How does White's test for heteroscedasticity differ from the Breusch-Pagan test?

## Both tests share a skeleton Square the OLS residuals, regress them on some set of variables, and ask whether that auxiliary regression explains anything. The statistic is `n` times the auxiliary R-squared, referred to a chi-square distribution whose degrees of freedom equal the number of auxiliary regressors excluding the intercept. The null in both cases is constant error variance, and a small p-value rejects it. The tests differ only in what goes on the right-hand side of the auxiliary regression - which is to say, in what shapes of heteroscedasticity they are capable of noticing. ## What each one looks for **Breusch-Pagan** puts the original predictors there. The implicit variance model is roughly linear: variance rising or falling steadily with a predictor. That covers the most common real pattern - spread growing with the scale of a driver - and it is cheap, needing only one degree of freedom per suspect. **White** puts the predictors, their squares, and all pairwise cross-products there. Because a smooth function of the predictors can be approximated by such a second-order expansion, the test does not commit to a functional form for the variance at all. It therefore catches: - **non-monotone variance** - large at both ends of a predictor's range and small in the middle, a pattern to which a linear auxiliary term is essentially blind because the positive and negative contributions cancel; - **interaction-driven variance** - noise that is high only for the combination of two conditions, invisible to either main effect alone. ## The degrees-of-freedom price With `k` original predictors, White's auxiliary regression has `k` linear terms, `k` squared terms and `k*(k-1)/2` cross-products, giving `k*(k+3)/2` regressors in total. For `k = 2` that is 5. For `k = 5` it is 20. For `k = 10` it is 65. Two problems arise as that number grows. First, spreading the test across 65 degrees of freedom dilutes it: if the real signal lives in one or two of those terms, the chi-square threshold for 65 degrees of freedom is far away and you fail to reject something that a targeted test would have caught easily. Second, you may not have the rows to fit the auxiliary regression stably at all - 65 terms on 300 observations is a fragile fit whose R-squared is inflated by chance. This makes the choice sample-dependent rather than a matter of one test being better. With few predictors and plenty of rows, White's generality is nearly free. With many predictors and a modest sample, Breusch-Pagan aimed at two or three genuine suspects is the more informative test. A middle path is to run Breusch-Pagan with hand-chosen terms - a specific predictor plus its square, for instance - which buys the shape flexibility exactly where you suspect it and spends only two degrees of freedom. ## Ambiguity of a rejection This is the subtlest point and the one worth raising unprompted. A squared residual can be inflated for two different reasons: the observation genuinely comes from a high-variance region, or the mean function is wrong there and the residual is absorbing systematic misfit. If the true relationship is curved but you fit a straight line, the residuals are systematically large at both ends and small in the middle - and that is a pattern in the *squares* that correlates with the squared predictor, which is precisely a term in White's auxiliary regression. So White's test is really a joint test against heteroscedasticity *and* certain kinds of specification error. A rejection tells you something is wrong with the second moments of your residuals; it does not tell you which of the two problems you have. The correct response is not to reach immediately for robust standard errors, because robust standard errors do nothing whatsoever about a mis-specified mean function - they would just give you correctly measured uncertainty around a wrong line. Check the functional form first, and only then treat the remaining unequal variance. Breusch-Pagan run on the linear predictors alone is somewhat less exposed to this, since it lacks the squared terms that curvature most readily loads on - though it is not immune either. ## When to use which - Concrete hypothesis about which variable drives the spread: **Breusch-Pagan** with that variable, possibly plus its square. - No hypothesis, few predictors, ample data: **White**, accepting the wide net. - Many predictors, modest data: neither test is very informative - reporting robust standard errors is the pragmatic route, since the correction costs little when the errors happen to be well behaved. In all three cases the test is diagnostic rather than decisive. Neither test changes a single number in the model, and neither answers the question you actually have, which is whether your conclusions survive a variance-agnostic standard error.