FRM Exam Part I · Regression Diagnostics

Heteroskedasticity in Regression: Tests and Fixes for FRM Part I

Updated 11 October 2026 · Fact-checked

Heteroskedasticity means the variance of regression errors is not constant across observations. OLS coefficients stay unbiased, but the usual standard errors are wrong, so t-tests and confidence intervals mislead. You detect it with residual plots or the Breusch-Pagan and White tests, and fix it with robust (White) standard errors.

Understand Heteroskedasticity

A basic OLS regression assumes homoskedasticity: the error term has the same variance for every observation. Var(εᵢ | X) = σ², a constant. If the spread of the errors changes with the regressors, you have heteroskedasticity.

A common example: you regress household spending on income. Rich households have far more room to vary their spending, so the errors fan out as income rises. In finance, return errors are often larger in volatile markets than in calm ones.

The key consequence is narrow. The OLS coefficient estimates remain unbiased and consistent (assuming the other assumptions hold). But OLS is no longer BLUE, because it is not the minimum-variance estimator. Worse, the usual formula for the standard errors is wrong. The standard errors can be too small or too large, depending on the pattern. If they are too small, t-statistics are inflated and you reject true null hypotheses too often (more Type I errors). The F-test is also unreliable.

There are two types. Unconditional heteroskedasticity is unrelated to the regressors, so it causes few problems. Conditional heteroskedasticity is correlated with the regressors, and it is the one that damages inference.

You handle it in two ways. The most common is robust standard errors (White, or heteroskedasticity-consistent), which fix the standard errors without changing the coefficients. The other is generalized least squares / weighted least squares, which reweights observations by the inverse of their error variance and can restore efficiency if you know the variance pattern.

Key formulas to remember

Homoskedasticity assumption: Var(εᵢ | X) = σ² for all i; Heteroskedasticity means Var(εᵢ | X) = σᵢ², which varies by observation.
Breusch-Pagan auxiliary regression: ε̂ᵢ² = α₀ + α₁X₁ + … + αₖXₖ + uᵢ; Regress squared residuals on the regressors. Null hypothesis: all α slopes = 0 (homoskedasticity).
Breusch-Pagan test statistic: LM = n × R² ~ χ²(k); R² is from the auxiliary regression, n is the sample size, k is the number of regressors in it. Reject the null if LM exceeds the critical value.
White test: ε̂ᵢ² regressed on the Xs, their squares and cross-products; LM = n × R² ~ χ²(q); q is the number of auxiliary regressors excluding the intercept. It detects nonlinear forms of heteroskedasticity.
Weighted least squares: Divide each observation by σᵢ, then run OLS; Needs a known or estimated variance pattern. Robust standard errors do not.

How to solve Heteroskedasticity questions

Use this sequence for any heteroskedasticity question, whether it asks about consequences, tests or remedies.

1Identify what is asked: the definition, the effect on estimates, a test calculation, or a remedy.
2Recall the effect: coefficients stay unbiased and consistent, but standard errors are biased, so t and F inferences are unreliable.
3If a plot is given, look at residuals against a regressor or fitted values. A fan or cone shape signals heteroskedasticity.
4If a test is asked, state the null as constant variance. Then compute LM = n × R² from the auxiliary regression.
5Compare LM with the chi-square critical value using the right degrees of freedom (number of auxiliary regressors excluding the intercept).
6Reject the null if LM is larger than the critical value, and conclude heteroskedasticity is present.
7Choose the remedy: robust standard errors for correct inference, or weighted least squares if the variance form is known.
8Check that your conclusion about direction is correct: understated standard errors inflate t-statistics.

Quickest way: Test with n × R² and pick the remedy

When to use it: Use it when a question gives the auxiliary regression R² and sample size, or asks which fix is best.

Multiply n by the auxiliary R² to get LM.
Count auxiliary regressors (not the intercept) for the degrees of freedom.
Compare LM to the chi-square critical value. Larger means reject homoskedasticity.
If asked how to fix it without knowing the variance form, answer robust (White) standard errors.
If the question asks about bias, remember the coefficients are still unbiased.

