We have seen Nearest Neighbor (NN) error rate as the number of samples approaches infinity is $ P=\int(1-\sum_{i=1}^c P^2(w_i|\vec{x}))p(\vec{x})d\vec{x} $
We would like to be able to answer two questions:
1) How good is that in terms of error rate?
2) How does it compare to Bayes, the best error rate we can achieve?
Recall error rate is $ P(e)=\int P(e|\vec{x})p(\vec{x})d\vec{x} $. For all x, Bayes rule yields minimum possible $ P(e|\vec{x})=:P^*(e|\vec{x}) $
Thus, we get the minimum $ P(e)=:P^*=\int P^*(e|\vec{x})p(\vec{x})d\vec{x} $
Claim 1: If $ P^* $ is low, then $ P\approx 2P^* $ (Assumes $ \infty $ number of samples.)
Justification: $ P^*(e|\vec{x})=1-P(w_{max}|\vec{x}) $, where $ w_{max} $ is such that $ P(w_{max}|\vec{x})\geq P(w_j|\vec{x}),\forall j $
So, $ P^* $ low => $ p^*(e|\vec{x}) $ is low for almost every x.
=> $ P(w_{max}|\vec{x}) $ is close to 1 for almost every x.
We have $ P=\int(1-\sum_{i=1}^cP^2(w_i|\vec{x}))p(\vec{x})d\vec{x} $ and for almost every x, $ 1-\sum_{i=1}^cP^2(w_i|\vec{x})\approx 1-P^2(w_{max}|\vec{x})\approx 2(1-P(w_{max}|\vec{x})) $, by Taylor expansion
$ =2(P^*(e|\vec{x}) $
=> $ P\approx\int 2P^*(e|\vec{x})p(\vec{x})d\vec{x}=2P^* $
Claim 2: (To be added later.)
