Appendix — Solutions — Problem Set 2.4 (Robust, Clustered, and HAC Standard Errors)
Full worked solutions to book/weeks/week-02/ps2.4.md. Notation follows CONVENTIONS §3 and Ch 2.4: the OLS estimator is \hat{\boldsymbol\beta}=(\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\mathbf{y} with sampling-error identity \hat{\boldsymbol\beta}=\boldsymbol\beta+(\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\boldsymbol\varepsilon; the error variance–covariance matrix is \boldsymbol\Omega=\operatorname{Var}(\boldsymbol\varepsilon\mid\mathbf{X}); the general variance is the sandwich (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\boldsymbol\Omega\mathbf{X}(\mathbf{X}'\mathbf{X})^{-1}; h_{ii} is the leverage (diagonal of the hat matrix \mathbf{H}=\mathbf{X}(\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'). Citations: Petersen, M. A. (2009). Estimating Standard Errors in Finance Panel Data Sets: Comparing Approaches. Review of Financial Studies, 22(1), 435–480. White (1980) and Newey & West (1987) referenced by name.
Problem 1 — Why \boldsymbol\Omega breaks the SE but not \hat\beta (12 pts)
(a) (4 pts) Take conditional expectations of the identity given \mathbf{X}: \mathbb{E}[\hat{\boldsymbol\beta}\mid\mathbf{X}] =\boldsymbol\beta+(\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\,\mathbb{E}[\boldsymbol\varepsilon\mid\mathbf{X}]. If \mathbb{E}[\boldsymbol\varepsilon\mid\mathbf{X}]=\mathbf{0} (exogeneity — the errors have mean zero given the regressors), the entire second term vanishes and \mathbb{E}[\hat{\boldsymbol\beta}\mid\mathbf{X}]=\boldsymbol\beta: unbiased. The only assumption used was mean independence of the errors, \mathbb{E}[\boldsymbol\varepsilon\mid\mathbf{X}]=\mathbf{0}. Nowhere did we touch \operatorname{Var}(\boldsymbol\varepsilon\mid\mathbf{X})=\boldsymbol\Omega — unbiasedness is a statement about the mean of the errors, and \boldsymbol\Omega is a statement about their spread and co-movement, a completely separate object.
(b) (4 pts) The general sandwich is \operatorname{Var}(\hat{\boldsymbol\beta}\mid\mathbf{X}) =(\mathbf{X}'\mathbf{X})^{-1}\,\mathbf{X}'\boldsymbol\Omega\mathbf{X}\,(\mathbf{X}'\mathbf{X})^{-1}. The piece that changes is the filling, \mathbf{X}'\boldsymbol\Omega\mathbf{X} (the “meat” of the sandwich); the two bread factors (\mathbf{X}'\mathbf{X})^{-1} depend only on the regressors and are untouched. Heteroskedasticity replaces the constant-\sigma^2 diagonal of \boldsymbol\Omega with varying \sigma_i^2, which changes the filling and hence the variance — but the variance is a property of the distribution of \hat{\boldsymbol\beta}, not of its center, so the center (\boldsymbol\beta) is left exactly where part (a) put it.
(c) (2 pts) No — \hat\beta_1=0.42 is not biased by the heteroskedasticity; it is still centered on the true effect. The thing that is wrong is the reported standard error 0.07, which was computed from the classical formula \hat\sigma^2(\mathbf{X}'\mathbf{X})^{-1} that silently assumed \boldsymbol\Omega=\sigma^2\mathbf{I}.
(d) (2 pts) If she trusts the 0.07 she loses the ability to say how confident to be: every t-statistic, p-value, and confidence interval she builds on top of it is wrong, so she may declare “significant” a result that an honest SE would call a coin-flip — being right about the point but wrong about the uncertainty is exactly how overconfident findings get published and then retracted.
Problem 2 — The sandwich by hand: classical vs robust (18 pts)
Data: x=(1,2,3,4,5), \hat\varepsilon=(+1,-2,+1,-4,+5), N=5, K=1.
(a) (3 pts) \sum_i x_i^2 = 1+4+9+16+25 = 55,\qquad \sum_i \hat\varepsilon_i^2 = 1+4+1+16+25 = 47.
(b) (4 pts) \hat\sigma^2=\frac{1}{N-K}\sum_i\hat\varepsilon_i^2=\frac{47}{4}=11.75. Then \widehat{\operatorname{Var}}(\hat\beta)_{\text{classical}}=\frac{\hat\sigma^2}{\sum_i x_i^2}=\frac{11.75}{55}=0.2136, \qquad \text{SE}_{\text{classical}}=\sqrt{0.2136}=\boxed{0.462}.
