% This chapter was modified on 5/17/03.
%\setcounter{chapter}{7}
\chapter{Law of Large Numbers}\label{chp 8} 

\section[Discrete Random Variables]{Law of Large Numbers for Discrete Random Variables}
\label{sec 8.1}
We are now in a position to prove our first fundamental theorem of probability.  
We have seen that an intuitive way to view the probability of a certain outcome is as the 
frequency with which that outcome occurs in the long run, when the experiment is repeated a large 
number of times.  We have also defined probability mathematically as a value of a distribution 
function for the random variable representing the experiment.   The Law of Large Numbers, which is
a theorem proved about the mathematical model of probability, shows that this model is consistent 
with the frequency interpretation of probability.  This theorem is sometimes called the \emx {law
of averages.}  To find out what would happen if this law were not true, see the article  by 
Robert M. Coates.\index{COATES, R. M.}\footnote{R.~M.~Coates, ``The Law," \emx {The World of
Mathematics,} ed. James R. Newman (New York: Simon and Schuster, 1956.}

\subsection*{Chebyshev Inequality}
To discuss the Law of Large Numbers, we first need an important inequality
called the \emx {Chebyshev Inequality.}

\begin{theorem}{\bf (Chebyshev Inequality)}\index{Chebyshev Inequality}
Let $X$ be a discrete random variable with expected value $\mu = E(X)$, and let $\epsilon
> 0$ be any positive real number.  Then
$$
P(|X - \mu| \geq \epsilon) \leq \frac {V(X)}{\epsilon^2}\ .
$$
\proof
Let $m(x)$ denote the distribution function of $X$.  Then the probability that $X$
differs from $\mu$ by at least $\epsilon$ is given by
$$P(|X - \mu| \geq \epsilon) = \sum_{|x - \mu| \geq \epsilon} m(x)\ .$$
We know that 
$$V(X) = \sum_x (x - \mu)^2 m(x)\ ,$$
and this is clearly at least as large as
$$\sum_{|x - \mu| \geq \epsilon} (x - \mu)^2 m(x)\ ,$$
since all the summands are positive and we have restricted the range of summation in the
second sum.  But this last sum is at least
\begin{eqnarray*}
\sum_{|x - \mu| \geq \epsilon} \epsilon^2 m(x) &=& 
\epsilon^2 \sum_{|x - \mu| \geq \epsilon} m(x) \\
&=& \epsilon^2 P(|X - \mu| \geq \epsilon)\ .\\
\end{eqnarray*}
So,
$$ P(|X - \mu| \geq \epsilon) \leq \frac {V(X)}{\epsilon^2}\ .
$$
\end{theorem}
Note that $X$ in the above theorem can be any discrete random variable, and
$\epsilon$ any positive number.

\begin{example}
Let $X$ by any random variable with $E(X) = \mu$ and $V(X) = \sigma^2$.  Then,
if $\epsilon = k\sigma$, Chebyshev's Inequality states that
$$
P(|X - \mu| \geq k\sigma) \leq \frac {\sigma^2}{k^2\sigma^2} = \frac 1{k^2}\ .
$$
Thus, for any random variable, the probability of a deviation from the mean of
more than~$k$ standard deviations is ${} \leq 1/k^2$.  If, for example,
$k = 5$, $1/k^2 = .04$.
\end{example}

Chebyshev's Inequality is the best possible inequality in the sense that, for
any $\epsilon > 0$, it is possible to give an example of a random variable for
which Chebyshev's Inequality is in fact an equality.  To see this, given
$\epsilon > 0$, choose $X$ with distribution
$$
p_X = \pmatrix{
-\epsilon & +\epsilon \cr
1/2 & 1/2 \cr}\ .
$$
Then $E(X) = 0$, $V(X) = \epsilon^2$, and
$$
P(|X - \mu| \geq \epsilon) = \frac {V(X)}{\epsilon^2} = 1\ .
$$

We are now prepared to state and prove the Law of Large Numbers.

\subsection*{Law of Large Numbers}
\begin{theorem}{\bf (Law of Large Numbers)}\index{Law of Large Numbers}
Let $X_1$,~$X_2$, \dots,~$X_n$ be an independent trials process, with
finite expected value $\mu = E(X_j)$ and finite variance $\sigma^2 =
V(X_j)$.  Let $S_n = X_1 + X_2 +\cdots+ X_n$.  Then for
any $\epsilon > 0$, 
$$ P\left( \left| \frac {S_n}n - \mu \right| \geq \epsilon
\right) \to 0
$$
as $n \rightarrow \infty$.
Equivalently,
$$
P\left( \left| \frac {S_n}n - \mu \right| < \epsilon \right) \to 1
$$
as $n \rightarrow \infty$.
\proof
Since $X_1$,~$X_2$, \dots,~$X_n$ are independent and have the same distributions,
we can apply Theorem~\ref{thm 6.9}.  We obtain
$$
V(S_n) = n\sigma^2\ ,
$$
and
$$
V (\frac {S_n}n) = \frac {\sigma^2}n\ .
$$
Also we know that
$$
E (\frac {S_n}n) = \mu\ .
$$
By Chebyshev's Inequality, for any $\epsilon > 0$,
$$
P\left( \left| \frac {S_n}n - \mu \right| \geq \epsilon \right) \leq \frac
{\sigma^2}{n\epsilon^2}\ .
$$
Thus, for fixed $\epsilon$,
$$
P\left( \left| \frac {S_n}n - \mu \right| \geq \epsilon \right) \to 0
$$
as $n \rightarrow \infty$, or equivalently,
$$
P\left( \left| \frac {S_n}n - \mu \right| < \epsilon \right) \to 1
$$
as $n \rightarrow \infty$.
\end{theorem}

\subsection*{Law of Averages}
Note that $S_n/n$ is an average of the individual outcomes, and one often
calls the Law of Large Numbers the ``law of averages." It is a striking fact
that we can start with a random experiment about which little can be predicted
and, by taking averages, obtain an experiment in which the outcome can be
predicted with a high degree of certainty.  The Law of Large Numbers, as we
have stated it, is often called the ``Weak Law of Large Numbers" to
distinguish it from the ``Strong Law of Large Numbers" described in
Exercise~\ref{exer 8.1.16}.

