Econometrics guide, Unit 3, step 12 of 20Previous step Next step

Step 12: Autocorrelation

By the EconTutorials team. 15-minute read with 6 practice questions.

SyllabusUndergraduate econometrics · Wooldridge, Introductory Econometrics, Ch. 12

In 30 seconds

Autocorrelation (serial correlation) means the errors of a time-series regression are correlated from one period to the next: $ \text{Corr}(u_t, u_s) \neq 0 $ for some $ t \neq s $.
The workhorse model is the AR(1): $ u_t = \rho u_{t-1} + e_t $. Errors $ s $ periods apart then have correlation $ \rho^s $.
Under strict exogeneity OLS stays unbiased, but the usual standard errors are wrong, usually too small, so “significant” results appear too often.
Detect it with residual plots, the correlogram, the Durbin–Watson statistic ($ d \approx 2(1-\hat\rho) $) and the Breusch–Godfrey test ($ TR^2 \sim \chi^2_p $).
Fix inference with Newey–West (HAC) standard errors, or quasi-difference the data by Cochrane–Orcutt / Prais–Winsten. Often the real cure is a better model of the dynamics.

Why time-series errors are correlated

In a cross-section of 1,000 workers, one worker’s error tells you nothing about the next: they were sampled independently. A time series is different. The observations come in a fixed order, and if consumption was unusually high relative to income last year, it is likely to be unusually high this year too. With time-series data we write $ t $ for the period and $ T $ for the number of observations:

\[ y_t = \beta_0 + \beta_1 x_{t1} + \dots + \beta_k x_{tk} + u_t, \qquad t = 1, \dots, T. \]

The classical assumptions for time series state explicitly what random sampling gave us for free: no serial correlation, $ \text{Corr}(u_t, u_s \mid \mathbf{X}) = 0 $ for all $ t \neq s $.

Key termThe errors are autocorrelated (or serially correlated) when $ \text{Corr}(u_t, u_s) \neq 0 $ for some $ t \neq s $. With positive autocorrelation a positive error tends to be followed by another positive one; with negative autocorrelation the errors tend to switch sign each period. Positive autocorrelation is by far the more common case in economics.

Chapter 12 of Econometrics Made Simple covers this in full: AR(1) errors, Durbin–Watson and Breusch–Godfrey, Newey–West and Prais–Winsten, with Python, R and Stata code.Get the textbook · $15

Four mechanisms produce it.

Inertia. Habits, contracts and expectations carry over from year to year. A burst of consumer optimism fades gradually.
Omitted persistent variables. If the error contains consumer confidence or credit conditions, which move slowly, successive errors share a common component.
Misspecified dynamics. If consumption depends on last year’s income too, the missing term sits in the error, and income is persistent. A straight line fitted to a curved trend leaves residuals on one side for long stretches.
Data construction. Smoothed, interpolated or overlapping-average data have autocorrelation built in.

The AR(1) error model

The simplest and most widely used model is the first-order autoregressive process:

\[ u_t = \rho\, u_{t-1} + e_t, \qquad |\rho| < 1, \]

where $ e_t $ is a fresh, unpredictable shock with variance $ \sigma_e^2 $. Each period’s error is a fraction $ \rho $ of last period’s plus something new. Substituting repeatedly shows that $ u_t = e_t + \rho e_{t-1} + \rho^2 e_{t-2} + \cdots $, a weighted sum of all past shocks. Two results follow:

\[ \text{Var}(u_t) = \frac{\sigma_e^2}{1-\rho^2}, \qquad \text{Corr}(u_t, u_{t-s}) = \rho^{s}. \]

The correlation decays geometrically with the distance between periods. The condition $ |\rho| < 1 $ makes the process stable: shocks die away instead of piling up.

Worked example

The errors in a quarterly model follow an AR(1) with $ \rho = 0.6 $ and $ \sigma_e = 1 $. Find the standard deviation of $ u_t $ and the correlation between errors one, two and four quarters apart.

Step 1$ \text{Var}(u_t) = 1/(1-0.36) = 1.5625 $, so the standard deviation is 1.25: 25% larger than the fresh shocks, because past shocks linger.

Step 2Correlations: $ 0.6 $ at one quarter, $ 0.6^2 = 0.36 $ at two, $ 0.6^4 = 0.1296 $ at four.

Errors a year apart are only weakly related, but neighbouring quarters are strongly linked.

