where
$L>0$ is a constant. A function satisfying condition
[link] is said to be Lipschitz on
$[0,1]$ . Notice
that such a function must be continuous, but it is not necessarilydifferentiable. An example of such a function is depicted in
[link] (a).
where
$\stackrel{i.i.d.}{\sim}$ means
independently and identically
distributed .
[link] (a) illustrates this
setup.
In many applications we can sample
$\mathcal{X}=[0,1]$ as we like, and not
necessarily at random. For example we can take
$n$ samples
uniformly on [0,1]
We will proceed with this setup (as in
[link] (b)) in the rest of the lecture.
Our goal is to find
${\widehat{f}}_{n}$ such that
$E[\parallel {f}^{*}-{\widehat{f}}_{n}{\parallel}^{2}]\to 0,\phantom{\rule{4.pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}n\to 0$ (here
$\parallel \xb7\parallel $ is the usual
${L}_{2}$ -norm; i.e.,
$\parallel {f}^{*}-{\widehat{f}}_{n}{\parallel}^{2}={\int}_{0}^{1}{|{f}^{*}\left(t\right)-{\widehat{f}}_{n}\left(t\right)|}^{2}dt$ ).
Let
$0<{m}_{1}\le {m}_{2}\le {m}_{3}\le \cdots $ be a sequence of integers
satisfying
${m}_{n}\to \infty $ as
$n\to \infty $ , and
${k}_{n}{m}_{n}=n$ for some integer
${k}_{n}>0$ . That is, for each value of
$n$ there is an associated integer value
${m}_{n}$ . Define the Sieve
${\mathcal{F}}_{1},\phantom{\rule{0.277778em}{0ex}}{\mathcal{F}}_{2},\phantom{\rule{0.277778em}{0ex}}{\mathcal{F}}_{3},\phantom{\rule{0.277778em}{0ex}}...,$
The above implies that
$\parallel {f}^{*}-{f}_{n}{\parallel}^{2}\to 0\phantom{\rule{4.pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}n\to \infty ,$ since
$m={m}_{n}\to \infty \phantom{\rule{4.pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}n\to \infty $ . In words, with
$n$ sufficiently large we can
approximate
${f}^{*}$ to arbitrary accuracy using models in
${\mathcal{F}}_{n}$ (even if the functions we are using to approximate
${f}^{*}$ are not
Lipschitz!).
For any
$f\in {\mathcal{F}}_{n},$$f={\sum}_{j=1}^{m}{c}_{j}\phantom{\rule{0.166667em}{0ex}}{\mathbf{1}}_{\{t\in {I}_{j,m}\}},$ we have
Note that
$E\left[{\widehat{c}}_{j}\right]={c}_{j}^{*}$ and therefore
$E\left[{\widehat{f}}_{n}\left(t\right)\right]={f}_{n}\left(t\right)$ . Lets analyze now the expected risk of
${\widehat{f}}_{n}$ :
where the final step follows from the fact that
$E\left[{\widehat{f}}_{n}\left(t\right)\right]={f}_{n}\left(t\right)$ . A couple of important remarks pertaining
the right-hand-side of equation
[link] : The
first term,
$\parallel {f}^{*}-{f}_{n}{\parallel}^{2}$ , corresponds to the approximation
error, and indicates how well can we approximate the function
${f}^{*}$ with a function from
${\mathcal{F}}_{n}$ . Clearly, the larger the class
${\mathcal{F}}_{n}$ is, the smallest we can make this term. This term is
precisely the squared bias of the estimator
${\widehat{f}}_{n}$ . The second
term,
$E[\parallel {f}_{n}-{\widehat{f}}_{n}{\parallel}^{2}]$ , is the estimation error, the
variance of our estimator. We will see that the estimation erroris small if the class of possible estimators
${\mathcal{F}}_{n}$ is also small.
The behavior of the first term in
[link] was
already studied. Consider the other term:
What is the best choice of
$m$ ? If
$m$ is small then the approximation
error (
i.e.,$O(1/{m}^{2})$ ) is going to be large, but the estimation error
(
i.e.,$O(m/n)$ ) is going to be small, and vice-versa. This two
conflicting goals provide a tradeoff that directs our choice of
$m$ (as a function of
$n$ ). In
[link] we depict
this tradeoff. In
[link] (a) we considered a
large
${m}_{n}$ value, and we see that the approximation of
${f}^{*}$ by a
function in the class
${\mathcal{F}}_{n}$ can be very accurate (that is, our
estimate will have a small bias), but when we use the measured dataour estimate looks very bad (high variance). On the other hand, as
illustrated in
[link] (b), using a very small
${m}_{n}$ allows our estimator to get very close to the best approximating
function in the class
${\mathcal{F}}_{n}$ , so we have a low variance estimator, but
the bias of our estimator (
i.e., the difference between
${f}_{n}$ and
${f}^{*}$ )
is quite considerable.
We need to balance the two terms in the right-hand-side of
[link] in order to maximize the rate of decay (with
$n$ ) of the expected risk. This implies that
$\frac{1}{{m}^{2}}=\frac{m}{n}$ therefore
${m}_{n}={n}^{1/3}$ and the Mean
Squared Error (MSE) is
produces a
$\mathcal{F}$ -consistent
estimator for
${f}^{*}=E\left[Y\right|X+x]\in \mathcal{F}$ .
It is interesting to note that the rate of decay of the MSE we obtainwith this strategy cannot be further improved by using more
sophisticated estimation techniques (that is,
${n}^{-2/3}$ is the
minimax MSE rate for this problem). Also, rather surprisingly, we
are considering classes of models
${\mathcal{F}}_{n}$ that are actually not
Lipschitz, therefore our estimator of
${f}^{*}$ is not a Lipschitz
function, unlike
${f}^{*}$ itself.
Questions & Answers
Is there any normative that regulates the use of silver nanoparticles?
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
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?