamma(0)\rho(h)$, the long-run variance is

\begin{aligned} \Omega &=4\left[1+2\sum_{h=1}^{\infty}0.5^h\right]\\ &=4\left[1+2\left(\frac{0.5}{1-0.5}\right)\right] =4(3)=12\ \text{index-points squared}. \end{aligned}

For $n=4$, the exact variance formula uses $\gamma(1)=2$, $\gamma(2)=1$, and $\gamma(3)=0.5$:

\begin{aligned} \operatorname{Var}(\bar X_4) &=\frac{1}{4}\left[4+2\left\{\left(1-\frac14\right)2+ \left(1-\frac24\right)1+\left(1-\frac34\right)0.5\right\}\right]\\ &=\frac{1}{4}\left[4+2(1.5+0.5+0.125)\right] =2.0625\ \text{index-points squared}. \end{aligned}

The large-sample approximation at $n=4$ is $\Omega/n=12/4=3$ index-points squared, which is not especially close for this small sample. At $n=100$, the approximation is $12/100=0.12$ index-points squared, giving an asymptotic standard error of $\sqrt{0.12}=0.3464$ index points. The approximation describes the leading large-$n$ behavior; it is not an exact finite-sample formula.

What the Long-Run Variance Says About Dependence

If observations are uncorrelated across distinct dates, then $\gamma(h)=0$ for $h\neq0$, and $\Omega=\gamma(0)$. The familiar variance $\gamma(0)/n$ is then recovered. Independence is sufficient for this covariance pattern, but it is not required: for the variance of the mean, it is the covariances that enter the calculation.

When autocovariances are positive overall, the mean varies more than it would under independent sampling with the same marginal variance. With negative autocovariances, the sum can be smaller than $\gamma(0)$. This is a comparison of uncertainty for the same number of observations and the same one-period variance; it does not say that a particular observed mean must be closer to $\mu$.

A central limit theorem supplies a distributional approximation in addition to the variance calculation. Covariance stationarity and absolute summability give the variance limit above, but by themselves do not guarantee a central limit theorem. Under suitable weak-dependence and moment conditions, the result established for weakly dependent series takes the form

\sqrt{n}(\bar X_n-\mu)\ \xrightarrow{d}\ N(0,\Omega), \qquad \Omega>0.

Here $N(0,\Omega)$ has mean zero and variance $\Omega$. Equivalently, for large $n$, $\bar X_n$ is approximately normal with mean $\mu$ and variance $\Omega/n$. The additional conditions behind the central limit theorem matter: the long-run variance formula alone is not enough to assert this normal approximation.

Long-Run Variance for a Stationary AR(1)

An AR(1) process provides a direct way to see how persistence changes the result. Consider

X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t, \qquad |\phi|<1,

where the innovations $\varepsilon_t$ have mean zero, variance $\sigma_\varepsilon^2$, are uncorrelated across time, and are uncorrelated with past values of the process. The restriction $|\phi|<1$ gives the stationary causal process. Iterating the equation yields $X_t-\mu=\sum_{j=0}^{\infty}\phi^j\varepsilon_{t-j}$. Therefore, for $h\geq0$, shared innovation terms give

\begin{aligned} \gamma(h) &=\operatorname{Cov}(X_t,X_{t-h})\\ &=\sigma_\varepsilon^2\sum_{j=0}^{\infty}\phi^{j+h}\phi^j =\frac{\sigma_\varepsilon^2\phi^h}{1-\phi^2}. \end{aligned}

By symmetry, $\gamma(-h)=\gamma(h)$. The geometric autocovariance sequence is absolutely summable, so

\begin{aligned} \Omega &=\frac{\sigma_\varepsilon^2}{1-\phi^2} \left(1+2\sum_{h=1}^{\infty}\phi^h\right)\\ &=\frac{\sigma_\varepsilon^2}{1-\phi^2} \left(1+\frac{2\phi}{1-\phi}\right) =\frac{\sigma_\varepsilon^2}{(1-\phi)^2}. \end{aligned}

The variance at an individual date is $\gamma(0)=\sigma_\varepsilon^2/(1-\phi^2)$, whereas the mean’s large-sample variance is governed by $\Omega/n$. Their ratio is

\frac{\Omega}{\gamma(0)}=\frac{1+\phi}{1-\phi}.

For positive $\phi$, this ratio exceeds 1; for negative $\phi$, it is below 1. As $\phi$ approaches 1, persistent positive movements make the mean substantially less precise than independent sampling would suggest.

