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"Pattern Classification" book I study that in nonparametric methods we need to estimate $p(x)$, and we don't want just the averaged version of it. They give an example of theoretical procedure to achieve estimation of $p(x)$ but don't understand it.

Suppose we use the following procedure. To estimate the density at x, we form a sequence of regions $R_1$, $R_2$,.. containing $x$ - the first region to be used with one sample, the second with two, and so on. Let $V_n$ be the volume of $R_n$, $k_n$ be the number of samples falling in $R_n$, and $p_n(x)$ be the $n$-th estimate for $p(x): p_n(x)=(k_n/n)/V_n$. [Eq.7]

If $p_n(x)$ is to converge to $p(x)$, three conditions appear to be required:

  • $\lim\limits_{n\to\infty} V_n = 0$;
  • $\lim\limits_{n\to\infty} k_n = \infty$;
  • $\lim\limits_{n\to\infty} k_n/n = 0$.

The first condition assures us that the space averaged $P/V$ will converge to $p(x)$, provided that the regions shrink uniformly and that $p(\cdot)$ is continuous at $x$. The second condition, which only makes sense if $p(x) = 0$, assures us that the frequency ratio will converge (in probability) to the probability $P$. The third condition is clearly necessary if $p_n(x)$ given by [Eq.7] is to converge at all. It also says that although a huge number of samples will eventually fall within the small region $R_n$, they will form a negligibly small fraction of the total number of samples.

My problem:

  1. I don't understand why do these conditions are necessary?
  2. If 2nd condition holds, how does it coexist with the 3rd? I mean that $k_n = \infty$, so $\infty/n = 0$ ?
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Your second point is a bit confused. $k_n$ is a sequence of $n \in \mathbb{N}$. So $\infty/n$ does not make sense. $k_n/n$ converges to zero means that the sequence $k_n$ goes to infinity less fast than $n$. –  Learner Dec 12 '12 at 8:44
I recommend reading chapter 1 (especially section 4) of the book "Elements of Large-Sample Theory" of Lehmann. It contains a very intuitive explanation of that aspect that is particularly relevant to statistics. –  Learner Dec 12 '12 at 8:48

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