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When I use 5-k cross validation, is the mean absolute error (MAE) equal to

$$\text{mean}(\text{MAEs of each 5 steps})=\frac{1}{5}(MAE_1 + \cdots + MAE_5)$$

or equal to

$$\text{mean}(\text{absolute errors of all predictions})$$?

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If there are an even number of training examples, wont this be exactly the same? With an odd number then they differ slightly. – B_Miner Sep 19 '11 at 15:20
@B_M How does that work? Perhaps you are thinking of medians? – whuber Sep 19 '11 at 19:02

1 Answer

up vote 2 down vote accepted

CV is a simulation of running testing algorithm on unseen data, so the first one.
The second is more an OOB error approximation of a bagging ensemble of your classifiers.

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Indeed, it makes more sense. wikidoc.org/index.php/Cross-validation : "The K results from the folds then can be averaged (or otherwise combined) to produce a single estimation" – hijack Sep 20 '11 at 14:26

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