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Introduction

In the last lecture , we studied bounds of the following form: for any δ > 0 , with probability at least 1 - δ ,

R ( f ) R ^ n ( f ) + log | F | + log 1 δ 2 n , f F

which led to upper bounds on the estimation error of the form

E [ R ( f ^ n ) ] - min f F R ( f ) log | F | + log ( n ) + 2 n .

The key assumptions made in deriving the error bounds were:

  • bounded loss function
  • finite collection of candidate functions

The bounds are valid for every P X Y and are called distribution-free.

Deriving bounds for countably infinite spaces

In this lecture we will generalize the previous results in a powerful way by developing bounds applicable to possibly infinitecollections of candidates. To start let us suppose that F is a countable, possibly infinite, collection of candidate functions.Assign a positive number c( f ) to each f F , such that

f F e - c ( f ) < .

The numbers c( f ) can be interpreted as

  • measures of complexity
  • -log of prior probabilities
  • codelengths

In particular, if P( f ) is the prior probability of f then

e - - log p ( f ) = p ( f )

so c ( f ) - log p ( f ) produces

f F e - c ( f ) = f F p ( f ) = 1 .

Now recall Hoeffding's inequality. For each f and every ϵ > 0

P R ( f ) - R ^ n ( f ) ϵ e - 2 n ϵ 2

or for every δ > 0

P R ( f ) - R ^ n ( f ) log 1 δ 2 n δ .

Suppose δ > 0 is specified. Using the values c( f ) for f F , define

δ ( f ) = e - c ( f ) δ .

Then we have

P R ( f ) - R ^ n ( f ) log 1 δ ( f ) 2 n δ ( f ) .

Furthermore we can apply the union bound as follows

P sup f F R ( f ) - R ^ n ( f ) - log ( 1 / δ ( f ) ) 2 n 0 P f F R ( f ) - R ^ n ( f ) log 1 δ ( f ) 2 n f F P R ( f ) - R ^ n ( f ) log 1 δ ( f ) 2 n f F δ ( f ) = f F e - c ( f ) δ = δ .

So for any δ > 0 with probability at least 1 - δ , we have that f F

R ( f ) R ^ n ( f ) + log 1 δ ( f ) 2 n = R ^ n ( f ) + c ( f ) + log 1 δ 2 n .

Special case

Suppose F is finite and c ( f ) = log | F | f F . Then

f F e - c ( f ) = f F e - log | F | = f F 1 | F | = 1

and

δ ( f ) = δ | F |

which implies that for any δ > 0 with probability at least 1 - δ , we have

R ( f ) R ^ n ( f ) + log | F | + log 1 δ ( f ) 2 n , f F .

Note that this is precisely the bound we derived in the last lecture .

Choosing c( f )

The generalized bounds allow us to handle countably infinite collections of candidate functions, but we require that

f F e - c ( f ) < .

Of course, if c ( f ) = - log p ( f ) where p ( f ) is a proper prior probability distribution then we have

f F e - c ( f ) = 1 .

However, it may be difficult to design a probability distribution over an infinite class of candidates. The coding perspectiveprovides a very practical means to this end.

Assume that we have assigned a uniquely decodable binary code to each f F , and let c( f ) denote the codelength for f . That is, the code for f is c( f ) bits long. A very useful class of uniquely decodable codes are called prefix codes.

Prefix Code
A code is called a prefix codeif no codeword is a prefix of any other codeword.

The kraft inequality

For any binary prefix code, the codeword lengths c 1 , c 2 , ... satisfy

i = 1 2 - c i 1 .

Conversely, given any c 1 , c 2 , ... satisfying the inequality above we can construct a prefix code with these codeword lengths.We will prove this result a bit later, but now let's see how this is useful in our learning problem.

Assume that we have assigned a binary prefix codeword to each f F , and let c( f ) denote the bit-length of the codeword for f . Set δ ( f ) = 2 - c ( f ) δ . Then

P f F R ( f ) - R ^ n ( f ) log 1 δ ( f ) 2 n f F P R ( f ) - R ^ n ( f ) log 1 δ ( f ) 2 n f F δ ( f ) = f F 2 - c ( f ) δ = δ .

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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salma
Commplementary angles
Idrissa Reply
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Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
hi
salma
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a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
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ramon Reply
Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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China
Cied
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abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
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Prasenjit
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Damian
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Damian
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Azam
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Uday
I'm interested in Nanotube
Uday
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Prasenjit
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
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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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