Random Forests: Question about R^2
Andy, thank you very much! One clarification question: If MSE = sum(residuals) / n, then in the formula (1 - mse / Var(y)) - shouldn't one square mse before dividing by variance? Dimitri
On Mon, Apr 13, 2009 at 10:52 AM, Liaw, Andy <andy_liaw at merck.com> wrote:
MSE is the mean squared residuals. ?For the training data, the OOB estimate is used (i.e., residual = data - OOB prediction, MSE = sum(residuals) / n, OOB prediction is the mean of predictions from all trees for which the case is OOB). ?It is _not_ the average OOB MSE of trees in the forest. I hope there's no question about how the pseudo R^2 is computed on a test set? ?If you understand how that's done, I assume the confusion is only how the OOB MSE is formed. Best, Andy From: Dimitri Liakhovitski
Dear Random Forests gurus, I have a question about R^2 provided by randomForest (for regression). I don't succeed in finding this information. In the help file for randomForest under "Value" it says: rsq: (regression only) - "pseudo R-squared'': 1 - mse / Var(y). Could someone please explain in somewhat more detail how exactly R^2 is calculated? Is "mse" mean squared error for prediction? Is "mse" an average of mse's for all trees run on out-of-bag holdout samples? In other words - is this R^2 based on out-of-bag samples? Thank you very much for clarification! -- Dimitri Liakhovitski MarketTools, Inc. Dimitri.Liakhovitski at markettools.com
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Dimitri Liakhovitski MarketTools, Inc. Dimitri.Liakhovitski at markettools.com