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In the last lecture we derived a risk (MSE) bound for regression problems; i.e., select an f F so that E [ ( f ( X ) - Y ) 2 ] - E [ ( f * ( X ) - Y ) 2 ] is small, where f * ( x ) = E [ Y | X = x ] . The result is summarized below.

Theorem

Complexity regularization with squared error loss

Let X = R d , Y = [ - b / 2 , b / 2 ] , { X i , Y i } i = 1 n iid, P X Y unknown, F = {collection of candidate functions},

f : R d Y , R ( f ) = E [ ( f ( X ) - Y ) 2 ] .

Let c ( f ) , f F , be positive numbers satisfying f F 2 - c ( f ) 1 , and select a function from F according to

f ^ n = arg min R ^ n ( f ) + 1 ϵ c ( f ) log 2 n ,

with ϵ 3 5 b 2 and R ^ n ( f ) = 1 n i = 1 n ( f ( X i ) - Y i ) 2 . Then,

E [ R ( f ^ n ) ] - R ( f * ) 1 + α 1 - α min f F R ( f ) - R ( f * ) + 1 ϵ c ( f ) log 2 n + O ( n - 1 )

where α = ϵ b 2 1 - 2 b 2 ϵ / 3 .

Maximum likelihood estimation

The focus of this lecture is to consider another approach to learning based on maximum likelihood estimation. Consider the classical signalplus noise model:

Y i = f i n + W i , i = 1 , , n

where W i are iid zero-mean noises. Furthermore, assume that W i P ( w ) for some known density P ( w ) . Then

Y i P y - f i n P f i ( y )

since Y i - f i n = W i .

A very common and useful loss function to consider is

R ^ n ( f ) = 1 n i = 1 n ( - log P f i ( Y i ) ) .

Minimizing R ^ n with respect to f is equivalent to maximizing

1 n i = 1 n log P f i ( Y i )

or

i = 1 n P f i ( Y i ) .

Thus, using the negative log-likelihood as a loss function leads to maximum likelihood estimation. If the W i are iid zero-mean Gaussian r.v.s then this is just the squared error loss weconsidered last time. If the W i are Laplacian distributed e.g. P ( w ) e - | w | , then we obtain the absolute error, or L 1 , loss function. We can also handle non-additive models such as thePoisson model

Y i P y | f i / n = e - f ( i / n ) [ f ( i / n ) ] y y ! .

In this case

- log P Y i | f i / n = f i / n - Y i log f i / n + constant

which is a very different loss function, but quite appropriate for many imaging problems.

Before we investigate maximum likelihood estimation for model selection, let's review some of the basic concepts. Let Θ denote a parameter space (e.g., Θ = R ), and assume we have observations

Y i i i d P θ * ( y ) , i = 1 , , n

where θ * Θ is a parameter determining the density of the { Y i }. The ML estimator of θ * is

θ ^ n = arg max θ Θ i = 1 n P θ ( Y i ) = arg max θ Θ i = 1 n log P θ ( Y i ) = arg min θ Θ i = 1 n - log P θ ( Y i ) .

θ ^ maximizes the expected log-likelihood. To see this, let's compare the expected log-likelihood of θ * with any other θ Θ .

E [ log P θ * ( Y ) - log P θ ( Y ) ] = E log P θ * ( Y ) P θ ( Y ) = log P θ * ( y ) P θ ( y ) P θ * ( y ) d y = K ( P θ , P θ * ) the KL divergence 0 with equality iff P θ * = P θ .

Why?

- E log P θ * ( y ) P θ ( y ) = E log P θ ( y ) P θ * ( y ) log E P θ ( y ) P θ * ( y ) = log P θ ( y ) d y = 0 K ( P θ , P θ * ) 0

On the other hand, since θ ^ n maximizes the likelihood over θ Θ , we have

i = 1 n log P θ * ( Y i ) P θ ^ n ( Y i ) = i = 1 n log P θ * ( Y i ) - log P θ ^ n ( Y i ) 0 .

Therefore,

1 n i = 1 n log P θ * ( Y i ) P θ ^ n ( Y i ) - K ( P θ ^ n , P θ * ) + K ( P θ ^ n , P θ * ) 0

or re-arranging

K ( P θ ^ n , P θ * ) 1 n i = 1 n log P θ * ( Y i ) P θ ^ n ( Y i ) - K ( P θ ^ n , P θ * ) .

Notice that the quantity

1 n i = 1 n log P θ * ( Y i ) P θ ( Y i )

is an empirical average whose mean is K ( P θ , P θ * ) . By the law of large numbers, for each θ Θ ,

1 n i = 1 n log P θ * ( Y i ) P θ ( Y i ) - K ( P θ , P θ * ) a . s . 0 .

If this also holds for the sequence { θ ^ n } , then we have

K ( P θ ^ n , P θ * ) 1 n log P θ * ( Y i ) P θ ^ n ( Y i ) - K ( P θ ^ n , P θ * ) 0 as n

which implies that

P θ ^ n P θ *

which often implies that

θ ^ n θ *

in some appropriate sense (e.g., point-wise or in norm).