(c) (5 pts) Build the numerator \sum_i x_i^2\,\hat\varepsilon_i^2 term by term:
| i | x_i^2 | \hat\varepsilon_i^2 | x_i^2\hat\varepsilon_i^2 |
|---|---|---|---|
| 1 | 1 | 1 | 1 |
| 2 | 4 | 4 | 16 |
| 3 | 9 | 1 | 9 |
| 4 | 16 | 16 | 256 |
| 5 | 25 | 25 | 625 |
| Σ | 907 |
So \widehat{\operatorname{Var}}(\hat\beta)_{\text{HC0}}=\frac{907}{55^2}=\frac{907}{3025}=0.2998, \qquad \text{SE}_{\text{HC0}}=\sqrt{0.2998}=\boxed{0.548}.
(d) (3 pts) Ratio =0.548/0.462=\boxed{1.18}. The robust SE is larger. Look at the table in (c): the biggest squared residuals (\hat\varepsilon_4^2=16, \hat\varepsilon_5^2=25) sit at the high-x, high-leverage observations, and the robust numerator multiplies each \hat\varepsilon_i^2 by x_i^2, so those points dominate. The classical formula instead spreads one average \hat\sigma^2 evenly and so undercounts the uncertainty contributed by exactly the points that matter most. Robust SEs catch the positive alignment of \hat\varepsilon_i^2 with x_i^2 (textbook heteroskedasticity) that classical SEs ignore.
(e) (3 pts) If every \hat\varepsilon_i^2 were the same constant c (here c=9), the robust numerator becomes \sum_i x_i^2\cdot c=c\sum_i x_i^2, so \text{Var}_{\text{HC0}}=c\sum x_i^2/(\sum x_i^2)^2=c/\sum x_i^2, while \text{Var}_{\text{classical}}=\frac{Nc}{N-K}\big/\sum x_i^2. The two now differ only by the degrees-of-freedom factor N/(N-K)=5/4: the HC0 SE is a touch smaller (ratio \sqrt{(N-K)/N}=\sqrt{4/5}\approx 0.89), and HC1 — which multiplies HC0 by exactly N/(N-K) — would match the classical SE on the nose. The point: under genuine homoskedasticity the robust and classical answers converge, so robust SEs cost essentially nothing; the gap in (d) appeared only because the residuals were not constant.
Problem 3 — HC0, HC1, HC2, HC3: which correction, and when (16 pts)
(a) (3 pts) N=60, K=4: \dfrac{N}{N-K}=\dfrac{60}{56}=\boxed{1.071}. It is a flat degrees-of-freedom inflation applied equally to every observation, correcting (crudely) for the fact that residuals are too small after fitting. It ignores leverage — it does not know that high-leverage points have their residuals shrunk more than others, so it under-inflates exactly those points.
(b) (5 pts) With h_{ii}=0.5, the multipliers on that observation’s \hat\varepsilon_i^2 are: \text{HC0: } 1,\qquad \text{HC2: } \frac{1}{1-h_{ii}}=\frac{1}{0.5}=2,\qquad \text{HC3: } \frac{1}{(1-h_{ii})^2}=\frac{1}{0.25}=4. Ranking by aggressiveness: \text{HC3 }(4)>\text{HC2 }(2)>\text{HC0 }(1). HC3 inflates this leveraged point’s contribution four-fold; HC0 not at all.
(c) (4 pts) OLS chooses \hat{\boldsymbol\beta} to minimize \sum_i\hat\varepsilon_i^2, which bends the fitted line toward each data point; the line is partly chasing the very noise it is supposed to leave in the residual, so \hat\varepsilon_i is on average smaller in magnitude than the true \varepsilon_i. Feeding too-small squared residuals into the sandwich makes HC0 standard errors too small — overconfident — in finite samples. The shrinkage is worst at high-leverage points (a point that pulls the line hardest toward itself has the most-shrunk residual), and the leverage factor 1-h_{ii} is precisely the amount of that shrinkage: dividing by 1-h_{ii} (HC2) or (1-h_{ii})^2 (HC3) undoes it observation-by-observation, restoring most weight to exactly the points the fitting deflated most.
(d) (4 pts) For N=9{,}000 loans with no extreme leverage, the four flavors are practically identical (they agree to several decimals — the corrections all \to 1 as N grows and leverage stays low), so it does not matter; report HC1. For N=25 with one big \mathbf{X}-outlier, the flavors diverge, and you should default to HC3: it inflates the high-leverage point’s residual the most via the (1-h_{ii})^2 denominator, it is never badly anti-conservative, and so it is the safe pick that keeps you honest when a single point could otherwise drive a spuriously tight SE.
