Module M06 Quiz: Heteroskedasticity

14 questions · Graduate (ECMT5001) · MCQ, calculation, short answer

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Question 1

Lesson L01 · Definition

Heteroskedasticity violates which Gauss-Markov assumption?

Type: Multiple Choice

A) MLR.1 — Linearity in parameters
B) MLR.4 — Zero conditional mean $E(u|\mathbf{x}) = 0$
C) MLR.5 — Homoskedasticity $\text{Var}(u|\mathbf{x}) = \sigma^2$
D) MLR.3 — No perfect multicollinearity

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Answer: C

Homoskedasticity (MLR.5) requires $\text{Var}(u_i | \mathbf{x}_i) = \sigma^2$ — constant error variance across all observations. Heteroskedasticity means $\text{Var}(u_i | \mathbf{x}_i) = \sigma_i^2$ varies with $i$ (depends on the regressors or observation characteristics). This leaves OLS unbiased and consistent (MLR.4 is intact), but violates the efficiency condition needed for Gauss-Markov.

Question 2

Lesson L01 · Consequences

When heteroskedasticity is present, OLS:

Type: Multiple Choice

A) Is biased and inconsistent
B) Is unbiased and consistent, but no longer BLUE; standard errors are incorrect, invalidating t and F tests
C) Is still BLUE — heteroskedasticity only affects efficiency, not inference
D) Produces coefficients that are too large in absolute value

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Answer: B

Since heteroskedasticity does not violate MLR.1–MLR.4, OLS remains unbiased ($E(\hat{\beta}|X) = \beta$) and consistent. However:

OLS is no longer BLUE — it is no longer the minimum-variance linear unbiased estimator (WLS with correct weights beats it).
The standard OLS standard error formulas assume homoskedasticity; with heteroskedasticity they are biased → t-statistics and F-statistics are invalid → confidence intervals and hypothesis tests are unreliable (typically too narrow → inflated Type I error).

Question 3

Lesson L01 · Detection

A residual plot shows residuals fanning out (increasing spread) as fitted values $\hat{y}$ increase. This pattern suggests:

Type: Multiple Choice

A) Serial correlation — errors are correlated across observations
B) Heteroskedasticity — error variance increases with the level of the fitted value
C) Omitted variable bias — a relevant variable is missing
D) Perfect multicollinearity among the regressors

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Answer: B

A fan pattern (residuals spread wider at higher fitted values) is the classic visual signature of heteroskedasticity where $\text{Var}(u_i) \propto \hat{y}_i$ or some function of the regressors. Common causes: wage data (high earners have more earnings variability), firm-level data (larger firms have larger absolute errors), cross-country data (larger economies have more variable growth). The remedy is robust standard errors or WLS.

Question 4

Lesson L02 · Breusch-Pagan Test

Describe the Breusch-Pagan (BP) test for heteroskedasticity step by step.

Type: Short Answer

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Steps:

Estimate the original regression by OLS: $y = X\beta + u$. Obtain residuals $\hat{u}_i$.
Compute squared residuals $\hat{u}_i^2$.
Regress $\hat{u}_i^2$ on the original regressors $x_1, \ldots, x_k$ (and a constant): $\hat{u}_i^2 = \delta_0 + \delta_1 x_{i1} + \cdots + \delta_k x_{ik} + v_i$.
Compute the LM statistic: $LM = n \cdot R^2_{\hat{u}^2}$ where $R^2_{\hat{u}^2}$ is the $R^2$ from the auxiliary regression.
Under $H_0$ (homoskedasticity), $LM \sim \chi^2_k$ asymptotically.
Reject $H_0$ if $LM > \chi^2_{k, \alpha}$.

$H_0$: $\delta_1 = \delta_2 = \cdots = \delta_k = 0$ (error variance does not depend on regressors).

The BP test assumes heteroskedasticity is a linear function of the regressors. If the true form is nonlinear, the White test (which adds squared and cross terms) has more power.

Question 5

Lesson L02 · BP Test Calculation

[Calculation] You run the auxiliary regression of $\hat{u}_i^2$ on $x_1$ and $x_2$ with $n = 200$ observations. The auxiliary regression gives $R^2 = 0.08$. Test for heteroskedasticity at the 5% level. $\chi^2_{2, 0.05} = 5.99$.

Type: Calculation

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\[LM = n \cdot R^2 = 200 \times 0.08 = \mathbf{16.0}\]

Since $LM = 16.0 > \chi^2_{2, 0.05} = 5.99$, reject $H_0$ at the 5% level. There is statistically significant evidence of heteroskedasticity. OLS standard errors are unreliable — use robust (HC) standard errors or WLS.

Note: $k = 2$ regressors (not counting the intercept in the auxiliary regression), so df = 2.

Question 6

Lesson L02 · White vs BP

What is the key difference between the White test and the Breusch-Pagan test? When would you prefer the White test?

