# 0.19 Applications of vc bound

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## Linear classifiers

Suppose $\mathcal{F}$ = {linear classifiers in ${\mathbf{R}}^{d}$ }, then we have

${V}_{\mathcal{F}}=d+1,\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{\stackrel{^}{f}}_{n}=arg\underset{f\in \mathcal{F}}{min}{\stackrel{^}{R}}_{n}\left(f\right)$
$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-\underset{f\in \mathcal{F}}{inf}R\left(f\right)\le 4\sqrt{\frac{\left(d+1\right)log\left(n+1\right)+log2}{n}}.$

## Generalized linear classifiers

Normally, we have a feature vector $X\in {\mathbf{R}}^{d}$ . A hyperplane in ${\mathbf{R}}^{d}$ provides a linear classifier in ${\mathbf{R}}^{d}$ . Nonlinear classifiers can be obtained by a straightforward generalization.

Let ${\phi }_{1},\cdots ,{\phi }_{{d}^{{}^{\text{'}}}}$ , ${d}^{{}^{\text{'}}}\ge d$ be a collection of functions mapping ${\mathbf{R}}^{d}\to \mathbf{R}$ . These functions, applied to a feature $X\in {\mathbf{R}}^{d}$ , produce a generalized set of features, $\phi ={\left({\phi }_{1}\left(X\right),{\phi }_{2}\left(X\right),\cdots ,{\phi }_{{d}^{\text{'}}}\left(X\right)\right)}^{\text{'}}$ . For example, if $X={\left({x}_{1},{x}_{2}\right)}^{\text{'}}$ , then we could consider ${d}^{\text{'}}=S$ and $\phi ={\left({x}_{1},{x}_{2},{x}_{1}{x}_{2},{x}_{1}^{2},{x}_{2}^{2}\right)}^{\text{'}}\in {\mathbf{R}}^{5}$ . We can then construct a linear classifier in the higher dimensional generalized feature space ${\mathbf{R}}^{{d}^{\text{'}}}$ .

The VC bounds immediately extend to this case, and we have for $\mathcal{F}$ ' = { generalized linear classifiers based on maps $\phi :{\mathbf{R}}^{d}\to {\mathbf{R}}^{{d}^{\text{'}}}$ },

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-\underset{f\in {\mathcal{F}}^{\text{'}}}{inf}R\left(f\right)\le 4\sqrt{\frac{\left({d}^{\text{'}}+1\right)log\left(n+1\right)+log2}{n}}.$

Theorem

## Steele '75, dudley '78

Let $\mathcal{G}$ be a finite-dimensional vector space of real-valued functions on ${\mathbf{R}}^{d}$ . The class of sets $\mathcal{A}=\left\{\left\{x:g\left(x\right)\ge 0\right\}:g\in \mathcal{G}\right\}$ has VC dimension $\ge$ dim( $\mathcal{G}$ ).

It is sufficient to show that no set of $n=dim\left(\mathcal{G}\right)+1$ points can be shattered by $\mathcal{A}$ . Take any $n$ points and for each $g\in \mathcal{G}$ , define the vector ${V}_{g}=\left(g\left({x}_{1}\right),\cdots ,g\left({x}_{n}\right)\right)$ .

The set $\left\{{V}_{g}:g\in \mathcal{G}\right\}$ is a linear subspace of ${\mathbf{R}}^{n}$ of dimension $\le$ dim ( $\mathcal{G}$ ) = $n-1$ . Therefore, there exists a non-zero vector $\alpha =\left({\alpha }_{1},\cdots ,{\alpha }_{n}\right)\in {\mathbf{R}}^{n}$ such that ${\sum }_{i=1}^{n}{\alpha }_{i}g\left({x}_{i}\right)=0$ . We can assume that at least one of these ${\alpha }_{i}^{S}$ is negative (if all are positive, just negate the sum). We can then re-arrange thisexpression as ${\sum }_{i:{\alpha }_{i}\ge 0}{\alpha }_{i}g\left({x}_{i}\right)={\sum }_{i:{\alpha }_{i}<0}-{\alpha }_{i}g\left({x}_{i}\right)$ .

Now suppose that there exists a $g\in \mathcal{G}$ such that the set $\left\{x:g\left(x\right)\ge 0\right\}$ selects precisely the ${x}_{i}^{S}$ on the left-hand side above. Then all terms on the left are non-negative and allthe terms on the right are non-positive. Since $\alpha$ is non-zero, this is a contradiction. Therefore, ${x}_{1},\cdots ,{x}_{n}$ cannot be shattered by sets in $\left\{x:g\left(x\right)\ge 0\right\}$ , $g\in \mathcal{G}$ .  6.375pt0.0pt6.375pt

Consider half-spaces in ${\mathbf{R}}^{d}$ of the form $\mathcal{A}=\left\{x\in {\mathbf{R}}^{d}:{x}_{i}\ge b,i\in \left\{1,\cdots ,d\right\},b\in R\right\}$ . Each half-space can be described by

$g\left(x\right)=\left[0,,,\cdots ,,,0,,,1,,,0,,,\cdots ,,,0\right]\left[\begin{array}{c}{x}_{1}\hfill \\ ⋮\hfill \\ {x}_{d}\hfill \end{array}\right]-b$
$⇒dim\left(\mathcal{G}\right)=d+1,\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{V}_{\mathcal{A}}\le d+1.$

## Tree classifiers

Let

${\mathcal{T}}_{k}=\left\{r,e,c,u,r,s,i,v,e,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},r,e,c,t,a,n,g,u,l,a,r,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},p,a,r,t,i,t,i,o,n,s,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},o,f,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},{\mathbf{R}}^{d},\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},w,i,t,h,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},k,+,1,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},c,e,l,l,s\right\}$

Let $T\in {\mathcal{T}}_{k}$ . Each cell of $T$ results from splitting a rectangular region into two smaller rectangles parallel to one ofthe coordinate axes.

