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I have a linear model with 12 regressors and sample size 42, which I want to test for heteroscedasticity. Hence, I applied a white test in the following way

  • regress $y=X\beta + e$
  • regress residuals $\hat e^2=\beta_1+\beta_2x_2+\cdots+\beta_{13}x_2^2+...+\beta_{25}x_{1}x_2+\cdots+u $
  • reject $H_0: \sigma_i^2=\sigma^2$ if $N\cdot R^2_{\hat e^2} > \chi^{2, 1-\alpha}_{s}$

Since I have $N=42$ and $s=90$* the test statistic is maximum $42$ and can thus never exceed $\chi^{2, 1-\alpha}_{s}$, which is $112$ for $\alpha=5%$.

What does that mean? Can there be no Heteroscedasticity in this case? This feels counterintuitive since I actually get an $R^2=1$ in step 2 and thus the $x$ terms have much explanatory power on the residuals. So am I doing something wrong?

Thanks in advance for your thoughts!

$*$ ($12$ variables, $12$ squared terms and $\frac{n*(n-1)}{2}=66$ crossterms)

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In your step 2 you are getting a perfect fit because you have more predictors than observations so $R^2=1$ is inevitable. I think most people would start the process here by plotting the residuals against the fitted values and looking at the spread of the residuals in that plot. Alternatively plot $\sqrt(| e |)$ against the fitted where $e$ is the residual in your notation, the so-called scale-location plot.

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  • $\begingroup$ thanks that makes perfect sense the regression in line 2 is obviously overfitted... Do you think it would be helpful do regularize it (ridge or lasso for example)? $\endgroup$
    – Jonasson
    Dec 9 '16 at 14:09
  • $\begingroup$ I would content myself with the plots. That is a personal view, others may have different opinions. $\endgroup$
    – mdewey
    Dec 9 '16 at 14:41

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