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K-Nearest Neighbors Density Estimation

A slecture by CIT student Raj Praveen Selvaraj

Partly based on the ECE662 Spring 2014 lecture material of Prof. Mireille Boutin.



Introduction

This slecture discusses about the K-Nearest Neighbors(k-NN) approach to estimate the density of a given distribution. The approach of K-Nearest Neighbors is very popular in signal and image processing for clustering and classification of patterns. It is an non-parametric density estimation technique which lets the region volume be a function of the training data. We will discuss the basic principle behind the k-NN approach to estimate density at a point X and then move on to building a classifier using the k-NN Density estimate.

Basic Principle

The general formulation for density estimation states that, for N Observations x1,x2,x3,...,xn the density at a point x can be approximated by the following function,

Knn1.jpg

where V is the volume of some neighborhood(say A) around x and k denotes the number of observations that are contained within the neighborhood. The basic idea of k-NN is to extend the neighborhood, until the k nearest values are included. If we consider the neighborhood around x as a sphere, the volume of the sphere is given by, $ v_{k}(x_{0}) = \underset{h}{min}\left \{ \frac{\pi ^{n/2}.h^{n}{\Gamma (\frac{n}{2} + 1)} \right \} $

where $ \Gamma (n) = (n - 1)! $

If xl is the kth closest sample point to x, then $ h_{k} = \parallel x_{l} - x_{0}\parallel $

$ v_{k}(x_{0}) = \frac{\pi ^{n/2}}{\Gamma (\frac{n}{2} + 1)} . h_{k}^{n} $

We approximate the density p(x) by,
$ \bar{\rho_{k}}(x_{0}) = \frac{k - \#s }{N.V_{k}(x_{0})} $
where #s is the number of samples on the boundary of circle wth radius hk

Most of the time this estimate is,
$ \bar{\rho_{k}}(x_{0}) = \frac{k - 1 }{N.V_{k}(x_{0})} $
The Estimated density at x0 is equal to the actual density x0
$ E(\bar{\rho_{k}}(x_{0})) = \rho(x_{0}) $

We will now prove the above claim that the estimated density is also the actual density at x0

The above claim is true only in cases where the window function $ \phi $ defines a region R with well defined boundaries.That is,

$ \phi(x) = \begin{Bmatrix}1 , x \ \epsilon \ R \\ 0, \ else \end{Bmatrix} $

The random variable here is Vk(x0)

Let, $ u = \int \rho(x)dx $ where u is a small band along the boundary of R

Observation : $ u \epsilon [0,1] $

Let G = event that "K-1 samples fall inside " $ R $
Let H = event that "1 sample falls inside the small band along the boundary"

Then,
$ Prob(G, H) = Prob(G).Prob(H\mid G) $

$ Prob(G) = \binom{N}{K-1}u^{k-1}(1-u)^{N-K+1} $

$ Prob(H\mid G) = \binom{N-K+1}{1} \left ( \frac{\Delta u}{1-u} \right )\left ( 1 - \frac{\Delta u}{1-u} \right )^{N-K} $

where $ {\Delta u} = \int \rho(x)dx $

$ Prob(G,H) = \binom{N}{K-1} \binom{N-K+1}{1} \left ( \frac{\Delta u}{1-u} \right ) u^{K-1}\left ( 1 - \frac{\Delta u}{1-u} \right )^{N-K}(1-u)^{N-K+1} $

$ Prob(G,H) = \frac{N!}{1!(N-K+1)!}.\frac{N-K+1}{1!(N-K)!} $

$ Prob(G,H) = \frac{N!}{(k-1)!(N-K)!}.\Delta u(1 - u)^{N-K}u^{K-1}\left ( 1- \frac{\Delta u}{1-u}\right )^{N-K} $

$ \left ( 1- \frac{\Delta u}{1-u}\right )^{N-K} = 1, when \ \Delta u \ is \ very \ small $

$ Prob(G,H)\cong \Delta u. \frac{N!}{(k-1)!(N-K)!}.u^{k-1}(1-u)^{N-K} $

So,
$ E(\bar{\rho }(x_{0})) = E(\frac{K-1}{N.V_{K}(x_{0})}) $

recall that
, $ u \cong \rho (x_{0}).V_{K}(x_{0}) $

$ \Rightarrow V_{K}(x_{0}) = \frac{u}{\rho (x_{0})} $

$ E(\bar{\rho }(x_{0})) = E\left ( \frac{K-1}{N.u}.\rho (x_{0}) \right ) $

$ E(\bar{\rho }(x_{0})) = \frac{K-1}{N}\rho (x_{0}) E\left ( \frac{1}{u}\right) $

