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ε ^ ( h i ) = 1 m j = 1 m Z j .

Thus, ε ^ ( h i ) is exactly the mean of the m random variables Z j that are drawn iid from a Bernoulli distribution with mean ε ( h i ) . Hence, we can apply the Hoeffding inequality, and obtain

P ( | ε ( h i ) - ε ^ ( h i ) | > γ ) 2 exp ( - 2 γ 2 m ) .

This shows that, for our particular h i , training error will be close to generalization error with high probability, assuming m is large. But we don't just want to guarantee that ε ( h i ) will be close to ε ^ ( h i ) (with high probability) for just only one particular h i . We want to prove that this will be true for simultaneously for all h H . To do so, let A i denote the event that | ε ( h i ) - ε ^ ( h i ) | > γ . We've already show that, for any particular A i , it holds true that P ( A i ) 2 exp ( - 2 γ 2 m ) . Thus, using the union bound, we have that

P ( h H . | ε ( h i ) - ε ^ ( h i ) | > γ ) = P ( A 1 A k ) i = 1 k P ( A i ) i = 1 k 2 exp ( - 2 γ 2 m ) = 2 k exp ( - 2 γ 2 m )

If we subtract both sides from 1, we find that

P ( ¬ h H . | ε ( h i ) - ε ^ ( h i ) | > γ ) = P ( h H . | ε ( h i ) - ε ^ ( h i ) | γ ) 1 - 2 k exp ( - 2 γ 2 m )

(The “ ¬ ” symbol means “not.”) So, with probability at least 1 - 2 k exp ( - 2 γ 2 m ) , we have that ε ( h ) will be within γ of ε ^ ( h ) for all h H . This is called a uniform convergence result, because this is a bound that holds simultaneously for all (as opposed to just one) h H .

In the discussion above, what we did was, for particular values of m and γ , give a bound on the probability that for some h H , | ε ( h ) - ε ^ ( h ) | > γ . There are three quantities of interest here: m , γ , and the probability of error; we can bound either one in terms of the other two.

For instance, we can ask the following question: Given γ and some δ > 0 , how large must m be before we can guarantee that with probability at least 1 - δ , training error will be within γ of generalization error? By setting δ = 2 k exp ( - 2 γ 2 m ) and solving for m , [you should convince yourself this is the right thing to do!], we find that if

m 1 2 γ 2 log 2 k δ ,

then with probability at least 1 - δ , we have that | ε ( h ) - ε ^ ( h ) | γ for all h H . (Equivalently, this shows that the probability that | ε ( h ) - ε ^ ( h ) | > γ for some h H is at most δ .) This bound tells us how many training examples we need in order makea guarantee. The training set size m that a certain method or algorithm requires in order to achieve a certain level of performance is also calledthe algorithm's sample complexity .

The key property of the bound above is that the number of training examples needed to make this guarantee is only logarithmic in k , the number of hypotheses in H . This will be important later.

Similarly, we can also hold m and δ fixed and solve for γ in the previous equation, and show [again, convince yourself that this is right!]that with probability 1 - δ , we have that for all h H ,

| ε ^ ( h ) - ε ( h ) | 1 2 m log 2 k δ .

Now, let's assume that uniform convergence holds, i.e., that | ε ( h ) - ε ^ ( h ) | γ for all h H . What can we prove about the generalization of our learning algorithm that picked h ^ = arg min h H ε ^ ( h ) ?

Define h * = arg min h H ε ( h ) to be the best possible hypothesis in H . Note that h * is the best that we could possibly do given that we are using H , so it makes sense to compare our performance to that of h * . We have:

ε ( h ^ ) ε ^ ( h ^ ) + γ ε ^ ( h * ) + γ ε ( h * ) + 2 γ

The first line used the fact that | ε ( h ^ ) - ε ^ ( h ^ ) | γ (by our uniform convergence assumption). The second used the fact that h ^ was chosen to minimize ε ^ ( h ) , and hence ε ^ ( h ^ ) ε ^ ( h ) for all h , and in particular ε ^ ( h ^ ) ε ^ ( h * ) . The third line used the uniform convergence assumption again, to show that ε ^ ( h * ) ε ( h * ) + γ . So, what we've shown is the following: If uniform convergence occurs,then the generalization error of h ^ is at most 2 γ worse than the best possible hypothesis in H !

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
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
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?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
Commplementary angles
Idrissa Reply
im all ears I need to learn
right! what he said ⤴⤴⤴
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
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?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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I'm interested in nanotube
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
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Sravani Reply
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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
good afternoon madam
what is system testing
what is the application of nanotechnology?
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
anybody can imagine what will be happen after 100 years from now in nano tech world
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
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
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
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Machine learning. OpenStax CNX. Oct 14, 2013 Download for free at http://cnx.org/content/col11500/1.4
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