Consider the important special case of Bernoulli trials with probability~$p$
for success.  Let $X_j = 1$ if the $j$th outcome is a success and~0 if it is a
failure.  Then $S_n = X_1 + X_2 +\cdots+ X_n$ is the number of successes in $n$
trials and $\mu = E(X_1) = p$.  The Law of Large Numbers states that for any
$\epsilon > 0$
$$
P\left( \left| \frac {S_n}n - p \right| < \epsilon \right) \to 1
$$
as $n \rightarrow \infty$.  The above statement says that, in a large number of repetitions
of a Bernoulli experiment, we can expect the proportion of times the event will
occur to be near $p$.  This shows that our mathematical model of probability
agrees with our frequency interpretation of probability.  

\subsection*{Coin Tossing}
Let us consider the special case of tossing a coin $n$ times with $S_n$ the
number of heads that turn up.  Then the random variable $S_n/n$ represents the
fraction of times heads turns up and will have values between 0~and~1.  The Law
of Large Numbers predicts that the outcomes for this random variable will, for
large~$n$, be near 1/2.

In Figure~\ref{fig 8.1}, we have plotted the distribution for this example for increasing values of
~$n$. We have marked the outcomes between .45~and~.55 by dots at the top of the
spikes.  We see that as
$n$ increases the distribution gets more and more concentrated around~.5 and a larger
and larger percentage of the total area is contained within the interval
$(.45,.55)$, as predicted by the Law of Large Numbers.

\putfig{5.0truein}{PSfig8-1}{Bernoulli trials distributions.}{fig 8.1} 


\subsection*{Die Rolling}
\begin{example}
Consider $n$ rolls of a die.  Let $X_j$ be the outcome of the $j$th roll.  Then
$S_n = X_1 + X_2 +\cdots+ X_n$ is the sum of the first $n$ rolls.  This is an
independent trials process with $E(X_j) = 7/2$.  Thus, by the Law of Large
Numbers, for any $\epsilon > 0$
$$
P\left( \left| \frac {S_n}n - \frac 72 \right| \geq \epsilon \right) \to 0
$$
as $n \rightarrow \infty$.  An equivalent way to state this is that, for any $\epsilon > 0$,
$$
P\left( \left| \frac {S_n}n - \frac 72 \right| < \epsilon \right) \to 1
$$
as $n \rightarrow \infty$.
\end{example}

\subsection*{Numerical Comparisons}
It should be emphasized that, although Chebyshev's Inequality proves the Law of Large
Numbers, it is actually a very crude inequality for the probabilities involved.  However,
its strength lies in the fact that it is true for any random variable at all, and it allows
us to prove a very powerful theorem.
\par
In the following example, we compare the estimates given by Chebyshev's Inequality with the
actual values.
\begin{example}
Let $X_1$,~$X_2$, \dots,~$X_n$ be a Bernoulli trials process with
probability~.3 for success and~.7 for failure.  Let $X_j = 1$ if the $j$th
outcome is a success and~0 otherwise.  Then, $E(X_j) = .3$ and $V(X_j) =
(.3)(.7) = .21$.  If
$$
A_n = \frac {S_n}n = \frac {X_1 + X_2 +\cdots+ X_n}n
$$
is the \emx {average} of the $X_i$, then $E(A_n) = .3$ and $V(A_n) =
V(S_n)/n^2 = .21/n$.  Chebyshev's Inequality states that if, for example,
$\epsilon = .1$,
$$
P(|A_n - .3| \geq .1) \leq \frac {.21}{n(.1)^2} = \frac {21}n\ .
$$
Thus, if $n = 100$,
$$
P(|A_{100} - .3| \geq .1) \leq .21\ ,
$$
or if $n = 1000$,
$$
P(|A_{1000} - .3| \geq .1) \leq .021\ .
$$
These can be rewritten as
 \begin{eqnarray*}
P(.2 < A_{100} < .4) &\geq& .79\ , \\
P(.2 < A_{1000} < .4) &\geq& .979\ .
\end{eqnarray*} 
These values should be compared with the actual values, which are (to six decimal
places)
\begin{eqnarray*}
P(.2 < A_{100} < .4) &\approx& .962549 \\
P(.2 < A_{1000} < .4) &\approx& 1\ .\\
\end{eqnarray*}
The program {\bf Law}\index{Law (program)} can be used to carry out the above calculations in a
systematic way.
\end{example}


\subsection*{Historical Remarks}
The Law of Large Numbers was first proved by the Swiss mathematician James
Bernoulli\index{BERNOULLI, J.|(} in the fourth part of his work \emx {Ars Conjectandi} published
posthumously in~1713.\footnote{J. Bernoulli, \emx {The Art of Conjecturing
IV,} trans.~Bing Sung, Technical Report No.~2, Dept.\ of Statistics, Harvard
Univ., 1966}  As often happens with a first proof, Bernoulli's proof was much
more difficult than the proof we have presented using Chebyshev's inequality. 
Chebyshev developed his inequality to prove a general form of the Law of Large
Numbers (see Exercise~\ref{exer 8.1.13}).  The inequality itself appeared
much earlier in a work by Bienaym\'e,\index{BIENAYM\'E, I.} and in discussing its history
Maistrov\index{MAISTROV, L.} remarks that it was referred to as the Bienaym\'e-Chebyshev
Inequality for a long time.\footnote{L. E. Maistrov, \emx {Probability Theory: A Historical
Approach,} trans.\ and ed.~Samual Kotz, (New York: Academic Press, 1974),
p.~202}

