# 0.13 Maximum likelihood estimation

 Page 1 / 1

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

Theorem

## Complexity regularization with squared error loss

Let $\mathcal{X}={\mathbb{R}}^{d}$ , $\mathcal{Y}=\left[-b/2,b/2\right]$ , ${\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{n}$ iid, ${P}_{XY}$ unknown, $\mathcal{F}$ = {collection of candidate functions},

$f:{\mathbb{R}}^{d}\to \mathcal{Y},\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\left(f\right)=E\left[{\left(f\left(X\right)-Y\right)}^{2}\right].$

Let $c\left(f\right)$ , $f\in \mathcal{F}$ , be positive numbers satisfying ${\sum }_{f\in \mathcal{F}}{2}^{-c\left(f\right)}\le 1$ , and select a function from $\mathcal{F}$ according to

${\stackrel{^}{f}}_{n}=argmin\left\{{\stackrel{^}{R}}_{n},\left(f\right),+,\frac{1}{ϵ},\phantom{\rule{0.166667em}{0ex}},\frac{c\left(f\right)log2}{n}\right\},$

with $ϵ\le \frac{3}{5{b}^{2}}$ and ${\stackrel{^}{R}}_{n}\left(f\right)=\frac{1}{n}{\sum }_{i=1}^{n}{\left(f\left({X}_{i}\right)-{Y}_{i}\right)}^{2}$ . Then,

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-R\left({f}^{*}\right)\phantom{\rule{4pt}{0ex}}\le \phantom{\rule{4pt}{0ex}}\left(\frac{1+\alpha }{1-\alpha }\right)\phantom{\rule{0.166667em}{0ex}}\underset{f\in \mathcal{F}}{min}\left\{R,\left(f\right),-,R,\left({f}^{*}\right),+,\frac{1}{ϵ},\phantom{\rule{0.166667em}{0ex}},\frac{c\left(f\right)log2}{n}\right\}+O\left({n}^{-1}\right)$

where $\alpha =\frac{ϵ{b}^{2}}{1-2{b}^{2}ϵ/3}$ .

## 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:

${Y}_{i}=f\left(\frac{i}{n}\right)+{W}_{i},i=1,\cdots ,n$

where ${W}_{i}$ are iid zero-mean noises. Furthermore, assume that ${W}_{i}\sim P\left(w\right)$ for some known density $P\left(w\right)$ . Then

${Y}_{i}\sim P\left(y,-,f,\left(\frac{i}{n}\right)\right)\equiv {P}_{{f}_{i}}\left(y\right)$

since ${Y}_{i}-f\left(\frac{i}{n}\right)={W}_{i}$ .

A very common and useful loss function to consider is

${\stackrel{^}{R}}_{n}\left(f\right)=\frac{1}{n}\sum _{i=1}^{n}\left(-log{P}_{{f}_{i}}\left({Y}_{i}\right)\right).$

Minimizing ${\stackrel{^}{R}}_{n}$ with respect to $f$ is equivalent to maximizing

$\frac{1}{n}\sum _{i=1}^{n}log{P}_{{f}_{i}}\left({Y}_{i}\right)$

or

$\prod _{i=1}^{n}{P}_{{f}_{i}}\left({Y}_{i}\right).$

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. $P\left(w\right)\propto {e}^{-|w|}$ , then we obtain the absolute error, or ${L}_{1}$ , loss function. We can also handle non-additive models such as thePoisson model

${Y}_{i}\sim P\left(y|f,\left(i,/,n\right)\right)={e}^{-f\left(i/n\right)}\frac{{\left[f\left(i/n\right)\right]}^{y}}{y!}.$

In this case

$-logP\left({Y}_{i},|f,\left(i,/,n\right)\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}f\left(i,/,n\right)-{Y}_{i}log\left(f,\left(i,/,n\right)\right)+\mathrm{constant}$

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 $\Theta$ denote a parameter space (e.g., $\Theta =R$ ), and assume we have observations

${Y}_{i}\stackrel{iid}{\sim }{P}_{{\theta }^{*}}\left(y\right),\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}i=1,\cdots ,n$

where ${\theta }^{*}\in \Theta$ is a parameter determining the density of the { ${Y}_{i}$ }. The ML estimator of ${\theta }^{*}$ is

$\begin{array}{ccc}\hfill {\stackrel{^}{\theta }}_{n}& =& arg\underset{\theta \in \Theta }{max}\prod _{i=1}^{n}{P}_{\theta }\left({Y}_{i}\right)\hfill \\ & =& arg\underset{\theta \in \Theta }{max}\sum _{i=1}^{n}log{P}_{\theta }\left({Y}_{i}\right)\hfill \\ & =& arg\underset{\theta \in \Theta }{min}\sum _{i=1}^{n}-log{P}_{\theta }\left({Y}_{i}\right).\hfill \end{array}$

$\stackrel{^}{\theta }$ maximizes the expected log-likelihood. To see this, let's compare the expected log-likelihood of ${\theta }^{*}$ with any other $\theta \in \Theta$ .

