For example, I have a model: $$Y= \beta_0 + \beta_1*X_1 + \beta_2*X_2 + \beta_3*X_3 + \beta_4*X_4 + \epsilon_1$$

I suppose that $X_1$ may be the factor causing heteroskedasticity, and I regress Y on $X_1$ $$Y= \alpha_0 + \alpha_1*X_1 + \epsilon_2 $$

Now, which residual, $\epsilon_1$ or $\epsilon_2$ could be used to regress on $X_1$ to test for heteroskedasticity ?


Park's original one-page paper (here) was more concerned with dealing with heteroskedasticity, rather than test for its existence. So given heteroskedasticity, Park assumes a specific form of it, namely a log-linear relationship between the variance of the error term and one regressor

$$\sigma^2_{\epsilon _i} = \sigma^2X_i^{\gamma}e^{u_i}$$ $$\Rightarrow \ln \sigma^2_{\epsilon _i} = \ln\sigma^2 + \gamma \ln X_i + u_i$$

To estimate this relationship, one needs to obtain a data series for $\ln \sigma^2_{\epsilon _i}$. Park suggested using the residuals from the original regression as a substitute, i.e.

$$\ln \sigma^2_{\epsilon _i} \approx \ln (\hat \epsilon^2_{1i})$$

assume $u_i$ is "nicely behaved" and estimate the regression

$$\ln (\hat \epsilon^2_{1i})= a + \gamma \ln X_i + u_i$$

Then, in order to deal with heteroskedasticity, one would transform the original equation by dividing by $X^{\hat \gamma/2}$

"Park's test" is to view instead the auxiliary regression as a test for heteroskedasticity, where if $\hat \gamma$ appears statistically significant, the null hypothesis of no-heteroskedasticity is rejected. In any case, I don't see where the second regression you mention in the question comes into play.

  • $\begingroup$ I want to do park test in R . So i wrote down the command lnu2 <- log(rs^2) lnX <- log(total) lm.p <- lm(lnu2~lnX) . But I am not understanding how can i add intercept coefficient $ln \sigma^2$ in the command ? $\endgroup$ – ABC May 23 '15 at 15:40
  • $\begingroup$ @ABC You are asking how you can add a constant term in a regression specification in R-environment? There have to be a million places online to obtain this basic information - which by the way I am not in a position to provide. $\endgroup$ – Alecos Papadopoulos May 23 '15 at 18:09

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