shahar slides on machine learning

7/23/2019 Shahar Slides on Machine learning

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A Geometric Approach to Statistical Learning

Theory

Shahar Mendelson

Centre for Mathematics and its ApplicationsThe Australian National University

Canberra, Australia


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What is a learning problem

A class of functions Fon a probability space (, ) A random variableYone wishes to estimate

A loss functional

The information we have: a sample(Xi, Yi)ni=1

Our goal: with high probability, find a good approximation to Y in F with

respect to, that is

Findf Fsuch that E(f(X), Y)) is almost optimal.

fis selected according to the sample (Xi, Yi)ni=1.Shahar Mendelson: A Geometric Approach to Statistical Learning Theory


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Example

Consider

The random variable Y is a fixed function T : [0, 1] (that is Yi =T(Xi)).

The loss functional is (u, v) = (u

v)2.

Hence, the goal is to find some f F for whichE(f(X), T(X)) = E(f(X) T(X))2 =

(f(t) T(t))2d(t)

is as small as possible.To selectfwe use the sampleX1,...,Xn and the valuesT(X1),...,T(Xn).

Shahar Mendelson: A Geometric Approach to Statistical Learning Theory


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A variation

Consider the following excess loss:

f(x) = (f(x) T(x))2 (f(x) T(x))2 =f(x) f(x),wherefminimizes Ef(x) = E(f(X) T(X))2 in the class.

The difference between the two cases:



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Our Goal

Given a fixed sample size, how close to the optimal can one get using em-pirical data?

How does the specific choice of the loss influence the estimate?

What parameters of the class Fare important? Although one has access to a random sample, the measure which generates

the data is not known.



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The algorithm

Given a sample (X1,...,Xn), select f Fwhich satisfies

argminfF1

n

ni=1

f(Xi),

that is, fis the best function in the class on the data.

The hope is thatwith high probability E(f|X1,...,Xn) =f(t)d(t)is close

to the optimal.

In other words, hopefully, with high probability, the empirical minimizer of theloss is almost the best function in the class with respect to the loss.



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Back to the squared loss

In the case of the squared excess loss -

f(x) = (f(x) T)2 (f(x) T(x))2,since the second term if the same for every f F, the empirical minimizationselects

argminfFn

i=1

(f(Xi) T(Xi))2

and the question is

how to relate thisempirical distance to the real distance we are interested

in.



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Analyzing the algorithm

For a second, lets forget the loss, and from here on, to simplify notation, denote

byGthe loss class.

We shall attempt to connect 1nn

i=1 g(Xi) (i.e. therandom, empirical structure

onG) to Eg.

We shall examine various notions of similarity of the structures.

Note: in the case of an excess loss, 0 Gand our aim is to be as close to 0 aspossible. Otherwise, our aim is to approachg

= 0.



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A road map

Asking the correct question - beware of loose methods of attack. Properties of the loss and their significance.

Estimating the complexity of a class.



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A little history

Originally, the study of

{0, 1

}-valued classes (e.g. Perceptrons) used the uni-

form law of large numbers:

Pr

g G

1

n

n

i=1g(Xi) Eg

,

which is a uniform measure of similarity.

If the probability of this is small, then for everyg G, the empirical structureis close to the real one. In particular, this is true for the empirical minimizer,

and thus, on the good event,

Eg 1n

ni=1

g(Xi) + .

In a minute: this approach is suboptimal!!!!Shahar Mendelson: A Geometric Approach to Statistical Learning Theory


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Why is this bad?

Consider the excess loss case.

We hope that the algorithm will get us close to 0...

So, it seems likely that we would only need to control the part ofGwhichis not too far from 0.

No need to control functions which are far away, while in the ULLN, wecontrol everyfunction inG.



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Why would this lead to a better bound?

Well, first, the set is smaller...

More important:

functions close to 0 in expectation are likely to have a small variance (undermild assumptions)...

On the other hand,

Because of the CLT, for every fixed function g G and n large enough,with probability 1/2,

1

n

ni=1

g(Xi) Eg

var(g)

n ,



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So?

Control over the entire class = control over functions with nonzero variance= rate of convergence cant be better than c/n.

Ifg = 0, we cant hope to get a faster rate than c/nusing this method.

This shows the statistical limitation of the loss.



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What does this tell us about the loss?

To get faster convergence rates one has to consider the excess loss. We also need a condition that would imply that if the expectation is small,

the variance is small (e.g E2f BEf - A Bernstein condition).

It turns out that this condition is connected to convexity properties ofat0.

One has to connect the richness ofG to that ofF (which follows from aLipshitz condition on).



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Localization - Excess loss

There are several ways to localize.

It is enough to bound

Pr

g G 1

n

ni=1

g(Xi) Eg 2

This event upper bounds the probability that the algorithm fails. If thisprobability is small , and sincen1

ni=1g(Xi) , then Eg 2.

Another (similar) option: relative bounds:

Prg G n

1/2ni=1(g(Xi)

E

g)var(g)



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Comparing structures

Suppose that one could find rn, for which, with high probability, for every g

G

with Eg rn,1

2Eg 1

n

ni=1

g(Xi) 32Eg

(here, 1/2 and 3/2 can be replaced by 1

and 1 + ).

Then if g was produced by the algorithm it can either

have a large expectation - Eg rn, =

The structures are similar and thus Eg

2

nni=1g(Xi),Or

have a small expectation = Eg rn,



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Comparing structures II

Thus, with high probability

Eg max

rn, 2

n

ni=1

g(Xi)

.

