2
$\begingroup$

I assume that this graph doesn't support the assumption of homoscedasticity. Am I right? Does it make sense to carry out another test to be sure?

y-axis: Regression Standardized Residual, x-axis: Regression Standardized Predicted Value, dependent variable: Score of a questionnaire measuring rage attacks, n=156

enter image description here

$\endgroup$
7
  • 1
    $\begingroup$ Welcome to the site. For those of us who don't speak German (I think it's German) could you translate the axis label? I think I understand most of it, but not "geschatzer Wert" and "think I understand" is not really good. $\endgroup$
    – Peter Flom
    Commented Dec 16, 2019 at 13:56
  • $\begingroup$ Sure, it's in my text: y-axis: Regression Standardized Residual, x-axis: Regression Standardized Predicted Value. $\endgroup$
    – user268744
    Commented Dec 16, 2019 at 14:05
  • $\begingroup$ You should have a look at this $\endgroup$
    – MGP
    Commented Dec 16, 2019 at 14:13
  • 2
    $\begingroup$ Does this answer your question? Chart indicates homoscedasticity but Breusch-Pagan test p<.001 $\endgroup$
    – Nick Cox
    Commented Dec 16, 2019 at 14:27
  • 3
    $\begingroup$ There is a presumably a constraint on the response, say no score is lower than zero. Hence, residuals must plot above some line that in spirit is residual = minimum observed $-$ fitted. (Standardization affects details only.) This alone inhibits or prohibits homoscedasticity. If there is an upper bound only, all the more reason to consider a generalized linear model with appropriate link rather than plain regression. For more, see the thread suggested just above as duplicate. $\endgroup$
    – Nick Cox
    Commented Dec 16, 2019 at 14:32

2 Answers 2

1
$\begingroup$

You have bigger problems than heteroskedasticity here

Based on the straight diagonal line forming the lower bound of the residuals on the lower-left side of the plot, it appears that you are using a linear regression to deal with a non-negative response variable (possibly a count variable?). If this is the case then it would be usual to either use a count regression model or at least use a logarithmic transformation on your response variable in the model. This would usually give you a more appropriate model for this kind of data and would typically deal with the heteroskedasticity at issue.

Before worrying about heteroskedasticity, I recommend that you reconsider whether you are using the correct model. If your response is a count variable then I would recommend starting with a negative-binomial count regression. If your response is not a count variable, but is some other non-negative variable, I would recommend that you consider a log-linear regression (with an adjustment for response values of zero). You can also consider variations on these models, but that is where I would start.

$\endgroup$
1
  • $\begingroup$ One thing I would like to add: the original poster DID look at the residuals plot, and decided that this is not how it should look. That is a good thing; I have seen many people not looking at it or ignoring it. $\endgroup$
    – W_vH
    Commented Oct 28, 2022 at 5:52
0
$\begingroup$

Two essential assumptions of regression are being unbiased and variance independence with observations. Mathematically, $\varepsilon \in \mathbb{R}^{n}$ is:

$$ \varepsilon \sim \mathcal{N} (0, \sigma^{2} I_{n}). $$

In your case, it seems that the estimated variance increase with observations. Two possibilities : the first one is a problem of heteroscedasticity and maybe a problem of sampling.

I hope it helps.

$\endgroup$
1
  • $\begingroup$ $\varepsilon \sim \mathcal N(0,sigma^2 I_n)$ is much stronger than being "unbiased" (this is not the usual terminology though, I'm guessing what you mean is that the errors are mean 0 conditional on regressors) and "variance independence with observations" (the usual terminology here should be "homoskedastic". Also, the answer is vague: what does it mean for those two assumptions to be "essential". For example, if the only issue is heteroskedasticity, OLS is still consistent (although conventional standard errors may be misleading). $\endgroup$ Commented Oct 28, 2022 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.