Problem 4 — Clustering and the Moulton factor (20 pts)
(a) (3 pts) Within a lender, the loans share a common pricing shock, so their errors are positively correlated: once you know one loan’s error you can partly predict its neighbors’. That means each additional loan from the same lender carries less than one observation’s worth of fresh information — it largely repeats the lender’s story. The naive formula treats all 9{,}000 as independent and so divides the error variance by too large an effective N, reporting a standard error that is too small. It is exactly like polling one household nine times and claiming a sample of nine: the nominal count overstates the real evidence.
(b) (5 pts) With \rho_x=\rho_\varepsilon=0.20 and n=200: 1+(n-1)\rho_x\rho_\varepsilon = 1+199\times(0.20)(0.20)=1+199\times 0.04 = 1+7.96 = \boxed{8.96}. The true variance is \approx 8.96\times the naive variance, so the true SE is \sqrt{8.96}\approx \boxed{2.99} times the naive one. A naive t-stat of 6.0 becomes an honest t-stat of \approx 6.0/2.99\approx\boxed{2.0} — still significant, but the megaphone has become a speaking voice.
(c) (4 pts) The inflation factor is 1+(n-1)\rho_x\rho_\varepsilon, driven by cluster size n, not by the total N. Option (i) — 9,000 more loans from the same 45 lenders — increases n (each lender now has \sim 400 loans), which makes the bracket larger, not smaller: more within-cluster repetition, no new independent evidence. Option (ii) — 45 new lenders — increases the number of clusters G while holding n fixed; the cluster-robust estimator’s precision lives in G, so new clusters are what actually buy evidence. Choose (ii). The slogan: pouring observations into existing clusters does not deliver the independence the naive formula assumes; more clusters does.
(d) (4 pts) The cluster-robust sandwich keeps the bread and rebuilds the filling cluster-by-cluster: \widehat{\operatorname{Var}}(\hat{\boldsymbol\beta})_{\text{cluster}} =(\mathbf{X}'\mathbf{X})^{-1}\left(\sum_{g=1}^{G}\mathbf{X}_g'\hat{\boldsymbol\varepsilon}_g\,\hat{\boldsymbol\varepsilon}_g'\mathbf{X}_g\right)(\mathbf{X}'\mathbf{X})^{-1}. White’s HC0 middle is \sum_i \hat\varepsilon_i^2\,\mathbf{x}_i\mathbf{x}_i' — a sum over single observations that keeps only the diagonal products \hat\varepsilon_i^2. The cluster version’s outer product \hat{\boldsymbol\varepsilon}_g\hat{\boldsymbol\varepsilon}_g' keeps the within-cluster cross-products \hat\varepsilon_i\hat\varepsilon_j for loans i,j in the same lender — exactly the off-diagonal covariances White’s version discards. Hence “White SEs are cluster SEs with clusters of size one”: when each cluster is a single observation there are no cross-products and the two formulas coincide.
(e) (4 pts) Clustering by state gives only G=9 clusters, and the cluster-robust estimator is consistent as G\to\infty, not as N\to\infty. With so few clusters the estimator is itself noisy and biased downward — overconfident, the very disease being treated — so the normal-approximation t-test is untrustworthy. The rule of thumb is to want \gtrsim 30–50 clusters. One concrete fix: use the wild cluster bootstrap (Cameron, Gelbach & Miller 2008, Review of Economics and Statistics, 90(3), 414–427) which resamples whole clusters; alternatively, use a t-distribution with G-1=8 degrees of freedom rather than the normal.
Problem 5 — HAC / Newey–West and serial correlation (16 pts)
(a) (4 pts) \boldsymbol\Omega is banded: nonzero entries along and near the diagonal, fading to zero away from it. The diagonal holds each period’s variance; the first off-diagonals hold the large \operatorname{Cov}(\varepsilon_t,\varepsilon_{t-1}), and as the lag \ell grows the covariance \operatorname{Cov}(\varepsilon_t,\varepsilon_{t-\ell}) shrinks toward zero (e.g. \operatorname{Cov}(\varepsilon_t,\varepsilon_{t-12})\approx 0). This differs from clustering: clustering gives block-diagonal \boldsymbol\Omega with dense blocks and clean group boundaries (correlation is all-or-nothing — full within a lender, zero across), whereas serial correlation gives a smear around the diagonal with no groups, the correlation decaying continuously with distance.
(b) (4 pts) Too big. Positive autocorrelation means each month’s residual partly echoes the previous month’s, so the 120 months contain fewer than 120 genuinely independent draws — the effective sample size is well below 120. The naive classical formula divides by the wrong (too large) effective T, producing a standard error that is too small and therefore a t-statistic that is too big.
(c) (4 pts) Switch to HAC / Newey–West standard errors. The one knob is the bandwidth L — the number of lags whose residual cross-products enter the middle of the sandwich (with declining Bartlett weights w_\ell=1-\ell/(L+1)). The rule of thumb: L\approx 4\left(\frac{T}{100}\right)^{2/9}=4\left(\frac{120}{100}\right)^{2/9}=4(1.2)^{0.2222}\approx 4.17 \;\Rightarrow\; \boxed{L=4}. Too small a bandwidth misses real persistence (it stops counting lagged correlation that is genuinely there, so the SE stays too small); too large adds noisy, near-zero lag terms that inflate the variance of the variance estimate, making the SE itself unstable.