Type: Short Answer

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Key difference: The Breusch-Pagan test assumes heteroskedasticity is a linear function of the regressors ($\text{Var}(u_i) = \sigma^2 h(\delta_0 + \delta_1 x_1 + \cdots + \delta_k x_k)$). The White test is more general — it adds squared terms and cross-products of all regressors to the auxiliary regression, allowing for nonlinear heteroskedasticity.

Prefer the White test when: - You suspect nonlinear heteroskedasticity (e.g. variance proportional to $x^2$) - You are unsure of the functional form of the heteroskedasticity - After a BP test that fails to reject, if the residual plot still suggests a problem

Cost of White test: Uses more degrees of freedom (with $k$ regressors: $2k + \binom{k}{2}$ terms in the auxiliary regression), reducing power in small samples. The BP test is preferred when the functional form of heteroskedasticity is known or suspected to be linear.

(Adapted from MIT 14.32 rec_4_20: "There is no universally better test — BP is better when the linear functional form is correct; White is better when nonlinear heteroskedasticity is present.")

Question 7

Lesson L02 · White Test Degrees of Freedom

[Calculation] With $k = 2$ original regressors, the White test auxiliary regression includes: the 2 original regressors, their squares ($x_1^2$, $x_2^2$), and their cross-product ($x_1 x_2$). How many regressors (excluding the intercept) are in the White auxiliary regression?

Type: Calculation

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Original regressors: $x_1$, $x_2$ → 2 terms
Squared terms: $x_1^2$, $x_2^2$ → 2 terms
Cross-product: $x_1 x_2$ → 1 term

Total = 5 terms → degrees of freedom for the $\chi^2$ test: $df = 5$.

General formula for $k$ regressors: $k + k + \binom{k}{2} = 2k + k(k-1)/2$ terms. For $k=3$: $6 + 3 = 9$ terms. This rapid growth in df explains why the White test loses power quickly as $k$ increases.

Question 8

Lesson L03 · Robust Standard Errors

Robust (heteroskedasticity-consistent / HC) standard errors fix the inference problem from heteroskedasticity but not the efficiency problem. Explain why.

Type: Short Answer

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Fixing inference: Robust (White/Huber-Eicker) SEs use a consistent estimator of $\text{Var}(\hat{\beta}|X)$ that remains valid under any form of heteroskedasticity: $\widehat{\text{Var}}(\hat{\beta}) = (X'X)^{-1} \left(\sum \hat{u}_i^2 x_i x_i'\right) (X'X)^{-1}$. Using these SEs for t and F tests gives asymptotically correct inference.

Not fixing efficiency: Robust SEs still use OLS point estimates $\hat{\beta}$. OLS is no longer BLUE under heteroskedasticity — WLS with weights $1/\sigma_i$ would give more precise (lower-variance) estimates. Robust SEs simply correct the reported uncertainty around the (inefficient) OLS estimates. They do not change $\hat{\beta}$ itself.

Practical implication: When heteroskedasticity is suspected, always report robust SEs. But if the form of heteroskedasticity is known, use WLS for efficiency gains.

Question 9

Lesson L03 · When to Use Robust SEs

You run a regression and suspect heteroskedasticity but are unsure of its form. Which SEs should you report?

Type: Multiple Choice

A) OLS SEs — they are always valid with large samples
B) Robust (HC) SEs — valid asymptotically under any form of heteroskedasticity
C) WLS SEs — always more efficient than OLS
D) Bootstrap SEs — always preferred over asymptotic SEs

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Answer: B

When the form of heteroskedasticity is unknown, robust (HC) SEs are the safe default. They are valid asymptotically regardless of the heteroskedasticity pattern. OLS SEs (Answer A) are invalid under heteroskedasticity even in large samples — they are biased, not just imprecise. WLS (C) requires knowing the form of heteroskedasticity to choose correct weights; misspecifying the weights can make things worse. Bootstrap SEs (D) are a good alternative but computationally intensive and not always needed.

Modern practice: Most applied econometricians report robust SEs by default, even when heteroskedasticity is not confirmed — the cost is negligible (slightly less power under homoskedasticity) and the protection against invalid inference is valuable.

Question 10

Lesson L04 · WLS Weights

If $\text{Var}(u_i) = \sigma^2 \cdot x_i$ (error variance proportional to $x_i$), what weight should observation $i$ receive in WLS?

Type: Multiple Choice

A) $w_i = x_i$
B) $w_i = 1/x_i$
C) $w_i = \sqrt{x_i}$
D) $w_i = 1/\sqrt{x_i}$

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Answer: B — $w_i = 1/x_i$

WLS multiplies each observation by $1/\sqrt{h_i}$ (transforming the equation), where $\text{Var}(u_i) = \sigma^2 h_i$. Here $h_i = x_i$, so the weight applied in WLS is $w_i = 1/h_i = 1/x_i$.

