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%%tth:\special{html: An introduction to information theory and entropy}
%Slide 1
\title{{\LARGE\bf An introduction to information theory and entropy}\newline \newline \newline}
\author{Tom Carter
\newline
\newline
\newline
\tthdump{\href{http://astarte.csustan.edu/\~tom/SFI-CSSS}{http://astarte.csustan.edu/\~\ tom/SFI-CSSS}}
%%tth:\href{http://astarte.csustan.edu/~tom/SFI-CSSS}{http://astarte.csustan.edu/\~tom/SFI-CSSS}
\vfill
Complex Systems Summer School
\newline
}
\date{June, 2006}
\maketitle
%Slide 2
\sectionhead{Our general topics:}
%%tth:\begin{itemize}
%%tth:\item
\tthdump{\hyperlink{Measuring complexity}
{\ $\circledcirc$ Measuring complexity\newline}}
%%tth:\makehyperlink{Measuring complexity}
%%tth:\item
\tthdump{\hyperlink{Some probability background}
{$\circledcirc$ Some probability background\newline}}
%%tth:\makehyperlink{Some probability background}
%%tth:\item
\tthdump{\hyperlink{Basics of information theory}
{$\circledcirc$ Basics of information theory\newline}}
%%tth:\makehyperlink{Basics of information theory}
%%tth:\item
\tthdump{\hyperlink{Some entropy theory}
{$\circledcirc$ Some entropy theory\newline}}
%%tth:\makehyperlink{Some entropy theory}
%%tth:\item
\tthdump{\hyperlink{The Gibbs inequality}
{$\circledcirc$ The Gibbs inequality\newline}}
%%tth:\makehyperlink{The Gibbs inequality}
%%tth:\item
\tthdump{\hyperlink{A simple physical example (gases)}
{$\circledcirc$ A simple physical example (gases)\newline}}
%%tth:\makehyperlink{A simple physical example (gases)}
%%tth:\item
\tthdump{\hyperlink{Shannon's communication theory}
{$\circledcirc$ Shannon's communication theory\newline}}
%%tth:\makehyperlink{Shannon's communication theory}
%%tth:\item
\tthdump{\hyperlink{Application to Biology (analyzing genomes)}
{$\circledcirc$ Application to Biology (analyzing\\\ \ \ \ \ genomes)\newline}}
%%tth:\makehyperlink{Application to Biology (analyzing genomes)}
%%tth:\item
\tthdump{\hyperlink{Some other measures}
{$\circledcirc$ Some other measures\newline}}
%%tth:\makehyperlink{Some other measures}
%%tth:\item
\tthdump{\hyperlink{Some additional material}
{$\circledcirc$ Some additional material\newline}}
%%tth:\makehyperlink{Some additional material}
%%tth:\item
\tthdump{\hyperlink{Examples using Bayes' Theorem}
{$\circledcirc$ Examples using Bayes' Theorem\newline}}
%%tth:\makehyperlink{Examples using Bayes' Theorem}
%%tth:\item
\tthdump{\hyperlink{Analog channels}
{$\circledcirc$ Analog channels\newline}}
%%tth:\makehyperlink{Analog channels}
%%tth:\item
\tthdump{\hyperlink{A Maximum Entropy Principle}
{$\circledcirc$ A Maximum Entropy Principle\newline}}
%%tth:\makehyperlink{A Maximum Entropy Principle}
%%tth:\item
\tthdump{\hyperlink{Application: Economics I (a Boltzmann Economy)}
{$\circledcirc$ Application: Economics I (a Boltzmann Economy)\newline}}
%%tth:\makehyperlink{Application: Economics I (a Boltzmann Economy)}
%%tth:\item
\tthdump{\hyperlink{Application: Economics II (a power law)}
{$\circledcirc$ Application: Economics II (a power law)\newline}}
%%tth:\makehyperlink{Application: Economics II (a power law)}
%%tth:\item
\tthdump{\hyperlink{Application to Physics (lasers)}
{$\circledcirc$ Application to Physics (lasers)\newline}}
%%tth:\makehyperlink{Application to Physics (lasers)}
%%tth:\item
\tthdump{\hyperlink{References}
{$\circledcirc$ References\newline}}
%%tth:\makehyperlink{References}
%%tth:\end{itemize}
\pagedone
\quotesection{The quotes}
%%tth:\begin{itemize}
%%tth:\item
\tthdump{\hyperlink{Science, wisdom, and counting}
{\ $\circledcirc$ Science, wisdom, and counting\newline}}
%%tth:\makehyperlink{Science, wisdom, and counting}
%%tth:\item
\tthdump{\hyperlink{Being different -- or random}
{$\circledcirc$ Being different -- or random\newline}}
%%tth:\makehyperlink{Being different -- or random}
%%tth:\item
\tthdump{\hyperlink{Surprise, information, and miracles}
{$\circledcirc$ Surprise, information, and miracles\newline}}
%%tth:\makehyperlink{Surprise, information, and miracles}
%%tth:\item
\tthdump{\hyperlink{Information (and hope)}
{$\circledcirc$ Information (and hope)\newline}}
%%tth:\makehyperlink{Information (and hope)}
%%tth:\item
\tthdump{\hyperlink{H (or S) for Entropy}
{$\circledcirc$ H (or S) for Entropy\newline}}
%%tth:\makehyperlink{H (or S) for Entropy}
%%tth:\item
\tthdump{\hyperlink{Thermodynamics}
{$\circledcirc$ Thermodynamics\newline}}
%%tth:\makehyperlink{Thermodynamics}
%%tth:\item
\tthdump{\hyperlink{Language, and putting things together}
{$\circledcirc$ Language, and putting things together\newline}}
%%tth:\makehyperlink{Language, and putting things together}
%%tth:\item
\tthdump{\hyperlink{Tools}
{$\circledcirc$ Tools}}
%%tth:\makehyperlink{Tools}
%%tth:\end{itemize}
%\thepage
\tthdump{\hyperlink{Our general topics:}{\hfil To topics $\leftarrow$}}
%%tth:{\special{html: Back to top of file}}
\pagedone
%Slide 3
\quotesection{Science, wisdom, and counting}
%%tth:\begin{quote}
``Science is organized knowledge. Wisdom is organized life.''
