Reference Bands In ACF Plots

Why ACF Plots Have Reference Bands

A sample autocorrelation function (ACF) plot displays estimated correlations at successive lags. Reference bands provide a quick benchmark: if observations have no serial dependence, how large might a sample autocorrelation appear just from sampling variation? Comparing plotted values with that benchmark can help flag patterns for further investigation.

The bands are approximate, not a dividing line between “dependent” and “independent.” They are usually calculated for each lag separately under a simple null model. Looking across many lags makes it more likely that at least one estimate will cross its band by chance. The right interpretation therefore considers the series length, the overall pattern, and how many lags were inspected.

Definition: For a series of length \(n\), a commonly drawn approximate 95% reference band at a nonzero lag is \(+1.96/\sqrt{n}\) and \(-1.96/\sqrt{n}\). Under the usual large-sample, no-serial-correlation benchmark, these are pointwise bounds: at any one prespecified lag, the sample autocorrelation falls between them with approximate probability 0.95.

Here, \(n\) is the number of observations used to calculate the sample ACF, not the number of lags displayed. The bounds are symmetric around zero, and their width decreases as the series gets longer. They apply to nonzero lags; the sample ACF at lag zero is one by definition, so it is not assessed using these bands.

Where the Approximation Comes From

The usual rule is motivated by the behavior of sample autocorrelations when observations are independent, have a common distribution, and have finite variance. This is a benchmark for understanding the bands, not a claim that every real series meets those conditions. For a fixed nonzero lag \(k\), the sample autocorrelation is approximately normally distributed with mean zero and variance \(1/n\) under this benchmark, for sufficiently large \(n\).

A short derivation explains the scale \(1/\sqrt{n}\). Suppose the observations are independent with mean zero and variance \(\sigma^2\), and write the lag-\(k\) sample autocovariance using a divisor of \(n\). Temporarily suppose the mean is known to be zero. Its numerator is a sum of products of observations separated by \(k\) time steps:

\widehat{\gamma}_k=\frac{1}{n}\sum_{t=k+1}^{n}X_tX_{t-k}.

Each product has mean zero. Its variance is \(\sigma^4\), since independence gives \(E[X_t^2X_{t-k}^2]=\sigma^2\sigma^2\). For distinct terms in the sum, the covariance is zero: even when two products share an observation, another mean-zero observation appears only once in their product, making its expected value zero. Thus, for fixed \(k\),

\operatorname{Var}(\widehat{\gamma}_k) =\frac{n-k}{n^2}\sigma^4 \approx \frac{\sigma^4}{n} \quad \text{when } n \text{ is large}.

The lag-zero sample autocovariance estimates \(\sigma^2\), so dividing the lag-\(k\) sample autocovariance by it gives the sample autocorrelation. A large-sample normal approximation then gives a standard error of about \(1/\sqrt{n}\). Estimating the mean rather than knowing it changes details in finite samples, but the same approximation is commonly used for the plotted bands.

For a central 95% normal interval, the critical value is approximately 1.96. Multiplying it by the approximate standard error gives the reference limits. This derivation describes why the rule is useful; it does not make the bounds exact for every series, sample size, or model.

\text{Approximate band at lag } k:\qquad -\frac{1.96}{\sqrt{n}} \ \text{to}\ \frac{1.96}{\sqrt{n}}, \qquad k>0.

Worked Examples

Worked Example: Two Crossings in a Short ACF Plot

A hypothetical analyst has 64 consecutive daily measurements and plots the sample ACF at lags 1 through 4. The plotted values are \(0.10\), \(-0.27\), \(0.22\), and \(-0.25\). Using the usual approximate 95% reference bands, identify which values cross the bands and state what that does, and does not, indicate.

Method and assumptions. Use the standard pointwise bands as a rough benchmark under the independent-observation null. The series length is \(n=64\). This calculation flags values outside the benchmark; it does not establish a cause for them.

Step 1: Calculate the standard-error approximation.

\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{64}}=\frac{1}{8}=0.125.

Step 2: Calculate the limits.

1.96(0.125)=0.245, \qquad \text{so the bands are } -0.245 \text{ and } 0.245.

Step 3: Compare each value with the limits. The value at lag 1, \(0.10\), is inside the bands. The value at lag 2, \(-0.27\), is below \(-0.245\), so it crosses. The value at lag 3, \(0.22\), is inside. The value at lag 4, \(-0.25\), is also below \(-0.245\), so it crosses.

Conclusion and check. There are two crossings among the four plotted lags. The arithmetic is consistent with the limits: \(-0.27<-0.245\) and \(-0.25<-0.245\), while the other two values lie between the limits. The crossings merit attention, but two pointwise flags do not by themselves prove that the underlying series has dependence. The common mistake is to treat every crossing as a definitive finding rather than a signal to consider the whole plot and the analysis context.

Worked Example: One Crossing Among Many Lags

For a hypothetical series of 144 observations, an analyst examines the first 20 nonzero-lag sample autocorrelations. Exactly one, at lag 7, is outside the usual approximate 95% bands; its value is \(0.18\). Calculate the bands and use a simple repeated-comparison calculation to explain why that single crossing needs cautious interpretation.