$\begin{array}{ccc}\hfill E\left[log{P}_{{\theta }^{*}}\left(Y\right)-log{P}_{\theta }\left(Y\right)\right]& =& E\left[log,\frac{{P}_{{\theta }^{*}}\left(Y\right)}{{P}_{\theta }\left(Y\right)}\right]\hfill \\ & =& \int log\frac{{P}_{{\theta }^{*}}\left(y\right)}{{P}_{\theta }\left(y\right)}{P}_{{\theta }^{*}}\left(y\right)dy\hfill \\ & =& K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)\phantom{\rule{1.em}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}\text{KL}\phantom{\rule{4.pt}{0ex}}\text{divergence}\hfill \\ & \ge & 0\phantom{\rule{1.em}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}\text{equality}\phantom{\rule{4.pt}{0ex}}\text{iff}\phantom{\rule{0.166667em}{0ex}}{P}_{{\theta }^{*}}={P}_{\theta }.\hfill \end{array}$

Why?

$\begin{array}{ccc}\hfill -E\left[log,\frac{{P}_{{\theta }^{*}}\left(y\right)}{{P}_{\theta }\left(y\right)}\right]& =& E\left[log,\frac{{P}_{\theta }\left(y\right)}{{P}_{{\theta }^{*}}\left(y\right)}\right]\hfill \\ & \le & logE\left[\frac{{P}_{\theta }\left(y\right)}{{P}_{{\theta }^{*}}\left(y\right)}\right]\hfill \\ & =& log\int {P}_{\theta }\left(y\right)dy=0\hfill \\ & ⇒& K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)\ge 0\hfill \end{array}$

On the other hand, since ${\stackrel{^}{\theta }}_{n}$ maximizes the likelihood over $\theta \in \Theta$ , we have

$\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)}=\sum _{i=1}^{n}log{P}_{{\theta }^{*}}\left({Y}_{i}\right)-log{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)\le 0.$

Therefore,

$\frac{1}{n}\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)}-K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)+K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\le 0$

or re-arranging

$K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\le \left|\frac{1}{n},\sum _{i=1}^{n},log,\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)},-,K,\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\right|.$

Notice that the quantity

$\frac{1}{n}\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{\theta }\left({Y}_{i}\right)}$

is an empirical average whose mean is $K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)$ . By the law of large numbers, for each $\theta \in \Theta$ ,

$\left|\frac{1}{n},\sum _{i=1}^{n},log,\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{\theta }\left({Y}_{i}\right)},-,K,\left({P}_{\theta },{P}_{{\theta }^{*}}\right)\right|\stackrel{a.s.}{\to }0.$

If this also holds for the sequence $\left\{{\stackrel{^}{\theta }}_{n}\right\}$ , then we have

$K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\le \left|\frac{1}{n},\sum ,log,\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)},-,K,\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\right|\to 0\phantom{\rule{0.166667em}{0ex}}\mathrm{as}\phantom{\rule{0.166667em}{0ex}}n\to \infty$

