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\begin{document}

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\lecture{10}{October 17, 1994}{Ron Rivest}{Mikael Goldmann}
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\section{Outline}

\begin{itemize}
\item A lower bound on sample complexity through VC-dimension.
\item Cover's coin problem.
\end{itemize}

\section{A lower bound on the sample complexity through VC-dimension}
In previous lectures it has been shown that the VC-dimension
of a concept class gives an upper bound on the number of
samples needed to PAC-learn concepts from the class. In 
particular it has been shown that for VC-dimension $d$ it
is possible to PAC-learn with parameters $\delta$ and $\epsilon$
given that the number of samples $m$ is at least
\begin{equation}\label{upper}
m\geq c_0\left(\frac{d}\epsilon \log\frac1\epsilon +
\frac1\epsilon\log\frac1\delta\right) 
\end{equation}
where $c_0$ is a fixed constant.
So, as long as $\mbox{VC-dim}({\cal C})$ is finite, it is
possible to PAC-learn concepts from ${\cal C}$ even though
$|{\cal C}|$ might be infinite. 

One might ask if there are concept classes
of infinite VC-dimension that can be learned from finite sample size,
or if at least the upper bound of (\ref{upper}) can be improved in any
substantial way to, say, $O(d\log(1/\epsilon))$. The answer
to both these questions is ``No.'' 
It is pretty clear that one cannot PAC-learn a concept class ${\cal
C}$ of
infinite VC-dimension. Say that we have an algorithm that makes $m$
queries to the examples oracle ($m$ depends only on $\delta$ and
$\epsilon$), and let $Y$ be a subset of the domain
of size $\geq 2m$ that is 
shattered by ${\cal C}$. After $m$ examples $A$ knows nothing about
at least half the points in $Y$, and therefore it cannot predict the
labels on those points better than a random coin toss.
In fact, for finite VC-dimension equation (\ref{upper}) is nearly
tight, and we will show that
for VC-dimension $d$, achieving error rate $\epsilon$ requires
$c_1 d/\epsilon$ samples for some fixed constant $c_1$. 

Let $X=\{x_1,\ldots,x_d\}$ and let ${\cal C}=2^X$. Obviously,
$\VC({\cal C})=d$. We will show that PAC-learning ${\cal C}$ requires 
$\Omega(d/\epsilon)$ examples. 
\begin{theorem}\label{VCD-LB}
Let $X=\{x_1,\ldots,x_d\}$ and let ${\cal C}=2^X$.
Any algorithm $A$ that PAC-learns  ${\cal C}$ with parameters $\epsilon$
and $\delta\leq 1/15$ must use more than $(d-1)/(64\epsilon)$
queries to the examples oracle.
\end{theorem}
This is in fact a general result since
PAC-learning ${\cal C}$ is a sub-problem of PAC-learning any concept class
of VC-dimension $d$.
\begin{corollary}
Let ${\cal C}$ be an arbitrary concept class and let $d=\VC({\cal
C})$. 
To PAC-learn ${\cal C}$ with parameters $\epsilon$ and
$\delta\leq1/15$ requires more than
$(d-1)/(64\epsilon)$ queries to the examples 
oracle.
\end{corollary}

\begin{proofof}{of Theorem \ref{VCD-LB}}
Let $m=(d-1)/(64\epsilon)$, and $A$ be an algorithm that makes at most
$m$ queries to the examples oracle
and then produces a hypothesis $h$.
We need to show that there are a distribution $p$ on $X$ 
and a concept $c\in{\cal C}$ such that $er(h)>\epsilon$ with
probability at least $1/15$. 

Let us first fix $p$ independently of $A$:
\begin{eqnarray*}
&&p(x_1) = 1-16\epsilon \\
&&p(x_2) = p(x_3) =\cdots =p(x_d)=\frac{16\epsilon}{d-1}
\end{eqnarray*}
In what follows $S$ is a random sample of $X$ that we get by $m$ 
independent draws from $p$. 
We want to establish that there is a $c$ so that
$\Pr_S[er(h)>\epsilon] > \frac1{15}$.

Let $X'=\{x_2,\ldots,x_d\}$. For any fixed $c\in{\cal C}$ and
hypothesis $h$, let
$er'(h)=\Pr[c(x)\neq h(x) \land x\in X']$. 
For technical reasons, it is easier to prove that
$\Pr_S[er'(h)>\epsilon] > 1/15$, which is enough 
since $er'(h)\leq er(h)$.

We pick a random $c\in{\cal C}$ and show that with positive
probability $c$ is hard to learn for $A$, thereby showing that there
must be some fixed $c$ that is hard to learn for $A$.

