Exercise 6.11

Let’s assume $|x-x_1|\le |x-x_2|\le \cdots \le |x-x_N|$, i.e. point $x_1$ is the nearest neighbor of $x$, and $x_N$ is the farest neighbor. Then we have $\varphi (\frac{|x-x_1|}{r})\ge \cdots \ge \varphi (\frac{|x-x_n|}{r})$, i.e. ${\alpha }_1(x)\ge \cdots \ge {\alpha }_N(x)$.

Then $\frac{{\alpha }_n(x)}{{\alpha }_1(x)}=e^{-\frac{1}{2}\frac{|x-x_n{\text{|}}^2-|x-x_1{\text{|}}^2}{r^2}}\to 0$ when $r\to 0$ if $n\ne 1$, and $\frac{{\alpha }_n(x)}{{\alpha }_1(x)}=1$ if $n=1$.

We thus have

This is the same as the nearest neighbor rule.