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ECON2515 · Intermediate Applied Econometrics Ii

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Chapter 9 of 10 · ECON 2515

Heteroskedasticity: Detection and Correction

Heteroskedasticity means the error variance is not constant across observations — Var(u | x) = σᵢ² ≠ σ² — which breaks the homoskedasticity assumption MLR.5. The result you must state precisely: OLS coefficients stay unbiased and linear, but they are no longer BLUE, and the usual standard errors are biased, so every t-test, F-test and confidence interval built on them becomes unreliable. This ECON 2515 Week 9 topic (and Quiz 4) teaches you to detect it — residual/û² plots plus the Breusch-Pagan and White LM tests, where LM = n·R²_aux ~ χ² — and to correct it with robust (White) standard errors or WLS/GLS. The one thing you never do is claim the coefficients are biased or 're-estimate the betas'.

In this chapter

What this chapter covers

  • 011. What it is — Var(u | x) = σᵢ² varies with the regressors, violating MLR.5 (homoskedasticity)
  • 022. Consequences — OLS stays unbiased and linear but is no longer BLUE; the usual SEs are biased
  • 033. What breaks vs survives — coefficients and R² look fine; t-tests, F-tests and CIs are invalid
  • 044. Detection by eye — residual/û² plots; the funnel/trumpet shape signals a non-constant variance
  • 055. Breusch-Pagan test — regress û² on the x's; LM = n·R² ~ χ²(df = number of auxiliary regressors)
  • 066. White test — add squares and cross-products (or use ŷ, ŷ²); catches non-linear variance forms
  • 077. Robust (White) standard errors — the easy fix; corrects the SEs and leaves coefficients unchanged
  • 088. WLS / GLS / FGLS — re-weight by 1/√h to restore efficiency (BLUE); interpret in original units
Worked example · free

Run a Breusch-Pagan test and prescribe the fix

Q [12 marks]. A house-price model price = β₀ + β₁size + β₂beds + β₃age + u is fitted by OLS on n = 250 homes, and a plot of the squared residuals against fitted price fans out as price rises. (a) Define heteroskedasticity and name the OLS assumption the fanning violates. (b) Write the Breusch-Pagan auxiliary regression and state H₀ and H₁. (c) The auxiliary regression returns R² = 0.048; form the LM statistic and test at the 5% level, given χ²₀.₀₅ with 3 df = 7.81. (d) What is the standard fix, and does it change the coefficients?
  • +2(a) Heteroskedasticity means the error variance is not constant across observations — here it rises with fitted price, so Var(u | x) = σᵢ² rather than a single σ². The fanning residual plot is evidence that the spread of the errors grows with the regressors, which violates MLR.5, the homoskedasticity assumption Var(u | x) = σ².
  • +4(b) Breusch-Pagan regresses the squared OLS residuals on the model's regressors: û² = δ₀ + δ₁size + δ₂beds + δ₃age + v. The hypotheses are H₀: δ₁ = δ₂ = δ₃ = 0 (the variance does not depend on the x's — homoskedasticity) versus H₁: at least one δ ≠ 0 (the variance depends on at least one regressor — heteroskedasticity).
  • +4(c) The test statistic is LM = n·R² = 250 × 0.048 = 12.0, which follows χ²(3) under H₀ — the df equal the number of regressors in the auxiliary regression. Since 12.0 > χ²₀.₀₅,₃ = 7.81, the statistic falls in the upper-tail rejection region, so reject H₀: there is significant evidence of heteroskedasticity at the 5% level.
  • +2(d) Report heteroskedasticity-robust (White) standard errors. The coefficients are unchanged — only the standard errors, and therefore the t-statistics, confidence intervals and p-values, are corrected so that inference is valid again. (An alternative fix is WLS/GLS, which re-weights the data to restore efficiency.)
Fanning = MLR.5 (homoskedasticity) violated; BP auxiliary regression is û² on size, beds and age with H₀: all δ = 0; LM = 250 × 0.048 = 12.0 > 7.81 = χ²₀.₀₅,₃ so reject H₀; the fix is robust (White) standard errors, which leave the coefficients unchanged.
Sia tip — The whole test collapses to one comparison: LM = n·R² versus the χ² critical value at df = number of auxiliary regressors, upper tail. Read LM against the χ² table — never the F or t table — and finish by naming the fix (robust SEs) and stressing that the coefficients do not move.
Glossary

