Lag Length, Serial Correlation Tests, and HAC Standard Errors
Jake Anderson
UCLA
Plan for Today
1
Lag length
How many past values of $y$ should an AR model use?
Compare AIC and BIC across lag lengths and pick the lowest.
Spring 2025 Midterm V4 (Boswell), Question 12 (modified)
2
Breusch-Godfrey test
Are the errors of the chosen model autocorrelated?
Write the auxiliary regression, then compare $LM = T \times R^2$ to a $\chi^2$ critical value.
Winter 2026 Midterm 1 (Rojas), Questions 23 and 24
3
HAC standard errors
The test says the errors are autocorrelated. What now?
Build a confidence interval with standard errors that allow for the autocorrelation.
Spring 2023 Midterm V1, Question 12
Problem 1 · Lag length
The Question: How Many Lags?
You fit AR(1) to AR(5) models to a stationary time series $y_t$ and get the following output. Based on AIC and BIC, what lag length would you choose?
Criterion
Lag 1
Lag 2
Lag 3
Lag 4
Lag 5
AIC
$-5.43$
$-5.88$
$-5.51$
$-5.50$
$-5.30$
BIC
$-5.20$
$-5.45$
$-5.02$
$-4.89$
$-4.67$
a1
b2
c3
dAIC 2; BIC 3
eNone of the given answer options are correct.
Source: Spring 2025 Midterm V4 (Boswell), Question 12, modified: the original asks about VARSelect output for two series; here the five models are AR models for one series.
Lag length · The idea
Each Extra Lag Improves the Fit a Little and Adds a Penalty
Up to $p = 3$ the fit falls faster than the penalty rises; after that, more slowly. SC is lowest at $p = 3$: pick the lag length with the lowest value.
AIC uses a smaller penalty, $2k/T$ (0.014 per coefficient here), and is lowest at $p = 4$. This series is not the one in the exam question.
A different series from the exam question, computed from okun5_aus (POE5): AR($p$) models for the quarterly change in Australian unemployment, the same 145 quarters for every $p$. The lecture’s formulas, $n\ln(SSE/n) + k\ln(n)$ and $2k - 2\ln(L)$, are $T$ times these (plus a constant for AIC), so they pick the same $p$.
Lag length · Step 1
Find the Lowest AIC and the Lowest BIC
Which is lower, $-5.43$ or $-5.88$?
AICLowest at lag 2: $-5.88$ is further below 0 than every other AIC value.
BICLowest at lag 2: $-5.45$. At every lag BIC is higher than AIC because its penalty is larger ($\ln T > 2$ once $T \ge 8$).
LAG 2Both criteria pick lag 2 $\implies$ AR(2).
When AIC and BIC disagreeChoose BIC. Its penalty per lag is larger, so it picks fewer lags, and extra lags that only fit noise are left out.
Lag length · Interpretation
Checking Each Answer Choice
a1AIC at lag 1 ($-5.43$) is higher than at lag 2 ($-5.88$): it is closer to 0, not lower.
c3Neither criterion is lowest at lag 3.
dAIC 2; BIC 3BIC at lag 3 ($-5.02$) is higher than at lag 2 ($-5.45$).
eNone of the given answer options are correct.Choice (b) is correct.
b2Answer: (b)
InterpretationUse an AR(2): $y_t$ depends on the two most recent periods. Next, check the residuals of the chosen model for autocorrelation (Problem 2).
When AIC and BIC disagree, choose BIC: its penalty per lag is larger, so it picks fewer lags.
Problem 2 · Breusch-Godfrey test
The Question: Which Auxiliary Regression?
Question 23. The R code below estimates an ARDL model (answer: ARDL(1,1)).
model <- dynlm(du_ts ~ L(du_ts, 1) +
L(g_ts, 0:1))
L(du_ts, 1)$du_{t-1}$: last quarter’s change in unemploymentL(g_ts, 0:1)$g_t$ and $g_{t-1}$: GDP growth this quarter and last quarter
Question 24. Based on question 23, below we test for serial correlation in an ARDL model’s residuals using the Breusch-Godfrey test of order up to 4 (bgtest(ardl model, order = 4)). Which auxiliary regression is used for the BG test? Let $\hat e_t$ be the OLS residual from the estimated ARDL model. $du_t$ is the change in unemployment and $g_t$ is GDP growth.
$\theta$: how much of last quarter’s change in unemployment carries into this quarter. OLS gives $\hat\theta = 0.379$.
It could be the case thatthis quarter’s error is related to last quarter’s error: $e_t = \rho\, e_{t-1} + v_t$ with $\rho \neq 0$. For example, something the model leaves out, such as a slowdown in one industry, raises unemployment for several quarters.
Then the regressor contains the errorOne quarter earlier, $du_{t-1} = \delta + \theta\, du_{t-2} + \beta_0 g_{t-1} + \beta_1 g_{t-2} + e_{t-1}$. So $du_{t-1}$ contains $e_{t-1}$, which is related to $e_t$.
