K-Nearest Neighbors Density Estimation

A slecture by CIT student Raj Praveen Selvaraj

Completely 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 any point x0 and then move on to building a classifier using the k-NN Density estimate.

For a background knowledge on Local Density estimation methods, check the slecture on Introduction to local density estimation methods

## K-Nearest Neighbor Density Estimation

The general formulation for density estimation states that, for N Observations x1,x2,x3,...,xn the density at a point x0 can be approximated by,

$\rho(x_{0}) = \frac{K}{NV}$

where V is the volume of some neighborhood around x0 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 where k ≥ 2. If we consider the neighborhood around x0 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 $\Gamma (n) = (n - 1)!$ There are K samples within the sphere(including the boundaries) of radius h.

If xl is the kth closest sample point to x0 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 ρ(x0) 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 with 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 thorough K-NN gives the actual density at 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 be a small band along the boundary of R given by,

$u = \int \rho(x)dx$

Observation : $u \ \epsilon \ [0,1]$

Let G be the event that "K-1 samples fall inside " $R$ and let H be the 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 ${\Delta u}$ is given by,
${\Delta u} = \int \rho(x)dx$

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

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

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

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

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

The Expected value of the density at x0 by k-NN is given by,
$E(\bar{\rho }(x_{0})) = E(\frac{K-1}{N.V_{K}(x_{0})})$

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 any given point x0 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 be a sample from Rn to be classified.

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

We now pick a "good" ki for each class and a window function, and we try to approximate the density at x0 for each class as discussed above.We then pick the class with the largest density at x0 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.
For more information on ROC curves check the slecture on ROC Curves

Method 2:

Given are samples xi1,xx2,..,xxn from a Gaussian Mixture. We choose a good value for k and and a window function,

We then approximate the density at x0 by,

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

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,

$Prob (\omega _{io}\mid x_{0})\geq Prob (\omega _{i}\mid x_{0}), \forall \ i = 1,2,..,c$

On applying Bayes Rule, the above equation becomes,

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

If we assume that the priors of the classes are equal then,

$\Rightarrow \rho(x_{0}, \omega _{i0}) \geq \rho(x_{0},\omega _{i}), \forall \ i = 1,2,..,c$

Since the density is based on the number of neighbors of each class we can rewrite the above equation as,

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

So classification using this method can also be thought of as assigning a class based on the majority vote of the k-Nearest Neighbors at any point x0

## Computational Complexity of k-NN

Suppose we have n samples in d dimensions, and we need to classify x0. Let's assume that we are using the Euclidean distance to compute the nearest neighbors. Then,

• It takes O(d) to compute distance to one sample
• O(nd) to find one nearest neighbor
• O(knd) to find k closest neighbors

The total complexity of k-NN is O(knd). We can see that this complexity is expensive for a large number of samples.But there are several pre-processing and optimization techniques to improve the efficiency of K-NN.

## Summary

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.

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

## References

[1] Mireille Boutin, "ECE662: Statistical Pattern Recognition and Decision Making Processes," Purdue University, Spring 2014.
[2] http://www.csd.uwo.ca/~olga/Courses/CS434a_541a/Lecture6.pdf
[3] http://research.cs.tamu.edu/prism/lectures/pr/pr_l8.pdf