Gaussian distributions

P θ * ( y ) = 1 π e - ( y - θ * ) 2
Θ = R , { Y i } i = 1 n i i d P θ * ( y )
K ( P θ , P θ * ) = log P θ * ( y ) P θ ( y ) P θ * ( y ) d y = [ ( y - θ ) 2 - ( y - θ * ) 2 ] P θ * ( y ) d y = E θ * [ ( y - θ ) 2 ] - E θ * [ ( y - θ * ) 2 ] = E θ * [ Y 2 - 2 Y θ + θ 2 ] - 1 / 2 = ( θ * ) 2 + 1 / 2 - 2 θ * θ + θ 2 - 1 / 2 = ( θ * - θ ) 2
θ * maximizes E [ log P θ ( Y ) ] wrt θ Θ
θ ^ n = arg max θ { - ( Y i - θ ) 2 } = arg min θ { ( Y i - θ ) 2 } = 1 n i = 1 n Y i

Hellinger distance

The KL divergence is not a distance function.

K ( P θ 1 , P θ 2 ) K ( P θ 2 , P θ 1 )

Therefore, it is often more convenient to work with the Hellinger metric,

H ( P θ 1 , P θ 2 ) = P θ 1 1 2 - P θ 2 1 2 2 d y 1 2 .

The Hellinger metric is symmetric, non-negative and

H ( P θ 1 , P θ 2 ) = H ( P θ 2 , P θ 1 )

and therefore it is a distance measure. Furthermore, the squared Hellinger distance lower bounds the KL divergence, so convergence in KL divergence implies convergence of the Hellinger distance.

Proposition 1

H 2 ( P θ 1 , P θ 2 ) K ( P θ 1 , P θ 2 )

Proof:

H ( P θ 1 , P θ 2 ) = P θ 1 ( y ) - P θ 2 ( y ) 2 d y = P θ 1 ( y ) d y + P θ 2 ( y ) d y - 2 P θ 1 ( y ) P θ 2 ( y ) d y = 2 - 2 P θ 1 ( y ) P θ 2 ( y ) d y , since P θ ( y ) d y = 1 θ = 2 1 - E θ 2 P θ 1 ( Y ) / P θ 2 ( Y ) 2 log E θ 2 P θ 2 ( Y ) / P θ 1 ( Y ) , since 1 - x - log x 2 E θ 2 log P θ 2 ( Y ) / P θ 1 ( Y ) , by Jensen's inequality = E θ 2 log ( P θ 2 ( Y ) / P θ 1 ( Y ) ) K ( P θ 1 , P θ 2 )

Note that in the proof we also showed that

H ( P θ 1 , P θ 2 ) = 2 1 - P θ 1 ( y ) P θ 2 ( y ) d y

and using the fact log x x - 1 again, we have

H ( P θ 1 , P θ 2 ) - 2 log P θ 1 ( y ) P θ 2 ( y ) d y .

The quantity inside the log is called the affinity between P θ 1 and P θ 2 :

A ( P θ 1 , P θ 2 ) = P θ 1 ( y ) P θ 2 ( y ) d y .

This is another measure of closeness between P θ 1 and P θ 2 .

Gaussian distributions

P θ ( y ) = 1 π e - ( y - θ ) 2
- 2 log P θ 1 ( y ) P θ 2 ( y ) d y = - 2 log 1 π e - ( y - θ 1 ) 2 1 2 1 π e - ( y - θ 2 ) 2 1 2 d y = - 2 log 1 π e - ( y - θ 1 ) 2 2 + ( y - θ 2 ) 2 2 d y = - 2 log 1 π e - y - ( θ 1 + θ 2 2 ) 2 + ( θ 1 - θ 2 2 ) 2 d y = - 2 log e - ( θ 1 - θ 2 2 ) 2 = 1 2 ( θ 1 - θ 2 ) 2
- 2 log A ( P θ 1 , P θ 2 ) = 1 2 ( θ 1 - θ 2 ) 2 for Gaussian distributions H ( P θ 1 , P θ 2 ) 1 2 ( θ 1 - θ 2 ) 2 for Gaussian .

Poisson distributions

If P θ ( y ) = e - θ θ y y ! , θ 0 , then

- 2 log A ( P θ 1 , P θ 2 ) = ( θ 1 - θ 2 ) 2 .

Summary

Y i i i d P θ *
  1. Maximum likelihood estimator maximizes the empirical average
    1 n i = 1 n log P θ ( Y i )
    (our empirical risk is negative log-likelihood)
  2. θ * maximizes the expectation
    E 1 n i = 1 n log P θ ( Y i )
    (the risk is the expected negative log-likelihood)
  3. 1 n i = 1 n log P θ ( Y i ) a . s . E 1 n i = 1 n log P θ ( Y i )
    so we expect some sort of concentration of measure.
  4. In particular, since
    1 n i = 1 n log P θ * ( Y i ) P θ ( Y i ) a . s . K ( P θ , P θ * )
    we might expect that K ( P θ ^ n , P θ * ) 0 for the sequence of estimates { P θ ^ n } n = 1 .

So, the point is that maximum likelihood estimator is just a special case of a loss function in learning. Due to its special structure, weare naturally led to consider KL divergences, Hellinger distances, and Affinities.

Questions & Answers

are nano particles real
Missy Reply
yeah
Joseph
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Lale Reply
no can't
Lohitha
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Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
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Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
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learn
Google
da
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Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
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ya I also want to know the raman spectra
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Crow Reply
what about nanotechnology for water purification
RAW Reply
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
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Alexandre
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Brian Reply
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?
LITNING Reply
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LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
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LITNING
scanning tunneling microscope
Sahil
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Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
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Mahi
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Rafiq
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Anam
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Anam
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Hafiz
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Bob Reply
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Bob
The nanotechnology is as new science, to scale nanometric
brayan
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Damian
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Damian Reply
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Renato
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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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