After weighting: $\tilde{y}_i = y_i/\sqrt{x_i}$, $\tilde{x}_i = x_i/\sqrt{x_i} = \sqrt{x_i}$, $\tilde{u}_i = u_i/\sqrt{x_i}$. Then $\text{Var}(\tilde{u}_i) = \sigma^2 x_i / x_i = \sigma^2$ — homoskedastic. OLS on the transformed model is equivalent to WLS and is now BLUE.

Question 11

Lesson L04 · FGLS

In Feasible GLS (FGLS), you estimate $\ln(\hat{u}_i^2) = \delta_0 + \delta_1 x_{i1} + \delta_2 x_{i2} + v_i$ and obtain fitted values $\hat{g}_i$. You then set $\hat{h}_i = e^{\hat{g}_i}$. What weight do you use in the subsequent WLS step?

Type: Multiple Choice

A) $w_i = \hat{h}_i$
B) $w_i = 1/\hat{h}_i$
C) $w_i = \ln(\hat{h}_i)$
D) $w_i = \sqrt{\hat{h}_i}$

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Answer: B — $w_i = 1/\hat{h}_i$

The auxiliary log-regression estimates $\ln(\text{Var}(u_i)) \approx \hat{g}_i$, so $\hat{h}_i = e^{\hat{g}_i}$ estimates $\text{Var}(u_i)/\sigma^2$. WLS requires weighting by the inverse of the estimated variance: $w_i = 1/\hat{h}_i$. This is FGLS — it is efficient when the model for $h_i$ is correct, and is asymptotically as efficient as GLS with known $h_i$. In practice, FGLS beats OLS (more efficient) but is less reliable than robust SEs when the heteroskedasticity model is misspecified.

Question 12

Lesson L02 · MIT-Style Comparison

[Short Answer — adapted from MIT 14.32 rec_4_20] "Which is better — the Breusch-Pagan test or the White test?" Give a balanced answer.

Type: Short Answer

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Neither is universally better — each has advantages:

BP test advantages: - More powerful when heteroskedasticity is actually a linear function of the regressors - Uses fewer degrees of freedom → more power in small samples - Simpler to implement and interpret

White test advantages: - More general — detects nonlinear heteroskedasticity that BP misses - Also detects certain specification errors (since the White test statistic is related to the general test for misspecification) - Does not require prior knowledge of the heteroskedasticity's functional form

Practical recommendation: Run BP first. If rejected, heteroskedasticity is confirmed. If not rejected but the residual plot looks suspicious, run the White test as a robustness check. Regardless of test results, in applied work it is now standard to report robust SEs — making the test results less critical for inference (though still important for understanding the data structure).

Question 13

Lesson L01 · Log Transformation

Wages are often modelled as $\ln(\text{wage})$ rather than wage levels. Why does taking logs reduce heteroskedasticity?

Type: Short Answer

Show Answer

Log transformation compresses the scale of the dependent variable, which tends to reduce heteroskedasticity when the variance of the outcome grows with its level.

Mechanism: If $\text{Var}(y_i) \propto \mu_i^2$ (i.e. a constant coefficient of variation), then $\text{Var}(\ln y_i) \approx \text{Var}(y_i)/\mu_i^2 \approx $ constant — homoskedastic. This is exactly the case for wages: high earners have higher absolute variance but similar proportional variability. The log transformation effectively converts absolute dispersion to proportional dispersion.

Additionally, log-wages are more normally distributed than levels (wages are right-skewed), which validates the normality assumption for exact finite-sample inference.

Question 14

Lesson L03 · ECMT5001-Style

[Short Answer + Calculation] A regression of house prices on income and household size gives:

	Coefficient	OLS SE	Robust SE
Intercept	120,000	35,000	48,000
Income	3,840	1,200	1,670
Household size	−8,000	2,500	3,200

(a) Calculate the t-statistics using OLS SEs and robust SEs for the income coefficient. (b) Is the income effect still significant at 5% under robust SEs? (Critical value $\approx 1.96$.) (c) What does the change in SEs suggest about the data?

Type: Short Answer + Calculation

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(a) t-statistics for income:

OLS: $t = 3840/1200 = \mathbf{3.20}$
Robust: $t = 3840/1670 = \mathbf{2.30}$

(b) Since robust $|t| = 2.30 > 1.96$, the income effect is still significant at 5% under robust SEs — though less convincingly so (it would not be significant at 1% with the robust SE).

(c) The robust SEs are substantially larger than OLS SEs (income SE: 1670 vs 1200, a 39% increase; household size: 3200 vs 2500, a 28% increase). This is strong evidence of heteroskedasticity: the OLS SEs understate the true sampling variability, making OLS t-tests too liberal (over-rejecting). In house price data, this is common — variance in prices tends to be higher for higher-income households (greater dispersion in luxury spending). Reports should use robust SEs; WLS could improve efficiency if the form of heteroskedasticity can be modelled.