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What I don't understand is if you can train a final layer on the reservoir's random representation, why is this better than just training the final layer on you
by jphoward 8y ago
What I don't understand is if you can train a final layer on the reservoir's random representation, why is this better than just training the final layer on your data directly? I assume the answer has something to do with dimensionality reduction?
- falcolas 8y agoIf I were to guess, I would say that the reservoir would amplify differences in data that may not be easily separated in the source data. Sort of acting like a hash function - small differences in the source become large differences in the output.
- breuderink 8y agoThe reservoir can compute non-linear functions of the input over time. So, with a reservoir, you can train a linear projection of the reservoir state that responds non-linearly to the input. Without the reservoir, the output projection can only be linearly related to the input.