IF SO$\hat\theta = 0.379$ is estimating $\theta$ plus the part of $e_t$ that moves with $e_{t-1}$ instead of $\theta$, the effect we want. More data does not fix this: OLS is inconsistent.
$\implies$ Before trusting $\hat\theta$, test whether the errors are related over time: the Breusch-Godfrey test.
Breusch-Godfrey · The idea
Are This Quarter’s Residuals Related to Earlier Ones?
The model in R, estimated on Australian data (151 quarters)
model <- dynlm(du_ts ~ L(du_ts, 1) + L(g_ts, 0:1))
R adds the intercept ($0.174$) without being asked.
Error and residual$e_t$: the true error, never observed. $\hat e_t = du_t - \widehat{du}_t$: the residual, our estimate of $e_t$. The test uses the residuals in place of the errors.
Residual correlogram$r_1 = -0.162$ and $r_2 = 0.163$, just outside the bounds $\pm 1.96/\sqrt{151} = \pm 0.160$. By eye, it is hard to decide.
The BG test checks lags 1 to 4 together$e_t = \rho_1 e_{t-1} + \rho_2 e_{t-2} + \rho_3 e_{t-3} + \rho_4 e_{t-4} + v_t$ $H_0$: $\rho_1 = \rho_2 = \rho_3 = \rho_4 = 0$ $\iff$ no autocorrelation up to lag 4 $H_1$: at least one $\rho_j \neq 0$ The auxiliary regression in Step 2 also includes the model’s regressors.
Breusch-Pagan (week 2) regresses $\hat e^2$ and tests heteroskedasticity. Breusch-Godfrey regresses $\hat e$ on lagged $\hat e$ and tests autocorrelation.
Computed from okun5_aus (POE5), the data behind this model: 1978 Q4 to 2016 Q2.
Breusch-Godfrey · Step 1
Save the Residuals and Make Four Lagged Copies
order = 4Four lagged copies of the residuals, $\hat e_{t-1}$ to $\hat e_{t-4}$, each moved down one more row (as in last week’s lag table). The first rows have no earlier quarter: NA.
fill = 0 (the default)bgtest() replaces each NA with 0 and keeps all $T = 151$ rows. 0 is the mean of the residuals, so the filled rows add no pattern of their own.
fill = NADrops the first 4 rows instead: $T - p = 151 - 4 = 147$ rows. This is the lecture’s $n = T - p$. Both versions give the same decision here (Step 3).
Breusch-Godfrey · Step 2
Regress $\hat e_t$ on the Original Regressors and the Lagged Residuals
Auxiliary regressionA second regression that we run only to compute the test statistic. (The lecture writes $\alpha$ where this question writes $\gamma$.)
Every regressor of the ARDL model, copied as it is: $du_{t-1}$, $g_t$, $g_{t-1}$ in levels, not differences.
order = 4 $\implies$ the residual at lags 1, 2, 3, and 4.
Why keep the regressors?$du_{t-1}$ contains last quarter’s error, so it is correlated with $\hat e_{t-1}$ (correlation 0.72 in these data). Without $du_{t-1}$, the test rejects too rarely. With these data it gives $LM = 8.42$ and does not reject (Step 3).
Breusch-Godfrey · Step 3
Compute $LM = T \times R^2$ and Compare to $\chi^2_{(4)}$
Choice (b), without the regressors: $R^2 = 0.05577$, $LM = 151 \times 0.05577 = 8.42 < 9.49$. That test does not reject $H_0$.
Breusch-Godfrey · Interpretation
Checking Each Answer Choice
a$\hat e_t = \gamma_0 + \gamma_1 du_{t-1} + \gamma_2 g_t + \gamma_3 g_{t-1} + \rho \hat e_{t-1} + v_t$One lagged residual: this tests lag 1 only, but order = 4.
b$\hat e_t = \gamma_0 + \rho_1 \hat e_{t-1} + \rho_2 \hat e_{t-2} + \rho_3 \hat e_{t-3} + \rho_4 \hat e_{t-4} + v_t$Leaves out $du_{t-1}$, $g_t$, $g_{t-1}$. With these data it gives $LM = 8.42$, which does not reject $H_0$.
d$\Delta\hat e_t = \gamma_0 + \gamma_1 \Delta du_{t-1} + \gamma_2 \Delta g_t + \gamma_3 \Delta g_{t-1} + \rho_1 \Delta\hat e_{t-1} + \cdots + \rho_4 \Delta\hat e_{t-4} + v_t$The test uses the residuals and regressors as they are in the model. Nothing is differenced.
The test rejects. Can we just switch to HAC standard errors?
InterpretationThe errors of this ARDL(1,1) are autocorrelated. Write the model one quarter earlier: $du_{t-1} = \delta + \theta\, du_{t-2} + \beta_0 g_{t-1} + \beta_1 g_{t-2} + e_{t-1}$. So the regressor $du_{t-1}$ contains $e_{t-1}$, and with autocorrelated errors $e_{t-1}$ is correlated with $e_t$. OLS is then inconsistent: its estimates do not get close to the true coefficients even with a very large sample. HAC standard errors do not change that.
