# Breusch–Pagan test for heteroscedasticity contradicts White's test?

Testing for heteroscedasticity I get these results:

Breusch–Pagan / Cook–Weisberg test for heteroskedasticity

• $$H_0$$: Constant variance

• $$H_a$$: Heteroskedasticity

  Variables: fitted values of log_expdu

chi2(1)      =    21.41
Prob > chi2  =   0.0000


White's test

• $$H_0$$: homoskedasticity

• against $$H_a$$: unrestricted heteroskedasticity

  chi2(47)     =     48.91
Prob > chi2  =    0.3964


The Breusch–Pagan test tells me that there is heteroskedasticity, while the White's test tells me the opposite. Which result should I use?

• I ran White Test and Breusch-Pagan Test on a data and both test returns contradictory outcomes which is the opposite of the earlier post. In the post above White test indicated heteroscedasticity while Breusch-pagan indicated the opposite. However in y own analysis the White’s test result was with a p-value of 0.006148 suggesting that there is no heteroscedasticity, while the Breusch-Pagan test with a p-value of 0.005277 indicates that there is heteroscedasticity in the model. Apr 23, 2021 at 15:51
• Welcome to the site. Was this intended as an answer to the OP's question, a comment requesting clarification from the OP or one of the answerers, or a new question of your own? Please only use the "Your Answer" field to provide answers to the original question. You will be able to comment anywhere when your reputation is >50. If you have a new question, click the blue ASK QUESTION at the top of the page & ask it there, then we can help you properly. Since you're new here, you may want to take our tour, which has information for new users. Apr 23, 2021 at 15:57

The Breusch-Pagan test only checks for the linear form of heteroskedasticity i.e. it models the error variance as $\sigma_i^2 = \sigma^2h(z_i'\alpha)$ where $z_i$ is a vector of your independent variables. It tests $H_0: \alpha = 0$ versus $H_a: \alpha \neq 0$.