Let $B$ be the event that $A$'s sample $S$ contains 
less than $(d-1)/2$
points in $X'$.
We establish three things:
\begin{eqnarray}
\Pr\nolimits_S[B] &\geq& 1/2 \label{prB}\\
\Exp_{c,S}[\,er'(h)\,\mid\,B\,] & > &4\epsilon\label{erhlower}\\
\Exp_{S}[\,er'(h)\,\mid\,er'(h)>\epsilon\,] &\leq& 16\epsilon 
\mbox{~~for any fixed $c$.} \label{erhupper}
\end{eqnarray}
Inequality
(\ref{erhupper}) is obvious since for any $h$ we have
$er'(h)\leq\Pr[x\in X'] = 16\epsilon$. However, if we consider $er(h)$
rather than $er'(h)$, then (\ref{erhupper}) does not necessarily hold,
which is the reason for introducing $er'(h)$. 
Both (\ref{prB}) and (\ref{erhlower}) make intuitive sense:
if $A$ only sees a few samples it is likely to miss a large fraction
of $X'$ (this is (\ref{prB})), and if $A$ knows nothing about a
large fraction of $X'$ then the error rate of the hypothesis is likely
to be high (this is (\ref{erhlower})).
Before formally proving (\ref{prB}) and (\ref{erhlower})
we show that they imply that there is a $c_*\in{\cal C}$ so that
$\Pr_S[er'(h)>\epsilon]>1/15$.

First we use (\ref{prB}) and (\ref{erhlower}) to get a lower bound on 
$\Exp_{c,S}[er'(h)]$.
\[
\Exp_{c,S}[er'(h)] \geq 
\Pr_S[B]\cdot\Exp_{c,S}[\,er'(h)\,\mid\,B\,]
>\frac12\cdot 4\epsilon =2\epsilon.
\]
In particular, there is some $c_*\in{\cal C}$
 such that $\Exp_{S}[er'(h)]> 2\epsilon$.
We take $c_*$ as the target concept and show that $A$ is likely to
produce a hypothesis with high error rate.
Using conditional expectation we get
\begin{eqnarray*}
2\epsilon &<& \Exp_{S}[er'(h)] \\
&=& \Pr\nolimits_S[er'(h)>\epsilon]\cdot 
\Exp_{S}[\,er'(h)\,\mid\,er'(h)>\epsilon\,] \\
& & +  (1-\Pr\nolimits_S[er'(h)>\epsilon] )\cdot
    \Exp_{S}[\,er'(h)\,\mid\,er'(h)\leq\epsilon\,]. 
\end{eqnarray*}
Next we apply (\ref{erhupper}) to get 
\begin{eqnarray*}
2\epsilon < \Exp_S[er'(h)]&\leq& 
\Pr\nolimits_S[er'(h)>\epsilon]\cdot 16\epsilon +
(1-\Pr\nolimits_S[er'(h)>\epsilon])\cdot\epsilon\\
&=& 15\epsilon\Pr\nolimits_S[er'(h)>\epsilon]+\epsilon,
\end{eqnarray*}
which implies $\Pr_S[er'(h)>\epsilon] > 1/15$.

It remains to prove (\ref{prB}) and
(\ref{erhlower}).

{\bf Proof of (\ref{prB}).}
Let $T$ be the number of
times $A$ gets a sample point in $X'$. Clearly, $\Exp[T] = 16\epsilon
m = (d-1)/4$. We have $\Pr_S[B]  \geq 1- \Pr[T\geq (d-1)/2]$, and
by Markov's inequality,\footnote{Markov's inequality: if $X$ is a
positive random variable, then $\Pr[X\geq t]\leq \Exp[X]/t$.}
$\Pr[T\geq (d-1)/2]\leq 1/2$.

{\bf Proof of (\ref{erhlower}).}
Let $S$ be the set of sample points that $A$ gets. Choosing a random
$c$ is equivalent to flipping a fair coin for each point in $X$ to
determine its label. Since $h$ is independent of the labeling of
$X'- S$, the contribution to $er'(h)$ is expected to be
$16\epsilon/(2(d-1))$ for each point in $X'- S$.
When $B$ occurs, we have $|X'- S|>(d-1)/2$ and thus the
expected value of $er'(h)$ given $B$ is strictly greater than
$4\epsilon$. 


The proof is now complete.
\end{proofof}

\section{Cover's coin problem}

Cover's coin problem is somewhat connected to the idea of ``learning
the bias of a coin.'' The precise statement of the problem is as
follows:

{\bf Cover's coin problem}\\
INPUT: A coin.\\
OUTPUT: ``rational'' if $\Pr[\mbox{heads}]$ is rational and
``irrational'' otherwise.

