# 0.5 Plug-in classifier and histogram classifier

 Page 1 / 2

We return to the topic of classification, and we assume an input (feature) space $\mathcal{X}$ and a binary output (label) space $\mathcal{Y}=\left\{0,1\right\}$ . Recall that the Bayes classifier (which minimizes the probability of misclassification) is defined by

${f}^{*}\left(x\right)=\left\{\begin{array}{cc}1,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & P\left(Y=1|X=x\right)\ge 1/2\hfill \\ 0,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & otherwise\hfill \end{array}\right).$

Throughout this section, we will denote the conditional probability function by

$\begin{array}{ccc}\hfill \eta \left(x\right)& \equiv & P\left(Y=1|X=x\right)\hfill \end{array}.$

## Plug-in classifiers

One way to construct a classifier using the training data ${\left\{{X}_{i},{Y}_{i}\right\}}_{\phantom{\rule{4pt}{0ex}}i=1}^{n}$ is to estimate $\eta \left(x\right)$ and then plug-it into the form of the Bayes classifier. That is obtain an estimate,

${\stackrel{^}{\eta }}_{n}\left(x\right)=\eta \left(x;{\left\{{X}_{i},{Y}_{i}\right\}}_{\phantom{\rule{4pt}{0ex}}i=1}^{n}\right)$

and then form the “plug-in" classification rule

$\stackrel{^}{f}\left(x\right)=\left\{\begin{array}{cc}1,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & \stackrel{^}{\eta }\left(x\right)\ge 1/2\hfill \\ 0,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & otherwise\hfill \end{array}\right).$
The function $\eta \left(x\right)$ is generally more complicated than the ultimate classification rule (binary-valued), as we cansee
$\begin{array}{ccc}\hfill \eta & :& \mathcal{X}\to \left[0,1\right]\hfill \\ \hfill f& :& \mathcal{X}\to \left\{0,1\right\}\hfill \end{array}.$

Therefore, in this sense plug-in methods are solving a more complicated problem than necessary. However, plug-in methods can perform well,as demonstrated by the next result.

Theorem

## Plug-in classifier

Let $\stackrel{˜}{\eta }$ be an approximation to $\eta$ , and consider the plug-in rule

$f\left(x\right)=\left\{\begin{array}{cc}1,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & \stackrel{˜}{\eta }\left(x\right)\ge 1/2\hfill \\ 0,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & otherwise\hfill \end{array}\right).$

Then,

$R\left(f\right)-{R}^{*}\le 2E\left[|\eta \left(x\right)-\stackrel{˜}{\eta }\left(x\right)|\right]$

where

$\begin{array}{ccc}\hfill R\left(f\right)& =& P\left(f\left(X\right)\ne Y\right)\hfill \\ \hfill {R}^{*}& =& R\left({f}^{*}\right)=\underset{f}{inf}R\left(f\right)\hfill \end{array}.$

Consider any $x\in {\mathbf{R}}^{d}$ . In proving the optimality of the Bayes classifier ${f}^{*}$ in Lecture 2 , we showed that

$\begin{array}{c}\hfill P\left(f,\left(,x,\right),\ne ,Y,|,X,=,x\right)-P\left({f}^{*},\left(x\right)\ne Y|X=x\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\left(2,\eta ,\left(,x,\right),-,1\right)\left[{\mathbf{1}}_{\left\{{f}^{*}\left(x\right)=1\right\}},-,{\mathbf{1}}_{\left\{f\left(x\right)=1\right\}}\right],\end{array}$

which is equivalent to

$\begin{array}{c}\hfill P\left(f,\left(,x,\right),\ne ,Y,|,X,=,x\right)-P\left({f}^{*},\left(x\right)\ne Y|X=x\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\left|2,\eta ,\left(,x,\right),-,1\right|\phantom{\rule{0.166667em}{0ex}}{\mathbf{1}}_{\left\{{f}^{*}\left(x\right)\ne f\left(x\right)\right\}},\end{array}$

