# 2.3 Machine learning lecture 4 course notes  (Page 5/5)

 Page 5 / 5

Let's put all this together into a theorem.

Theorem. Let $|\mathcal{H}|=k$ , and let any $m,\delta$ be fixed. Then with probability at least $1-\delta$ , we have that

$\epsilon \left(\stackrel{^}{h}\right)\le \left(\underset{h\in \mathcal{H}}{min},\epsilon ,\left(h\right)\right)+2\sqrt{\frac{1}{2m}log\frac{2k}{\delta }}.$

This is proved by letting $\gamma$ equal the $\sqrt{·}$ term, using our previous argument that uniform convergence occurs with probability at least $1-\delta$ , and then noting that uniform convergence implies $\epsilon \left(h\right)$ is at most $2\gamma$ higher than $\epsilon \left({h}^{*}\right)={min}_{h\in \mathcal{H}}\epsilon \left(h\right)$ (as we showed previously).

This also quantifies what we were saying previously saying about the bias/variance tradeoff in model selection. Specifically, suppose we have some hypothesis class $\mathcal{H}$ , and are considering switching to some much larger hypothesis class ${\mathcal{H}}^{\text{'}}\supseteq \mathcal{H}$ . If we switch to ${\mathcal{H}}^{\text{'}}$ , then the first term ${min}_{h}\epsilon \left(h\right)$ can only decrease (since we'd then be taking a min over a larger set of functions). Hence,by learning using a larger hypothesis class, our “bias” can only decrease. However, if k increases, then the second $2\sqrt{·}$ term would also increase. This increase corresponds to our “variance” increasingwhen we use a larger hypothesis class.

By holding $\gamma$ and $\delta$ fixed and solving for $m$ like we did before, we can also obtain the following sample complexity bound:

Corollary. Let $|\mathcal{H}|=k$ , and let any $\delta ,\gamma$ be fixed. Then for $\epsilon \left(\stackrel{^}{h}\right)\le {min}_{h\in \mathcal{H}}\epsilon \left(h\right)+2\gamma$ to hold with probability at least $1-\delta$ , it suffices that

$\begin{array}{ccc}\hfill m& \ge & \frac{1}{2{\gamma }^{2}}log\frac{2k}{\delta }\hfill \\ & =& O\left(\frac{1}{{\gamma }^{2}},log,\frac{k}{\delta }\right),\hfill \end{array}$

## The case of infinite $\mathcal{H}$

We have proved some useful theorems for the case of finite hypothesis classes. But many hypothesis classes, including any parameterized by real numbers(as in linear classification) actually contain an infinite number of functions. Can we prove similar results for this setting?

Let's start by going through something that is not the “right” argument. Better and more general arguments exist , but this will be useful for honing our intuitions about the domain.

Suppose we have an $\mathcal{H}$ that is parameterized by $d$ real numbers. Since we are using a computer to represent real numbers, and IEEE double-precision floating point ( double 's in C) uses 64 bitsto represent a floating point number, this means that our learning algorithm, assuming we're using double-precision floating point, is parameterized by $64d$ bits. Thus, our hypothesis class really consists of at most $k={2}^{64d}$ different hypotheses. From the Corollary at the end of the previous section, we therefore find that, to guarantee $\epsilon \left(\stackrel{^}{h}\right)\le \epsilon \left({h}^{*}\right)+2\gamma$ , with to hold with probability at least $1-\delta$ , it suffices that $m\ge O\left(\frac{1}{{\gamma }^{2}},log,\frac{{2}^{64d}}{\delta }\right)=O\left(\frac{d}{{\gamma }^{2}},log,\frac{1}{\delta }\right)={O}_{\gamma ,\delta }\left(d\right)$ . (The $\gamma ,\delta$ subscripts are to indicate that the last big- $O$ is hiding constants that may depend on $\gamma$ and $\delta$ .) Thus, the number of training examples needed is at most linear in the parameters of the model.

The fact that we relied on 64-bit floating point makes this argument not entirely satisfying, but the conclusion is nonetheless roughly correct: If what we're going to do is try to minimize training error,then in order to learn “well” using a hypothesis class that has $d$ parameters, generally we're going to need on the order of a linear number of training examples in $d$ .

(At this point, it's worth noting that these results were proved for an algorithm that uses empirical risk minimization. Thus, while the linear dependence of samplecomplexity on $d$ does generally hold for most discriminative learning algorithms that try to minimize trainingerror or some approximation to training error, these conclusions do not always apply as readily to discriminative learning algorithms. Giving good theoreticalguarantees on many non-ERM learning algorithms is still an area of active research.)