which implies that

${P}_{{\stackrel{^}{\theta }}_{n}}\to {P}_{{\theta }^{*}}$

which often implies that

${\stackrel{^}{\theta }}_{n}\to {\theta }^{*}$

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

## Gaussian distributions

${P}_{{\theta }^{*}}\left(y\right)=\frac{1}{\sqrt{\pi }}{e}^{-{\left(y-{\theta }^{*}\right)}^{2}}$
$\Theta =\mathbb{R},\phantom{\rule{1.em}{0ex}}{\left\{{Y}_{i}\right\}}_{i=1}^{n}\stackrel{iid}{\sim }{P}_{{\theta }^{*}}\left(y\right)$
$\begin{array}{ccc}\hfill K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)& =& \int log\frac{{P}_{{\theta }^{*}}\left(y\right)}{{P}_{\theta }\left(y\right)}{P}_{{\theta }^{*}}\left(y\right)dy\hfill \\ & =& \int \left[{\left(y-\theta \right)}^{2}-{\left(y-{\theta }^{*}\right)}^{2}\right]{P}_{{\theta }^{*}}\left(y\right)dy\hfill \\ & =& {E}_{{\theta }^{*}}\left[{\left(y-\theta \right)}^{2}\right]-{E}_{{\theta }^{*}}\left[{\left(y-{\theta }^{*}\right)}^{2}\right]\hfill \\ & =& {E}_{{\theta }^{*}}\left[{Y}^{2}-2Y\theta +{\theta }^{2}\right]-1/2\hfill \\ & =& {\left({\theta }^{*}\right)}^{2}+1/2-2{\theta }^{*}\theta +{\theta }^{2}-1/2\hfill \\ & =& {\left({\theta }^{*}-\theta \right)}^{2}\hfill \end{array}$
$⇒{\theta }^{*}\phantom{\rule{0.166667em}{0ex}}\text{maximizes}\phantom{\rule{0.166667em}{0ex}}E\left[log{P}_{\theta }\left(Y\right)\right]\phantom{\rule{0.166667em}{0ex}}\text{wrt}\phantom{\rule{0.166667em}{0ex}}\theta \in \Theta$
$\begin{array}{ccc}\hfill {\stackrel{^}{\theta }}_{n}& =& arg\underset{\theta }{max}\left\{-\sum {\left({Y}_{i}-\theta \right)}^{2}\right\}\hfill \\ & =& arg\underset{\theta }{min}\left\{\sum {\left({Y}_{i}-\theta \right)}^{2}\right\}\hfill \\ & =& \frac{1}{n}\sum _{i=1}^{n}{Y}_{i}\hfill \end{array}$

## Hellinger distance

The KL divergence is not a distance function.

$K\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\ne K\left({P}_{{\theta }_{2}},{P}_{{\theta }_{1}}\right)$

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

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)={\left(\int ,{\left({P}_{{\theta }_{1}}^{\frac{1}{2}},-,{P}_{{\theta }_{2}}^{\frac{1}{2}}\right)}^{2},d,y\right)}^{\frac{1}{2}}.$

The Hellinger metric is symmetric, non-negative and

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=H\left({P}_{{\theta }_{2}},{P}_{{\theta }_{1}}\right)$

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

${H}^{2}\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\le K\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)$

## Proof:

$\begin{array}{ccc}\hfill H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)& =& \int {\left(\sqrt{{P}_{{\theta }_{1}}\left(y\right)},-,\sqrt{{P}_{{\theta }_{2}}\left(y\right)}\right)}^{2}dy\hfill \\ & =& \int {P}_{{\theta }_{1}}\left(y\right)dy+\int {P}_{{\theta }_{2}}\left(y\right)dy-2\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy\hfill \\ & =& 2-2\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy,\phantom{\rule{1.em}{0ex}}\text{since}\phantom{\rule{0.166667em}{0ex}}\int {P}_{\theta }\left(y\right)dy=1\phantom{\rule{0.166667em}{0ex}}\forall \theta \hfill \\ & =& 2\left(1,-,{E}_{{\theta }_{2}},\left[\sqrt{{P}_{{\theta }_{1}}\left(Y\right)/{P}_{{\theta }_{2}}\left(Y\right)}\right]\right)\hfill \\ & \le & 2log\left({E}_{{\theta }_{2}},\left[\sqrt{{P}_{{\theta }_{2}}\left(Y\right)/{P}_{{\theta }_{1}}\left(Y\right)}\right]\right),\phantom{\rule{1.em}{0ex}}\text{since}\phantom{\rule{0.166667em}{0ex}}1-x\le -logx\hfill \\ & \le & 2{E}_{{\theta }_{2}}\left[log,\sqrt{{P}_{{\theta }_{2}}\left(Y\right)/{P}_{{\theta }_{1}}\left(Y\right)}\right],\phantom{\rule{1.em}{0ex}}\text{by}\phantom{\rule{4.pt}{0ex}}\text{Jensen's}\phantom{\rule{4.pt}{0ex}}\text{inequality}\hfill \\ & =& {E}_{{\theta }_{2}}\left[log,\left(,{P}_{{\theta }_{2}},\left(Y\right),/,{P}_{{\theta }_{1}},\left(Y\right),\right)\right]\phantom{\rule{4pt}{0ex}}\equiv \phantom{\rule{4pt}{0ex}}K\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\hfill \end{array}$

