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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 A be a collection of subsets of R d , definition: The n t h shatter coefficient of A is defined by

S A ( n ) = m a x x 1 , ... , x n ϵ R d { { x 1 , ... , x n } A , A ϵ A } .

The shatter coefficients are a measure of the richness of the collection A . S A ( n ) is the largest number of different subsets of a set of n points that can be generated by intersecting the set with elements of A .

In 1-d, Let A = { - , t , t ϵ R } Possible subsets of { x 1 , ... , x n } generated by intersecting with sets of the form - , t are { x 1 , ... , x n } , { x 1 , ... , x n - 1 } , ... , { x 1 } , φ . Hence S d ( n ) = n + 1 .

In 2-d, Let A = { all rectangles in R 2 }

Consider a set { x 1 , x 2 , x 3 , x 4 } 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 R 2 to cover any subsets of the four points as the above picture, i.e. S A ( 4 ) = 2 4 = 16 .

Clearly, S A ( n ) = 2 n , n = 1 , 2 , 3 as well.

However, for n = 5 , S A ( n ) < 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 S A 2 n .

If { { x 1 , ... , x n } A , A ϵ A } = 2 n then we say that A shatters x 1 , ... , x n .

Vc dimension

The VC dimension
V A of a collection of sets A is defined as the largest interger n such that S A ( n ) = 2 n .

Sauer's lemma:

Let A be a collection of set with VC dimension V A < . Then n , S A ( n ) i = 0 V A n i , also S A ( n ) ( n + 1 ) V A , n .

Vc dimension and classifiers

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

linear (hyperplane) classifiers in R d

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

When n = 2 , we can also assign 1 to all possible subsets { { x 1 , x 2 } , { x 1 } , { x 2 } , φ } and 0 to their complements, and vice versa. Hence S F ( 2 ) = 4 = 2 2 .

When n = 3 , we can arrange arrange the point x 1 , x 2 , x 3 (non-colinear) so that the set of linear classifiers shatters the three points, hence S F ( 3 ) = 8 = 2 3

When n = 4 , no matter where the points x 1 , x 2 , x 3 , x 4 and what designated binary values y 1 , y 2 , y 3 , y 4 are. It's clear that 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 S F ( 4 ) < 16 = 2 4 and V F = 3 .

We state here without proving it that in general the class of linear classifiers in R d has V F = d + 1 .

The vc inequality

Let X 1 , , ... , X n be i.i.d. R d -valued random variables. Denote the common distribution of X i , 1 i n by μ ( A ) = P ( X 1 ϵ A ) for any subset A R d . Similarly, define the empirical distribution μ n ( A ) = 1 n 1 n 1 { X i ϵ A } .

Theorem

Vc '71

For any probablilty measure μ and collection of subsets A , and for any ϵ > 0 .

P s u p A ϵ A μ n ( A ) - μ ( A ) > ϵ 8 S A ( n ) e - n ϵ 2 / 32

and

E s u p A ϵ A μ n ( A ) - μ ( A ) 2 log 2 S A ( n ) n

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

Corollary

Let F be a collection of classifiers of the form f : R d { 0 , 1 } with VC dimension V F < ,  Let R ( f ) = P ( f ( X ) Y ) and R ^ n ( f ) = 1 n 1 n 1 { f ( X i ) Y i } , where X i , Y i , 1 i n are i.i.d. with joint distribution P X Y .

Define

f ^ n = a r g m i n f ϵ F R ^ n ( f ) .

Then

E [ R ( f ^ n ) ] - i n f f ϵ F R ( f ) 4 V F log n + 1 + log 2 n .

Let A = { { x : f ( x ) = 1 } × { 0 } { x : f ( x ) = 0 } × { 1 } , f ϵ F }

Note that

P ( f ( X ) Y ) = P ( ( X , Y ) ϵ A ) : = μ ( A )

where A = { x : f ( x ) = 1 } × { 0 } { x : f ( x ) = 0 } × { 1 } .

Similarly,

1 n 1 n 1 { f ( X i ) Y i } = 1 n 1 n 1 { ( X i , Y i ) ϵ A } : = μ ( A ) .

Therefore, according to the VC theorem.

E s u p f ϵ F R ^ n ( f ) - R ( f ) = E s u p A ϵ A μ n ( A ) - μ ( A ) 2 log 2 S A ( n ) n = 2 log 2 S F ( n ) n

Since V F < , S F ( n ) ( n + 1 ) V F and

E s u p f ϵ F R ^ n ( f ) - R ( f ) 2 V F log ( n + 1 ) + log 2 n .

Next, note that

R f ^ n - i n f f ϵ F R ( f ) = R f ^ n - R ^ n f ^ n + R ^ n f ^ n - i n f f ϵ F R ( f ) = R f ^ n - R ^ n f ^ n + s u p f ϵ F R ^ n ( f ^ n ) - R ( f ) R f ^ n - R ^ n f ^ n + s u p f ϵ F R ^ n ( f ) - R ( f ) 2 s u p f ϵ F R ^ n ( f ) - R ( f ) .

Therefore,

E R f ^ n - i n f f ϵ F R ( f ) 2 E s u p f ϵ F R ^ n ( f ) - R ( f ) 4 V F log ( n + 1 ) + log 2 n .

Questions & Answers

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Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
Maira Reply
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
Google
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Bhagvanji
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Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
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Application of nanotechnology in medicine
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please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
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I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
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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
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LITNING Reply
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LITNING Reply
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LITNING Reply
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LITNING
scanning tunneling microscope
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Santosh
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Rafiq
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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
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Anam
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Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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Bob Reply
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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
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biomolecules are e building blocks of every organics and inorganic materials.
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Source:  OpenStax, Statistical learning theory. OpenStax CNX. Apr 10, 2009 Download for free at http://cnx.org/content/col10532/1.3
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