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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?

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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 $H_a: \alpha \neq 0$.

The White test on the other hand is more generic. It relies on the intuition that if there is no heteroskedasticity the classical error variance esitmator should gives you standard error estimates close enough to those estimated by the robust estimator. Therefore, it is able to detect more general form of heteroskedasticity than the Breusch-Pagan test.

A shortcoming of the White test is that it can lose its power very quickly particularly if the model has many regressors. This could be the reason for the results such as yours.

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Breusch-Pagan / Cook-Weisberg tests the null hypothesis that the error variances are all equal versus the alternative that the error variances are a multiplicative function of one or more variables. For example, in the default form of the hettest command shown above, the alternative hypothesis states that the error variances increase (or decrease) as the predicted values of Y increase, e.g. the bigger the predicted value of Y, the bigger the error variance is. A large chi-square would indicate that heteroscedasticity was present. In your example, the chi-square value was large and the p-value is small, indicating heteroskedasticity was present.

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    $\begingroup$ Thank you for answering--that information is quite clear. But what is your explanation for why the BP/CW tests appear (strongly) to contradict the White test, which would also satisfy your description? $\endgroup$ – whuber Apr 2 '13 at 13:00

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