In \emx {Ars Conjectandi} Bernoulli provides his reader with a long discussion
of the meaning of his theorem with lots of examples.  In modern notation he has
an event that occurs with probability~$p$ but he does not know $p$.  He wants
to estimate $p$ by the fraction $\bar{p}$ of the times the event occurs
when the experiment is repeated a number of times.  He discusses in detail the
problem of estimating, by this method, the proportion of white balls in an urn
that contains an unknown number of white and black balls.  He would do this by
drawing a sequence of balls from the urn, replacing the ball drawn after each
draw, and estimating the unknown proportion of white balls in the urn by the
proportion of the balls drawn that are white.  He shows that, by choosing $n$
large enough he can obtain any desired accuracy and reliability for the
estimate.  He also provides a lively discussion of the applicability of his
theorem to estimating the probability of dying of a particular disease, of
different kinds of weather occurring, and so forth.

In speaking of the number of trials necessary for making a judgement, Bernoulli
observes that the ``man on the street" believes the ``law of averages."

\begin{quote}
Further, it cannot escape anyone that for judging in this way about any event
at all, it is not enough to use one or two trials, but rather a great number of
trials is required.  And sometimes the stupidest man---by some instinct of
nature \emx {per se} and by no previous instruction (this is truly amazing)---
knows for sure that the more observations of this sort that are taken, the less
the danger will be of straying from the mark.\footnote{Bernoulli, op.\ cit., p.~38.}
\end{quote}

\noindent But he goes on to say that he must contemplate another possibility.

\begin{quote}
Something futher must be contemplated here which perhaps no one has thought
about till now.  It certainly remains to be inquired whether after the number
of observations has been increased, the probability is increased of attaining
the true ratio between the number of cases in which some event can happen and
in which it cannot happen, so that this probability finally exceeds any given
degree of certainty; or whether the problem has, so to speak, its own
asymptote---that is, whether some degree of certainty is given which one can
never exceed.\footnote{ibid., p.~39.}
\end{quote}

\noindent Bernoulli recognized the importance of this theorem, writing:

\begin{quote}
Therefore, this is the problem which I now set forth and make known after I
have already pondered over it for twenty years.  Both its novelty and its very
great usefullness, coupled with its just as great difficulty, can exceed in
weight and value all the remaining chapters of this thesis.\footnote{ibid.,
p.~42.}
\end{quote}

\noindent Bernoulli concludes his long proof with the remark:

\begin{quote}
Whence, finally, this one thing seems to follow: that if observations of all
events were to be continued throughout all eternity, (and hence the ultimate
probability would tend toward perfect certainty), everything in the world would
be perceived to happen in fixed ratios and according to a constant law of
alternation, so that even in the most accidental and fortuitous occurrences we
would be bound to recognize, as it were, a certain necessity and, so to speak,
a certain fate.

I do now know whether Plato wished to aim at this in his doctrine of the
universal return of things, according to which he predicted that all things
will return to their original state after countless ages have
past.\footnote{ibid., pp.~65--66.}
\end{quote}\index{BERNOULLI, J.|)}

\exercises
\begin{LJSItem}

\i\label{exer 8.1.1} A fair coin is tossed 100 times.  The expected number of
heads is~50, and the standard deviation for the number of heads is $(100 \cdot
1/2 \cdot 1/2)^{1/2} = 5$.  What does Chebyshev's Inequality tell you about the
probability that the number of heads that turn up deviates from the expected
number 50 by three or more standard deviations (i.e., by at least 15)?

\i\label{exer 8.1.100} Write a program that uses the function 
$\mbox {binomial}(n,p,x)$ to compute the exact probability that you estimated in
Exercise~\ref{exer 8.1.1}.  Compare the two results.

\i\label{exer 8.1.101} Write a program to toss a coin 10{,}000 times.  Let $S_n$ be the
number of heads in the first $n$ tosses.  Have your program print out, after every
1000 tosses, $S_n - n/2$.  On the basis of this simulation, is it correct to
say that you can expect heads about half of the time when you toss a coin a
large number of times?

\i\label{exer 8.1.102} A 1-dollar bet on craps has an expected winning of $-.0141$.  What
does the Law of Large Numbers say about your winnings if you make a large number of
1-dollar bets at the craps table?  Does it assure you that your losses will be
small?  Does it assure you that if $n$ is very large you will lose?

\i\label{exer 8.1.103} Let $X$ be a random variable with $E(X) =0$ and $V(X) = 1$.  What
integer value~$k$ will assure us that $P(|X| \geq k) \leq .01$?

\i\label{exer 8.1.6} Let $S_n$ be the number of successes in $n$ Bernoulli
trials with probability~$p$ for success on each trial.  Show, using Chebyshev's
Inequality, that for any $\epsilon > 0$
$$
P\left( \left| \frac {S_n}n - p \right| \geq \epsilon \right) \leq \frac {p(1 -
p)}{n\epsilon^2}\ .
$$

\i\label{exer 8.1.7} Find the maximum possible value for $p(1 - p)$ if $0 < p
< 1$.  Using this result and Exercise~\ref{exer 8.1.6}, show that the estimate
$$
P\left( \left| \frac {S_n}n - p \right| \geq \epsilon \right) \leq \frac
1{4n\epsilon^2}
$$
is valid for any $p$.

\i\label{exer 8.1.104} A fair coin is tossed a large number of times.  Does the Law of Large
Numbers assure us that, if $n$ is large enough, with $\mbox {probability} >
.99$ the number of heads that turn up will not deviate from $n/2$ by more than
100?