Common mistakes in Heteroskedasticity

Saying heteroskedasticity makes OLS coefficients biased.
Students link any violated assumption to bias.
Fix: Remember that only the standard errors and efficiency suffer. Bias comes from omitted variables or correlation between errors and regressors.
Assuming standard errors are always too small.
Textbook examples usually show understated errors.
Fix: Say the standard errors are unreliable. They are often too small, but the direction depends on the pattern.
Thinking robust standard errors change the coefficient estimates.
Confusing robust errors with weighted least squares.
Fix: Robust errors change only the standard errors. Weighted least squares changes the coefficients.
Using the wrong degrees of freedom in the chi-square test.
Counting the intercept or using the original model's variables.
Fix: Use the number of slope regressors in the auxiliary regression. For White, include squares and cross-products.
Using the original regression R² in n × R².
Two regressions are involved and both report R².
Fix: Use the R² from the regression of squared residuals on the regressors.
Treating a significant test as a reason to discard the model.
Confusing a diagnostic with model failure.
Fix: A positive test calls for corrected standard errors, not abandoning the regression.

Worked examples

Example 1

You regress returns on 3 factors using 200 observations. You regress the squared residuals on the same 3 factors and obtain R² = 0.06. Using the Breusch-Pagan test at 5% (chi-square critical value with 3 degrees of freedom = 7.81), what do you conclude?

Show the solution

Null hypothesis: homoskedasticity (all slope coefficients in the auxiliary regression are zero).
LM = n × R² = 200 × 0.06 = 12.0.
Degrees of freedom = 3, critical value = 7.81.
12.0 > 7.81, so reject the null.

Answer: LM = 12.0 exceeds 7.81, so you reject homoskedasticity. Heteroskedasticity is present, and you should use robust standard errors.

Example 2

A regression shows a coefficient of 0.80 with a conventional standard error of 0.30. Residuals are found to be conditionally heteroskedastic, and the robust standard error is 0.50. Is the coefficient significant at 5% (critical |t| about 1.96) under each standard error?

Show the solution

Conventional t = 0.80 ÷ 0.30 = 2.67, which exceeds 1.96, so it looks significant.
Robust t = 0.80 ÷ 0.50 = 1.60, which is below 1.96, so it is not significant.
The coefficient 0.80 is the same in both cases, because robust errors do not change it.
The conventional standard error was understated, which inflated the t-statistic.

Answer: Conventionally t = 2.67 (significant), but with robust errors t = 1.60 (not significant at 5%). The conclusion should rely on the robust result.

Exam tips

Expect conceptual questions on consequences: unbiased coefficients, wrong standard errors, inflated t-statistics.
Know the Breusch-Pagan and White tests by their auxiliary regression and n × R² statistic.
When asked for the simplest correction, choose robust standard errors.
Distinguish conditional from unconditional heteroskedasticity. Only conditional causes real inference problems.
Read option wording carefully. Statements saying OLS is biased under heteroskedasticity are wrong.

Practice questions from Regression Diagnostics

Heteroskedasticity: frequently asked questions

What is the difference between homoskedasticity and heteroskedasticity?

Under homoskedasticity, the error variance is the same for every observation. Under heteroskedasticity, it changes, often with the level of a regressor. Only the second one distorts the usual standard errors.

How do the Breusch-Pagan and White tests differ?

Both regress squared residuals on explanatory variables and use n × R² as a chi-square statistic. Breusch-Pagan uses the regressors themselves, so it finds linear patterns. White adds squares and cross-products, so it detects more general forms.

What are robust standard errors?

They are standard errors computed in a way that stays valid when error variance is not constant. The coefficients are the same as OLS, but the t-statistics and confidence intervals become reliable.

Does heteroskedasticity bias the OLS estimates?

No. OLS coefficients remain unbiased and consistent. What fails is the efficiency of OLS and the validity of its standard errors.