Using $R^2$ for RF Can the $R^2$ measure be used to measure the performance of Random Forest model? My explanatory and dependent variables are linearly dependent.
 A: There is no advantage over using MSE or RMSE. Any model that has better $R^2$ than another model also will have better MSE and RMSE (assuming the same data). In that sense, all three are equivalent loss functions and measures of performance.
A common reason for wanting to use $R^2$ over MSE or RMSE is the desire to say that $R^2=0.94$ means that $94\%$ of the variation in the data is explained by the model, and $94\%$ is an $\text{A}$ grade in school. This interpretation of $R^2$ fails for nonlinear models like random forest, as the residuals and predictions are not orthogonal. (That there is a linear relationship between your variables is not relevant to this point.)
So you can use $R^2$ when MSE or RMSE would be viable loss functions, but I don’t see a reason to do so.
A: If you take the definition of $R^2$ coming as a log-likelihood difference ratio (see this answer) then it is a perfectly valid choice.
Under this framework, you can interpret $R^2$ as the fraction of explained deviance.
