Are the model residuals from GEE interpretable as residuals from simple linear regression, so that I may plot a residual versus fitted plot to determine whether there's heteroskedasticity?

It is known that residuals of the model from a generalized linear mixed model are not useful by themselves to determine heteroskedasticity (https://mirrors.nics.utk.edu/cran/web/packages/DHARMa/vignettes/DHARMa.html).


1 Answer 1


Okay, so keep in mind that the GEE is estimating the marginal effect of the parameters across the groups, since the score equation is setting the parameter estimates to solve the following score equation, where the variables below are either matrix or scalar.

$$ \sum_{i=1}^G D_i^T V_i^{-1} (Y_i-\mu_i)=0 $$

Even if there is heteroskedasticity, the GEE isn't really trying to directly model it. It just assumes an approximate covariance matrix, and estimates using that working structure. The variance-covariance matrix is not of inferential interest in the GEE (GEE1) typically ... (there are various extensions such as GEE2).

For a GEE the point estimates should be identical to the standard estimates from a GLM, under the independence working assumption, but they only differ with respect to the standard errors. Liang and Zeger pointed out the equation above reduces to the standard Quasi-Likelihood equations for Weederburn in the case of working independence ...

So the residuals are just about as meaningful as running a standard linear regression since the point estimates themselves should be unchanged, and they will still tell you if there are any patterns in the fitted values. If there is heteroskedasticity in your original regression, it shouldn't disappear after you apply the GEE.

  • $\begingroup$ Is the goal to find $D_i^T$ or $V_i$? @user1848065 $\endgroup$
    – user318514
    Commented Aug 25, 2022 at 12:10
  • $\begingroup$ The direct goal is to find the $\beta$ that solve the equation and enter into the score via $D_i^T=\frac{\partial \mu_i}{\partial\beta}$, where $\mu$ is the working response under the specified model. $V_i$ depends on $\beta$ through $A_i$ and is also dependent on the working correlation matrix $R$, and it enters the equation as follows $V_i=A_i^{1/2}R(\alpha)A_i^{1/2}$. $A_i$ contains the working variance as a function of the $\beta$ on the diagonal. Updates for $\beta,\alpha$ are performed on each scoring iteration, so in some sense both are needed @Germania $\endgroup$ Commented Aug 25, 2022 at 13:15

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