[R] Random Forests: Question about R^2

[Dimitri Liakhovitski]

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