The other part of our previous argument that's slightly unsatisfying is that it relies on the parameterization of $\mathcal{H}$ . Intuitively, this doesn't seem like it should matter: We had written the classof linear classifiers as ${h}_{\theta }\left(x\right)=1\left\{{\theta }_{0}+{\theta }_{1}{x}_{1}+\cdots {\theta }_{n}{x}_{n}\ge 0\right\}$ , with $n+1$ parameters ${\theta }_{0},...,{\theta }_{n}$ . But it could also be written ${h}_{u,v}\left(x\right)=1\left\{\left({u}_{0}^{2}-{v}_{0}^{2}\right)+\left({u}_{1}^{2}-{v}_{1}^{2}\right){x}_{1}+\cdots \left({u}_{n}^{2}-{v}_{n}^{2}\right){x}_{n}\ge 0\right\}$ with $2n+2$ parameters ${u}_{i},{v}_{i}$ . Yet, both of these are just defining the same $\mathcal{H}$ : The set of linear classifiers in $n$ dimensions.

To derive a more satisfying argument, let's define a few more things.

Given a set $S=\left\{{x}^{\left(i\right)},...,{x}^{\left(d\right)}\right\}$ (no relation to the training set) of points ${x}^{\left(i\right)}\in \mathcal{X}$ , we say that $\mathcal{H}$ shatters $S$ if $\mathcal{H}$ can realize any labeling on $S$ . I.e., if for any set of labels $\left\{{y}^{\left(1\right)},...,{y}^{\left(d\right)}\right\}$ , there existssome $h\in \mathcal{H}$ so that $h\left({x}^{\left(i\right)}\right)={y}^{\left(i\right)}$ for all $i=1,...d$ .

Given a hypothesis class $\mathcal{H}$ , we then define its Vapnik-Chervonenkis dimension , written $\mathrm{VC}\left(\mathcal{H}\right)$ , to be the size of the largest set that is shattered by $\mathcal{H}$ . (If $\mathcal{H}$ can shatter arbitrarily large sets, then $\mathrm{VC}\left(\mathcal{H}\right)=\infty$ .)

For instance, consider the following set of three points:

Can the set $\mathcal{H}$ of linear classifiers in two dimensions ( $h\left(x\right)=1\left\{{\theta }_{0}+{\theta }_{1}{x}_{1}+{\theta }_{2}{x}_{2}\ge 0\right\}$ ) can shatter the set above? The answer is yes. Specifically, we see that, for any of the eight possiblelabelings of these points, we can find a linear classifier that obtains “zero training error” on them:

Moreover, it is possible to show that there is no set of 4 points that this hypothesis class can shatter. Thus, the largest set that $\mathcal{H}$ can shatter is of size 3, and hence $\mathrm{VC}\left(\mathcal{H}\right)=3$ .

Note that the VC dimension of $\mathcal{H}$ here is 3 even though there may be sets of size 3 that it cannot shatter. For instance, if we had a set of three pointslying in a straight line (left figure), then there is no way to find a linear separator for the labeling of the three points shown below (right figure):

In order words, under the definition of the VC dimension, in order to prove that $\mathrm{VC}\left(\mathcal{H}\right)$ is at least $d$ , we need to show only that there's at least one set of size $d$ that $\mathcal{H}$ can shatter.

The following theorem, due to Vapnik, can then be shown. (This is, many would argue, the most important theorem in all of learning theory.)

Theorem. Let $\mathcal{H}$ be given, and let $d=\mathrm{VC}\left(\mathcal{H}\right)$ . Then with probability at least $1-\delta$ , we have that for all $h\in \mathcal{H}$ ,

$|\epsilon \left(h\right)-\stackrel{^}{\epsilon }\left(h\right)|\le O\left(\sqrt{\frac{d}{m}log\frac{m}{d}+\frac{1}{m}log\frac{1}{\delta }}\right).$

Thus, with probability at least $1-\delta$ , we also have that:

$\epsilon \left(\stackrel{^}{h}\right)\le \epsilon \left({h}^{*}\right)+O\left(\sqrt{\frac{d}{m}log\frac{m}{d}+\frac{1}{m}log\frac{1}{\delta }}\right).$

In other words, if a hypothesis class has finite VC dimension, then uniform convergence occurs as $m$ becomes large. As before,this allows us to give a bound on $\epsilon \left(h\right)$ in terms of $\epsilon \left({h}^{*}\right)$ . We also have the following corollary:

Corollary. For $|\epsilon \left(h\right)-\stackrel{^}{\epsilon }\left(h\right)|\le \gamma$ to hold for all $h\in \mathcal{H}$ (and hence $\epsilon \left(\stackrel{^}{h}\right)\le \epsilon \left({h}^{*}\right)+2\gamma$ ) with probability at least $1-\delta$ , it suffices that $m={O}_{\gamma ,\delta }\left(d\right)$ .

In other words, the number of training examples needed to learn “well” using $\mathcal{H}$ is linear in the VC dimension of $\mathcal{H}$ . It turns out that, for “most” hypothesis classes, the VC dimension (assuming a “reasonable” parameterization) is also roughly linear in the number of parameters.Putting these together, we conclude that (for an algorithm that tries to minimize training error) the number of trainingexamples needed is usually roughly linear in the number of parameters of $\mathcal{H}$ .

what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
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
what is the actual application of fullerenes nowadays?
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
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
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.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!