0.4 An example of the use of sieves for complexity regularization

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Consider the following setting. Let

$Y={f}^{*}\left(X\right)+W,$

where X is a random variable (r.v.) on $\mathcal{X}=\left[0,1\right]$ , $W$ is a r.v. on $\mathcal{Y}=\mathbf{R},$ independent of $X$ and satisfying

$E\left[W\right]=0\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}E\left[{W}^{2}\right]={\sigma }^{2}<\infty .$

Finally let ${f}^{*}:\left[0,1\right]\to \mathbf{R}$ be a function satisfying

$|{f}^{*}\left(t\right)-{f}^{*}\left(s\right)|\le L|t-s|,\phantom{\rule{4pt}{0ex}}\forall t,s\in \left[0,1\right],$

where $L>0$ is a constant. A function satisfying condition [link] is said to be Lipschitz on $\left[0,1\right]$ . Notice that such a function must be continuous, but it is not necessarilydifferentiable. An example of such a function is depicted in [link] (a).

Note that

$\begin{array}{ccc}\hfill E\left[Y|X=x\right]& =& E\left[{f}^{*}\left(X\right)+W|X=x\right]\hfill \\ & =& E\left[{f}^{*}\left(x\right)+W|X=x\right]\hfill \\ & =& {f}^{*}\left(x\right)+E\left[W\right]={f}^{*}\left(x\right).\hfill \end{array}$

Consider our usual setup: Estimate ${f}^{*}$ using $n$ training examples

$\begin{array}{c}\hfill {\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{n}\stackrel{i.i.d.}{\sim }{P}_{XY},\\ \hfill {Y}_{i}={f}^{*}\left({X}_{i}\right)+{W}_{i},\phantom{\rule{4pt}{0ex}}i=\left\{1,...,n\right\},\end{array}$

where $\stackrel{i.i.d.}{\sim }$ means independently and identically distributed . [link] (a) illustrates this setup.

In many applications we can sample $\mathcal{X}=\left[0,1\right]$ as we like, and not necessarily at random. For example we can take $n$ samples uniformly on [0,1]

$\begin{array}{ccc}\hfill {x}_{i}& =& \frac{i}{n},\phantom{\rule{4pt}{0ex}}i=1,...,n\phantom{\rule{0.277778em}{0ex}},\hfill \\ \hfill {Y}_{i}& =& f\left({x}_{i}\right)+{W}_{i}\hfill \\ & =& f\left(\frac{i}{n}\right)+{W}_{i}.\hfill \end{array}$

We will proceed with this setup (as in [link] (b)) in the rest of the lecture.

Our goal is to find ${\stackrel{^}{f}}_{n}$ such that $E\left[\parallel {f}^{*}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]\to 0,\phantom{\rule{4.pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}n\to 0$ (here $\parallel ·\parallel$ is the usual ${L}_{2}$ -norm; i.e., $\parallel {f}^{*}-{\stackrel{^}{f}}_{n}{\parallel }^{2}={\int }_{0}^{1}{|{f}^{*}\left(t\right)-{\stackrel{^}{f}}_{n}\left(t\right)|}^{2}dt$ ).

Let

$\mathcal{F}=\left\{f:\phantom{\rule{4.pt}{0ex}}f\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}\text{Lipschitz}\phantom{\rule{4.pt}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}\text{constant}\phantom{\rule{4.pt}{0ex}}L\right\}.$

The Riskis defined as

$R\left(f\right)=\parallel {f}^{*}{-f\parallel }^{2}={\int }_{0}^{1}{|{f}^{*}\left(t\right)-f\left(t\right)|}^{2}dt.$

The Expected Risk (recall that ourestimator ${\stackrel{^}{f}}_{n}$ is based on $\left\{{x}_{i},{Y}_{i}\right\}$ and hence is a r.v.) is defined as

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]=E\left[\parallel {f}^{*}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right].$

Finally the Empirical Riskis defined as

${\stackrel{^}{R}}_{n}\left(f\right)=\frac{1}{n}\sum _{i=1}^{n}{\left(f,\left(\frac{i}{n}\right),-,{Y}_{i}\right)}^{2}.$

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 ${\mathcal{F}}_{1},\phantom{\rule{0.277778em}{0ex}}{\mathcal{F}}_{2},\phantom{\rule{0.277778em}{0ex}}{\mathcal{F}}_{3},\phantom{\rule{0.277778em}{0ex}}...,$

${\mathcal{F}}_{n}=\left\{f,:,f,\left(t\right),=,\sum _{j=1}^{{m}_{n}},{c}_{j},\phantom{\rule{0.166667em}{0ex}},{\mathbf{1}}_{\left\{\frac{j-1}{{m}_{n}}\le t<\frac{j}{{m}_{n}}\right\}},,,\phantom{\rule{4pt}{0ex}},{c}_{j},\in ,\mathbf{R}\right\}.$