\i\label{exer 8.1.105} In Exercise~\ref{sec 6.2}.\ref{exer 6.2.16}, you showed that, for the
hat check problem, the number $S_n$ of people who get their own hats back has $E(S_n) =
V(S_n) = 1$.  Using Chebyshev's Inequality, show that $P(S_n \geq 11) \leq .01$
for any $n \geq 11$.

\i\label{exer 8.1.106} Let $X$ by any random variable which takes on values 0,~1,~2,
\dots,~$n$ and has $E(X) = V(X) = 1$.  Show that, for any positive integer $k$,
$$
P(X \geq k + 1) \leq \frac 1{k^2}\ .
$$

\i\label{exer 8.1.107} We have two coins: one is a fair coin and the other is a coin that
produces heads with probability 3/4.  One of the two coins is picked at random,
and this coin is tossed $n$ times.  Let $S_n$ be the number of heads that turns
up in these $n$ tosses.  Does the Law of Large Numbers allow us to predict the
proportion of heads that will turn up in the long run?  After we have observed
a large number of tosses, can we tell which coin was chosen?  How many tosses
suffice to make us 95~percent sure?

\i\label{exer 8.1.13} (Chebyshev\index{CHEBYSHEV, P. L.}\footnote{P. L. Chebyshev, ``On Mean
Values," \emx {J.\ Math.\ Pure.\ Appl.,} vol.~12 (1867), pp.~177--184.})  Assume
that $X_1$,~$X_2$, \dots,~$X_n$ are independent random variables with possibly
different distributions and let $S_n$ be their sum.  Let $m_k = E(X_k)$,
$\sigma_k^2 = V(X_k)$, and $M_n = m_1 + m_2 +\cdots+ m_n$.  Assume that
$\sigma_k^2 < R$ for all~$k$.  Prove that, for any $\epsilon > 0$,
$$
P\left( \left| \frac {S_n}n - \frac {M_n}n \right| < \epsilon \right) \to 1
$$
as $n \rightarrow \infty$.

\i\label{exer 8.1.108}  A fair coin is tossed repeatedly.  Before each toss, you are allowed
to decide whether to bet on the outcome.  Can you describe a betting system with infinitely
many bets which will enable you, in the long run, to win more than half of your bets? 
(Note that we are disallowing a betting system that says to bet until you are ahead, then
quit.) Write a computer program that implements this betting system.  As stated above, your
program must decide whether to bet on a particular outcome before that outcome is determined.
For example, you might select only outcomes that come after
there have been three tails in a row.  See if you can get more than 50\% heads
by your ``system."

\istar\label{exer 8.1.109} Prove the following analogue of Chebyshev's Inequality:
$$
P(|X - E(X)| \geq \epsilon) \leq \frac 1\epsilon E(|X - E(X)|)\ .
$$

\istar\label{exer 8.1.16} We have proved a theorem often called the ``Weak Law of
Large Numbers."  Most people's intuition and our computer simulations suggest
that, if we toss a coin a sequence of times, the proportion of heads will really
approach 1/2; that is, if $S_n$ is the number of heads in $n$ times, then we
will have
$$
A_n = \frac {S_n}n \to \frac 12
$$
as $n \to \infty$.  Of course, we cannot be sure of this since we are not able
to toss the coin an infinite number of times, and, if we could, the coin could
come up heads every time.  However, the ``Strong Law of Large Numbers,"\index{Strong Law of 
Large\\ Numbers} proved in more advanced courses, states that
$$
P\left( \frac {S_n}n \to \frac 12 \right) = 1\ .
$$
Describe a sample space $\Omega$ that would make it possible for us to talk
about the event
$$
E = \left\{\, \omega : \frac {S_n}n \to \frac 12\, \right\}\ .
$$
Could we assign the equiprobable measure to this space?  
\choice{}{(See
Example~\ref{exam 2.2.12}.)} 

\istar\label{exer 8.1.16.5} 
In this exercise, we shall construct an example of a sequence of random
variables that satisfies the weak law of large numbers, but not the strong
law. The distribution of $X_i$ will have to depend on $i$, because
otherwise both laws would be satisfied.  (This problem was communicated to us
by David Maslen.)
\vskip .1in
Suppose we have an infinite sequence of mutually independent events $A_1,
A_2, \ldots$. Let $a_i = P(A_i)$, and let $r$ be a positive integer.

\begin{enumerate}
\item Find an expression of the probability that none of the $A_i$ with
$i>r$ occur.

\item Use the fact that $x-1 \leq e^{-x}$ to show that
$$
P(\mbox{No\ $A_i$\ with\ $i > r$\ occurs}) \leq e^{-\sum_{i=r}^{\infty}
a_i}
$$
\item (The first Borel-Cantelli lemma)  Prove that if $\sum_{i=1}^{\infty} a_i$ diverges, then
$$
P(\mbox{infinitely\ many\ $A_i$\ occur}) = 1.
$$
\vskip .1in
\noindent
Now, let $X_i$ be a sequence of mutually independent random variables such
that for each positive integer $i \geq 2$,
$$
P(X_i = i) = \frac{1}{2i\log i}, \quad P(X_i = -i) =
\frac{1}{2i\log i}, \quad P(X_i =0) = 1 - \frac{1}{i \log i}.
$$
When $i=1$ we let $X_i=0$ with probability $1$.  As usual we let $S_n = X_1
+ \cdots + X_n$.  Note that the mean of each $X_i$ is $0$.
\item Find the variance of $S_n$.
\item Show that the sequence $\langle X_i \rangle$ satisfies the Weak
Law of Large Numbers, i.e. prove that for any $\epsilon > 0$
$$
P\biggl(\biggl|{\frac{S_n}{n}}\biggr| \geq \epsilon\biggr) \rightarrow 0\ ,
$$
as $n$ tends to infinity.
\vskip .1in
\noindent
We now show that $\{ X_i \}$ does not satisfy the Strong Law of
Large Numbers.  Suppose that $S_n / n \rightarrow 0$. Then because
$$
\frac{X_n}{n} = \frac{S_n}{n} - \frac{n-1}{n} \frac{S_{n-1}}{n-1}\ ,
$$
we know that $X_n / n \rightarrow 0$. From the definition of limits, we
conclude that the inequality $|X_i| \geq \frac{1}{2} i$ can only be
true for finitely many $i$.
\item Let $A_i$ be the event $|X_i| \geq \frac{1}{2} i$. Find
$P(A_i)$. Show that $\sum_{i=1}^{\infty} P(A_i)$ diverges (use the 
Integral Test).
\item Prove that $A_i$ occurs for infinitely many $i$.
\item Prove that 
$$
P\biggl(\frac{S_n}{n} \rightarrow 0\biggr) = 0,
$$
and hence that the Strong Law of Large Numbers fails for the sequence
$\{ X_i \}$.

