# 0.5 Plug-in classifier and histogram classifier

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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.

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 $\left\{{Q}_{i}\right\},\phantom{\rule{4pt}{0ex}}i=1,...,M$ .

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.

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