# 0.10 Complexity regularization

 Page 1 / 1

## 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.

how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!