Exploratory search for interactions in linear regression with normal distr. of residuals but non-normally distr. DV I am conducting a very exploratory analysis (I.e., few to no specific predictions about which predictors/interactions are most relevant) assessing whether ten IVs (all binary) predict one DV (continuous). Preliminary analyses showed that the DV was non-normally distributed, but when running the full regression model, the residuals ARE normally distributed. I take this to mean it’s fine to use linear regression as the strategy for the overall model, but I am also interested in finding out whether any of my predictors interact with gender to predict the DV.
In other papers, I have seen people search for interaction terms of interest (again, this is very exploratory work) by running a series of ANOVAs. So, for example, I would test whether gender interacts with predictor 1 to predict the DV, then whether gender interacts with predictor 2 to predict the DV, and so on. Significant interaction terms are then added to the full regression model to see if they significantly improve model fit.
This leads to my question: given that my DV is non-normally distributed, can I use ANOVAs (which rely on normality) to look for possible interactions? I am thinking yes because I end up using linear regression anyway (and ANOVA is after all, a form of regression with similar assumptions). But, is there any reason I should actually look for interactions to include in my regression using some non-parametric technique?
 A: When you say that the residuals are normally distributed, did you test that formally? If so, then it's probably reasonable to use a typical linear model. The key is that $r = y - X\beta$ is normal, since least squares is assuming that $y$ is a combination of a deterministic component from the regressors and an iid normal distribution error term that, crucially, is independent of $X$ and $y$. So the dependent variable not having a normal marginal distribution is fine.
Now, looking at the residuals' marginal distribution alone can obscure the possibility that they aren't iid, depending potentially on y. OLS and ANOVA both make the assumption that the error is homoskedastic- the variance of the error is constant and errors between samples are independent. It's quite common for that not to be the case. The result is that the Gauss-Markov theorem no longer applies and the OLS estimate of the coefficients is no longer the BLUE. If the errors are still normal, then OLS is unbiased still, just not efficient. A bigger consequence for your purposes is that the OLS/ANOVA inferences about the variances in relationship to independent variables will be biased, and in general it's not possible to say whether they will be high or low estimates. Essentially, if you don't have homoskedastic variance, you can't rely on anything from ANOVA without dealing with it first or using another method. Iteratively reweighted least squares with Tikhonov regularization can fit a linear model, estimate an error covariance matrix, use that to improve the model, and that to improve the error estimate repeatedly. The regularization helps to avoid the model using regression coefficients to "cover up" relationships between independent variables and the variance of the error- if, for example, the errors have variance that is a function of the mean (a very common case) OLS will penalize every residual equally, and even strong associations between independent variables and the response in smaller-mean situations can disappear as the larger variance samples dominate the objective. IRLS can decouple that, and it'll also tell you directly whether the errors are homoskedastic or not. If OLS is sufficient, then the regularization will mean that IRLS will converge quickly and estimate the error covariance as the identity matrix. If not, it just requires fitting an OLS model prepared in a particular way each iteration, so it's not much more complex. I'd start there and verify that your assumptions are good. If the response variable is significantly non normal but the residuals are, you might be fine, or it your assumptions could be violated wildly. Better to know for sure.