$T\in {\mathcal{T}}_{3}$ , $d=2$ .

Each additional split is analogous to a half-space set. Therefore, each additional split can potentially shatter $d+1$ points. This implies that

${V}_{{\mathcal{T}}_{k}}\le \left(d+1\right)k.$

$d=1$ .

$k=1$ split shatters two points.

$k=2$ splits shatters three points $<4$ .

## Vc bound for tree classifiers

${\mathcal{F}}_{k}=\left\{tree\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}classifiers\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}with\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}k+1\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}leafs\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}on\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{\mathbf{R}}^{d}\right\}$
$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-\underset{f\in {\mathcal{F}}_{k}}{inf}R\left(f\right)\le 4\sqrt{\frac{\left(d+1\right)klogn+log2}{n}}.$

How can we decide what dimension to choose for a generalized linear classifier?

How many leafs should be used for a classification tree?

Complexity Regularization using VC bounds!

## Structural risk minimization (srm)

SRM is simply complexity regularization using VC type bounds in place of Chernoff's bound or other concentration inequalities.

The basic idea is to consider a sequence of sets of classifiers ${\mathcal{F}}_{1},{\mathcal{F}}_{2},...,$ of increasing VC dimensions ${V}_{{\mathcal{F}}_{1}}\le {V}_{{\mathcal{F}}_{2}}\le ...$ . Then for each $k=1,2,...$ we find the minimum empirical risk classifier

${\stackrel{^}{f}}_{n}^{\left(k\right)}=arg\underset{f\in {\mathcal{F}}_{k}}{min}{\stackrel{^}{R}}_{n}\left(f\right)$

and then select the final classifier according to

$\stackrel{^}{k}=arg\underset{k\ge 1}{min}\left\{{\stackrel{^}{R}}_{n},\left({\stackrel{^}{t}}_{n}^{\left(k\right)}\right),+,\sqrt{\frac{32{V}_{{\mathcal{F}}_{k}}\left(logn+1\right)}{n}}\right\}$

and ${\stackrel{^}{f}}_{n}\equiv {\stackrel{^}{f}}_{n}^{\left(\stackrel{^}{k}\right)}$ is the final choice.

The basic rational is that we know

${R}_{n}\left({\stackrel{^}{f}}_{n}^{\left(k\right)}\right)-\underset{f\in {\mathcal{F}}_{k}}{inf}R\left(f\right)\le {C}^{\text{'}}\sqrt{\frac{{V}_{{\mathcal{F}}_{k}}logn}{n}}$

where ${C}^{\text{'}}$ is a constant.

The end result is that

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\le \underset{k\ge 1}{min}\left\{\underset{f\in {\mathcal{F}}_{k}}{min},R,\left(f\right),+,16,\sqrt{\frac{{V}_{{\mathcal{F}}_{k}}logn+4}{2n}}\right\}$

analogous to our pervious complexity regularization results, except thatcodelengths are replaced by VC dimensions.

In order to prove the result we use the VC probability concentration bound and assume that $△={\sum }_{k\ge 1}{V}_{{\mathcal{F}}_{k}}<\infty$ . This enables a union bounding argument and leads to a risk bound of the form given above.

## Key point of vc theory

Complexity of classes depends on richness (shattering capability) relative to a set of $n$ arbitrary points. This allows us to effectively “quantize" collections of functions in a slightlydata-dependent manner.

## Application to trees

Let

${\mathcal{F}}_{k}=\left\{k,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},l,e,a,f,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},d,e,c,i,s,i,o,n,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},t,r,e,e,s,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},i,n,\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}},{\mathbf{R}}^{d}\right\},\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{V}_{{\mathcal{F}}_{k}}\le \left(d+1\right)\left(k+1\right)$
${\stackrel{^}{f}}_{n}^{\left(k\right)}=arg\underset{f\in {\mathcal{F}}_{k}}{min}{\stackrel{^}{R}}_{n}\left(f\right)$
$\stackrel{^}{k}=arg\underset{k\ge 1}{min}\left(\underset{f\in {\mathcal{F}}_{k}}{min},R,\left(f\right),+,\sqrt{\frac{32\left(d+1\right)\left(k-1\right)\left(logn+1\right)}{n}}\right)$

Then

${\stackrel{^}{f}}_{n}={\stackrel{^}{f}}_{n}^{\left(\stackrel{^}{k}\right)}$

satisfies

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\le \underset{k\ge 1}{min}\left(\underset{f\in {\mathcal{F}}_{k}}{min},R,\left(f\right),+,16,\sqrt{\frac{\left(d+1\right)\left(k-1\right)logn+4}{2n}}\right)$

compare with

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\le \underset{k\ge 1}{min}\left(\underset{f\in dyadic\phantom{\rule{4pt}{0ex}}k\phantom{\rule{4pt}{0ex}}leaf\phantom{\rule{4pt}{0ex}}trees}{min},R,\left(f\right),+,\sqrt{\frac{\left(3k-1\right)log2+\frac{1}{2}logn}{2n}}\right)$

from Lecture 11 .

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