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
what is variations in raman spectra for nanomaterials
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
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?