What a Sample ACF Plot Shows
A sample autocorrelation function (sample ACF) plot summarizes estimated linear association between observations separated by different time lags. Each bar belongs to a particular lag: lag 1 compares observations one time step apart, lag 2 compares observations two time steps apart, and so on. Reading a plot carefully means identifying both the bar’s lag and its position relative to zero.
For an observed series \(x_1,x_2,\ldots,x_n\), let \(\bar{x}\) be its sample mean. Using the sample autocovariance convention with divisor \(n\), the sample ACF at lag \(k\) is the sample autocovariance at that lag divided by the sample autocovariance at lag 0.
The plot’s horizontal axis gives the lag, and the vertical position of a bar gives the estimated autocorrelation. A bar above zero is positive, a bar below zero is negative, and a bar close to zero has a small estimated magnitude. The sign describes the direction of the estimated association, while the absolute value \(|\hat{\rho}_k|\) describes its size. Thus, \(-0.7\) has greater magnitude than \(+0.3\), even though one estimate is negative and the other positive.
Why the Sample ACF Is a Ratio
The lag-\(k\) sample autocovariance measures whether deviations from the series mean tend to have the same or opposite signs when separated by \(k\) time steps. Dividing by the lag-0 sample autocovariance expresses that quantity relative to the series’ overall sample variance. This removes the scale of measurement: converting the observations from dollars to thousands of dollars multiplies every sample autocovariance by the same squared conversion factor, which cancels in the ratio.
The second equality follows because both numerator and denominator contain the same factor \(1/n\). The denominator is positive for a nonconstant series, so the sign of \(\hat{\rho}_k\) is the sign of the lag-\(k\) product sum. For the convention above, the absolute value cannot exceed 1: by the Cauchy–Schwarz inequality, the absolute product sum is at most the square root of the product of the sums of squared deviations in the two lagged portions. Each of those sums is no greater than the full lag-0 sum of squares. Consequently, \(|\hat{\rho}_k|\leq 1\).
Reading Bars by Lag, Sign, and Size
Consider a hypothetical sample ACF plot for weekly service requests at a small, fictional roadside-assistance program. The bars are summarized below by their plotted approximate heights. The service-request setting is only an illustration; the values are invented.
| Lag | Approximate bar height | How to read it |
|---|---|---|
| 0 | +1.00 | The defining lag-0 value |
| 1 | +0.65 | Positive estimate; largest nonzero magnitude shown |
| 2 | -0.20 | Negative estimate; smaller magnitude than lag 1 |
| 3 | +0.10 | Small positive estimate |
| 4 | -0.50 | Negative estimate; second-largest nonzero magnitude shown |
The lag-1 bar is above zero, so adjacent observations have positive estimated association in this sample. The lag-4 bar is below zero, so observations four time steps apart have negative estimated association. Comparing absolute values gives \(0.65>0.50>0.20>0.10\): among the nonzero lags displayed, lag 1 has the largest estimated magnitude and lag 3 the smallest. This is a comparison of estimates, not by itself evidence that one underlying population correlation is different from another.
Worked Example: Comparing Bars in a Plotted ACF
A hypothetical plot of monthly equipment-repair counts has bars at lag 1 of \(+0.42\), lag 2 of \(-0.68\), lag 3 of \(+0.15\), and lag 4 of \(-0.31\). Identify the sign at each lag, rank the nonzero lags by magnitude, and state the largest-magnitude estimate.
Method. Read the sign directly from whether each bar is above or below zero. For the size comparison, take absolute values: \(0.42,\ 0.68,\ 0.15,\ 0.31\).
Lag 1 is positive, with magnitude \(0.42\). Lag 2 is negative, with magnitude \(0.68\). Lag 3 is positive, with magnitude \(0.15\). Lag 4 is negative, with magnitude \(0.31\).
Ordering the magnitudes from largest to smallest gives \(0.68>0.42>0.31>0.15\), so the order is lag 2, lag 1, lag 4, lag 3. The largest-magnitude estimate is the negative lag-2 estimate, \(-0.68\).
