0.11 Decision trees

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Minimum complexity penalized function

Recall the basic results of the last lectures: let $\mathcal{X}$ and $\mathcal{Y}$ denote the input and output spaces respectively. Let $X\in \mathcal{X}$ and $Y\in \mathcal{X}$ be random variables with unknown joint probability distribution ${P}_{XY}$ . We would like to use $X$ to “predict” $Y$ . Consider a loss function $0\le \ell \left({y}_{1},{y}_{2}\right)\le 1,\phantom{\rule{4pt}{0ex}}\forall {y}_{1},{y}_{2}\in \mathcal{Y}$ . This function is used to measure the accuracy of our prediction. Let $\mathcal{F}$ be a collection of candidate functions (models), $f:\mathcal{X}\to \mathcal{Y}$ . The expected risk we incur is given by $R\left(f\right)\equiv {E}_{XY}\left[\ell \left(f\left(X\right),Y\right)\right]$ . We have access only to a number of i.i.d. samples, ${\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{n}$ . These allow us to compute the empirical risk ${\stackrel{^}{R}}_{n}\left(f\right)\equiv \frac{1}{n}{\sum }_{i=1}^{n}\ell \left(f\left({X}_{i}\right),{Y}_{i}\right)$ .

Assume in the following that $\mathcal{F}$ is countable. Assign a positive number $c\left(f\right)$ to each $f\in \mathcal{F}$ such that ${\sum }_{f\in \mathcal{F}}{2}^{-c\left(f\right)}\le 1$ . If we use a prefix code to describe each element of $\mathcal{F}$ and define $c\left(f\right)$ to be the codeword length (in bits) for each $f\in \mathcal{F}$ , the last inequality is automatically satisfied.

We define the minimum complexity penalized estimator as

${\stackrel{^}{f}}_{n}\equiv arg\underset{f\in \mathcal{F}}{min}\left\{{\stackrel{^}{R}}_{n},\left(f\right),+,\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}}\right\}.$

As we showed previously we have the bound

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\le \underset{f\in \mathcal{F}}{min}\left\{R,\left(f\right),+,\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}},+,\frac{1}{\sqrt{n}}\right\}.$

The performance (risk) of ${\stackrel{^}{f}}_{n}$ is on average better than

$R\left({f}_{n}^{*}\right)+\sqrt{\frac{c\left({f}_{n}^{*}\right)log2+\frac{1}{2}logn}{2n}}+\frac{1}{\sqrt{n}},$

where

${f}_{n}^{*}=arg\underset{f\in \mathcal{F}}{min}\left\{R,\left(f\right),+,\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}}\right\}.$

If it happens that the optimal function, that is

${f}^{*}=arg\underset{f\phantom{\rule{4.pt}{0ex}}\text{measurable}}{min}R\left(f\right),$

is close to an $f\in \mathcal{F}$ with a small $c\left(f\right)$ , then ${\stackrel{^}{f}}_{n}$ will perform almost as well as the optimal function.

Suppose ${f}^{*}\in \mathcal{F}$ , then

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\le R\left({f}^{*}\right)+\sqrt{\frac{c\left({f}^{*}\right)log2+\frac{1}{2}logn}{2n}}+\frac{1}{\sqrt{n}}.$

Furthermore if $c\left({f}^{*}\right)=O\left(logn\right)$ then

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\le R\left({f}^{*}\right)+O\left(\sqrt{\frac{logn}{n}}\right),$

that is, only within a small $O\left(\sqrt{\frac{logn}{n}}\right)$ offset of the optimal risk.

In general, we can also bound the excess risk $E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-{R}^{*}$ , where ${R}^{*}$ is the Bayes risk,

${R}^{*}=\underset{f\phantom{\rule{4.pt}{0ex}}\text{measurable}}{inf}R\left(f\right).$

By subtracting ${R}^{*}$ (a constant) from both sides of the inequality

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\le \underset{f\in \mathcal{F}}{min}\left\{R,\left(f\right),+,\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}},+,\frac{1}{\sqrt{n}}\right\}$

we obtain

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-{R}^{*}\le \underset{f\in \mathcal{F}}{min}\left\{R,\left(f\right),-,{R}^{*},+,\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}},+,\frac{1}{\sqrt{n}}\right\}.$

Note that two terms in this upper bound: $R\left(f\right)-{R}^{*}$ is a bound on the approximation error of a model $f$ , and remainder is a bound on the estimation error associated with $f$ . Thus, we see that complexity regularization automatically optimizes a balance between approximation and estimationerrors. In other words, complexity regularization is adaptive to the unknown tradeoff between approximation and estimation.

Classification

Consider the particularization of the above to a classification scenario. Let $\mathcal{X}={\left[0,1\right]}^{d}$ , $\mathcal{Y}=\left\{0,1\right\}$ and $\ell \left(\stackrel{^}{y},y\right)\equiv {\mathbf{1}}_{\left\{\stackrel{^}{y}\ne y\right\}}$ . Then $R\left(f\right)={E}_{XY}\left[{\mathbf{1}}_{\left\{f\left(X\right)\ne Y\right\}}\right]=P\left(f\left(X\right)\ne Y\right)$ . The Bayes risk is given by

${R}^{*}=\underset{f\phantom{\rule{4.pt}{0ex}}\text{measurable}}{inf}R\left(f\right).$

As it was observed before, the Bayes classifier ( i.e., a classifier that achieves the Bayes risk) is given by

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

This classifier can be expressed in a different way. Consider the set ${G}^{*}=\left\{x:\phantom{\rule{4pt}{0ex}}P\left(Y=1|X=x\right)\ge 1/2\right\}$ . The Bayes classifier can written as ${f}^{*}\left(x\right)={\mathbf{1}}_{\left\{x\in {G}^{*}\right\}}$ . Therefore the classifier is characterized entirely by the set ${G}^{*}$ , if $X\in {G}^{*}$ then the “best” guess is that $Y$ is one, and vice-versa. The boundary of this set corresponds to the points where the decision is harder.The boundary of ${G}^{*}$ is called the Bayes Decision Boundary . In [link] (a) this concept is illustrated. If $\eta \left(x\right)=P\left(Y=1|X=x\right)$ is a continuous function then the Bayes decision boundary is simply given by $\left\{x:\phantom{\rule{4pt}{0ex}}P\left(Y=1|X=x\right)=1/2\right\}$ . Clearly the structure of the decision boundary provides importantinformation on the difficulty of the problem.

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