Key terms

Heteroskedasticity
A non-constant error variance, Var(u | x) = σᵢ² ≠ σ², so the typical size of the error changes with the regressors. It violates MLR.5 and makes the usual OLS standard errors wrong.
Homoskedasticity (MLR.5)
The Gauss-Markov assumption that the conditional error variance is a single constant, Var(u | x) = σ², for every observation. It is needed for OLS efficiency (BLUE) and for correct standard errors.
Consequence for OLS
Under heteroskedasticity OLS stays unbiased, linear and consistent, but is no longer BLUE (efficient), and the conventional standard errors are biased — so t-tests, F-tests, CIs and p-values are unreliable.
Breusch-Pagan (LM) test
Regress the squared residuals û² on the original regressors; the statistic LM = n·R²_aux follows χ²(df = number of auxiliary regressors). A large LM rejects homoskedasticity. It assumes a linear variance form.
White test
A more flexible test that regresses û² on the regressors, their squares and cross-products (or, to save df, on the fitted values ŷ and ŷ²). Same LM = n·R² ~ χ² rule, but it catches non-linear variance forms.
Robust (White) standard errors
Also Huber/Eicker or 'sandwich' SEs. They give a variance estimate valid under arbitrary, unknown heteroskedasticity. The coefficients are unchanged; only the standard errors — and hence t, F, CI, p — are corrected.
Weighted least squares (WLS) / GLS
If Var(u | x) = σ²h(x) with h known, divide every term (including the intercept) by √hᵢ so the transformed error is homoskedastic; OLS on the transformed model is GLS = BLUE, giving less weight to high-variance observations.
Feasible GLS (FGLS)
GLS when the variance form is unknown: run OLS, form ln(û²), regress it on the x's to get fitted ĝ, set ĥ = exp(ĝ), then run WLS with weights 1/ĥ. Inference is on the transformed model; interpret in original units.
FAQ

Heteroskedasticity: Detection and Correction FAQ

Does heteroskedasticity bias the OLS coefficients?

No — this is the single most common wrong answer. The point estimates stay unbiased and consistent; what breaks are the standard errors, and therefore all inference (t-tests, F-tests, confidence intervals, p-values). Contrast this with omitted-variable bias, which does bias the coefficients. Confusing the two diagnostics loses easy marks in both the MCQ and Part B.

What is the difference between the Breusch-Pagan and White tests?

Both form LM = n·R² from an auxiliary regression of the squared residuals and compare it to a χ² critical value. Breusch-Pagan regresses û² on the original regressors only, so it assumes a linear variance form. The White test adds the squares and cross-products of the regressors (or uses ŷ and ŷ²), so it can detect arbitrary, non-linear heteroskedasticity — at the cost of many more degrees of freedom.

What degrees of freedom does the LM statistic use, and which table?

df equals the number of regressors in the auxiliary regression — not n and not n − k − 1. And LM = n·R² is a chi-square statistic, so you read it against the χ² table in the upper tail, never the F or t table. Getting the df or the table wrong flips the decision on a borderline statistic.

How do I fix heteroskedasticity once I have found it?

The easy default is heteroskedasticity-robust (White) standard errors: keep the OLS coefficients and just replace the SE formula with one valid under unknown heteroskedasticity, then do t/F inference as usual. If you know the variance form you can use WLS/GLS, which divides every term by √hᵢ to restore efficiency (BLUE); when the form is unknown, estimate it first with feasible GLS.

The White test uses ŷ or y?

The fitted values ŷ (and ŷ²), never the raw outcome y. The df-saving version of the White test regresses û² on ŷ and ŷ², because the fitted values are a compact summary of all the regressors. Using y or y² instead is a marked error.

After a WLS transformation, how do I interpret the coefficients?

Do the t-tests and F-tests on the transformed (re-weighted) model, because that is where the errors are homoskedastic and the SEs are valid. But interpret the coefficients in the original units of the model — the transformation is only a device to fix the variance, not a change in what the parameters mean. Remember also to divide the intercept term (the 1) by √hᵢ when doing WLS by hand.

Study strategy

Exam move

Memorise the one-line consequence verbatim and be able to defend every word: 'under heteroskedasticity OLS is still unbiased and linear but no longer BLUE, and the usual standard errors — and every t, F and CI built on them — are wrong.' Then drill the detect-and-fix routine until it is automatic: describe the funnel/trumpet residual plot and name MLR.5; write the auxiliary regression of û² and state H₀ (all δ = 0) versus H₁; form LM = n·R² and compare it to χ² at df = number of auxiliary regressors in the upper tail; decide; and prescribe the fix (robust SEs, which leave the coefficients unchanged, or WLS/GLS). Keep a one-page reference with the two tests side by side (Breusch-Pagan on the original x's, White adding squares and cross-products or ŷ and ŷ²), the LM = n·R² ~ χ² rule, and the WLS transform (divide every term by √hᵢ). Finally, practise the traps that examiners plant: never say the coefficients are biased, never read LM against the F table, never regress û² on y in the White test, and always separate this diagnostic (SEs break) from omitted-variable bias (coefficients break) and multicollinearity (coefficients fine but imprecise).

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