Check. Lag 2 is the largest bar by distance from zero, not because it is the most positive. The common mistake is to rank signed values as if a larger number always means a larger association; magnitude comparisons require absolute values.
Connecting a Short Series to Its ACF
When the bars are not supplied, the same reading starts by calculating the sample mean, then the deviations, then the lagged products. The lag-0 sum of squared deviations provides the common denominator for the ACF ratios. Only pairs separated by the lag being calculated contribute to its product sum.
Worked Example: Interpreting Lag 1 for the Series 2, 4, 2
A hypothetical three-day series records \(2,4,2\) service alerts per day. Calculate the lag-1 sample autocorrelation and interpret its sign. Use the divisor-\(n\) sample autocovariance convention.
Step 1: Find the mean and deviations. $$ \bar{x}=\frac{2+4+2}{3}=\frac{8}{3}, \qquad x_t-\bar{x}=\left(-\frac{2}{3},\frac{4}{3},-\frac{2}{3}\right). $$
Step 2: Calculate the lag-0 sum of squares. $$ S_0=\left(-\frac{2}{3}\right)^2+ \left(\frac{4}{3}\right)^2+ \left(-\frac{2}{3}\right)^2 =\frac{4}{9}+\frac{16}{9}+\frac{4}{9} =\frac{24}{9}=\frac{8}{3}. $$ Thus, \(\hat{\gamma}_0=S_0/3=8/9\).
Step 3: Calculate the lag-1 product sum. The lag-1 pairs use the first and second deviations, then the second and third deviations: $$ S_1=\left(-\frac{2}{3}\right)\left(\frac{4}{3}\right) +\left(\frac{4}{3}\right)\left(-\frac{2}{3}\right) =-\frac{8}{9}-\frac{8}{9} =-\frac{16}{9}. $$ Therefore, \(\hat{\gamma}_1=S_1/3=-16/27\).
Step 4: Divide to obtain the sample ACF. $$ \hat{\rho}_1=\frac{-16/27}{8/9} =-\frac{16}{27}\cdot\frac{9}{8} =-\frac{2}{3}\approx -0.667. $$
Interpretation and check. The lag-1 estimate is negative: the two adjacent pairs in this series have deviations with opposite signs. It lies between \(-1\) and \(1\), as required. The common mistake is to omit one of the two lag-1 products or to miscalculate their sum; the correct product sum is \(-16/9\), not \(-4/9\).
Comparing Several Lags
A plot may show several bars of alternating signs. That does not change the reading procedure: identify each lag first, then note the sign, then compare absolute values. Calculating a few values by hand is useful because it shows precisely which observation pairs contribute to each bar.
Worked Example: Comparing All Nonzero Lags in a Short Series
A hypothetical sequence of five weekly audit adjustments, in thousands of dollars, is \(3,5,4,6,5\). Using the divisor-\(n\) convention, calculate the sample ACF at lags 1 through 4, identify the largest-magnitude lag, and describe the signs.
Step 1: Calculate deviations and the lag-0 sum. $$ \bar{x}=\frac{3+5+4+6+5}{5}=\frac{23}{5}=4.6. $$ The deviations are \(-1.6,\ 0.4,\ -0.6,\ 1.4,\ 0.4\), and $$ S_0=(-1.6)^2+(0.4)^2+(-0.6)^2+(1.4)^2+(0.4)^2 =2.56+0.16+0.36+1.96+0.16=5.20. $$ Thus, \(\hat{\gamma}_0=5.20/5=1.04\).