Worked Example: Positive Persistence Raises Uncertainty

Suppose a hypothetical stationary AR(1) process measures an outcome in millimetres, with $\phi=0.6$ and innovation variance $\sigma_\varepsilon^2=4$ square millimetres. For $n=100$, calculate the long-run variance and the asymptotic standard error of the mean, then compare with independent sampling having the same marginal variance.

The stationarity condition holds because $|0.6|<1$. Its marginal variance is

\gamma(0)=\frac{4}{1-0.6^2} =\frac{4}{0.64}=6.25\ \text{square millimetres}.

The long-run variance and resulting asymptotic standard error are

\Omega=\frac{4}{(1-0.6)^2} =\frac{4}{0.16}=25\ \text{square millimetres}, \qquad SE(\bar X_{100})\approx\sqrt{\frac{25}{100}}=0.5000\ \text{millimetres}.

Under independent sampling with marginal variance $6.25$, the standard error would be $\sqrt{6.25/100}=0.2500$ millimetres. Thus positive serial dependence doubles the asymptotic standard error in this example. The variance ratio is $25/6.25=4$, consistent with $(1+0.6)/(1-0.6)=4$. This comparison concerns the uncertainty of the mean, not a causal effect of persistence.

Worked Example: Negative Persistence Reduces Uncertainty

A second hypothetical stationary AR(1) outcome is measured in seconds and has $\phi=-0.5$ and innovation variance $3$ square seconds. Find its marginal and long-run variances, then calculate the asymptotic standard error of its mean for $n=100$.

The stationarity condition holds because $|-0.5|<1$. The marginal variance is

\gamma(0)=\frac{3}{1-(-0.5)^2} =\frac{3}{0.75}=4\ \text{square seconds}.

Using the AR(1) long-run variance formula,

\Omega=\frac{3}{(1-(-0.5))^2} =\frac{3}{2.25}=1.333\ \text{square seconds}.

Hence the asymptotic variance of the sample mean is $1.333/100=0.01333$ square seconds, and its asymptotic standard error is $\sqrt{0.01333}=0.1155$ seconds. For independent observations with the same marginal variance, the standard error would be $\sqrt{4/100}=0.2000$ seconds. Here alternating deviations tend to offset one another, so the long-run variance is smaller than the marginal variance. It remains positive, as required for this nondegenerate approximation.

Common Mistake and Practical Meaning

Wrong reasoning: “The variance of the mean is the one-period variance divided by $n$, so serial correlation does not matter.” Correction: That formula omits all cross-date covariance terms. Under absolute summability, the relevant large-sample quantity is $\Omega/n$, with $\Omega=\gamma(0)+2\sum_{h\geq1}\gamma(h)$.
Wrong reasoning: “Positive autocorrelation means the long-run variance is always just the marginal variance multiplied by a fixed factor greater than one.” Correction: The adjustment depends on the entire autocorrelation sequence. For an AR(1), it is $(1+\phi)/(1-\phi)$, but other processes have different patterns.
Wrong reasoning: “Because the variance limit exists, a normal approximation automatically follows.” Correction: A central limit theorem requires additional dependence and moment conditions. The variance limit is a second-moment result; the normal approximation is a distributional result.

In applications, the long-run variance is the quantity needed to quantify the sampling uncertainty of a time-series mean when observations are dependent. Its value depends on persistence across all lags, not just on the variance at one date. A later step in practical inference is to estimate this theoretical quantity from data; the derivation here clarifies what such an estimate must represent.

Key takeaway: For an absolutely summable covariance-stationary process, $n\operatorname{Var}(\bar X_n)$ converges to the sum of all autocovariances. Serial dependence changes the mean’s uncertainty through that sum: positive persistence can inflate it, while negative dependence can reduce it.

Check Your Understanding

Answer each question using the definitions and formulas above.

State the long-run variance in terms of $\gamma(0)$ and the positive-lag autocovariances. Why is there a factor of 2?
If observations are uncorrelated across distinct dates, what is $\Omega$, and what large-sample variance does this imply for the sample mean?
For a stationary AR(1) with $\phi=0.2$ and innovation variance $1$, calculate its marginal variance and long-run variance.
Why does absolute summability of autocovariances justify a variance limit but not, by itself, guarantee a central limit theorem?
A colleague uses $\gamma(0)/n$ for a series with positive autocovariances at several lags. Identify the omitted terms and explain the likely direction of the error.