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.
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
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Rafiq
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analytical skills graphene is prepared to kill any type viruses .
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Damian
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