10.3 Heteroskedasticity

• Using roust standard errors are always valid

• If there is significant evidence of heteroskedasticity implying A4 does not hold

  • Gauss-Markov Theorem no longer holds, OLS is not BLUE.
  • Should consider using a better linear unbiased estimator (Weighted Least Squares or Generalized Least Squares)

10.3.1 Breusch-Pagan test

A4 implies

$$E(\epsilon_i^2|\mathbf{x_i})=\sigma^2$$

$$\epsilon_i^2 = \gamma_0 + x_{i1}\gamma_1 + ... + x_{ik -1}\gamma_{k-1} + error$$

and determining whether or not $\mathbf{x}_i$ has any predictive value

• if $\mathbf{x}_i$ has predictive value, then the variance changes over the levels of $\mathbf{x}_i$ which is evidence of heteroskedasticity
• if $\mathbf{x}_i$ does not have predictive value, the variance is constant for all levels of $\mathbf{x}_i$

The Breusch-Pagan test for heteroskedasticity would compute the F-test of total significance for the following model

$$e_i^2 = \gamma_0 + x_{i1}\gamma_1 + ... + x_{ik -1}\gamma_{k-1} + error$$

A low p-value means we reject the null of homoskedasticity

However, Breusch-Pagan test cannot detect heteroskedasticity in non-linear form

10.3.2 White test

test heteroskedasticity would allow for a non-linear relationship by computing the F-test of total significance for the following model (assume there are three independent random variables)

$$\begin{aligned} e_i^2 &= \gamma_0 + x_i \gamma_1 + x_{i2}\gamma_2 + x_{i3}\gamma_3 \\ &+ x_{i1}^2\gamma_4 + x_{i2}^2\gamma_5 + x_{i3}^2\gamma_6 \\ &+ (x_{i1} \times x_{i2})\gamma_7 + (x_{i1} \times x_{i3})\gamma_8 + (x_{i2} \times x_{i3})\gamma_9 + error \end{aligned}$$

A low p-value means we reject the null of homoskedasticity

Equivalently, we can compute LM as $LM = nR^2_{e^2}$ where the $R^2_{e^2}$ come from the regression with the squared residual as the outcome

• The LM statistic has a [$\chi_k^2$][Chi-squared] distribution