\end{enumerate}
\istar\label{exer 8.1.110} Let us toss a biased coin that comes up heads with probability~$p$
and assume the validity of the Strong Law of Large Numbers as described
in Exercise~\ref{exer 8.1.16}.  Then, with probability~1, $$
\frac {S_n}n \to p
$$
as $n \to \infty$.  If $f(x)$ is a continuous function on the unit interval,
then we also have
$$
f\left( \frac {S_n}n \right) \to f(p)\ .
$$

Finally, we could hope that
$$
E\left(f\left( \frac {S_n}n \right)\right) \to E(f(p)) = f(p)\ .
$$
Show that, if all this is correct, as in fact it is, we would have proven that
any continuous function on the unit interval is a limit of polynomial
functions.  This is a sketch of a probabilistic proof of an important theorem 
in mathematics called the \emx {Weierstrass approximation theorem.}\index{Weierstrass
Approximation Theorem}
\end{LJSItem}


\choice{}{\section[Continuous Random Variables]{Law of Large Numbers for Continuous Random
Variables}
\label{sec 8.2}
In the previous section we discussed in some detail the Law of Large Numbers
for discrete probability distributions.  This law has a natural analogue for
continuous probability distributions, which we consider somewhat more briefly
here.

\subsection*{\bf Chebyshev Inequality} \hfill\break\index{Chebyshev Inequality}
Just as in the discrete case, we begin our discussion with the Chebyshev
Inequality.

\begin{theorem}{\bf (Chebyshev Inequality)} \index{Chebyshev Inequality}
Let $X$ be a continuous random variable with density function $f(x)$.  Suppose $X$ has a 
finite expected value $\mu = E(X)$ and finite variance $\sigma^2 = V(X)$.  Then for any
positive number $\epsilon > 0$ we have
$$
P(|X - \mu| \geq \epsilon) \leq \frac {\sigma^2}{\epsilon^2}\ .
$$
\end{theorem}

The proof is completely analogous to the proof in the discrete case, and we omit it.
\par
Note that this theorem says nothing if $\sigma^2 = V(X)$ is infinite.


\begin{example}
Let $X$ be any continuous random variable with $E(X) = \mu$ and $V(X) =
\sigma^2$.  Then, if $\epsilon = k\sigma = k$ standard deviations
for some integer~$k$, then
$$
P(|X - \mu| \geq k\sigma) \leq \frac {\sigma^2}{k^2\sigma^2} = \frac 1{k^2}\ ,
$$
just as in the discrete case.
\end{example}

\subsection*{Law of Large Numbers}
With the Chebyshev Inequality we can now state and prove the Law of Large
Numbers for the continuous case.

\begin{theorem}{\bf (Law of Large Numbers)}\index{Law of Large Numbers}
Let $X_1$,~$X_2$, \dots,~$X_n$ be an independent trials process with a
continuous density function~$f$, finite expected value~$\mu$, and finite
variance~$\sigma^2$.  Let $S_n = X_1 + X_2 +\cdots+ X_n$ be the sum of the
$X_i$.  Then for any real number $\epsilon > 0$ we have
$$
\lim_{n \to \infty} P\left( \left| \frac {S_n}n - \mu \right| \geq \epsilon
\right) = 0\ ,
$$
or equivalently,
$$
\lim_{n \to \infty} P\left( \left| \frac {S_n}n - \mu \right| < \epsilon
\right) = 1\ .
$$
\end{theorem}

Note that this theorem is not necessarily true if $\sigma^2$ is infinite
(see Example~\ref{exam 8.2.5}).

As in the discrete case, the Law of Large Numbers says that the average value
of $n$ independent trials tends to the expected value as $n \to \infty$, in the
precise sense that, given $\epsilon > 0$, the probability that the average
value and the expected value differ by more than $\epsilon$ tends to~0 as $n
\to \infty$.

