Questions: Hausman Test: Fixed Effects vs. Random Effects

Question 1

A researcher runs a Hausman test on a panel dataset and obtains a p-value of 0.02. What should she conclude?

• A. The random effects estimator is more efficient and should be preferred
• B. There is significant evidence that unit effects are correlated with the regressors; use fixed effects
• C. The fixed effects estimator is biased and should be replaced with random effects
• D. Neither FE nor RE is valid; the model needs to be respecified from scratch

Answer: (B)

A p-value of 0.02 means we reject H₀: Cov(αᵢ, X) = 0 at the 5% level. Rejection indicates that the unit-specific effects are correlated with the regressors, which invalidates the random effects assumption. Under this condition, RE is inconsistent (suffers omitted variable bias), while FE remains consistent by sweeping out the unit effects through the within transformation. The researcher should use fixed effects. Option A has the conclusion backwards; options C and D misread what rejection implies.

Question 2

Under the null hypothesis of the Hausman test, why does the variance of (β̂_FE − β̂_RE) simplify to Var(β̂_FE) − Var(β̂_RE)?

• A. Because under H₀, FE and RE produce numerically identical coefficient estimates
• B. Because under H₀, RE is the efficient estimator, so the covariance between β̂_RE and (β̂_FE − β̂_RE) is zero — a property of efficient estimators
• C. Because FE always has higher variance than RE by algebraic construction, regardless of the null
• D. Because the test requires the two estimators to be run on independent subsamples of the data

Answer: (B)

This is an application of the Hausman (1978) efficiency result. Under H₀, RE is the efficient estimator in the class of linear unbiased estimators. A general property of efficient estimators is that their covariance with any other consistent estimator's deviation from them is zero: Cov(β̂_RE, β̂_FE − β̂_RE) = 0. From this, Var(β̂_FE − β̂_RE) = Var(β̂_FE) − Var(β̂_RE). This simplification is convenient because it means you only need the two variance-covariance matrices to construct the test statistic, with no need to estimate the cross-covariance.

Question 3

Failing to reject the null hypothesis in a Hausman test proves that the random effects model is correctly specified.

• T. True
• F. False

Answer: False

Failure to reject H₀ means the data are *consistent with* the RE assumption (Cov(αᵢ, X) = 0) — it does not prove that assumption is true. The test may have low power in small samples, so mild endogeneity might go undetected. Additionally, the Hausman test assumes FE is correctly specified; it does not test for omitted time-varying confounders, which would bias both estimators. 'Failure to reject' should be read as 'insufficient evidence against RE,' not as 'RE is validated.'

Question 4

Under the null hypothesis of the Hausman test, both FE and RE are consistent estimators of the coefficients, but FE is less efficient than RE.

• T. True
• F. False

Answer: True

Under H₀ (Cov(αᵢ, X) = 0), the unit effects are uncorrelated with regressors, so both FE and RE consistently estimate the true parameters. However, FE discards between-unit variation by demeaning — it uses only within-unit changes over time. RE exploits both within and between variation, leading to smaller standard errors. The efficiency gain of RE under H₀ is precisely why we might prefer it when the null holds, and why the Hausman test is worth running: it lets you use the more efficient estimator when it is valid.

Question 5

Explain the core logic of the Hausman test: what properties of FE and RE does it exploit, and what does a large discrepancy between their coefficient estimates indicate?

Model answer: The test exploits the fact that FE and RE respond differently to endogeneity. FE is consistent whether or not Cov(αᵢ, X) = 0, because it eliminates unit effects through demeaning. RE is only consistent when Cov(αᵢ, X) = 0. Under H₀, both estimators are consistent and should produce similar estimates (apart from sampling noise). Under H₁ (endogeneity), RE is biased while FE is not — so their estimates diverge systematically. A large, statistically significant gap indicates that RE is picking up the correlation between unit effects and regressors, producing biased estimates. The test statistic quantifies this divergence and compares it to a χ² distribution.

The key insight is that the test does not pick the 'better' model in any general sense — it specifically tests whether RE's efficiency gains come at the cost of consistency. FE is always the safe choice for consistency; RE is the efficient choice when safe. The Hausman test tells you whether the safe and the efficient choice agree closely enough to trust the efficient one.