0.18 The vapnik-chervonenkis inequality

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The vapnik-chervonenkis inequality

The VC inequality is a powerful generalization of the bounds we obtained for the hyperplane classifier in the previous lecture . The basic idea of the proof is quite similar. Before starting the inequality, we need to introduce theconcept of shatter coefficients and VC dimension.

Shatter coefficients

Let $\mathcal{A}$ be a collection of subsets of ${\mathcal{R}}^{d}$ , definition: The ${n}^{th}$ shatter coefficient of $\mathcal{A}$ is defined by

${\mathcal{S}}_{\mathcal{A}}\left(n\right)=\begin{array}{c}max\\ {x}_{1},...,{x}_{n}\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}{\mathcal{R}}^{d}\end{array}\left|\left\{,\left\{{x}_{1},...,{x}_{n}\right\},\bigcap ,A,,,A,\phantom{\rule{0.166667em}{0ex}},ϵ,\phantom{\rule{0.166667em}{0ex}},\mathcal{A},\right\}\right|.$

The shatter coefficients are a measure of the richness of the collection $\mathcal{A}$ . ${\mathcal{S}}_{\mathcal{A}}\left(n\right)$ is the largest number of different subsets of a set of $n$ points that can be generated by intersecting the set with elements of $\mathcal{A}$ .

In 1-d, Let $\mathcal{A}=\left\{\left(-,\infty ,,,t\right],t\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{R}\right\}$ Possible subsets of $\left\{{x}_{1},...,{x}_{n}\right\}$ generated by intersecting with sets of the form $\left(-,\infty ,,,t\right]$ are $\left\{{x}_{1},...,{x}_{n}\right\},\phantom{\rule{0.166667em}{0ex}}\left\{{x}_{1},\phantom{\rule{0.166667em}{0ex}}...,{x}_{n-1}\right\},\phantom{\rule{0.166667em}{0ex}}...,\phantom{\rule{0.166667em}{0ex}}\left\{{x}_{1}\right\},\phantom{\rule{0.166667em}{0ex}}\phi$ . Hence ${\mathcal{S}}_{d}\left(n\right)=n+1$ .

In 2-d, Let $\mathcal{A}$ = $\left\{$ all rectangles in ${\mathcal{R}}^{2}$ $\right\}$

Consider a set $\left\{{x}_{1},\phantom{\rule{0.166667em}{0ex}}{x}_{2},\phantom{\rule{0.166667em}{0ex}}{x}_{3},\phantom{\rule{0.166667em}{0ex}}{x}_{4}\right\}$ of training points. If we arrange the four points into the corner of a diamond shape. It's easyto see that we can find a rectangle in ${\mathcal{R}}^{2}$ to cover any subsets of the four points as the above picture, i.e. ${\mathcal{S}}_{\mathcal{A}}\left(4\right)={2}^{4}=16$ .

Clearly, ${\mathcal{S}}_{\mathcal{A}}\left(n\right)={2}^{n},n=1,\phantom{\rule{0.166667em}{0ex}}2,\phantom{\rule{0.166667em}{0ex}}3$ as well.

However, for $n=5,{\mathcal{S}}_{\mathcal{A}}\left(n\right)<{2}^{5}$ . This is because we can always select four points such that the rectangle, which just contains fourof them, contains the other point. Consequently, we cannot find a rectangle classifier which contains the four outer points and does not contain the innerpoint as shown above.

Note the ${\mathcal{S}}_{\mathcal{A}}\le {2}^{n}$ .

If $\left|\left\{,\left\{{x}_{1},...,{x}_{n}\right\},\bigcap ,A,,,A,\phantom{\rule{0.166667em}{0ex}},ϵ,\phantom{\rule{0.166667em}{0ex}},\mathcal{A},\right\}\right|={2}^{n}$ then we say that $\mathcal{A}$ shatters ${x}_{1},\phantom{\rule{0.166667em}{0ex}}...,\phantom{\rule{0.166667em}{0ex}}{x}_{n}$ .