Note that in the proof we also showed that

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=2\left(1,-,\int ,\sqrt{{P}_{{\theta }_{1}}\left(y\right)},\sqrt{{P}_{{\theta }_{2}}\left(y\right)},d,y\right)$

and using the fact $logx\le x-1$ again, we have

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\le -2log\left(\int ,\sqrt{{P}_{{\theta }_{1}}\left(y\right)},\sqrt{{P}_{{\theta }_{2}}\left(y\right)},d,y\right).$

The quantity inside the log is called the affinity between ${P}_{{\theta }_{1}}$ and ${P}_{{\theta }_{2}}$ :

$A\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy.$

This is another measure of closeness between ${P}_{{\theta }_{1}}$ and ${P}_{{\theta }_{2}}$ .

## Gaussian distributions

${P}_{\theta }\left(y\right)=\frac{1}{\pi }{e}^{-{\left(y-\theta \right)}^{2}}$
$\begin{array}{ccc}& & -2log\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy\hfill \\ & =& -2log\int {\left(\frac{1}{\sqrt{\pi }},{e}^{-{\left(y-{\theta }_{1}\right)}^{2}}\right)}^{\frac{1}{2}}{\left(\frac{1}{\sqrt{\pi }},{e}^{-{\left(y-{\theta }_{2}\right)}^{2}}\right)}^{\frac{1}{2}}dy\hfill \\ & =& -2log\left(\int ,\frac{1}{\sqrt{\pi }},{e}^{-\left[\frac{{\left(y-{\theta }_{1}\right)}^{2}}{2},+,\frac{{\left(y-{\theta }_{2}\right)}^{2}}{2}\right]},d,y\right)\hfill \\ & =& -2log\left(\int ,\frac{1}{\sqrt{\pi }},{e}^{-\left[{\left(y,-,\left(,\frac{{\theta }_{1}+{\theta }_{2}}{2},\right)\right)}^{2},+,{\left(\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)}^{2}\right]},d,y\right)\hfill \\ & =& -2log{e}^{-{\left(\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)}^{2}}\hfill \\ & =& \frac{1}{2}{\left({\theta }_{1}-{\theta }_{2}\right)}^{2}\hfill \end{array}$
$\begin{array}{ccc}& ⇒& -2logA\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=\frac{1}{2}{\left({\theta }_{1}-{\theta }_{2}\right)}^{2}\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{Gaussian}\phantom{\rule{4.pt}{0ex}}\text{distributions}\hfill \\ & ⇒& H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\le \frac{1}{2}{\left({\theta }_{1}-{\theta }_{2}\right)}^{2}\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{Gaussian}.\hfill \end{array}$

## Poisson distributions

If ${P}_{\theta }\left(y\right)={e}^{-\theta }\frac{{\theta }^{y}}{y!},\theta \ge 0$ , then

$-2logA\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)={\left(\sqrt{{\theta }_{1}}-\sqrt{{\theta }_{2}}\right)}^{2}.$

## Summary

${Y}_{i}\stackrel{iid}{\sim }{P}_{{\theta }^{*}}$
1. Maximum likelihood estimator maximizes the empirical average
$\frac{1}{n}\sum _{i=1}^{n}log{P}_{\theta }\left({Y}_{i}\right)$
(our empirical risk is negative log-likelihood)
2. ${\theta }^{*}$ maximizes the expectation
$E\left[\frac{1}{n},\sum _{i=1}^{n},log,{P}_{\theta },\left({Y}_{i}\right)\right]$
(the risk is the expected negative log-likelihood)
3. $\frac{1}{n}\sum _{i=1}^{n}log{P}_{\theta }\left({Y}_{i}\right)\stackrel{a.s.}{\to }E\left[\frac{1}{n},\sum _{i=1}^{n},log,{P}_{\theta },\left({Y}_{i}\right)\right]$
so we expect some sort of concentration of measure.
4. In particular, since
$\frac{1}{n}\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{\theta }\left({Y}_{i}\right)}\stackrel{a.s.}{\to }K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)$
we might expect that $K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\to 0$ for the sequence of estimates ${\left\{{P}_{{\stackrel{^}{\theta }}_{n}}\right\}}_{n=1}^{\infty }$ .

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.

are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
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
nanocopper obvius
Alexandre
what is the stm
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?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
Got questions? Join the online conversation and get instant answers!

#### Get Jobilize Job Search Mobile App in your pocket Now! By OpenStax By Marion Cabalfin By Mary Cohen By OpenStax By By Robert Morris By Madison Christian By OpenStax By OpenStax By Mackenzie Wilcox