${\mathcal{F}}_{n}$ is the space of functions that are constant on intervals

${\text{I}}_{j,{m}_{n}}\equiv \left[\frac{j-1}{{m}_{n}},,,\frac{j}{{m}_{n}}\right),\phantom{\rule{4pt}{0ex}}j=1,...,{m}_{n}.$

From here on we will use $m$ and $k$ instead of ${m}_{n}$ and ${k}_{n}$ (dropping the subscript $n$ ) for notational ease. Define

${f}_{n}\left(t\right)=\sum _{j=1}^{m}{c}_{j}^{*}\phantom{\rule{0.166667em}{0ex}}{\mathbf{1}}_{\left\{t\in {I}_{j,m}\right\}},\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{where}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}{c}_{j}^{*}=\frac{1}{k}\sum _{i:\frac{i}{n}\in {I}_{j,m}}{f}^{*}\left(\frac{i}{n}\right).$

Note that ${f}_{n}\in {\mathcal{F}}_{n}$ .

Exercise 1

Upper bound $\parallel {f}^{*}-{f}_{n}{\parallel }^{2}$ .

$\begin{array}{ccc}\hfill \parallel {f}^{*}{-f\parallel }^{2}& =& {\int }_{0}^{1}{|{f}^{*}\left(t\right)-{f}_{n}\left(t\right)|}^{2}dt\hfill \\ & =& \sum _{j=1}^{m}{\int }_{{I}_{j,m}}{|{f}^{*}\left(t\right)-{f}_{n}\left(t\right)|}^{2}dt\hfill \\ & =& \sum _{j=1}^{m}{\int }_{{I}_{j,m}}{|{f}^{*}\left(t\right)-{c}_{j}^{*}|}^{2}dt\hfill \\ & =& \sum _{j=1}^{m}{\int }_{{I}_{j,m}}{\left|{f}^{*},\left(t\right),-,\frac{1}{k},\sum _{i:\frac{i}{n}\in {I}_{j,m}},{f}^{*},\left(\frac{i}{n}\right)\right|}^{2}dt\hfill \\ & =& \sum _{j=1}^{m}{\int }_{{I}_{j,m}}{\left(\frac{1}{k},\left|\sum _{i:\frac{i}{n}\in {I}_{j,m}},\left({f}^{*},\left(t\right),-,{f}^{*},\left(\frac{i}{n}\right)\right)\right|\right)}^{2}dt\hfill \\ & \le & \sum _{j=1}^{m}{\int }_{{I}_{j,m}}{\left(\frac{1}{k},\sum _{i:\frac{i}{n}\in {I}_{j,m}},\left|{f}^{*},\left(t\right),-,{f}^{*},\left(\frac{i}{n}\right)\right|\right)}^{2}dt\hfill \\ & \le & \sum _{j=1}^{m}{\int }_{{I}_{j,m}}{\left(\frac{1}{k},\sum _{i:\frac{i}{n}\in {I}_{j,m}},\frac{L}{m}\right)}^{2}dt\hfill \\ & =& \sum _{j=1}^{m}{\int }_{{I}_{j,m}}{\left(\frac{L}{m}\right)}^{2}dt\hfill \\ & =& \sum _{j=1}^{m}\frac{1}{m}{\left(\frac{L}{m}\right)}^{2}={\left(\frac{L}{m}\right)}^{2}.\hfill \end{array}$

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 $f\in {\mathcal{F}}_{n},$ $f={\sum }_{j=1}^{m}{c}_{j}\phantom{\rule{0.166667em}{0ex}}{\mathbf{1}}_{\left\{t\in {I}_{j,m}\right\}},$ we have

${\stackrel{^}{R}}_{n}\left(f\right)=\frac{1}{n}\sum _{j=1}^{m}\left(\sum _{i:\frac{i}{n}\in {I}_{j,m}},{\left({c}_{j}-{Y}_{i}\right)}^{2}\right).$

Let ${\stackrel{^}{f}}_{n}=arg{min}_{f\in {\mathcal{F}}_{n}}{\stackrel{^}{R}}_{n}\left(f\right).$ Then

${\stackrel{^}{f}}_{n}\left(t\right)=\sum _{j=1}^{m}{\stackrel{^}{c}}_{j}\phantom{\rule{0.166667em}{0ex}}{\mathbf{1}}_{\left\{t\in {I}_{j,m}\right\}},\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\text{where}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4.pt}{0ex}}{\stackrel{^}{c}}_{j}=\frac{1}{k}\sum _{i:\frac{i}{n}\in {I}_{j,m}}{Y}_{i}$