Vc dimension

The VC dimension
${V}_{\mathcal{A}}$ of a collection of sets $\mathcal{A}$ is defined as the largest interger $n$ such that ${S}_{\mathcal{A}}\left(n\right)={2}^{n}$ .

Sauer's lemma:

Let $\mathcal{A}$ be a collection of set with VC dimension ${V}_{\mathcal{A}}<\infty$ . Then $\forall n,{\mathcal{S}}_{\mathcal{A}}\left(n\right)\le {\sum }_{i=0}^{{V}_{\mathcal{A}}}\left(\begin{array}{c}n\\ i\end{array}\right)$ , also ${\mathcal{S}}_{\mathcal{A}}\left(n\right)\le {\left(n+1\right)}^{{V}_{\mathcal{A}}},\forall n$ .

Vc dimension and classifiers

Let $\mathcal{F}$ be a collection of classifiers of the form $f:{\mathcal{R}}^{d}\to \left\{0,1\right\}$ Define $\mathcal{A}=\left\{\left\{x:f\left(x\right)=1\right\}×\left\{0\right\}\bigcup \left\{x:f\left(x\right)=0\right\}×\left\{1\right\},f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\right\}$ In words, this is collection of subsets of $\mathcal{X}×\mathcal{Y}$ for which on $fϵ\mathcal{F}$ maps the features $x$ to a label opposite of $y$ .  The size of $\mathcal{A}$ expresses the richness of $\mathcal{F}$ .  The larger $\mathcal{A}$ is the more likely it is that there exists an $fϵ\mathcal{F}$ for which $R\left(f\right)=P\left(f\left(X\right)\ne Y\right)$ is close to the Bayes risk ${R}^{*}=P\left({f}^{*}\left(X\right)\ne Y\right)$ where ${f}^{*}$ is the Bayes classifier. The ${n}^{th}$ shatter coefficient of $\mathcal{F}$ is defined as ${\mathcal{S}}_{\mathcal{F}}\left(n\right)={\mathcal{S}}_{\mathcal{A}}\left(n\right)$ and the VC dimesion of $\mathcal{F}$ is defined as ${V}_{\mathcal{F}}={V}_{\mathcal{A}}$ .

linear (hyperplane) classifiers in ${\mathcal{R}}^{d}$

Consider $d$ = 2. Let $n$ be the number of training points, it is easy to see that when $n=1$ , let $\mathcal{A}$ be as above. By using linear classifiers in ${\mathcal{R}}^{2}$ , it is easy to see that we can assign 1 to all possible subsets $\left\{\left\{{x}_{1}\right\},\phi \right\}$ and 0 to their complements. Hence ${\mathcal{S}}_{\mathcal{F}}\left(1\right)=2$ .

When $n=2$ , we can also assign 1 to all possible subsets $\left\{\left\{{x}_{1},{x}_{2}\right\},\phantom{\rule{0.166667em}{0ex}}\left\{{x}_{1}\right\},\phantom{\rule{0.166667em}{0ex}}\left\{{x}_{2}\right\},\phantom{\rule{0.166667em}{0ex}}\phi \right\}$ and 0 to their complements, and vice versa. Hence ${\mathcal{S}}_{\mathcal{F}}\left(2\right)=4={2}^{2}$ .

When $n=3$ , we can arrange arrange the point ${x}_{1},\phantom{\rule{0.166667em}{0ex}}{x}_{2},\phantom{\rule{0.166667em}{0ex}}{x}_{3}$ (non-colinear) so that the set of linear classifiers shatters the three points, hence ${\mathcal{S}}_{\mathcal{F}}\left(3\right)=8={2}^{3}$