\smallskip
At first glance the question might look strange, especially
since the maximum likelihood estimate $\bar p$ of
$p=\Pr[\mbox{heads}]$ 
after
a finite number of coin flips is always rational ($\bar
p=\mbox{\#heads}/\mbox{\#flips}$). On the face of it, the problem
appears to be impossible. In order to get a better understanding,
let us look at a simpler problem and see in what sense it can be solved.

{\bf Fair coin problem}\\
INPUT: A coin.\\
OUTPUT: ``fair'' if $\Pr[\mbox{heads}]=1/2$ and
``biased'' otherwise.

It is possible to devise an algorithm $F$ for the ``Fair coin
problem'' that flips the coin a few times, guesses ``fair'' or
``biased,'' then
flips again and guesses, and so on. With probability~1, the infinite 
sequence of guesses will contain only a finite number of mistakes.
\begin{center}
\fbox{\begin{minipage}{\textwidth}
\begin{tabbing}
F:~~\=\\
\> For \=$i=1,2,3,\ldots$\\
\>\>$\delta_i = 2^{-i}$\\
\>\> $N_i=4^i i$\\
\>\>Flip $N_i$ times and let $H$ be the number of ``heads'' observed\\
\>\>If $|H-N_i/2|>\delta_iN_i$ then output ``fair'' else output ``biased''\\
\> End for
\end{tabbing}
\end{minipage}}
\end{center}
Consider first the case that the coin is fair, and let $B_i$ be the
event that the $i^{\mbox{th}}$ output is ``biased.'' Using the
additive version of the Chernoff bounds we get
\[
\Pr[B_i]=\Pr\bigl[|H-N_i/2|>\delta_iN_i\bigr] \leq
2e^{-2N_i\delta_i^2} = 2e^{-2i}.
\]
The following result is very useful when studying the behavior of $F$.
\begin{theorem}[Borel--Cantelli lemmas]
Let $A_1,A_2,\ldots$ be a countable sequence of events, and let 
$A=\cap_{n\geq 1}\cup_{m\geq n} A_i$ be the event that infinitely many
of the $A_i$'s occur. Then
\begin{eqnarray*}
\Pr[A] = 0 &\mbox{if}& \sum_{i\geq 1}\Pr[A_i]<\infty,\\
\Pr[A] = 1 &\mbox{if}& \sum_{i\geq 1}\Pr[A_i]<\infty
\mbox{~~and $A_1,A_2,\ldots$ are independent events.}
\end{eqnarray*}
\end{theorem}
In our case $\sum_{i\geq
1}\Pr[B_i]<\infty$, which implies that $F$ ouputs ``fair'' all
but finitely often.

The second case is when the probability of heads is $p\neq1/2$.
In this case, with probability~$1$, it only happens finitely many times
that $H/N_i \not\in [p-\delta_j,p+\delta_j]$. When $j$ is sufficiently
large, the intervals $[p-\delta_j,p+\delta_j]$ and 
$[\frac12-\delta_j,\frac12+\delta_j]$ are disjoint, so $F$ will say
``fair'' only finitely many times with probability~$1$.

Let us now return to Cover's coin problem. We will construct an
algorithm with the following properties. If $p=\Pr[\mbox{heads}]$ is
rational, then with probability~$1$ the algorithm will output
``irrational'' only a finite number of times. For all irrational $p$,
except for a set of measure~0, the algorithm will output ``rational''
only finitely many times with probability~$1$. 

What we would like to do
is to put shrinking intervals around all rational numbers, but these
intervals would of course cover $[0,1]$ and any irrational number
would always be close to {\em some\/} rational number. The key is to
consider a slowly growing, but always finite, set of rationals.
If $p$ is rational, then it will eventually become a member of the
set, and with probability~$1$ the empiric average $\bar p$ will be
close to $p$ all but finitely often. 
On the other hand, If $p$ is
irrational, the union of the intervals around the rationals in the
finite set shrinks in size, and eventually $\bar p$ should always fall
outside this union. This is almost what happens. The problem is that
while the union of intervals has smaller and smaller measure, we keep
adding more rationals to our finite set, and therefore an irrational
number that is not in an interval at some point may be too close to a
rational number that is about to be added to the set.