$ E(\bar{\rho }(x_{0})) = \frac{K-1}{N}\rho (x_{0}) \int_{0}^{1}\frac{1}{u}\rho_{u}du $

$ E(\bar{\rho }(x_{0})) = \frac{K-1}{N}\rho (x_{0}) \int_{0}^{1}\frac{1}{u}\frac{N!}{(K-1)!(N-K)!}u^{K-1}(1-u)^{N-K}du $

$ E(\bar{\rho }(x_{0})) = \frac{K-1}{N}\rho (x_{0}) \int_{0}^{1}\frac{1}{u}\frac{N(N-1)!}{(K-1)(K-Z)!(N-K)!}u^{K-Z}(1-u)^{N-K}du $

$ E(\bar{\rho }(x_{0})) = \frac{\rho (x_{0}).(N-1)!}{(K-Z)!}.\frac{1}{(N-K)!} \int_{0}^{1}u^{K-Z}(1-u)^{N-K}du $

Now, $ \int_{0}^{1}u^{K-Z}(1-u)^{N-K}du = \frac{\Gamma (k-1)\Gamma (N-K+1)}{\Gamma (N)} $ and recall $ \Gamma (n) = (n-1)! $. Substituting these in the above equation we get,

$ E(\bar{\rho }(x_{0})) = \rho(x_{0}) $ as claimed.

How to classify data using k-NN Density Estimate

Having seen how the density at a given point x is estimated based on the value of k and the given observations x1,x2,x3,...,xn, let's discuss about using the k-NN density estimate for classification. </br>

Method 1:

Let x0 from Rn be a point to classify.

Given are samples xi1,xx2,..,xxn for i classes.

We now pick a ki for each class and a window function, and we try to approximate the density at x0 for each class and then pick the class with the largest density based on,

$ \rho(x_{0}\mid \omega _{i}) = \frac{k_{i} - 1 }{N_{i}.V_{k_{i}}(x_{0})}.Prob(\omega _{i}) $

If the priors of the classes are unknown, we use ROC curves to estimate the priors.

Method 2: </br>

Given are samples xi1,xx2,..,xxn from a Gaussian Mixture. We choose a single value of k and and one window function,

We then approximate the density at x0 by,
$ \rho(x_{0}\mid \omega _{i}) = \frac{k_{i} - 1 }{N_{i}.V_{k_{i}}(x_{0})} $</br>

where Vi is the volume of the smallest window that contains k samples and ki is the number of samples among these k that belongs to class i.

We pick a class i0 such that,
$ Prob (\omega _{io}\mid x_{0})\geq Prob (\omega _{i}\mid x_{0}), \forall \ i = 1,2,..,c $
$ \Rightarrow \rho(x_{0}\mid \omega _{i0}).Prob (\omega _{io})\geq \rho(x_{0}\mid \omega _{i}).Prob (\omega _{i}), \forall \ i = 1,2,..,c $
$ \Rightarrow \rho(x_{0}, \omega _{i0}) \geq \rho(x_{0},\omega _{i}), \forall \ i = 1,2,..,c $

$ k_{i0} \geq k_{i}, \forall \ i = 1,2,...,c $

It boils down to assigning a class based on the majority vote of the k-nearest neighbors.

Computational Complexity

k-NN Algorithm stores all the training data samples. Suppose we have n samples each of dimension k, k-NN takes,

  • O(d) time to calculate a single distance
  • O(nd) time to find one closest neighbor
  • O(knd) time to find k closest neighbors
  • Giving a total complexity of O(knd)

Advantages and DisAdvantages of k-NN

K-NN is a simple and intuitive algorithm that can be applied to any kind of distribution. It gives a very good classification rate when the number of samples is large enough. But choosing the best "k" for the classifier may be difficult. The time and space complexity of the algorithm is very high, and we need to make several optimizations for efficiently running the algorithm.

Neverthless, it's one among the most popular techniques used for classification.





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