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