What does it look like? With $ \rho = 0.6 $ a series moves in smooth waves, staying above or below zero for several periods at a time. With $ \rho = -0.6 $ it zig-zags. Counting sign changes is a quick informal check: for independent errors the sign changes about half the time.

What autocorrelation does to OLS

The proof that OLS is unbiased never used the correlation between errors. For time series it needs strict exogeneity: $ E(u_t \mid \mathbf{X}) = 0 $ for every $ t $, so each error is unrelated to the regressors in all periods, past and future. Under that assumption OLS is unbiased whether or not the errors are autocorrelated. Strict exogeneity rules out feedback, such as a central bank raising rates because consumption was unexpectedly strong. It always fails when $ y_{t-1} $ is a regressor, which matters later.

The damage is to the variance. With $ \tilde{x}_t = x_t – \bar{x} $, the variance of the slope includes every covariance between the errors:

\[ \text{Var}(\hat\beta_1) = \frac{\sigma^2}{\text{SST}_x} + \frac{2}{\text{SST}_x^2} \sum_{t < s} \tilde{x}_t \tilde{x}_s\, \text{Cov}(u_t, u_s). \]

The usual formula keeps only the first term. If the errors are AR(1) with parameter $ \rho $ and $ x $ is also persistent, with autocorrelation roughly $ \phi $, the true variance is approximately

\[ \text{Var}(\hat\beta_1) \approx \frac{\sigma^2}{\text{SST}_x} \times \frac{1+\rho\phi}{1-\rho\phi}. \]

When both are positive the factor exceeds 1: the usual standard errors are too small, $ t $ statistics too large, confidence intervals too narrow, and true null hypotheses are rejected too often. Intuitively, fifty years in which the error stays positive for a decade at a time carry less information than fifty independent draws. OLS is also no longer BLUE.

A 5% test of a true zero slope, T = 50 2,000 simulated samples

usual OLS seNewey–West se, L = 3

Figure 1. When both the regressor and the errors are persistent, the usual standard errors are too small and a nominal 5% test rejects a true null up to a third of the time. Newey–West standard errors help but, with only 50 observations, only partly. Numbers from the Monte Carlo experiment in Econometrics Made Simple, Chapter 12.

Figure 1 comes from a Monte Carlo experiment in Econometrics Made Simple, Chapter 12: 2,000 samples of $ T = 50 $, a true slope of zero, a regressor with $ \phi = 0.8 $. The mean of $ \hat\beta_1 $ is essentially zero in every design, so OLS stays unbiased. But with $ \rho = 0.55 $ the estimates have a standard deviation of 0.178 while OLS reports 0.114 on average, so a 5% test rejects 20.3% of the time. The formula predicts this well: with $ \rho\phi = 0.44 $ the variance factor is $ 1.44/0.56 = 2.57 $, so the true standard deviation should be $ \sqrt{2.57} = 1.60 $ times the usual one; the simulation gives $ 0.178/0.114 = 1.56 $.

Common mistakeSaying that autocorrelation “biases the coefficients”. Under strict exogeneity it does not. The problem is the standard errors, and so every $ t $ test, $ F $ test and confidence interval. The exception is a model with a lagged dependent variable, where autocorrelated errors do make OLS biased and inconsistent.

The running example: a consumption function

The book’s file consumption.csv holds annual data for 1975–2024 ($ T = 50 $) on real consumption, disposable income and household wealth per person, plus a real interest rate. The data are simulated with AR(1) errors and $ \rho = 0.55 $, so we can check each method against the truth. OLS gives

\[ \widehat{\text{cons}}_t = \underset{(387.4)}{768.06} + \underset{(0.0254)}{0.6978}\,\text{income}_t + \underset{(0.0035)}{0.0310}\,\text{wealth}_t – \underset{(26.58)}{109.21}\,\text{interest}_t, \qquad R^2 = 0.9985. \]

An extra dollar of income goes with about 70 cents more consumption, a dollar of wealth with 3.1 cents, and a one-point rise in the interest rate with $109 less per person. The very high $ R^2 $ is typical of trending series and says little on its own.

Consumption-function residuals in time order, 1975–2024 T = 50

positive residualnegative residual

Figure 2. Residuals from the regression of consumption on income, wealth and the interest rate. There are runs on one side of zero (negative in 1982–83 and 1999–2001, positive in 2012–14 and 2020–23), and the sign changes 21 times in 49 years against about 24.5 expected for independent errors. Data: consumption.csv, recomputed.

