# 0.3 Introduction to complexity regularization  (Page 3/3)

 Page 3 / 3

The number of unique labellings of the training data that can be achieved with linear classifiers is, in fact, finite. A line can bedefined by picking any pair of training points, as illustrated in [link] . Two classifiers can be defined from each such line: one that outputs a label “1” for everything on or abovethe line, and another that outputs “0” for everything on or above. There exist $\left(\genfrac{}{}{0pt}{}{n}{2}\right)$ such pairs of training points, and these define all possible unique labellings of the training data.Therefore, there are at most $2\left(\genfrac{}{}{0pt}{}{n}{2}\right)$ unique linear classifiers for any random set of $n$ 2-dimensional features (the factor of 2 is due to the fact that for each linear classifier thereare 2 possible assignments of the labelling). Fitting a linear classifier to 2-dimensional data. There are an infinite number of such classifiers. We can generate alinear classifier by choosing two data points, drawing a line with both points on one side, and declaring all points on or above theline to be “ + 1 ” (or “ - 1 ”) and all points below the line to be “ - 1 ” (or “ + 1 ”). From the discussion in the previous figure, we see that the two linear classifiers depicted in this figure are equivalent for this setof data points, and hence relative to the set of n training data there are only on the order of n 2 unique linear classifiers.

Thus, instead of infinitely many linear classifiers, we realize that as far as a random sample of $n$ training data is concerned, there are at most

$\begin{array}{ccc}\hfill 2\left(\genfrac{}{}{0pt}{}{n}{2}\right)& =& \frac{2n!}{\left(n-2\right)!2!}\hfill \\ & =& n\left(n-1\right)\hfill \end{array}$

unique linear classifiers. That is, using linear classification rules, there are at most $n\left(n-1\right)\approx {n}^{2}$ unique label assignments for $n$ data points. If we like, we can encode each possibility with ${log}_{2}n\left(n-1\right)\approx 2{log}_{2}n$ bits. In $d$ dimensions there are $2\left(\genfrac{}{}{0pt}{}{n}{d}\right)$ hyperplane classification rules which can be encoded in roughly $d{log}_{2}n$ bits. Roughly speaking, the number of bits required for encoding each model is the VC dimension. Theremarkable aspect of the VC dimension is that it is often finite even when $\mathcal{F}$ is infinite (as in this example).

If $\mathcal{X}$ has $d$ dimensions in total, we might consider linear classifiers based on $1,2,\cdots ,d$ features at a time. Lower dimensional hyperplanes are less complex than higher dimensionalones. Suppose we set

$\begin{array}{ccc}\hfill {\mathcal{F}}_{1}& =& \text{linear}\phantom{\rule{4.pt}{0ex}}\text{classifiers}\phantom{\rule{4.pt}{0ex}}\text{using}\phantom{\rule{4.pt}{0ex}}\text{1}\phantom{\rule{4.pt}{0ex}}\text{feature}\hfill \\ \hfill {\mathcal{F}}_{2}& =& \text{linear}\phantom{\rule{4.pt}{0ex}}\text{classifiers}\phantom{\rule{4.pt}{0ex}}\text{using}\phantom{\rule{4.pt}{0ex}}\text{2}\phantom{\rule{4.pt}{0ex}}\text{features}\hfill \\ \hfill \cdots & & \text{and}\phantom{\rule{4.pt}{0ex}}\text{so}\phantom{\rule{4.pt}{0ex}}\text{on}\phantom{\rule{4.pt}{0ex}}\hfill \end{array}.$

These spaces have increasing VC dimensions, and we can try to balance the empirical risk and a cost function depending on the VC dimension.Such procedures are often referred to as Structural Risk Minimization . This gives you a glimpse of what the VC dimension is all about. In future lectures we will revisit this topic in greaterdetail.

## Hold-out methods

The basic idea of “hold-out” methods is to split the $n$ samples $D\equiv {\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{n}$ into a training set, ${D}_{T}$ , and a test set, ${D}_{V}$ .

$\begin{array}{ccc}\hfill {D}_{T}={\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{m},& & {D}_{V}={\left\{{X}_{i},{Y}_{i}\right\}}_{i=m+1}^{n}\hfill \end{array}.$

Now, suppose we have a collection of different model spaces $\left\{{\mathcal{F}}_{\lambda }\right\}$ indexed by $\lambda \in \Lambda$ (e.g., ${\mathcal{F}}_{\lambda }$ is the set of polynomials of degree $d$ , with $\lambda =d$ ), or suppose that we have a collection of complexity penalization criteria ${L}_{\lambda }\left(f\right)$ indexed by $\lambda$ ( e.g., let ${L}_{\lambda }\left(f\right)=\stackrel{^}{R}\left(f\right)+\lambda c\left(f\right)$ , with $\lambda \in {\mathbf{R}}^{+}$ ). We can obtain candidate solutions using the training set as follows. Define

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

and take

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

or

$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{\lambda }& =& arg\underset{f\in \mathcal{F}}{min}\phantom{\rule{0.166667em}{0ex}}\left\{{\stackrel{^}{R}}_{m},\left(f\right),+,\lambda ,c,\left(f\right)\right\}\hfill \end{array}.$

This provides us with a set of candidate solutions $\left\{{\stackrel{^}{f}}_{\lambda }\right\}$ . Then we can define the hold-out error estimate using the test set:

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

and select the “best” model to be $\stackrel{^}{f}={\stackrel{^}{f}}_{\stackrel{^}{\lambda }}$ where

$\begin{array}{ccc}\hfill \stackrel{^}{\lambda }& =& arg\underset{\lambda }{min}{\stackrel{^}{R}}_{V}\left({\stackrel{^}{f}}_{\lambda }\right)\hfill \end{array}.$

This type of procedure has many nice theoretical guarantees, provided both the training and test set grow with $n$ .