When $n=4$ , no matter where the points ${x}_{1},\phantom{\rule{0.166667em}{0ex}}{x}_{2},\phantom{\rule{0.166667em}{0ex}}{x}_{3},\phantom{\rule{0.166667em}{0ex}}{x}_{4}$ and what designated binary values ${y}_{1},\phantom{\rule{0.166667em}{0ex}}{y}_{2},\phantom{\rule{0.166667em}{0ex}}{y}_{3},\phantom{\rule{0.166667em}{0ex}}{y}_{4}$ are. It's clear that $\mathcal{A}$ does not shatter the four points. To see the claim, first observe that the four points will form a 4-gon (if the four points are co-linear, or if the three points are co-linear then clearly linear classifiers cannot shatter the points). The two points that belong to the same diagonal lines form 2 groups and no linear classifier can assign different values to the 2 groups. Hence ${\mathcal{S}}_{\mathcal{F}}\left(4\right)<16={2}^{4}$ and ${V}_{\mathcal{F}}=3$ .

We state here without proving it that in general the class of linear classifiers in ${\mathcal{R}}^{d}$ has ${V}_{\mathcal{F}}=d+1$ .

The vc inequality

Let ${X}_{1},\phantom{\rule{0.166667em}{0ex}},...,\phantom{\rule{0.166667em}{0ex}}{X}_{n}$ be i.i.d. ${\mathcal{R}}^{d}$ -valued random variables. Denote the common distribution of ${X}_{i},1\le i\le n$ by $\mu \left(A\right)=P\left({X}_{1}\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}A\right)$ for any subset $A\subset {\mathcal{R}}^{d}$ . Similarly, define the empirical distribution ${\mu }_{n}\left(A\right)=\frac{1}{n}{\sum }_{1}^{n}{1}_{\left\{{X}_{i}ϵA\right\}}$ .

Theorem

Vc '71

For any probablilty measure $\mu$ and collection of subsets $\mathcal{A}$ , and for any $ϵ>0$ .

$P\left(\begin{array}{c}sup\\ Aϵ\mathcal{A}\end{array},\left|{\mu }_{n},\left(A\right),-,\mu ,\left(A\right)\right|,>,ϵ\right)\le 8{\mathcal{S}}_{\mathcal{A}}\left(n\right){e}^{-n{ϵ}^{2}/32}$

and

$E\left[\begin{array}{c}sup\\ Aϵ\mathcal{A}\end{array},\left|{\mu }_{n},\left(A\right),-,\mu ,\left(A\right)\right|\right]\le 2\sqrt{\frac{log2{\mathcal{S}}_{\mathcal{A}}\left(n\right)}{n}}$

Before giving a proof to the theorem. We present a Corollary.

Corollary

Let $\mathcal{F}$ be a collection of classifiers of the form $f:{\mathcal{R}}^{d}\to \left\{0,1\right\}$ with VC dimension ${V}_{\mathcal{F}}<\infty$ ,  Let $R\left(f\right)=P\left(f\left(X\right)\ne Y\right)$ and ${\stackrel{^}{R}}_{n}\left(f\right)=\frac{1}{n}{\sum }_{1}^{n}{1}_{\left\{f\left({X}_{i}\right)\ne {Y}_{i}\right\}}$ , where ${X}_{i},{Y}_{i},1\le i\le n$ are i.i.d. with joint distribution ${P}_{XY}$ .

Define

${\stackrel{^}{f}}_{n}=\begin{array}{c}argmin\\ fϵ\mathcal{F}\end{array}{\stackrel{^}{R}}_{n}\left(f\right)$ .

Then

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-\begin{array}{c}inf\\ fϵ\mathcal{F}\end{array}R\left(f\right)\le 4\sqrt{\frac{{\mathcal{V}}_{\mathcal{F}}log\left(n,+,1\right)+log2}{n}}.$

Let $\mathcal{A}=\left\{\left\{x:f\left(x\right)=1\right\}×\left\{0\right\}\bigcup \left\{x:f\left(x\right)=0\right\}×\left\{1\right\},f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\right\}$

Note that

$P\left(f\left(X\right)\ne Y\right)=P\left(\left(X,Y\right)\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}A\right):=\mu \left(A\right)$

where $A=\left\{x:f\left(x\right)=1\right\}×\left\{0\right\}\bigcup \left\{x:f\left(x\right)=0\right\}×\left\{1\right\}$ .

