Question About AdaBoost algorithm

From: DCR <d.cranshaw@xxxxxxxxx>
Date: Mon, 30 Nov 2009 17:34:39 -0800 (PST)

We know that in Adaboost algorithm, the alpha (a(t)) is typically:

1/2 * ln (1 -error(t) / error(t))

And the error bound is at most: e^(-2 * sum(t=1 to T: Y(t)^2))

What if we make a(t) = 1/3 * ln (1 -error(t) / error(t))

What would be the error bound (derivation appreciated)?

Thanks.
Dave

.

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