Detecting autocorrelation

The correlogram and the residual regression

Always plot the residuals in time order first, as in Figure 2. Then compute the sample autocorrelations $ r_j $ of the residuals at lags $ j = 1, 2, \dots $ and plot them: the correlogram. If the errors are uncorrelated, each $ r_j $ has a standard deviation of about $ 1/\sqrt{T} $, so values outside $ \pm 2/\sqrt{T} $ are individually significant at roughly 5%. For the consumption function that band is $ \pm 2/\sqrt{50} = \pm 0.283 $. The first two autocorrelations are 0.225 and −0.256: both inside the band.

Under AR(1) errors, $ u_t = \rho u_{t-1} + e_t $ is itself a regression. Regress $ \hat{u}_t $ on $ \hat{u}_{t-1} $ and test $ \rho = 0 $ with an ordinary $ t $ test (valid in large samples when the regressors are strictly exogenous). Here $ \hat\rho = 0.227 $, $ t = 1.62 $, $ p = 0.112 $: not significant at 5%.

The Durbin–Watson statistic

The oldest test, still printed by most software, is due to Durbin and Watson (1950):

\[ d = \frac{\sum_{t=2}^{T} (\hat{u}_t – \hat{u}_{t-1})^2}{\sum_{t=1}^{T} \hat{u}_t^2} \;\approx\; 2(1-\hat\rho). \]

Expand the numerator and the approximation follows. So $ d $ near 2 means no first-order autocorrelation, values towards 0 mean positive autocorrelation, and values towards 4 mean negative autocorrelation. For the consumption function $ d = 1.509 $; the approximation gives $ 2(1-0.227) = 1.546 $.

The exact distribution of $ d $ depends on the regressors, so printed tables give only a lower bound $ d_L $ and an upper bound $ d_U $ for the critical value. Against positive autocorrelation: reject if $ d < d_L $, do not reject if $ d > d_U $, and the test is inconclusive in between. With $ T = 50 $ and three regressors, the 5% bounds are $ d_L = 1.421 $ and $ d_U = 1.674 $. Our 1.509 is in between. Modern software can compute the exact $ p $-value: R’s lmtest::dwtest gives 0.018, so at 5% the test does detect positive autocorrelation. Our article on the Durbin–Watson test works through the bounds procedure in detail.

Common mistakeUsing Durbin–Watson when the model contains a lagged dependent variable $ y_{t-1} $. Strict exogeneity then fails and $ d $ is biased towards 2, so the test tends to miss autocorrelation that is there. DW also needs an intercept, tests only first order and has its inconclusive region. Use Breusch–Godfrey instead.

The Breusch–Godfrey test

The Breusch–Godfrey (BG) test, from Breusch (1978) and Godfrey (1978), is an LM test for autocorrelation up to any order $ p $, with $ H_0: \rho_1 = \dots = \rho_p = 0 $.

Step	What you do
1	Estimate the model by OLS and save the residuals $ \hat{u}_t $.
2	Regress $ \hat{u}_t $ on all the original regressors and on $ \hat{u}_{t-1}, \dots, \hat{u}_{t-p} $ (missing early lags set to zero). Record $ R^2_{\hat{u}} $.
3	$ LM = T \times R^2_{\hat{u}} $, approximately $ \chi^2_p $ under $ H_0 $. Reject if it exceeds the critical value.

Including the original regressors in step 2 is what keeps the test valid when they are not strictly exogenous, for example when $ y_{t-1} $ is among them. Choose $ p $ to match the data: one or two lags for annual data, four for quarterly, twelve for monthly.

Worked example

For the consumption function, the BG auxiliary regression with $ p = 1 $ has $ R^2_{\hat{u}} = 0.0523 $. With $ p = 2 $, $ LM = 7.825 $. Carry out both tests at 5%.

Step 1$ p = 1 $: $ LM = 50 \times 0.0523 = 2.615 $. The 5% critical value of $ \chi^2_1 $ is 3.84, so do not reject ($ p = 0.106 $).

Step 2$ p = 2 $: $ LM = 7.825 $ against the $ \chi^2_2 $ critical value 5.99. Reject ($ p = 0.020 $).

Step 3The second-order test rejects because the second autocorrelation, −0.256, adds to the evidence.

The evidence is mixed: the residual regression and BG(1) do not reject; the exact DW test and BG(2) do.

