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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

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Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
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Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
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there is no specific books for beginners but there is book called principle of nanotechnology
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s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
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Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
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CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
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s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
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SUYASH Reply
for screen printed electrodes ?
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s. Reply
of graphene you mean?
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in general
s.
Graphene has a hexagonal structure
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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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