# 0.10 Complexity regularization

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## Review: pac bounds

Consider a finite collection of models $\mathcal{F}$ , and recall the basic PAC bound: for any $\delta >0$ , with probability at least $1-\delta$

$\begin{array}{c}\hfill R\left(f\right)\le {\stackrel{^}{R}}_{n}\left(f\right)+\sqrt{\frac{log|\mathcal{F}|+log\left(1/\delta \right)}{2n}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\forall f\in \mathcal{F}\end{array}$

where

$\begin{array}{ccc}\hfill {\stackrel{^}{R}}_{n}\left(f\right)& =& \frac{1}{n}\sum _{i=1}^{n}\ell \left(f\left({X}_{i}\right),{Y}_{i}\right)\hfill \\ \hfill R\left(f\right)& =& E\left[\ell ,\left(,f,\left(,X,\right),,,Y,\right)\right]\hfill \end{array}$

and the loss $\ell$ is assumed to be bounded between 0 and 1. Note that we can write the inequality above as:

$\begin{array}{c}\hfill R\left(f\right)\le {\stackrel{^}{R}}_{n}\left(f\right)+\sqrt{\frac{log\left(\frac{|\mathcal{F}|}{\delta }\right)}{2n}}\end{array}$

Letting ${\delta }_{f}=\frac{\delta }{|\mathcal{F}|}$ , we have:

$\begin{array}{c}\hfill R\left(f\right)\le {\stackrel{^}{R}}_{n}\left(f\right)+\sqrt{\frac{log\left(1/{\delta }_{f}\right)}{2n}}\end{array}$

This is precisely the form of Hoeffding's inequality, with ${\delta }_{f}$ in place of the usual $\delta$ . In effect, in order to have Hoeffding's inequality hold with probability $1-\delta$ for all $f\in \mathcal{F}$ , we must distribute the “ $\delta$ -budget” or “confidence-budget” over all $f\in \mathcal{F}$ (in this case, evenly distributed):

$\begin{array}{ccc}\hfill \sum _{f\in \mathcal{F}}{\delta }_{f}& =& \sum _{f\in \mathcal{F}}\frac{\delta }{|\mathcal{F}|}\hfill \\ & =& \delta \hfill \end{array}$

However, to apply the union bound, we do not need to distribute $\delta$ evenly among the candidate models. We only require:

$\begin{array}{ccc}\hfill \sum _{f\in \mathcal{F}}{\delta }_{f}& =& \delta \hfill \end{array}$

So, if $p\left(f\right)$ are positive numbers satisfying ${\sum }_{f\in \mathcal{F}}p\left(f\right)=1$ , then we can take ${\delta }_{f}=p\left(f\right)\delta$ . This provides two advantages:

1. By choosing $p\left(f\right)$ larger for certain $f$ , we can preferentially treat those candidates
2. We do not need $\mathcal{F}$ to be finite and we only require ${\sum }_{f\in \mathcal{F}}p\left(f\right)=1$

Prefix codes are one way to achieve this. If we assign a binary prefix code of length $c\left(f\right)$ to each $f\in \mathcal{F}$ , then the values $p\left(f\right)={2}^{-c\left(f\right)}$ satisfy ${\sum }_{f\in \mathcal{F}}p\left(f\right)\le 1$ according to the Kraft inequality.

The main point of this lecture is to examine how PAC bounds of the form w.p. $\ge 1-\delta$

$\begin{array}{c}\hfill R\left(f\right)\le {\stackrel{^}{R}}_{n}\left(f\right)+\sqrt{\frac{c\left(f\right)log2+log\left(1/\delta \right)}{2n}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\forall f\in \mathcal{F}\end{array}$

can be used to select a model that comes close to achieving the best possible performace

$\begin{array}{c}\hfill \underset{f\in \mathcal{F}}{inf}R\left(f\right)\end{array}$

Let ${\stackrel{^}{f}}_{n}$ be the model selected from $\mathcal{F}$ using the training data ${\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{n}$ . We will specify this model in a moment, but keep in mind that it is notnecessarily the model with minimum empirical risk as before. We would like to have