Exam tipWhy do the residuals show $ \hat\rho \approx 0.23 $ when the errors were built with $ \rho = 0.55 $? With only 50 observations $ \hat\rho $ is imprecise (its standard error is 0.140), and when the regressors trend, OLS bends the fitted line towards slow swings in the error, so the residuals look less autocorrelated than the errors are. Tests for autocorrelation have low power in short series. Failing to reject is not proof of no autocorrelation.

Remedy 1: Newey–West (HAC) standard errors

Since OLS is still unbiased, the simplest response keeps the coefficients and fixes the standard errors. This is the time-series version of the robust standard errors of Step 11. Newey and West (1987) proposed, for simple regression with $ \hat{v}_t = \tilde{x}_t\hat{u}_t $ and lag length $ L $,

\[ \widehat{\text{Var}}_{\text{NW}}(\hat\beta_1) = \frac{1}{\text{SST}_x^2}\left[\sum_{t=1}^{T} \hat{v}_t^2 + 2\sum_{j=1}^{L} w_j \sum_{t=j+1}^{T} \hat{v}_t \hat{v}_{t-j}\right], \qquad w_j = 1 – \frac{j}{L+1}. \]

The first sum alone is the White variance. The second adds estimated covariances up to lag $ L $, with Bartlett weights that shrink as the lag grows and guarantee a positive variance. The result is valid under heteroskedasticity and autocorrelation of unknown form: HAC standard errors.

Key termA common rule of thumb for the lag length, from Stock and Watson’s Introduction to Econometrics, is $ L \approx 0.75\,T^{1/3} $, rounded. For $ T = 50 $, $ 0.75 \times 50^{1/3} = 2.76 $, so $ L = 3 $. Check that conclusions survive neighbouring values of $ L $.

In Python: smf.ols(...).fit(cov_type="HAC", cov_kwds={"maxlags": 3}). In R: coeftest(m, vcov = NeweyWest(m, lag = 3, prewhite = FALSE)). In Stata: newey y x1 x2, lag(3) after tsset.

For the consumption function the HAC(3) standard errors are slightly smaller than the usual ones, by 3–4%: 0.0246 against 0.0254 for income. That fits the correlogram: the positive first autocorrelation is roughly offset by the negative second one. Here the correction changes nothing. In real US quarterly data (1959–2009), a static regression of consumption on income gives a usual standard error of 0.00287 and a Newey–West one, with $ L = 4 $, of 0.00645: the $ t $ statistic falls, and the confidence interval widens, by a factor of 2.25.

Remedy 2: feasible GLS by quasi-differencing

If the errors really are AR(1), the model can be transformed so its errors are uncorrelated. Write the model for period $ t $, subtract $ \rho $ times the model for $ t-1 $, and use $ u_t – \rho u_{t-1} = e_t $:

\[ \underbrace{y_t – \rho y_{t-1}}_{\tilde{y}_t} = \beta_0(1-\rho) + \beta_1 \underbrace{(x_t – \rho x_{t-1})}_{\tilde{x}_t} + e_t, \qquad t = 2, \dots, T. \]

This quasi-differencing leaves the slope unchanged and gives a serially uncorrelated error, so OLS on the transformed data is GLS. In practice $ \rho $ is estimated first, from the OLS residuals, which makes it feasible GLS. Cochrane–Orcutt drops the first observation; Prais–Winsten keeps it, multiplied by $ \sqrt{1-\hat\rho^2} $. Both can be iterated until $ \hat\rho $ settles.

Worked example

Consumption was 15,297 in 1975 and 15,592 in 1976; income was 18,401 and 18,912. With $ \hat\rho = 0.2262 $, compute the transformed 1976 values and the Prais–Winsten 1975 consumption.

Step 1$ \tilde{y}_{1976} = 15{,}592 – 0.2262 \times 15{,}297 = 12{,}131.8 $.

Step 2$ \tilde{x}_{1976} = 18{,}912 – 0.2262 \times 18{,}401 = 14{,}749.7 $. The intercept column becomes $ 1 – 0.2262 = 0.7738 $.

Step 31975: $ \sqrt{1-0.2262^2} = 0.9741 $, so $ 0.9741 \times 15{,}297 = 14{,}900.8 $, and the intercept column in that row is 0.9741.

Iterated to convergence, both estimators give $ \hat\rho \approx 0.24 $, an income coefficient of about 0.70 and an interest-rate effect of about −120 instead of −109 (from Econometrics Made Simple, Chapter 12).

