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Describe how to check whether heteroskedasticity affects a regression.

FRM® Part I · Quantitative Analysis · Regression Diagnostics · FRM1.QA.9.a

The idea

Heteroskedasticity means the variance of the regression errors is not constant but depends on the regressors or the fitted value. OLS coefficients stay unbiased, but the usual standard errors are wrong, so t-tests and confidence intervals mislead. Check first by eye. Plot the residuals against the fitted values and against each regressor, and look for a funnel or fan where the spread changes. Then test formally. The Breusch-Pagan test regresses the squared residuals on the regressors, and White's test also adds their squares and cross-products. In both the null is constant variance. The statistic n × R² from the auxiliary regression is compared with a chi-square whose degrees of freedom equal the number of auxiliary regressors, excluding the intercept. A large statistic rejects homoskedasticity. The common slip is comparing R² itself, not n × R², with the critical value.

Formulas

Lagrange multiplier statistic for heteroskedasticity

LM = n × R²

n
the number of observations
R²
the R² of the auxiliary regression of squared residuals on the regressors (and, for White, their squares and cross-products)

Compare with chi-square, degrees of freedom equal to the number of auxiliary regressors excluding the intercept. The trap is comparing R² itself with the critical value.

A practice question

From the question bank, with the reason each wrong option is wrong.

A risk analyst regresses monthly fund returns on a market factor and plots the residuals against the fitted values. The residuals form a funnel: tightly clustered around zero for small fitted values and widely scattered for large ones. What does this pattern indicate?

  1. Heteroskedasticity: the error variance changes with the level of the fitted value. - correct
  2. Serial correlation: residuals from neighbouring observations move together.Why not: Serial correlation shows up in a time-ordered plot or an ACF as runs of same-sign residuals, not as a funnel against fitted values.
  3. Multicollinearity among the regressors, which widens the spread of the residuals.Why not: Multicollinearity inflates coefficient standard errors; it does not produce a spread that grows with the fitted value.
  4. Normally distributed errors, because the residuals are centred on zero.Why not: Centring on zero is guaranteed by an OLS fit with an intercept and says nothing about whether the variance is constant.

A funnel in the residual-versus-fitted plot means the dispersion of the errors depends on the fitted value, which is the definition of heteroskedasticity. Constant spread would look like a horizontal band.

A harder one

An analyst runs a Breusch-Pagan test on a regression with 2 regressors and 150 observations. The auxiliary regression of squared residuals on the regressors has an R² of 0.06. The 5% chi-square critical value with 2 degrees of freedom is 5.99. What is the correct conclusion?

  1. LM = 0.06 is below 5.99, so fail to reject the null of homoskedasticity.Why not: Compares the auxiliary R² itself with the critical value instead of n x R².
  2. LM = 9.0 exceeds 5.99, so reject the null of homoskedasticity. - correct
  3. LM = 9.0 exceeds 5.99, so fail to reject the null of homoskedasticity.Why not: Reverses the hypotheses: constant variance is the null, so a statistic above the critical value rejects it.
  4. LM = 9.0 exceeds 5.99, so reject, which shows the OLS coefficients are biased.Why not: Draws the wrong implication: heteroskedasticity leaves the coefficients unbiased and distorts the standard errors.

The statistic is n x R² = 150 x 0.06 = 9.0. It exceeds the chi-square critical value of 5.99, so homoskedasticity is rejected; the consequence is unreliable standard errors, not biased coefficients.

The hardest tier

A regression has 3 regressors and 500 observations. White's auxiliary regression of squared residuals on the 3 regressors, their 3 squares and their 3 cross-products gives an R² of 0.03. Chi-square 5% critical values: 3 df 7.81; 6 df 12.59; 9 df 16.92; 10 df 18.31. Which conclusion is correct?

  1. LM = 15.0 with 3 degrees of freedom exceeds 7.81, so reject homoskedasticity.Why not: Counts only the original regressors; White's auxiliary regression also contains the squares and cross-products.
  2. LM = 15.0 with 9 degrees of freedom is below 16.92, so do not reject homoskedasticity. - correct
  3. LM = 15.0 with 6 degrees of freedom exceeds 12.59, so reject homoskedasticity.Why not: Counts the squares and cross-products but forgets the three original regressors in the auxiliary regression.
  4. LM = 15.0 with 10 degrees of freedom is below 18.31, so do not reject homoskedasticity.Why not: Counts the intercept of the auxiliary regression as a restriction; only the slope terms are tested, giving 9 degrees of freedom.

The auxiliary regression has 3 + 3 + 3 = 9 slope terms, so LM = n x R² = 500 x 0.03 = 15.0 is compared with the 9 df critical value 16.92. 15.0 is smaller, so homoskedasticity is not rejected.

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