This result is based on a ratio limit theorem, because we would like to show

thatsup

gG,Egrn

n1

ni=1 g(Xi)

Eg 1

.This normalization is possible if Eg2 can be bounded using Eg (which is aproperty of the loss). Otherwise, one needs a slightly different localization.



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Star shape

IfG is star-shaped, its relative richness increases as r becomes smaller.



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Why is this better?

Thanks to a star-shape assumption, our aim is to find the smallest rnsuch that

with high probability,

supgG, Eg=r

1

n

n

i=1g(Xi) Eg

r

2.

This would imply that the error of the algorithm is at mostr.

For the non-localized result, to obtain the same error, one needs to show

supgG1n

ni=1

g(Xi) Eg r,where the supremum is on a much larger set, and includes functions with alarge variance.



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BTW, even this is not optimal....

It turns out that a structural approach (uniform or localized) does not give the

best result that one could get on the error of the EM algorithm.

A sharp bound follows from a direct analysis of the algorithm (under mild

assumptions) and depends on the behavior of the (random) function

n(r) = supgG, Eg=r

1

n

ni=1

g(Xi) Eg.



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Application of concentration

Suppose that we can show that with high probability, for every r,

supgG, Eg=r

1nn

i=1

g(Xi) Eg E supgG, Eg=r

1nn

i=1

g(Xi) Eg n(r)

i.e. that the expectation is a good estimation of the random variable. Then,

The uniform estimate: the error is close to n(1). The localized estimate: the error is close to the fixed point: n(r) =r/2.

The direct analysis ......



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A short summary

bounding the right quantity - one needs to understand

Pr

supgA

1nn

i=1

g(Xi) Eg t

.

For an estimate on EM, A G. The smaller we can take A - the betterthe bound!

loss class vs. excess loss: being close to 0 (hopefully) implies small variance(property of the loss).

One has to connect the complexity ofA

G to the complexity of the

subset of the base classFthat generated it.

If one considers excess loss (better statistical error), there is a question ofthe approximation error.



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Estimating the complexity

Concentration:supgA

1nn

i=1

g(Xi) Eg E supgA

1nn

i=1

g(Xi) Eg .

symmetrization

E supgA

1nn

i=1

g(Xi) Eg EXE supgA

1nn

i=1

ig(Xi)

=Rn(A). iare independent,P r(= 1) =P r(= 1) = 1/2.



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Estimating the Complexity II

For= (X1,...,Xn) considerPA= {(g(X1),...,g(Xn))}. Then

Rn(A) = EX

E sup

vPA

1

n

n

i=1ivi

.

For every (random) coordinate projection V = PA, the complexity pa-rameter E supvPA

n1ni=1 ivi measures the correlation ofV with arandom noise.

The noise model: a random point in {1, 1}n (then-dimensional combina-torial cube), i.e., (1,...,n). Rn(A) measures howVis correlated with this

noise.



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Example - A large set

If the class of functions is bounded by 1 then for any sample (X1,...,Xn),PA [1, 1]n.

ForV = [1, 1]n (orV = {1, 1}n),

1nE sup

vPA

ni=1

ivi = 1,

(If there are many large coordinate projections, Rn does not converge to

0 asn !).

Question: what subsets of [1, 1]n are big in the context of this complexityparameter?



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Combinatorial dimensions

Consider a class of

{1, 1

}-valued functions. Define the Vapnik-Chervonenkis

dimension ofAby

vc(A) = sup|| | PA= {1, 1}||

.

In other words, vc(A) is the largest dimension of a coordinate projection ofA

which is the entire (combinatorial) cube.

There is a real-valued analog of the Vapnik-Chervonenkis dimension, which is

called thecombinatorial dimension.



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Combinatorial dimensions II

The combinatorial dimension:

For every, it measures the largest dimension|| of a cube of side length that can be found in a coordinate projectionPA.



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Some connection between the parameters

Ifvc(A) dthenRn(A) c

d/n.

Ifvc(A, ) is the combinatorial dimension ofAat scale, then

Rn(A)

C

n

0vc(A, )d.

Note: These bounds on Rn are (again) not optimal and can be improved in

various ways. For example:

The bounds take into account the worst case projection - not the average

projection.

The bound does not take into account the diameter ofA.Shahar Mendelson: A Geometric Approach to Statistical Learning Theory


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Important

The vc dimension (combinatorial dimension) and other related complexity pa-

rameters (e.g covering numbers) are only ways of upper bounding Rn.

Sometimes such a bound is good, but at times it is not.

Although the connections between the various complexity parameters are very

interesting and nontrivial, for SLT it is always best to try and bound Rndirectly.

Again, the difficulty of the learning problem is captured by the richness of a

random coordinate projection of the loss class.



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Example: Error rates for VC classes

Let

Fbe a class of{0, 1}-valued functions withvc(F) dandT F. (Properlearning)

is the squared loss andGis the loss class. Note, 0

G!

H= star(G, 0) = {g| g G} is the star-shaped hull ofG.



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Example: Error rates for VC classes II

Then:

Sincef(x) 0 and functions in F are {0, 1}-valued, then Eh2 Eh.

The error rate is upper bounded by the fixed point of

E suphH, Eh=r

n1n

i=1ih(Xi)

=Rn(Hr),i.e. when

Rn(Hr) =r

2.

The next step is to relate the complexity ofHr to the complexity ofF.



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Bounding Rn(Hr)

Fis small in the appropriate sense. is a Lipschitz function, and thus G= (F) is not much larger than F.

The star-shaped hull ofGis not much larger than G.In particular, for every n d, with probability larger than 1 edn cd, ifEng infgG Eng+ , then

Eg c maxd

nlog

n

ed , .


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