$\begin{array}{c}\hfill E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-\underset{f\in \mathcal{F}}{inf}R\left(f\right)\end{array}$

as small as possible. First, for any $\delta >0$ , define

$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{n}^{\delta }& =& arg\underset{f\in \mathcal{F}}{min}\left\{{\stackrel{^}{R}}_{n},\left(f\right),+,C,\left(f,n,\delta \right)\right\}\hfill \end{array}$

where

$\begin{array}{c}\hfill C\left(f,n,\delta \right)\equiv \sqrt{\frac{c\left(f\right)log2+log\left(1/\delta \right)}{2n}}\end{array}$

Then w.p. $\ge 1-\delta$

$\begin{array}{c}\hfill R\left(f\right)\le {\stackrel{^}{R}}_{n}\left(f\right)+C\left(f,n,\delta \right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}},\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\forall f\in \mathcal{F}\end{array}$

and in particular,

$\begin{array}{c}\hfill R\left({\stackrel{^}{f}}_{n}^{\delta }\right)\le {\stackrel{^}{R}}_{n}\left({\stackrel{^}{f}}_{n}^{\delta }\right)+C\left({\stackrel{^}{f}}_{n}^{\delta },n,\delta \right)\phantom{\rule{4pt}{0ex}},\end{array}$

so, by the definition of ${\stackrel{^}{f}}_{n}^{\delta }$ , $\forall f\in \mathcal{F}$

$\begin{array}{c}\hfill R\left({\stackrel{^}{f}}_{n}^{\delta }\right)\le {\stackrel{^}{R}}_{n}\left(f\right)+C\left(f,n,\delta \right)\phantom{\rule{4pt}{0ex}}.\end{array}$

We will make use of the inequality above in a moment. First note that $\forall f\phantom{\rule{4pt}{0ex}}\in \mathcal{F}$

$\begin{array}{ccc}\hfill E\left[R\left({\stackrel{^}{f}}_{n}^{\delta }\right)\right]-R\left(f\right)& =& E\left[R\left({\stackrel{^}{f}}_{n}^{\delta }\right)-{\stackrel{^}{R}}_{n}\left(f\right)\right]+E\left[{\stackrel{^}{R}}_{n}\left(f\right)-R\left(f\right)\right]\hfill \end{array}$

The second term is exactly 0, since $E\left[{\stackrel{^}{R}}_{n}\left(f\right)\right]=R\left(f\right)$ .

Now consider the first term $E\left[R\left({\stackrel{^}{f}}_{n}^{\delta }\right)-{\stackrel{^}{R}}_{n}\left(f\right)\right]$ . Let $\Omega$ be the set of events on which

$\begin{array}{c}\hfill R\left({\stackrel{^}{f}}_{n}^{\delta }\right)\le {\stackrel{^}{R}}_{n}\left(f\right)-C\left(f,n,\delta \right),\phantom{\rule{3.33333pt}{0ex}}\forall \phantom{\rule{3.33333pt}{0ex}}f\in \mathcal{F}\end{array}$

From the bound above, we know that $P\left(\Omega \right)\ge 1-\delta$ . Thus,

$\begin{array}{ccc}\hfill E\left[R\left({\stackrel{^}{f}}_{n}^{\delta }\right)-{\stackrel{^}{R}}_{n}\left(f\right)\right]& =& E\left[R\left({\stackrel{^}{f}}_{n}^{\delta }\right)-{\stackrel{^}{R}}_{n}\left(f\right)|\Omega \right]P\left(\Omega \right)+E\left[R\left({\stackrel{^}{f}}_{n}^{\delta }\right)-{\stackrel{^}{R}}_{n}\left(f\right)|{\Omega }^{c}\right]\left(1-P\left(\Omega \right)\right)\hfill \\ & \le & C\left(f,n,\delta \right)+\delta \phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\left(\text{since}\phantom{\rule{3.33333pt}{0ex}}0\le R,\phantom{\rule{3.33333pt}{0ex}}\stackrel{^}{R}\le 1,\phantom{\rule{3.33333pt}{0ex}}P\left(\Omega \right)\le 1\phantom{\rule{3.33333pt}{0ex}}\text{and}\phantom{\rule{3.33333pt}{0ex}}1-P\left(\Omega \right)\le \delta \right)\hfill \\ & =& \sqrt{\frac{c\left(f\right)log2+log\left(1/\delta \right)}{2n}}+\delta \hfill \\ & =& \sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}}+\frac{1}{\sqrt{n}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\left(\text{by}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{setting}\phantom{\rule{3.33333pt}{0ex}}\delta =\frac{1}{\sqrt{n}}\right)\hfill \end{array}$

We can summarize our analysis with the following theorem.

