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Can the $R^2$ measure be used to measure the performance of Random Forest model? My explanatory and dependent variables are linearly dependent.

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  • $\begingroup$ This has been written about before on this site. Check out, for instance, stats.stackexchange.com/q/13869/40036 $\endgroup$ – josliber Dec 11 '20 at 19:39
  • $\begingroup$ Why are you using a random forest of the relationship between your variables is known to be linear? $\endgroup$ – Dave Dec 11 '20 at 19:58
  • $\begingroup$ @Dave At some points it might not yield a linear model. $\endgroup$ – Agi Dec 11 '20 at 20:29
  • $\begingroup$ @josliberI did actually but it doesn't give a straight answer of if it is theoretically wrong to do it. $\endgroup$ – Agi Dec 11 '20 at 20:29
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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.

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  • $\begingroup$ Are you trying to compare the RF performance to the linear regression $R^2& value? $\endgroup$ – Dave Dec 11 '20 at 19:59
  • $\begingroup$ Yes. I want to compare the performance of different models with one another. The concern that I have is that RF might build a non-linear model for some training sets (I'm using rolling time window), that's why I wonder if it makes sense to use rf for non-linear mdoels. $\endgroup$ – Agi Dec 11 '20 at 20:25
  • $\begingroup$ I do not follow your concerns. Perhaps post them as a new question. It looks like it will be interesting to discuss. $\endgroup$ – Dave Dec 11 '20 at 20:28
  • $\begingroup$ Could you provide the formula for MSE to be used for non-linear models? Is it $MSE=\frac{1}{n}\Sigma_{i=1}^n(y_i-\hat{y_i})^2$ for all sort of models? $\endgroup$ – Agi Dec 13 '20 at 18:15
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    $\begingroup$ @Agi The formula for MSE does not depend on the model, so what you wrote applies in general. $\endgroup$ – Dave Dec 13 '20 at 18:29
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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.

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    $\begingroup$ Ben’s answer seems to be specific to GLMs, though I should keep it in mind if I want some kind of scaled performance metric for a logistic regression. $\endgroup$ – Dave Dec 11 '20 at 20:10
  • $\begingroup$ @Dave it's actually valid to any likelihood of choice, since it's simply a comparison of likelihoods. GLMs do, however, maximize that quantity directly, if the respective distribution conforms to the GLM framework $\endgroup$ – Firebug Dec 11 '20 at 20:52
  • $\begingroup$ @Firebug Do non-linear models fit into GLM category? Cause the way it computes the r-squared leads down to the same traditional r-squared definition of the linear models. $\endgroup$ – Agi Dec 11 '20 at 21:06
  • $\begingroup$ @Agi Your random forest is not a GLM, no. $\endgroup$ – Dave Dec 11 '20 at 23:11
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    $\begingroup$ @Agi you’re mixing up two questions. One is if random forest is a generalized linear model (GLM), and the answer is that it is not. The other is about how proportion of deviance explained works in general, and I will let Firebug address that. $\endgroup$ – Dave Dec 12 '20 at 3:55

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