FGLS needs more than OLS does. The transformed error must be uncorrelated with $ x_t – \rho x_{t-1} $, which requires strict exogeneity. If interest rates respond to last year’s consumption surprise, FGLS is inconsistent while OLS, which needs only contemporaneous exogeneity, is still consistent. That is why OLS with HAC standard errors is the modern default, with FGLS reserved for short samples with strongly autocorrelated errors and credibly exogenous regressors.

Re-thinking the dynamics

Autocorrelated residuals often signal that something evolving over time has been left out. Rearranging the quasi-differenced model shows that AR(1) errors are a dynamic model in disguise:

\[ y_t = \beta_0(1-\rho) + \rho\, y_{t-1} + \beta_1 x_t – \rho\beta_1 x_{t-1} + e_t . \]

Estimating $ y_t $ on $ y_{t-1} $, $ x_t $ and $ x_{t-1} $ without the restriction lets the data decide how $ y $ responds over time. In the consumption data the lags do not help ($ t $ statistics below 0.5), and BG(2) still rejects ($ p = 0.023 $): these data really are a static relationship with autocorrelated errors. With real data you rarely know that, so it is worth trying.

Common mistakeTreating HAC standard errors or FGLS as a cure for a spurious regression. The US levels regression of consumption on income has $ R^2 = 0.998 $ and $ d = 0.253 $. Granger and Newbold (1974) warned that when $ R^2 $ exceeds $ d $, two trending series may only appear related. That calls for differences or a test for unit roots and cointegration (Step 18; see also our article on the augmented Dickey–Fuller test), not a mechanical fix.

Drill Durbin–Watson, Breusch–Godfrey and HAC questions with full worked answers in the Practice Questions book.1,150+ questions · $9

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Errors follow $ u_t = 0.7u_{t-1} + e_t $ with $ \sigma_e = 2 $. Find the standard deviation of $ u_t $ and the correlation between $ u_t $ and $ u_{t-3} $. Show worked answer

$ \text{Var}(u_t) = \sigma_e^2/(1-\rho^2) = 4/(1-0.49) = 7.84 $, so the standard deviation is $ \sqrt{7.84} = 2.80 $.
$ \text{Corr}(u_t, u_{t-3}) = 0.7^3 = 0.343 $.

EasyQ2. A regression on $ T = 60 $ years with two regressors gives $ d = 1.10 $. The 5% bounds are $ d_L = 1.51 $ and $ d_U = 1.65 $. Estimate $ \hat\rho $ and carry out the test against positive autocorrelation. Show worked answer

$ \hat\rho \approx 1 – d/2 = 1 – 0.55 = 0.45 $.
$ d = 1.10 < d_L = 1.51 $, so reject $ H_0: \rho = 0 $: there is evidence of positive first-order autocorrelation.
(Had $ d $ been between 1.51 and 1.65, the bounds test would be inconclusive.)

MediumQ3. A model on 80 quarters is tested for autocorrelation up to order 4. The Breusch–Godfrey auxiliary regression has $ R^2 = 0.15 $. Compute LM, state the distribution under $ H_0 $ and conclude at 5%. Why does the auxiliary regression include the original regressors? Show worked answer

$ LM = T R^2 = 80 \times 0.15 = 12.0 $, approximately $ \chi^2_4 $ under $ H_0: \rho_1 = \dots = \rho_4 = 0 $.
The 5% critical value is 9.49; $ 12.0 > 9.49 $, so reject ($ p = 0.017 $).
Including the regressors keeps the test valid when they are not strictly exogenous (for example a lagged dependent variable), because it allows for any correlation between the residuals and the regressors.

MediumQ4. Using the rule $ L \approx 0.75T^{1/3} $, choose the Newey–West lag length for $ T = 120 $ monthly observations, and write down the Bartlett weights $ w_j $. Show worked answer

$ 120^{1/3} = 4.93 $, so $ 0.75 \times 4.93 = 3.70 $: $ L = 4 $.
$ w_j = 1 – j/(L+1) = 1 – j/5 $: $ w_1 = 0.8 $, $ w_2 = 0.6 $, $ w_3 = 0.4 $, $ w_4 = 0.2 $.
More distant covariances get less weight. With monthly data and persistent errors, also check a larger $ L $ such as 12.