Once again, we suppress the proof, as it is identical to the proof in the discrete case.
\subsection*{Uniform Case}
\begin{example}
Suppose we choose at random $n$ numbers from the interval $[0,1]$ with
uniform distribution.  Then if $X_i$ describes the $i$th choice, we have
\begin{eqnarray*}
          \mu & = & E(X_i) = \int_0^1 x\, dx = \frac 12\ , \\
     \sigma^2 & = & V(X_i) = \int_0^1 x^2\, dx - \mu^2 \\
              & = & \frac 13 - \frac 14 = \frac 1{12}\ .
\end{eqnarray*}
Hence,
\begin{eqnarray*}
E \left( \frac {S_n}n \right)  & = & \frac 12\ , \\
V \left( \frac {S_n}n \right)  & = & \frac 1{12n}\ ,
\end{eqnarray*}
and for any $\epsilon > 0$,
$$
P \left( \left| \frac {S_n}n - \frac 12 \right| \geq \epsilon \right) \leq \frac
1{12n \epsilon^2}\ .
$$

This says that if we choose $n$ numbers at random from $[0,1]$, then the
chances are better than $1 - 1/(12n\epsilon^2)$ that the difference $|S_n/n -
1/2|$ is less than~$\epsilon$.  Note that $\epsilon$ plays the role of the
amount of error we are willing to tolerate: If we choose $\epsilon = 0.1$, say,
then the chances that $|S_n/n - 1/2|$ is less than~0.1 are better than $1 -
100/(12n)$.  For $n = 100$, this is about .92, but if $n = 1000$, this is better
than .99 and if $n = 10{,}000$, this is better than .999.
\putfig{5.0truein}{PSfig8-2}{Illustration of Law of Large Numbers --- uniform case.}{fig 8.2}

We can illustrate what the Law of Large Numbers says for this example
graphically.  The density for $A_n = S_n/n$ is determined by
$$
f_{A_n}(x) = nf_{S_n}(nx)\ .
$$

We have seen in Section~\ref{sec 7.2}, that we can compute the density
$f_{S_n}(x)$ for the sum of $n$ uniform random variables.  In Figure~\ref{fig 8.2} we have
used this to plot the density for $A_n$ for various values of~$n$.  We have
shaded in the area for which $A_n$ would lie between .45~and~.55.  We see that as
we increase $n$, we obtain more and more of the total area inside the shaded
region.  The Law of Large Numbers tells us that we can obtain as much of the
total area as we please inside the shaded region by choosing $n$ large enough
(see also Figure~\ref{fig 8.1}).
\end{example}

\subsection*{Normal Case}
\begin{example}
Suppose we choose $n$ real numbers at random,
using a normal distribution with mean~0 and variance~1.  Then
\begin{eqnarray*}
         \mu &=& E(X_i) = 0\ , \\
    \sigma^2 &=& V(X_i) = 1\ .
\end{eqnarray*}
Hence,
\begin{eqnarray*}
E \left( \frac {S_n}n \right) &=& 0\ , \\
V \left( \frac {S_n}n \right) &=& \frac 1n\ ,
\end{eqnarray*}
and, for any $\epsilon > 0$,
$$
P\left( \left| \frac {S_n}n - 0 \right| \geq \epsilon \right) \leq \frac
1{n\epsilon^2}\ .
$$
In this case it is possible to compare the Chebyshev estimate for $P(|S_n/n -
\mu| \geq \epsilon)$ in the Law of Large Numbers with exact values, since we
know the density function for $S_n/n$ exactly (see Example~\ref{exam 7.12}). 
The comparison is shown in Table~\ref{table 8.1}, for $\epsilon = .1$.  The data
in this table was produced by the program {\bf LawContinuous}.\index{LawContinuous (program)}
\begin{table}
\centering
\begin{tabular}{r|r|r}
$n$ & $P(|S_n/n| \ge .1)$ & Chebyshev \\
\hline
100 & .31731 & 1.00000 \\
200 & .15730 & .50000 \\
300 & .08326 & .33333 \\
400 & .04550 & .25000 \\
500 & .02535 & .20000 \\
600 & .01431 & .16667 \\
700 & .00815 & .14286 \\
800 & .00468 & .12500 \\
900 & .00270 & .11111 \\
1000 & .00157 & .10000 \\
\hline
\end{tabular}
\caption{Chebyshev estimates.}
\label{table 8.1}
\end{table}
We see here that the Chebyshev estimates are in general \emx {not} very 
accurate.
\end{example}

\subsection*{Monte Carlo Method}
Here is a somewhat more interesting example.

\begin{example}
Let $g(x)$ be a continuous function defined for $x \in [0,1]$ with values in
$[0,1]$.  In Section~\ref{sec 2.1}, we showed how to estimate the area of
the region under the graph of $g(x)$ by the Monte Carlo method, that is, by
choosing a large number of random values for $x$~and~$y$ with uniform
distribution and seeing what fraction of the points $P(x,y)$ fell inside the
region under the graph (see Example~\ref{exam 2.1.2}).
\par
Here is a better way to estimate the same area (see Figure~\ref{fig 8.3}).  Let us choose a
large number of independent values $X_n$ at random from $[0,1]$ with uniform
density, set $Y_n = g(X_n)$, and find the average value of the $Y_n$.  Then
this average is our estimate for the area.  To see this, note that if the
density function for~$X_n$ is uniform,
\begin{eqnarray*}
\mu & = & E(Y_n) = \int_0^1 g(x) f(x)\, dx \\
    & = & \int_0^1 g(x)\, dx \\
    & = & \mbox {average\ value\ of\ }  g(x)\ ,
\end{eqnarray*}
while the variance is
$$
\sigma^2 = E((Y_n - \mu)^2) = \int_0^1 (g(x) - \mu)^2\, dx < 1\ ,
$$
since for all $x$ in $[0, 1]$, $g(x)$ is in $[0, 1]$, hence $\mu$ is in $[0, 1]$, and
so $|g(x) - \mu| \le 1$.  Now let $A_n = (1/n)(Y_1 + Y_2 +\cdots+ Y_n)$.  Then by Chebyshev's
Inequality, we have
$$
P(|A_n - \mu| \geq \epsilon) \leq \frac {\sigma^2}{n\epsilon^2} < \frac
1{n\epsilon^2}\ .
$$

\putfig{3truein}{PSfig8-3}{Area problem.}{fig 8.3} 

This says that to get within $\epsilon$ of the true value for $\mu = \int_0^1
g(x)\, dx$ with probability at least $p$, we should choose $n$ so that
$1/n\epsilon^2 \leq 1 - p$ (i.e., so that $n \geq 1/\epsilon^2(1 - p)$).  Note
that this method tells us how large to take $n$ to get a desired accuracy.
\end{example}

The Law of Large Numbers requires that the variance $\sigma^2$ of the original
underlying density be finite: $\sigma^2 < \infty$.  In cases where this fails
to hold, the Law of Large Numbers may fail, too.  An example follows.

