Auxiliary Model in the White Test

Question

Why in the white test, we estimate auxiliary regression model of the squared residuals (in the original model) and not just plain residuals?

Answer 1

Note that $\text{Var}(\epsilon | x_i) = E[\epsilon^2 | x_i] - \left(E[\epsilon | x_i]\right)^2.$ Recall a typical Gauss-Markov assumption that $E[\epsilon | x_i]=0$. Hence, $\text{Var}(\epsilon | x_i) = E[\epsilon^2 | x_i].$ Since we don't observe $\epsilon^2$, we have to use some estimate. The best estimate that we have is $\hat{e}^2$, where $\hat{e}$ is our residual. The variance is the expectation of the squared error and our best estimate of this is the squared residuals; hence, they are the outcome of our auxiliary regression.

Now, we suppose that $\sigma^2_i$ varies across units $i$. One way to check this is to determine whether the squared residuals are correlated with other stuff that varies across units, namely, the variables in the regression that we care about, plus their squares, and the interactions, as suggested by White. If the squared residuals can't be predicted by stuff that we observe that varies across units, then we cannot reject the null hypothesis that the expected squared residuals and don't vary across units, which suggests that the variance doesn't vary across units and the errors are homoskedastic.

Answer 2

Because the White's test examines for heteroscedasticity by assuming that:

$\sigma^2 = \beta_0 + x_1 \beta_1 + x_2 \beta_2 + ...$

(include squares and cross-product of covariates on the right hand side)

In other words, White's test investigates if the error variance is heteroscedastic by regressing an estimate of $\sigma^2$ against the available regressors. An estimate of $\sigma^2$ is given by the squared residuals as the mean of the residuals is zero.