Similarly,

$\frac{1}{n}\sum _{1}^{n}{1}_{\left\{f\left({X}_{i}\right)\phantom{\rule{0.166667em}{0ex}}\ne \phantom{\rule{0.166667em}{0ex}}{Y}_{i}\right\}}=\frac{1}{n}\sum _{1}^{n}{1}_{\left\{\left({X}_{i},{Y}_{i}\right)\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}A\right\}}:=\mu \left(A\right).$

Therefore, according to the VC theorem.

$\begin{array}{ccc}\hfill E\left[\begin{array}{c}sup\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array},\left|{\stackrel{^}{R}}_{n},\left(f\right),-,R,\left(f\right)\right|\right]=E\left[\begin{array}{c}sup\\ A\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{A}\end{array},\left|{\mu }_{n},\left(A\right),-,\mu ,\left(A\right)\right|\right]& \le & 2\sqrt{\frac{log2{\mathcal{S}}_{\mathcal{A}}\left(n\right)}{n}}\hfill \\ & =& 2\sqrt{\frac{log2{\mathcal{S}}_{\mathcal{F}}\left(n\right)}{n}}\hfill \end{array}$

Since ${V}_{\mathcal{F}}<\infty ,{\mathcal{S}}_{\mathcal{F}}\left(n\right)\le {\left(n+1\right)}^{{V}_{\mathcal{F}}}$ and

$E\left[\begin{array}{c}sup\\ fϵ\mathcal{F}\end{array},\left|{\stackrel{^}{R}}_{n},\left(f\right),-,R,\left(f\right)\right|\right]\le 2\sqrt{\frac{{V}_{\mathcal{F}}log\left(n+1\right)+log2}{n}}.$

Next, note that

$\begin{array}{ccc}\hfill R\left({\stackrel{^}{f}}_{n}\right)-\begin{array}{c}inf\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array}R\left(f\right)& =& \left[R,\left({\stackrel{^}{f}}_{n}\right),-,{\stackrel{^}{R}}_{n},\left({\stackrel{^}{f}}_{n}\right)\right]+\left[{\stackrel{^}{R}}_{n},\left({\stackrel{^}{f}}_{n}\right),-,\begin{array}{c}inf\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array},R,\left(f\right)\right]\hfill \\ & =& \left[R,\left({\stackrel{^}{f}}_{n}\right),-,{\stackrel{^}{R}}_{n},\left({\stackrel{^}{f}}_{n}\right)\right]+\left[\begin{array}{c}sup\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array},\left({\stackrel{^}{R}}_{n},\left({\stackrel{^}{f}}_{n}\right),-,R,\left(f\right)\right)\right]\hfill \\ & \le & \left[R,\left({\stackrel{^}{f}}_{n}\right),-,{\stackrel{^}{R}}_{n},\left({\stackrel{^}{f}}_{n}\right)\right]+\left[\begin{array}{c}sup\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array},\left({\stackrel{^}{R}}_{n},\left(f\right),-,R,\left(f\right)\right)\right]\hfill \\ & \le & 2\begin{array}{c}sup\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array}\left|{\stackrel{^}{R}}_{n},\left(f\right),-,R,\left(f\right)\right|\hfill \end{array}.$

Therefore,

$\begin{array}{ccc}\hfill E\left[R,\left({\stackrel{^}{f}}_{n}\right)\right]-\begin{array}{c}inf\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array}R\left(f\right)& \le & 2E\left[\begin{array}{c}sup\\ f\phantom{\rule{0.166667em}{0ex}}ϵ\phantom{\rule{0.166667em}{0ex}}\mathcal{F}\end{array},\left|{\stackrel{^}{R}}_{n},\left(f\right),-,R,\left(f\right)\right|\right]\hfill \\ & \le & 4\sqrt{\frac{{V}_{\mathcal{F}}log\left(n+1\right)+log2}{n}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\hfill \end{array}.$

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