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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 n 2 such pairs of training points, and these define all possible unique labellings of the training data.Therefore, there are at most 2 n 2 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

2 n 2 = 2 n ! ( n - 2 ) ! 2 ! = n ( n - 1 )

unique linear classifiers. That is, using linear classification rules, there are at most n ( n - 1 ) n 2 unique label assignments for n data points. If we like, we can encode each possibility with log 2 n ( n - 1 ) 2 log 2 n bits. In d dimensions there are 2 n d 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 F is infinite (as in this example).

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

F 1 = linear classifiers using 1 feature F 2 = linear classifiers using 2 features and so on .

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 { X i , Y i } i = 1 n into a training set, D T , and a test set, D V .

D T = { X i , Y i } i = 1 m , D V = { X i , Y i } i = m + 1 n .

Now, suppose we have a collection of different model spaces { F λ } indexed by λ Λ (e.g., F λ is the set of polynomials of degree d , with λ = d ), or suppose that we have a collection of complexity penalization criteria L λ ( f ) indexed by λ ( e.g., let L λ ( f ) = R ^ ( f ) + λ c ( f ) , with λ R + ). We can obtain candidate solutions using the training set as follows. Define

R ^ m ( f ) = i = 1 m ( f ( X i ) , Y i )

and take

f ^ λ = arg min f F λ R ^ m ( f )


f ^ λ = arg min f F R ^ m ( f ) + λ c ( f ) .

This provides us with a set of candidate solutions { f ^ λ } . Then we can define the hold-out error estimate using the test set:

R ^ V ( f ) = 1 n - m + 1 i = m + 1 n ( f ( X i ) , Y i ) ,

and select the “best” model to be f ^ = f ^ λ ^ where

λ ^ = arg min λ R ^ V f ^ λ .

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 λ we compute

f ^ λ ( k ) = arg min f F 1 n i k i = 1 n ( f ( X i ) , Y i ) + λ C ( f )


f ^ λ ( k ) = arg min f F λ 1 n i k i = 1 n ( f ( X i ) , Y i ) .

Then we have cross-validation function

V ( λ ) = 1 n k = 1 n ( f λ ( k ) ( X k ) , Y k ) λ * = arg min λ V ( λ ) .


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 = { X i , Y i } i = 1 n , our overall goal is to find

f * = arg min f F R ( f )

from some collection of functions, F . Because we do not know the true distribution P X Y underlyingthe data points D n , it is difficult to get an exact handle on the risk, R ( f ) . If we only focus on minimizing the empirical risk R ^ ( f ) 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 { F λ } λ Λ such that the complexity of F λ increases as λ increases, and
    lim λ F λ = F .
    A solution is given by
    f ^ λ * = arg min f F λ * R ^ n ( f )
    where either λ * is a function which increases with n ,
    λ * = λ ( n ) ,
    or λ * 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 { C λ } λ Λ satisfying the following properties:
    1. C λ : F R + ;
    2. For each f F and λ 1 < λ 2 we have C λ 1 ( f ) C λ 2 ( f ) ;
    3. There exists λ 0 Λ such that C λ 0 ( f ) = 0 for all f F .
    In this formulation we find a solution
    f ^ λ * = arg min f F R ^ n ( f ) + C λ * ( f ) ,
    where either λ * = λ ( n ) , a function growing the number of data samples n , or λ * is selected by hold-out validation.


If an estimator or classifier f ^ λ * satisfies

E R ( f ^ λ * ) inf f F R ( f ) as n ,

then we say that f ^ λ * is F -consistent with respect to the risk R . When the context is clear, we will simply say that f ^ is consistent.

Questions & Answers

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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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