Exercise 2

Note that $E\left[{\stackrel{^}{c}}_{j}\right]={c}_{j}^{*}$ and therefore $E\left[{\stackrel{^}{f}}_{n}\left(t\right)\right]={f}_{n}\left(t\right)$ . Lets analyze now the expected risk of ${\stackrel{^}{f}}_{n}$ :

$\begin{array}{ccc}\hfill E\left[\parallel {f}^{*}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]& =& E\left[\parallel {f}^{*}-{f}_{n}+{f}_{n}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]\hfill \\ & =& \parallel {f}^{*}-{f}_{n}{\parallel }^{2}+E\left[\parallel {f}_{n}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]+2E\left[⟨{f}^{*}-{f}_{n},{f}_{n}-{\stackrel{^}{f}}_{n}⟩\right]\hfill \\ & =& \parallel {f}^{*}-{f}_{n}{\parallel }^{2}+E\left[\parallel {f}_{n}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]+2⟨{f}^{*}-{f}_{n},E\left[{f}_{n}-{\stackrel{^}{f}}_{n}\right]⟩\hfill \\ & =& \parallel {f}^{*}-{f}_{n}{\parallel }^{2}+E\left[\parallel {f}_{n}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right],\hfill \end{array}$

where the final step follows from the fact that $E\left[{\stackrel{^}{f}}_{n}\left(t\right)\right]={f}_{n}\left(t\right)$ . 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 ${\stackrel{^}{f}}_{n}$ . The second term, $E\left[\parallel {f}_{n}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]$ , 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:

$\begin{array}{ccc}\hfill E\left[\parallel {f}_{n}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]& =& E\left[{\int }_{0}^{1},{|{f}_{n}\left(t\right)-{\stackrel{^}{f}}_{n}\left(t\right)|}^{2},d,t\right]\hfill \\ & =& E\left[\sum _{j=1}^{m},{\int }_{{I}_{j,m}},{|{c}_{j}^{*}-{\stackrel{^}{c}}_{j}|}^{2},d,t\right]\hfill \\ & =& \sum _{j=1}^{m}{\int }_{{I}_{j,m}}E\left[|{c}_{j}^{*}-{\stackrel{^}{c}}_{j}{|}^{2}\right]\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& \sum _{j=1}^{m}{\int }_{{I}_{j,m}}\frac{E\left[{W}^{2}\right]}{k}\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & \le & \sum _{j=1}^{m}{\int }_{{I}_{j,m}}\frac{{\sigma }^{2}}{k}\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& \sum _{j=1}^{m}\frac{1}{m}\frac{{\sigma }^{2}}{k}=\frac{{\sigma }^{2}}{k}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\frac{m}{n}\phantom{\rule{0.166667em}{0ex}}{\sigma }^{2}\hfill \end{array}.$

Combining all the facts derived we have

$E\left[\parallel {f}^{*}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]\phantom{\rule{4pt}{0ex}}\le \phantom{\rule{4pt}{0ex}}\frac{{L}^{2}}{{m}^{2}}\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}\frac{m}{n}\phantom{\rule{0.166667em}{0ex}}{\sigma }^{2}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}O\left(max,\left\{\frac{1}{{m}^{2}},,,\frac{m}{n}\right\}\right).$

This equation used Big-O notation.

What is the best choice of $m$ ? If $m$ is small then the approximation error ( i.e., $O\left(1/{m}^{2}\right)$ ) is going to be large, but the estimation error ( i.e., $O\left(m/n\right)$ ) 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

$E\left[\parallel {f}_{n}-{\stackrel{^}{f}}_{n}{\parallel }^{2}\right]=O\left({n}^{-2/3}\right).$

So the sieve ${\mathcal{F}}_{1},\phantom{\rule{0.277778em}{0ex}}{\mathcal{F}}_{2},\phantom{\rule{0.277778em}{0ex}}\cdots$ with

${\mathcal{F}}_{n}=\left\{f,:,f,\left(t\right),=,\sum _{j=1}^{{m}_{n}},{c}_{j},\phantom{\rule{0.166667em}{0ex}},{\mathbf{1}}_{\left\{\frac{j-1}{{m}_{n}}\le t<\frac{j}{{m}_{n}}\right\}},,,\phantom{\rule{4pt}{0ex}},{c}_{j},\in ,\mathbf{R}\right\},$

produces a $\mathcal{F}$ -consistent estimator for ${f}^{*}=E\left[Y|X+x\right]\in \mathcal{F}$ .

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.

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