\subsection*{Cauchy Case}
\begin{example}\label{exam 8.2.5}
Suppose we choose $n$ numbers from $(-\infty,+\infty)$ with a Cauchy density
with parameter $a = 1$.  We know that for the Cauchy density the expected value
and variance are undefined (see Example~\ref{exam 6.23}).  In this case, the
density function for
$$
A_n = \frac {S_n}n
$$
is given by (see Example~\ref{exam 7.9})
$$
f_{A_n}(x) = \frac 1{\pi(1 + x^2)}\ ,
$$
that is, \emx {the density function for $A_n$ is the same for all $n$.}  In this
case, as $n$ increases, the density function does not change at all, and the
Law of Large Numbers does not hold.
\end{example}

\exercises
\begin{LJSItem}

\i\label{exer 8.2.1} Let $X$ be a continuous random variable with mean $\mu =
10$ and variance $\sigma^2 = 100/3$.  Using Chebyshev's Inequality, find an upper
bound for the following probabilities.
\begin{enumerate}
\item $P(|X - 10| \geq 2)$.

\item $P(|X - 10| \geq 5)$.

\item $P(|X - 10| \geq 9)$.

\item $P(|X - 10| \geq 20)$.
\end{enumerate}

\i\label{exer 8.2.2} Let $X$ be a continuous random variable with values
unformly distributed over the interval $[0,20]$.
\begin{enumerate}
\item Find the mean and variance of $X$.

\item Calculate $P(|X - 10| \geq 2)$, $P(|X - 10| \geq 5)$, $P(|X - 10| \geq
9)$, and $P(|X - 10| \geq 20)$ exactly.  How do your answers compare with those
of Exercise~\ref{exer 8.2.1}?  How good is Chebyshev's Inequality in this case?
\end{enumerate}

\i\label{exer 8.2.3} Let $X$ be the random variable of Exercise~\ref{exer
8.2.2}.
\begin{enumerate}
\item Calculate the function $f(x) = P(|X - 10| \geq x)$.

\item Now graph the function $f(x)$, and on the same axes, graph the
Chebyshev function $g(x) = 100/(3x^2)$.  Show that $f(x) \leq g(x)$ for all~$x >
0$, but that $g(x)$ is not a very good approximation for~$f(x)$.
\end{enumerate}

\i\label{exer 8.2.100} Let $X$ be a continuous random variable with values exponentially
distributed over $[0,\infty)$ with parameter $\lambda = 0.1$.
\begin{enumerate}
\item Find the mean and variance of $X$.

\item Using Chebyshev's Inequality, find an upper bound for the following
probabilities: $P(|X - 10| \geq 2)$, $P(|X - 10| \geq 5)$, $P(|X - 10| \geq
9)$, and $P(|X - 10| \geq 20)$.

\item Calculate these probabilities exactly, and compare with the bounds in
(b).
\end{enumerate}
\i\label{exer 8.2.101} Let $X$ be a continuous random variable with values normally
distributed over $(-\infty,+\infty)$ with mean $\mu = 0$ and variance $\sigma^2 = 1$.
\begin{enumerate}
\item Using Chebyshev's Inequality, find upper bounds for the following
probabilities: $P(|X| \geq 1)$, $P(|X| \geq 2)$, and $P(|X| \geq 3)$.

\item The area under the normal curve between $-1$~and~1 is .6827, between
$-2$~and~2 is .9545, and between $-3$~and~3 it is .9973 (see the table in
Appendix~A).  Compare your bounds in (a) with
these exact values.  How good is Chebyshev's Inequality in this case?
\end{enumerate}

\i\label{exer 8.2.102} If $X$ is normally distributed, with mean~$\mu$ and
variance~$\sigma^2$, find an upper bound for the following probabilities, using Chebyshev's
Inequality.
\begin{enumerate}
\item $P(|X - \mu| \geq \sigma)$.

\item $P(|X - \mu| \geq 2\sigma)$.

\item $P(|X - \mu| \geq 3\sigma)$.

\item $P(|X - \mu| \geq 4\sigma)$.
\end{enumerate}
\noindent Now find the exact value using the program {\bf NormalArea}\index{NormalArea (program)}
or the normal table in Appendix~A, and compare.

\i\label{exer 8.2.103} If $X$ is a random variable with mean~$\mu \ne 0$ and variance~$\sigma^2$,
define the \emx {relative deviation} $D$ of $X$ from its mean by
$$
D = \left| \frac {X - \mu}\mu \right|\ .
$$
\begin{enumerate}
\item Show that $P(D \geq a) \leq \sigma^2/(\mu^2a^2)$.

\item If $X$ is the random variable of Exercise~\ref{exer 8.2.1}, find an
upper bound for $P(D \geq .2)$, $P(D \geq .5)$, $P(D \geq .9)$, and $P(D \geq
2)$.
\end{enumerate}

\i\label{exer 8.2.104} Let $X$ be a continuous random variable and define the {\em
standardized version} $X^*$ of~$X$ by:
$$
X^* = \frac {X - \mu}\sigma\ .
$$
\begin{enumerate}
\item Show that $P(|X^*| \geq a) \leq 1/a^2$.