Theorem

## Complexity regularized model selection

Let $\mathcal{F}$ be a countable collection of models, and assign a positive number $c\left(f\right)$ to each $f\in \mathcal{F}$ such that ${\sum }_{f\in \mathcal{F}}{2}^{-c\left(f\right)}\le 1$ . Define the minimum complexity regularized risk model

$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{n}& =& arg\underset{f\in \mathcal{F}}{min}\left\{{\stackrel{^}{R}}_{n},\left(f\right),+,\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}}\right\}\hfill \end{array}$

Then,

$\begin{array}{c}\hfill E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\phantom{\rule{4pt}{0ex}}\le \phantom{\rule{4pt}{0ex}}\underset{f\in \mathcal{F}}{inf}\left\{R,\left(f\right),+,\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}},+,\frac{1}{\sqrt{n}}\right\}\end{array}$

This shows that

$\begin{array}{c}\hfill {\stackrel{^}{R}}_{n}\left(f\right)+\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}}\end{array}$

is a reasonable surrogate for

$\begin{array}{c}\hfill R\left(f\right)+\sqrt{\frac{c\left(f\right)log2+\frac{1}{2}logn}{2n}}\end{array}$

## Histogram classifiers

Let $\mathcal{X}={\left[0,1\right]}^{d}$ be the input space and $\mathcal{Y}=\left\{0,1\right\}$ be the output space. Let ${\mathcal{F}}_{k}$ , k = 1, 2, ...  denotes the collection of histogram classification rules withk equal volume bins. One choice of prefix code for this example is: $k=1⇒\text{code}=0,k=3⇒\text{code}=10,k=3⇒\text{code}=110$ and so on .... Then, if first code is corresponding to k $⇒f\in {\mathcal{F}}_{k}$ , followed by $k={log}_{2}|{\mathcal{F}}_{k}|$ bits to indicate which of the ${2}^{k}$ histogram rules in ${\mathcal{F}}_{k}$ is under consideration, we have

$\begin{array}{c}\hfill f\in {\mathcal{F}}_{k}⇒c\left(f\right)=2k\phantom{\rule{3.33333pt}{0ex}}bits\end{array}$

Let ${\stackrel{^}{f}}_{n}$ be the model that solves the minimization i.e.,

$\begin{array}{c}\hfill \underset{k\ge 1}{min}\left\{\underset{f\in {\mathcal{F}}_{k}}{min},{\stackrel{^}{R}}_{n},\left(f\right),+,\sqrt{\frac{2klog2+\frac{1}{2}logn}{2n}}\right\}\end{array}$

That is, for each k, let

$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{n}^{\left(k\right)}& =& arg\underset{f\in {\mathcal{F}}_{k}}{min}{\stackrel{^}{R}}_{n}\left(f\right)\hfill \end{array}$

Then select the best $k$ according to

$\begin{array}{ccc}\hfill \stackrel{^}{k}& =& arg\underset{k\ge 1}{min}\left\{{\stackrel{^}{R}}_{n},\left({\stackrel{^}{f}}_{n}^{\left(k\right)}\right),+,\sqrt{\frac{2klog2+\frac{1}{2}logn}{2n}}\right\}\hfill \end{array}$

and set

$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{n}& =& {\stackrel{^}{f}}_{n}^{\left(\stackrel{^}{k}\right)}\hfill \end{array}$

Then,

$\begin{array}{c}\hfill E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\phantom{\rule{4pt}{0ex}}\le \phantom{\rule{4pt}{0ex}}\underset{k\ge 1}{inf}\left\{\underset{f\in {\mathcal{F}}_{k}}{min},R,\left(f\right),+,\sqrt{\frac{2klog2+\frac{1}{2}logn}{2n}},+,\frac{1}{\sqrt{n}}\right\}\end{array}$

It is a simple exercise to show that if $d=2$ and the Bayes decision boundary is a 1-d curve, then by setting $k=\sqrt{n}$ and selecting the best $f$ from ${\mathcal{F}}_{\sqrt{n}}$ we have

$\begin{array}{c}\hfill E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}O\left({n}^{-1/4}\right)\end{array}$
The complexity regularized classifier ${\stackrel{^}{f}}_{n}$ adaptively achieves this rate, without user intervention.

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