= One-nearest-neighbour asymptotic risk
{title2=$R_\infty=\mathbb E[2\eta(X)(1-\eta(X))]$}
For <binary classification>, write $\eta(x)=\mathbb P(Y=1\mid X=x)$ and assume <feature-measurable nearest-neighbour tie-breaking>. If $\mathbb E|\eta(X_{(1)}(X))-\eta(X)|\to0$, then the expected <conditional misclassification risk> of the <one-nearest-neighbour classifier> tends to $\mathbb E[2\eta(X)(1-\eta(X))]$. Couple each label and an oracle label using one uniform variable. Their disagreement has the preceding expected absolute difference. The oracle and the test label are conditionally <independent> <Bernoulli random variables> of parameter $\eta(X)$, so their disagreement <probability> is exactly $2\eta(X)(1-\eta(X))$. The difference of <classification in statistical learning> error probabilities is at most the coupling disagreement.
Back to article page