Method and assumptions. First apply the pointwise reference-band rule. Then, only as an illustration, suppose 20 separate tests each had a 0.05 chance of a crossing under the null and were independent of one another. Actual sample autocorrelations at different lags need not be independent, so this calculation is not an exact familywise probability for the plot.

Step 1: Calculate the bands. Since \(\sqrt{144}=12\),

\frac{1.96}{\sqrt{144}}=\frac{1.96}{12}=0.1633 \quad \text{(to 4 decimal places)}.

The approximate limits are therefore \(-0.1633\) and \(0.1633\). The observed value \(0.18\) exceeds the upper limit because \(0.18>0.1633\).

Step 2: Calculate the illustrative chance of at least one crossing. Under the simplifying independence assumption, the probability of no crossing at any of the 20 lags is \(0.95^{20}\). Therefore,

P(\text{at least one crossing}) =1-(0.95)^{20} \approx 1-0.3585 =0.6415.

Conclusion and check. The lag-7 value crosses its pointwise band, but the illustrative probability of at least one crossing across 20 tests is about 0.6415, or 64.15%. The multiplication uses the stated simplifying assumption; it should not be presented as the exact probability for real ACF values. The common mistake is to read a 95% pointwise band as though it gives a 95% guarantee that the entire plot will stay inside the limits.

Worked Example: The Same Correlation Estimate at Different Sample Sizes

Two hypothetical, separate series each have a sample autocorrelation of \(0.30\) at lag 2. The first series has 25 observations and the second has 100. Calculate the usual approximate bands for each series and explain why the crossing result differs even though the sample autocorrelations match.

Method and assumptions. Calculate the pointwise reference bands separately using each series’ own length. Do not apply one series’ bands to the other.

Step 1: Calculate the bands for \(n=25\).

\frac{1.96}{\sqrt{25}}=\frac{1.96}{5}=0.392, \quad \text{giving limits } -0.392 \text{ and } 0.392.

The estimate \(0.30\) is inside these limits because \(-0.392<0.30<0.392\).

Step 2: Calculate the bands for \(n=100\).

\frac{1.96}{\sqrt{100}}=\frac{1.96}{10}=0.196, \quad \text{giving limits } -0.196 \text{ and } 0.196.

The estimate \(0.30\) is above the upper limit because \(0.30>0.196\).

Conclusion and check. The estimate is inside the band for the shorter series and outside for the longer one. The longer series has narrower approximate bands because \(1.96/\sqrt{100}<1.96/\sqrt{25}\). The common mistake is to interpret \(0.30\) as having the same reference-band status regardless of sample size. A crossing is a comparison with a sampling-variation benchmark, not a direct measure of how important a relationship is.

How to Interpret Crossings

The usual bands are pointwise: at a particular lag, they provide an approximate 95% benchmark under the stated null model. They are not a simultaneous 95% envelope that is intended to contain every sample ACF value in a plot. As more lags are examined, there are more opportunities for one estimate to fall outside its pointwise limits by chance.

A single crossing is therefore not definitive proof of dependence. It may reflect chance variation, a feature of the data that needs further investigation, or a mismatch between the series and the assumptions behind the approximation. Conversely, values inside the bands do not prove that the series has no dependence. The bands are an aid to interpretation, not a complete decision procedure.

Consider the pattern as well as the individual values. Several notable values, a sustained sequence of values on one side of zero, or a recognizable pattern may prompt closer analysis. But do not replace that analysis with a mechanical count of crossings: the pointwise rule does not account for all features of the series or for the dependence that can occur among estimates at different lags.

Key takeaway: Calculate the usual bands as \(\pm1.96/\sqrt{n}\), compare each nonzero-lag sample autocorrelation with them, and interpret crossings as approximate pointwise flags. Neither an isolated crossing nor the absence of crossings establishes a complete conclusion about the series.

Check Your Understanding

For each problem, use the usual approximate 95% pointwise bands unless the question says otherwise.

A hypothetical series has 81 observations. Calculate its upper and lower reference limits, rounding to 4 decimal places. (Answer: \(-0.2178\) and \(0.2178\))
A series of 64 observations has a sample autocorrelation of \(-0.24\) at lag 3. Does it cross the usual approximate bands? (Answer: No; the limits are \(-0.245\) and \(0.245\))
An analyst inspects 15 lags, and 2 values cross their pointwise bands. Under the simplifying assumption of independent 0.05 crossing chances at each lag, how many crossings are expected on average? (Answer: \(15(0.05)=0.75\); this is an illustrative expectation, not a proof that either observed crossing is or is not dependence)
A series has 400 observations and a lag-5 sample autocorrelation of \(0.11\). Calculate the upper reference limit to 3 decimal places and state whether the value crosses it. (Answer: The upper limit is \(1.96/20=0.098\); yes, \(0.11>0.098\))
Two separate series have a lag-1 sample autocorrelation of \(0.30\), with lengths 25 and 100. For which series does this value cross the usual bands? (Answer: It crosses for the length-100 series only; the limits are \(\pm0.196\) for \(n=100\) and \(\pm0.392\) for \(n=25\))

What You'll Learn