\item If $X$ is the random variable of Exercise~\ref{exer 8.2.1}, find
bounds for $P(|X^*| \geq 2)$, $P(|X^*| \geq 5)$, and $P(|X^*| \geq 9)$.
\end{enumerate}

\i\label{exer 8.2.105}
\begin{enumerate} 
\item Suppose a number $X$ is chosen at random from $[0,20]$ with uniform
probability.  Find a lower bound for the probability that $X$ lies between
8~and~12, using Chebyshev's Inequality.

\item Now suppose 20 real numbers are chosen independently from $[0,20]$
with uniform probability.  Find a lower bound for the probability that their
average lies between 8~and~12.

\item Now suppose 100 real numbers are chosen independently from
$[0,20]$.  Find a lower bound for the probability that their average lies
between 8~and~12.
\end{enumerate}

\i\label{exer 8.2.106} A student's score on a particular calculus final is a random variable
with values of $[0,100]$, mean~70, and variance~25.
\begin{enumerate}
\item Find a lower bound for the probability that the student's score will
fall between 65~and~75.

\item If 100 students take the final, find a lower bound for the probability
that the class average will fall between 65~and~75.
\end{enumerate}

\i\label{exer 8.2.107} The Pilsdorff beer company runs a fleet of trucks along the 100~mile
road from Hangtown to Dry Gulch, and maintains a garage halfway in between. 
Each of the trucks is apt to break down at a point $X$~miles from Hangtown,
where $X$ is a random variable uniformly distributed over $[0,100]$.
\begin{enumerate}
\item Find a lower bound for the probability $P(|X - 50| \leq 10)$.

\item Suppose that in one bad week, 20 trucks break down.  Find a lower bound
for the probability $P(|A_{20} - 50| \leq 10)$, where $A_{20}$ is the average
of the distances from Hangtown at the time of breakdown.
\end{enumerate}

\i\label{exer 8.2.12} A share of common stock in the Pilsdorff beer company
has a price $Y_n$ on the $n$th business day of the year.  Finn observes that the
price change $X_n = Y_{n + 1} - Y_n$ appears to be a random variable with mean
$\mu = 0$ and variance $\sigma^2 =1/4$.  If $Y_1 = 30$, find a lower bound for
the following probabilities, under the assumption that the $X_n$'s are mutually independent.
\begin{enumerate}
\item $P(25 \leq Y_2 \leq 35)$.

\item $P(25 \leq Y_{11} \leq 35)$.

\item $P(25 \leq Y_{101} \leq 35)$.
\end{enumerate}

\i\label{exer 8.2.108} Suppose one hundred numbers $X_1$,~$X_2$, \dots,~$X_{100}$ are chosen
independently at random from $[0,20]$.  Let $S = X_1 + X_2 +\cdots+ X_{100}$
be the sum, $A = S/100$ the average, and $S^* = (S - 1000)/(10/\sqrt3)$ the
standardized sum.  Find lower bounds for the probabilities
\begin{enumerate}
\item $P(|S - 1000| \leq 100)$.

\item $P(|A - 10| \leq 1)$.

\item $P(|S^*| \leq \sqrt3)$.
\end{enumerate}

\i\label{exer 8.2.14} Let $X$ be a continuous random variable normally
distributed on $(-\infty,+\infty)$ with mean~0 and variance~1.  Using the normal
table provided in Appendix~A, or the program {\bf NormalArea},
find values for the function $f(x) = P(|X| \geq x)$ as $x$ increases from
0~to~4.0 in steps of~.25.  Note that for $x \geq 0$ the table gives
$ NA(0,x) = P(0 \leq X \leq x)$ and thus $P(|X| \geq x) = 2(.5 - NA(0,x)$.
Plot by hand
the graph of~$f(x)$ using these values, and the graph of the Chebyshev function
$g(x) = 1/x^2$, and compare (see Exercise~\ref{exer 8.2.3}).

\i\label{exer 8.2.109} Repeat Exercise~\ref{exer 8.2.14}, but this time with mean~10 and
variance~3.  Note that the table in Appendix~A presents values for
a standard normal variable.  Find the standardized version $X^*$ for~$X$, find
values for $f^*(x) = P(|X^*| \geq x)$ as in Exercise~\ref{exer 8.2.14}, and
then rescale these values for $f(x) = P(|X -10| \geq x)$.  Graph and compare
this function with the Chebyshev function $g(x) = 3/x^2$.

\i\label{exer 8.2.110} Let $Z = X/Y$ where $X$~and~$Y$ have normal densities with mean~0 and
standard deviation~1.  Then it can be shown that $Z$ has a Cauchy density.
\begin{enumerate}
\item Write a program to illustrate this result by plotting a bar graph of
1000 samples obtained by forming the ratio of two standard normal outcomes. 
Compare your bar graph with the graph of the Cauchy density.  Depending upon which
computer language you use, you may or may not need to tell the computer how to simulate a
normal random variable.  A method for doing this was described in Section~\ref{sec 5.2}.

\item We have seen that the Law of Large Numbers does not apply to the
Cauchy density (see Example~\ref{exam 8.2.5}).  Simulate a large number of
experiments with Cauchy density and compute the average of your results.  Do
these averages seem to be approaching a limit?  If so can you explain why this
might be?
\end{enumerate}

\i\label{exer 8.2.111} Show that, if $X \geq 0$, then $P(X \geq a) \leq E(X)/a$.

\i\label{exer 8.2.112} (Lamperti\footnote{Private communication.})
\index{LAMPERTI, J.} Let $X$ be a non-negative random
variable.  What is the best upper bound you can give for $P(X \geq a)$ if you know
\begin{enumerate}
\item $E(X) = 20$.

\item $E(X) = 20$ and $V(X) = 25$.

\item $E(X) = 20$, $V(X) = 25$, and $X$ is symmetric about its mean.
\end{enumerate}
\end{LJSItem}}