Next stepAdd lags of $du$ and $g$, choose how many with SC (Problem 1), and run the BG test again.
The posted solutions for this exam list (d) but describe the regression in (c) (“the original regressors plus lagged residuals up to order 4”). Answer letters can differ across exam versions; in the version shown here, the answer is (c).
Problem 3 · HAC standard errors
The Question: Taxes and the Top 1% Income Share
We are interested in the following model to determine the effect of the tax rate on inequality, where SHARE is the percentage income share held by those with the top 1% of incomes:
$SHARE_t = \alpha_1 + \alpha_2 TAX_t + \alpha_3 YEAR_t + \alpha_4 YEAR_t^2 + e_t$
bgtest(reg.mod, order=3, fill=0)
Breusch-Godfrey test for serial correlation of order up to 3
LM test = 56.568, df = 3, p-value = 3.177e-12
What is the confidence interval for the marginal effect of taxes on inequality, given the critical value, $t_{0.975,76} = 1.99$? Is the use of HAC errors justified?
aThe confidence interval is $[-0.23, 0.02]$ and the use of HAC errors is justified
bThe confidence interval is $[0.23, 23.02]$ and the use of HAC errors is justified
cThe confidence interval is $[-0.23, 0.02]$ and the use of HAC errors is not justified
dThe confidence interval is $[-.44, 0.02]$ and the use of HAC errors is not justified
eNone of the above
Source: Spring 2023 Midterm V1, Question 12
HAC standard errors · An aside
Ten Friends, One Study Guide
The questionIs the midterm curved? You ask 10 friends, and 9 say yes.
Ten separate answersEach friend found out on their own. That is 10 independent answers, and 9 out of 10 is strong evidence.
One shared sourceAll 10 read the same study guide, written by one student. That is about one answer repeated 10 times. Counting it as 10 makes you too confident.
Back to regressionThe usual OLS standard error treats the 80 yearly errors as 80 independent pieces of information. If each year’s error is close to last year’s, the data contain less information than that, and the usual standard error is too small.
HAC standard errors · The idea
This Year’s Error Is Close to Last Year’s
The dataCanada, 1921 to 2000: $T = 80$ years. $SHARE$: income share of the top 1% (%). $TAX$: median marginal tax rate of the top 1% (%). The plot shows the OLS residuals.
Long runs12 years in a row above 0, then later 13 years in a row below 0. The sign changes only 14 times in 80 years; independent errors would change sign about 40 times.
Correlogram$r_1 = 0.81$, $r_2 = 0.72$, $r_3 = 0.58$: far outside the bounds $\pm 1.96/\sqrt{80} = \pm 0.22$.
HAC standard errorsHeteroskedasticity and autocorrelation consistent (HAC) standard errors, computed by NeweyWest() in R. Same OLS estimates; standard errors that allow the errors to be correlated over time.
Computed from inequality (POE5): the variables ca_share and ca_tax in the exam’s R output.
Hypotheses$H_0$: no autocorrelation up to lag 3 vs $H_1$: autocorrelation at lag 1, 2, or 3. order = 3 $\implies$ $\chi^2_{(3)}$, 5% critical value $7.81$.
Statistic$LM = 56.568$, far above $7.81$.
Is $p = 3.177\text{e-}12$ small or large?
p-value$3.177\text{e-}12 = 3.177 \times 10^{-12} = 0.000000000003177$. Not 3.177.
REJECT $H_0$The errors are autocorrelated $\implies$ HAC standard errors are justified.
Unlike Problem 2, this model has no lagged $SHARE$ on the right side. If $TAX_t$ is uncorrelated with $e_t$, the OLS estimates are consistent, and only the standard errors need correcting.
The interval includes 0. This agrees with the output: $p = 0.098 > 0.05$.
HAC standard errors · Step 3
Usual Standard Error vs HAC Standard Error
If we drop HAC and use the usual standard errors, will $-0.107$ change?
Usual OLS standard error$0.0177$: interval $[-0.142,\ -0.072]$ does not include 0.
HAC standard error$0.064$, about 3.6 times larger: interval $[-0.234,\ 0.020]$ includes 0.
With the usual SE the effect looks significant. That SE assumes no autocorrelation, and the BG test rejects that assumption.
Usual OLS standard error computed from inequality (POE5) with coeftest(reg.mod); it is not in the exam output.
HAC standard errors · Interpretation
Checking Each Answer Choice
b$[0.23, 23.02]$, justifiedThe interval must contain the estimate $-0.107$; this one does not.
c$[-0.23, 0.02]$, not justifiedRight interval. But the BG test rejects, so HAC is justified.
d$[-.44, 0.02]$, not justifiedIts midpoint is $-0.21$, not $-0.107$. And HAC is justified.
eNone of the aboveChoice (a) holds.
a$[-0.23, 0.02]$, justifiedAnswer: (a)
InterpretationA tax rate 1 percentage point higher goes with a top 1% share 0.107 percentage points lower, holding $YEAR$ and $YEAR^2$ fixed. With HAC standard errors the interval includes 0: the effect is not statistically significant at the 5% level ($p = 0.098$).