Step 2: Calculate the lagged product sums. $$ S_1=(-1.6)(0.4)+(0.4)(-0.6)+(-0.6)(1.4)+(1.4)(0.4) =-0.64-0.24-0.84+0.56=-1.16, $$ $$ S_2=(-1.6)(-0.6)+(0.4)(1.4)+(-0.6)(0.4) =0.96+0.56-0.24=1.28, $$ $$ S_3=(-1.6)(1.4)+(0.4)(0.4) =-2.24+0.16=-2.08, \qquad S_4=(-1.6)(0.4)=-0.64. $$
Step 3: Divide by the lag-0 sum. The common divisor \(1/5\) cancels in each ratio: $$ \hat{\rho}_1=\frac{-1.16}{5.20}\approx -0.2231,\qquad \hat{\rho}_2=\frac{1.28}{5.20}\approx 0.2462, $$ $$ \hat{\rho}_3=\frac{-2.08}{5.20}=-0.4000,\qquad \hat{\rho}_4=\frac{-0.64}{5.20}\approx -0.1231. $$
Interpretation and check. Lags 1, 3, and 4 have negative estimates; lag 2 has a positive estimate. Their absolute magnitudes are approximately \(0.2231,\ 0.2462,\ 0.4000,\ 0.1231\), respectively, so lag 3 has the largest magnitude. The signs and magnitudes agree with the signs and sizes of the corresponding product sums, since the common denominator \(5.20\) is positive. Rounding is to four decimal places.
Common Reading Traps
- Confusing lag with time. Lag 3 means a separation of three observation steps. If observations are weekly, it means three weeks; if they are monthly, it means three months.
- Comparing signed values instead of magnitudes. A bar at \(-0.8\) is farther from zero than a bar at \(+0.5\), because \(0.8>0.5\).
- Forgetting that lag 0 is special. For a nonconstant series, the lag-0 sample ACF is 1 by definition. It is not evidence of a pattern at a positive time separation.
- Overstating what a bar establishes. The sample ACF describes estimated correlations in the observed series. A bar’s sign or size alone does not establish causation, explain a mechanism, or determine whether the estimate is distinguishable from zero in a statistical test.
- Mixing conventions in a calculation. If sample autocovariances are computed under one divisor convention, use that same convention for the lag-0 denominator. Under the definition used here, the shared divisor cancels in the ratio.
Check Your Understanding
For each problem, identify the lag-specific information before interpreting or comparing the estimates.
- A plotted sample ACF has values \(0.35\) at lag 1, \(-0.70\) at lag 2, and \(0.10\) at lag 3. Which lag has the largest magnitude, and which estimate is negative? (Answer: Lag 2 has the largest magnitude, \(0.70\); the negative estimate is \(-0.70\) at lag 2.)
- A nonconstant series has \(\hat{\gamma}_0=4\) and \(\hat{\gamma}_1=-2\). Calculate \(\hat{\rho}_1\) and state its sign. (Answer: \(\hat{\rho}_1=-2/4=-0.5\), negative.)
- For the series \(1,3,2\), calculate the lag-1 sample ACF using the divisor-\(n\) convention. Show the mean, lag-0 sum of squares, and lag-1 product sum. (Answer: \(\bar{x}=2\); lag-0 sum \(=2\); lag-1 product sum \(=-1\); \(\hat{\rho}_1=-1/2=-0.5\).)
- A plot gives lag-1 through lag-4 estimates of \(+0.30,-0.55,+0.20,-0.10\). Rank the lags by decreasing magnitude and identify the largest-magnitude sign. (Answer: Lag 2, lag 1, lag 3, lag 4; the largest-magnitude estimate is negative, \(-0.55\).)
- A hypothetical series of four daily measurements is \(4,2,5,3\). Using the divisor-\(n\) convention, calculate \(\hat{\rho}_1\) and \(\hat{\rho}_2\), then identify which has greater magnitude. (Answer: \(\bar{x}=3.5\), deviations \(0.5,-1.5,1.5,-0.5\), and lag-0 sum \(=5\). The lag-1 product sum is \(-3.75\), giving \(\hat{\rho}_1=-0.75\); the lag-2 product sum is \(1.50\), giving \(\hat{\rho}_2=0.30\). Lag 1 has greater magnitude.)