MediumQ5. With $ \hat\rho = 0.5 $, the data for 2023 and 2024 are $ y = 200, 212 $ and $ x = 150, 160 $. Compute the quasi-differenced 2024 values, the 2024 value of the intercept column, and the Prais–Winsten factor for the first observation. Show worked answer

$ \tilde{y}_{2024} = 212 – 0.5 \times 200 = 112 $.
$ \tilde{x}_{2024} = 160 – 0.5 \times 150 = 85 $.
Intercept column: $ 1 – 0.5 = 0.5 $.
Prais–Winsten factor: $ \sqrt{1-0.25} = 0.866 $, applied to every variable (and the intercept) in the first year.

ExamQ6. A study regresses annual GDP growth on a persistent policy index ($ \phi \approx 0.6 $) and reports $ t = 2.5 $ using the usual OLS standard error. The residuals suggest AR(1) errors with $ \rho \approx 0.5 $. (a) Explain why the usual standard error is likely to be too small. [2] (b) Using $ \text{Var} \approx (\sigma^2/\text{SST}_x)(1+\rho\phi)/(1-\rho\phi) $, estimate the corrected $ t $ statistic and say whether the result is still significant at 5%. [3] (c) Recommend a remedy and say what it does and does not change. [2] Show worked answer

(a) Positively autocorrelated errors and a persistent regressor make the cross-products $ \tilde{x}_t\tilde{x}_s\text{Cov}(u_t,u_s) $ positive. [B1] The usual formula ignores them, so it understates the variance. [B1]
(b) $ \rho\phi = 0.3 $; factor $ = 1.3/0.7 = 1.857 $. [M1] Standard errors scale by $ \sqrt{1.857} = 1.363 $. [A1] Corrected $ t \approx 2.5/1.363 = 1.83 < 1.96 $: no longer significant at 5%. [A1]
(c) Newey–West (HAC) standard errors with $ L \approx 0.75T^{1/3} $. [B1] They fix the standard errors and tests but leave the OLS coefficients unchanged. [B1]
Total: 7 marks.

Check yourself

1. A Durbin–Watson statistic close to 2 indicates

$ d \approx 2(1-\hat\rho) $, so $ d \approx 2 $ means $ \hat\rho \approx 0 $.

2. With strictly exogenous regressors and AR(1) errors, the OLS coefficients are

Unbiasedness does not use the correlation between errors; the variance formula does.

3. The model includes $ y_{t-1} $ as a regressor. Which test for autocorrelation is valid?

DW assumes strict exogeneity, which a lagged dependent variable breaks. BG includes all regressors in its auxiliary regression and stays valid.

Your score: 0 out of 3

The short version

Time-series errors are often autocorrelated: inertia, omitted persistent variables, missing dynamics or the way data are built.
AR(1): $ u_t = \rho u_{t-1} + e_t $, $ \text{Var}(u_t) = \sigma_e^2/(1-\rho^2) $, $ \text{Corr}(u_t,u_{t-s}) = \rho^s $.
Under strict exogeneity OLS is unbiased, but usual standard errors are too small when errors and regressors are both persistent.
Detect with plots, the correlogram, $ d \approx 2(1-\hat\rho) $ (bounds, inconclusive region, invalid with $ y_{t-1} $) and Breusch–Godfrey $ TR^2 \sim \chi^2_p $.
Newey–West HAC standard errors fix inference ($ L \approx 0.75T^{1/3} $); Cochrane–Orcutt and Prais–Winsten quasi-difference the data but need strict exogeneity.
Very high $ R^2 $ with very low $ d $ is a warning of a spurious regression, not a job for HAC.
The full chapter in Econometrics Made Simple has the code in Python, R and Stata, the real US data and a complete set of exercises with answers.

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Step	What you do
1	Estimate the model by OLS and save the residuals \( \hat{u}_t \).
2	Regress \( \hat{u}_t \) on all the original regressors and on \( \hat{u}_{t-1}, \dots, \hat{u}_{t-p} \) (missing early lags set to zero). Record \( R^2_{\hat{u}} \).
3	\( LM = T \times R^2_{\hat{u}} \), approximately \( \chi^2_p \) under \( H_0 \). Reject if it exceeds the critical value.

Step 12: Autocorrelation

Why time-series errors are correlated

The AR(1) error model

What autocorrelation does to OLS

The running example: a consumption function

Detecting autocorrelation

The correlogram and the residual regression

The Durbin–Watson statistic

The Breusch–Godfrey test

Remedy 1: Newey–West (HAC) standard errors

Remedy 2: feasible GLS by quasi-differencing

Re-thinking the dynamics

Practise: 6 questions

Check yourself

The short version

Further reading on EconTutorials

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