since ${f}^{*}\left(x\right)=1$ whenever $2\eta \left(x\right)-1>0$ . Thus,

$\begin{array}{ccc}\hfill P\left(f\left(X\right)\ne Y\right)-{R}^{*}& =& {\int }_{{\mathbf{R}}^{d}}2|\eta \left(x\right)-1/2|\phantom{\rule{0.166667em}{0ex}}{\mathbf{1}}_{\left\{{f}^{*}\left(x\right)\ne f\left(x\right)\right\}}{p}_{X}\left(x\right)dx\hfill \\ & & \text{where}\phantom{\rule{4.pt}{0ex}}{p}_{X}\left(x\right)\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}\text{marginal}\phantom{\rule{4.pt}{0ex}}\text{density}\phantom{\rule{4.pt}{0ex}}\text{of}\phantom{\rule{4.pt}{0ex}}X\hfill \\ & \le & {\int }_{{\mathbf{R}}^{d}}2|\eta \left(x\right)-\stackrel{˜}{\eta }\left(x\right)|{\mathbf{1}}_{\left\{{f}^{*}\left(x\right)\ne f\left(x\right)\right\}}{p}_{X}\left(x\right)dx\hfill \\ & \le & {\int }_{{\mathbf{R}}^{d}}2|\eta \left(x\right)-\stackrel{˜}{\eta }\left(x\right)|{p}_{X}\left(x\right)dx\hfill \\ & =& 2E\left[|\eta \left(X\right)-\stackrel{˜}{\eta }\left(X\right)|\right]\hfill \end{array}$

where the first inequality follows from the fact

$\begin{array}{ccc}\hfill f\left(x\right)\ne {f}^{*}\left(x\right)& ⇒& |\eta \left(x\right)-\stackrel{˜}{\eta }\left(x\right)|\ge |\eta \left(x\right)-1/2|\hfill \end{array}$

and the second inequality is simply a result of the fact that ${\mathbf{1}}_{\left\{{f}^{*}\left(x\right)\ne f\left(x\right)\right\}}$ is either 0 or 1. Pictorial illustration of | η ( x ) - η ˜ ( x ) | ≥ | η ( x ) - 1 / 2 | when f ( x ) ≠ f * ( x ) . Note that the inequality P ( f ( X ) ≠ Y ) - R * ≤ ∫ R d 2 | η ( x ) - η ˜ ( x ) | 1 { f * ( x ) ≠ f ( x ) } p X ( x ) d x shows that the excess risk is at most twice the integral over the setwhere f * ( x ) ≠ f ( x ) . The difference | η ( x ) - η ˜ ( x ) | may be arbitrarily large away from this set without effecting the error rate of the classifier. Thisillustrates the fact that estimating η well everywhere (i.e., regression) is unnecessary for the design of a good classifier (weonly need to determine where η crosses the 1 / 2 -level). In other words, “classification is easier than regression.”

The theorem shows us that a good estimate of $\eta$ can produce a good plug-in classification rule. By “good" estimate, we mean an estimator $\stackrel{˜}{\eta }$ that is close to $\eta$ in expected ${L}_{1}\text{-norm}$ .

## The histogram classifier

Let's assume that the (input) features are randomly distributed over theunit hypercube $\mathcal{X}={\left[0,1\right]}^{d}$ (note that by scaling and shifting any set of bounded features we can satisfy this assumption),and assume that the (output) labels are binary, i.e., $\mathcal{Y}=\left\{0,1\right\}$ . A histogram classifier is based on a partition the hypercube ${\left[0,1\right]}^{d}$ into $M$ smaller cubes of equal size.

## Partition of hypercube in 2 dimensions

Consider the unit square ${\left[0,1\right]}^{2}$ and partition it into $M$ subsquares of equal area (assuming $M$ is a squared integer). Let the subsquares be denoted by $\left\{{Q}_{i}\right\},\phantom{\rule{4pt}{0ex}}i=1,...,M$ . Example of hypercube [ 0 , 1 ] 2 in M equally sized partition

Define the following piecewise-constant estimator of $\eta \left(x\right)$ :

${\stackrel{^}{\eta }}_{n}\left(x\right)=\sum _{j=1}^{M}{\stackrel{^}{P}}_{j}{\mathbf{1}}_{\left\{x\in {Q}_{j}\right\}}$

where

${\stackrel{^}{P}}_{j}=\frac{{\sum }_{i=1}^{n}{\mathbf{1}}_{\left\{{X}_{i}\in {Q}_{j},{Y}_{i}=1\right\}}}{{\sum }_{i=1}^{n}{\mathbf{1}}_{\left\{{X}_{i}\in {Q}_{j}\right\}}}.$

Like our previous denoising examples, we expect that the bias of ${\stackrel{^}{\eta }}_{n}$ will decrease as $M$ increases, but the variance will increase as $M$ increases.

where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers! By Saylor Foundation By Brooke Delaney By Subramanian Divya By By Karen Gowdey By OpenStax By Rohini Ajay By Jonathan Long By OpenStax By Anh Dao