## Leaving-one-out cross-validation

A very popular hold-out method is the so call “leaving-one-out cross-validation” studied in depth by Grace Wahba (UW-Madison,Statistics). For each $\lambda$ we compute

$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{\lambda }^{\left(k\right)}& =& arg\underset{f\in \mathcal{F}}{min}\frac{1}{n}\sum _{\stackrel{i=1}{i\ne k}}^{n}\ell \left(f\left({X}_{i}\right),{Y}_{i}\right)+\lambda C\left(f\right)\hfill \end{array}$

or

$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{\lambda }^{\left(k\right)}& =& arg\underset{f\in {\mathcal{F}}_{\lambda }}{min}\frac{1}{n}\sum _{\stackrel{i=1}{i\ne k}}^{n}\ell \left(f\left({X}_{i}\right),{Y}_{i}\right).\hfill \end{array}$

Then we have cross-validation function

$\begin{array}{ccc}\hfill V\left(\lambda \right)& =& \frac{1}{n}\sum _{k=1}^{n}\ell \left({f}_{\lambda }^{\left(k\right)}\left({X}_{k}\right),{Y}_{k}\right)\hfill \\ \hfill {\lambda }^{*}& =& arg\underset{\lambda }{min}V\left(\lambda \right).\hfill \end{array}$

## Summary

To summarize, this lecture gave a brief and incomplete survey of different methods for dealing with the issues of overfitting and modelselection. Given a set of training data, ${D}_{n}={\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{n}$ , our overall goal is to find

${f}^{*}=arg\underset{f\in \mathcal{F}}{min}R\left(f\right)$

from some collection of functions, $\mathcal{F}$ . Because we do not know the true distribution ${P}_{XY}$ underlyingthe data points ${D}_{n}$ , it is difficult to get an exact handle on the risk, $R\left(f\right)$ . If we only focus on minimizing the empirical risk $\stackrel{^}{R}\left(f\right)$ we end up overfitting to the training data. Two general approaches were presented.

1. In the first approach we consider an indexed collection of spaces ${\left\{{\mathcal{F}}_{\lambda }\right\}}_{\lambda \in \Lambda }$ such that the complexity of ${\mathcal{F}}_{\lambda }$ increases as $\lambda$ increases, and
$\underset{\lambda \to \infty }{lim}{\mathcal{F}}_{\lambda }=\mathcal{F}.$
A solution is given by
$\begin{array}{c}\hfill {\stackrel{^}{f}}_{{\lambda }^{*}}=arg\underset{f\in {\mathcal{F}}_{{\lambda }^{*}}}{min}{\stackrel{^}{R}}_{n}\left(f\right)\end{array}$
where either ${\lambda }^{*}$ is a function which increases with $n$ ,
$\begin{array}{ccc}\hfill {\lambda }^{*}& =& \lambda \left(n\right),\hfill \end{array}$
or ${\lambda }^{*}$ is chosen by hold-out validation.
2. The alternative approach is to incorporate a penalty term into the risk minimization problem formulation. Here we consideran indexed collection of penalties ${\left\{{C}_{\lambda }\right\}}_{\lambda \in \Lambda }$ satisfying the following properties:
1. ${C}_{\lambda }:\mathcal{F}\to {\mathbf{R}}^{+}$ ;
2. For each $f\in \mathcal{F}$ and ${\lambda }_{1}<{\lambda }_{2}$ we have ${C}_{{\lambda }_{1}}\left(f\right)\le {C}_{{\lambda }_{2}}\left(f\right)$ ;
3. There exists ${\lambda }_{0}\in \Lambda$ such that ${C}_{{\lambda }_{0}}\left(f\right)=0$ for all $f\in \mathcal{F}$ .
In this formulation we find a solution
$\begin{array}{ccc}\hfill {\stackrel{^}{f}}_{{\lambda }^{*}}& =& arg\underset{f\in \mathcal{F}}{min}{\stackrel{^}{R}}_{n}\left(f\right)+{C}_{{\lambda }^{*}}\left(f\right),\hfill \end{array}$
where either ${\lambda }^{*}=\lambda \left(n\right)$ , a function growing the number of data samples $n$ , or ${\lambda }^{*}$ is selected by hold-out validation.

## Consistency

If an estimator or classifier ${\stackrel{^}{f}}_{{\lambda }^{*}}$ satisfies

$E\left[R,\left(,{\stackrel{^}{f}}_{{\lambda }^{*}},\right)\right]\to \underset{f\in \mathcal{F}}{inf}R\left(f\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}n\to \infty ,$

then we say that ${\stackrel{^}{f}}_{{\lambda }^{*}}$ is $\mathcal{F}$ -consistent with respect to the risk $R$ . When the context is clear, we will simply say that $\stackrel{^}{f}$ is consistent.

#### Questions & Answers

Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Statistical learning theory. OpenStax CNX. Apr 10, 2009 Download for free at http://cnx.org/content/col10532/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Statistical learning theory' conversation and receive update notifications?  By By  By     By