While the ideas are the same as in the ``Fair coin problem,'' the
technical details are a bit more involved.
Let $q_1,q_2,\ldots$ be an arbitrary enumeration of $\Q\cap[0,1]$, and
$\foo_k$ be $\{q_1,q_2,\ldots,q_k\}$. 
Define
\begin{eqnarray*}
\foo_k(\delta) &=& \cup_{i=1}^k [q_i-\delta,q_i+\delta]\\
N_0 &=& 0\\
N_k &=& 2^k\\
\delta_k &=&\sqrt{(\ln k)/2^k}
\end{eqnarray*}


\begin{center}
\fbox{\begin{minipage}{\textwidth}
\begin{tabbing}
G:~~\=\\
\> $h=0$\\
\> $n=0$ \\
\> $\mbox{\it guess}=\mbox{\rm``rational''}$\\
\> For\=~$k=1$ to $\infty$\\
\>\> For\=~$n=N_{k-1}+1$ to $N_k$\\
\>\>\> $h_n= \mbox{\rm flip}$\\
\>\>\> $h=h+h_n$\\
\>\>\> $\bar p_n = h/n$\\
\>\>\> output \mbox{\it guess}\\
\>\> End for\\
\>\>If $\bar p_{N_k} \in \foo_{k}(\delta_{k})$ 
then $\mbox{\it guess}=\mbox{\rm``rational''}$
else $\mbox{\it guess}=\mbox{\rm``irrational''}$\\
%\>\>\>then $\mbox{\it guess}=\mbox{\rm``rational''}$\\
%\>\>\>else $\mbox{\it guess}=\mbox{\rm``irrational''}$\\
\>End for
\end{tabbing}
\end{minipage}}
\end{center}

The following lemma is useful when analyzing $G$.
\begin{lemma}\label{p-delta}
If $p=\Pr[\mbox{\rm heads}]$ then 
$\bar p_{N_k}\in [p-\delta_{k},p+\delta_{k}]$ for all but finitely
many $k$ with probability~$1$.
\end{lemma}
\begin{proof}
Let $A_k$ be the event that 
$p_{N_k}\not\in [p-\delta_{N_k},p+\delta_{N_k}]$. Using the additive
version of the Chernoff bounds we 
\[
\Pr[A_k] \leq 2e^{-2N_k\delta^2_{k}} = 2/k^2.
\]
Therefore $\sum_{k\geq 1}\Pr[A_k]<\infty$, and the lemma follows
by the Borel--Cantelli lemmas.
\end{proof}

We consider three cases.


{\bf\boldmath Case~1: $p$ is rational.} There is an
$l$ such that $p=q_{l}\in\foo_k$ for all $k\geq l$. By
Lemma~\ref{p-delta} $\bar p_k \in [q_l-\delta_k,q_l+\delta_k]$ for all
but finitely many $k$ with probability~$1$, and
$[q_l-\delta_k,q_l+\delta_k]\subset \foo_k(\delta_k)$ for $k\geq l$. This
implies that $G$ will output ``rational'' all but finitely often with
probability~$1$. 


{\bf\boldmath Case~2: $p\in \foo_k(2\delta_k)$ for finitely many $k$.}
In this case there is some $l$ so that 
$[p-\delta_k,p+\delta_k]\cap \foo_k(\delta_k)=\emptyset$ for $k\geq l$.
Together with Lemma~\ref{p-delta} this implies that the output will be
``irrational'' all but finitely often with probability~$1$.


{\bf\boldmath Case~3: $p$ is irrational, but $p\in \foo_k(2\delta_k)$ for
infinitely many $k$.} This case is bad since the output may be 
``rational'' infinitely often. However, we will show that the Lebesgue
measure of the set of such $p$ is~0.

First note that
\[
\{\,p\,\mid\,\mbox{$p\in \foo_k(2\delta_k)$ for infinitely many $k$}\,\}
\subset\bigcup_{k\geq m} \foo_k(2\delta_k) \mbox{~~for all $m\geq0$.}
\]
Since $\lambda(\foo_k(2\delta_k))\leq 4k\delta_k$, we have
\[
\lambda\left(\bigcup_{k\geq m} \foo_k(2\delta_k)\right) \leq
\sum_{k\geq m} 4k\delta_k.
\]
The sum $\sum_{k\geq 1} k\delta_k$ converges, so
$\sum_{k\geq m}4k\delta_k$ 
tends to~$0$ as $m$ tends to $\infty$. This implies that
$\lambda\{\,p\,\mid\,
\mbox{$p\in \foo_k(2\delta_k)$ for infinitely many $k$}\,\} = 0$.

\begin{thebibliography}{99}
\bibitem{Cov73}
T.~M.~Cover, On determining the irrationality of the mean of a random
variable. {\em The Annals of Satistics}, 1(5):862--871, 1973.

\bibitem{EHKV33}
A.~Ehrenfeucht, D.~Haussler, M.~Kearns, and L.~Valiant.
A general lower bound on the number of examples needed for learning. 
{\em Information and Computation}, 82(3):247--251, 1989.

\bibitem{KeaVaz94} M.~Kearns and U.~Vazirani.  {\em An Introduction to 
Computational Learning Theory}, 52--64, 1994.
\end{thebibliography}

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