(d) (4 pts) For the high-frequency colleague with negative autocorrelation, naive t-stats are too small (too conservative). The general principle: the sign of the autocorrelation determines the direction of the naive SE’s error. Positive autocorrelation (the usual finance case — shocks persist) shrinks the effective sample and makes naive SEs too small / t-stats too big; negative autocorrelation (e.g. bid–ask bounce, where an up-tick tends to be followed by a down-tick) does the reverse, making naive SEs too large / t-stats too small. HAC gets either case right because it estimates the actual band of off-diagonal covariances rather than assuming the band is empty.
Problem 6 — Petersen (2009): which clustering dimension? (18 pts)
(a) (4 pts) A persistent firm effect dominates. Petersen’s diagnostic: whichever clustering moves the SE more away from the White number points at the bigger source of correlation. Here firm-clustering moves the SE from 0.021 to 0.044 (roughly doubling) while time-clustering barely moves it (0.021\to 0.024) — so the residuals are correlated across time within a firm (a firm mispriced this year tends to be mispriced next year) far more than across firms within a year. The correct choice is clustering by firm, which asserts that \boldsymbol\Omega’s off-diagonal blocks are organized by firm: one dense block per firm spanning all its years, zeros across firms.
(b) (3 pts) White (HC1) SEs assume the off-diagonals of \boldsymbol\Omega are zero — errors uncorrelated across observations — and only ever look at one observation’s own squared residual. They correct for heteroskedasticity (the diagonal) but are completely blind to the within-firm correlation across years (the off-diagonal blocks). That is why White’s 0.021 is barely above classical’s 0.018 and far below the honest firm-clustered 0.044.
(c) (4 pts) That opposite pattern — firm-clustered \approx White but time-clustered roughly double — would imply a common time effect dominates: a market-wide shock in each period hits all firms’ residuals together (correlation across firms within a period), with little persistence within a firm. You would cluster by time (or absorb it with time fixed effects and cluster on any residual common shock). The lesson restated: the correlation structure of the residuals — hence the right clustering — depends on what you put on the left-hand side, because different outcomes carry different shocks; the same panel of firms and years can demand firm-clustering for one regression and time-clustering for another. Clustering is an economic judgment about \boldsymbol\Omega, not a software default.
(d) (4 pts) They do different jobs. A firm fixed effect (a dummy per firm, the demeaning of Ch 2.3) removes the firm’s average level — the permanent, time-invariant part of the firm’s outcome. Clustering by firm handles correlation in the leftover residuals over time — the co-movement that remains after the mean is removed, which a fixed effect cannot touch because it is about the mean, not the covariance. So “I have firm FEs, so I don’t need to cluster by firm” conflates a mean with a covariance and leaves you overconfident. Yes — you typically want both at once: firm fixed effects to absorb permanent firm differences and clustering by firm to honor the within-firm residual correlation that survives the demeaning.
(e) (3 pts) Two-way (firm + time) clustered variance is built from three sandwiches you already know: \widehat{\operatorname{Var}}_{\text{two-way}} =\widehat{\operatorname{Var}}_{\text{cluster by firm}} +\widehat{\operatorname{Var}}_{\text{cluster by time}} -\widehat{\operatorname{Var}}_{\text{White (heterosk.-only)}}. The White (single-observation) sandwich is subtracted because each observation’s own variance is counted twice — once inside its firm block and once inside its time block — so the overlap (the pure diagonal, captured by White) must be removed once to avoid double-counting (Cameron, Gelbach & Miller 2011, Journal of Business & Economic Statistics, 29(2), 238–249).
End of solutions for PS 2.4. Cross-references: Ch 2.4 (sandwich, White/HC, HC0–HC3, clustering, Moulton factor, HAC/Newey–West, Petersen taxonomy), Ch 2.1 (hat matrix and leverage h_{ii}), Ch 2.2 (classical variance and Gauss–Markov), Ch 2.3 (demeaning = fixed effects). Numeric keys: P2 classical SE 0.462, HC0 SE 0.548, ratio 1.18; P3 HC1 factor 1.071, HC2/HC3 multipliers 2 and 4 at h=0.5; P4 bracket 8.96, SE factor 2.99, honest t ≈ 2.0; P5 NW bandwidth L=4; P6 firm effect dominates → cluster by firm, two-way = firm + time − White. Looks ahead to Weeks 3–4 (panel/fixed-effects methods) and Ch 5.4 (full Petersen 2009 reader’s guide).