- Immanuel Kant
``My own suspicion is that the universe is not only stranger than we suppose,
but stranger than we can suppose.''
- John Haldane
``Not everything that can be counted counts, and not everything that counts can be counted.''
- Albert Einstein (1879-1955)
``The laws of probability, so true in general, so fallacious in particular .''
- Edward Gibbon
%%tth:\end{quote}
\pagedone
\sectionhead{Measuring complexity}
\begin{itemize}
\item Workers in the field of complexity face a classic problem: how can we tell that
the system we are looking at is actually a complex system? (i.e., should we even
be studying this system? :-)
Of course, in practice, we will study the systems that interest us, for whatever reasons,
so the problem identified above tends not to be a real problem. On the other hand,
having chosen a system to study, we might well ask ``How complex is this system?''
In this more general context, we probably want at least to be able to compare two
systems, and be able to say that system A is more complex than system B. Eventually,
we probably would like to have some sort of numerical rating scale.
\pagedone
\item Various approaches to this task have been
proposed, among them:
\begin{enumerate}
\item Human observation and (subjective) rating
% (i.e., use
% the human perceptual system as a measurement instrument)
\item Number of parts or distinct elements (what counts as a distinct part?)
\item Dimension (measured how?)
\item Number of parameters controlling the system
\item Minimal description (in which language?)
\item Information content (how do we define/measure information?)
\item Minimal generator/constructor (what machines/methods can we use?)
\item Minimum energy/time to construct (how would evolution count?)
\end{enumerate}
\pagedone
\item Most (if not all) of these measures will actually be measures associated
with a {\em model} of a phenomenon. Two observers (of the same phenomenon?)
may develop or use very different models, and thus disagree in their assessments
of the complexity. For example, in a very simple case, counting the number of
parts is likely to depend on the scale at which the phenomenon is viewed (counting
atoms is different from counting molecules, cells, organs, etc.).
We shouldn't expect to be able to come up with a single universal measure
of complexity. The best we are likely to have is a measuring system useful
by a particular observer, in a particular context, for a particular purpose.
My first focus will be on measures related to how surprising or unexpected
an observation or event is. This approach has been described as {\em
information theory}.
\end{itemize}
\pagedone
\quotesection{Being different -- or random}
%%tth:\begin{quote}
``The man who follows the crowd will usually get no further than the crowd. The man who walks alone is likely to find himself in places no one has ever been before. Creativity in living is not without its attendant difficulties, for peculiarity breeds contempt. And the unfortunate thing about being ahead of your time is that when people finally realize you were right, they'll say it was obvious all along. You have two choices in life: You can dissolve into the mainstream, or you can be distinct. To be distinct is to be different. To be different, you must strive to be what no one else but you can be. ''
-Alan Ashley-Pitt
``Anyone who considers arithmetical methods of producing random digits is, of course, in a state of sin.''
- John von Neumann (1903-1957)
%%tth:\end{quote}
\pagedone
\sectionhead{Some probability background}
\begin{itemize}
\item At various times in what follows, I may float between two notions of the {\em probability}
of an event happening. The two general notions are:
\begin{enumerate}
\item A {\em frequentist} version of probability:
In this version, we assume we have a set of possible events, each of which
we assume occurs some number of times. Thus, if there are $N$ distinct possible
events $(x_1, x_2, \ldots , x_N),$ no two of which can occur simultaneously,
and the events occur with frequencies $(n_1, n_2, \ldots , n_N)$,
we say that the probability of event $x_i$ is given by
$$P(x_i) = \frac{n_i}{\sum_{j=1}^{N}{n_j}}$$
This definition has the nice property that
$$\sum_{i=1}^{N}P(x_i) = 1$$
\item An {\em observer relative} version of probability:
In this version, we take a statement of {\em probability} to be an
assertion about the belief that a specific observer has of the occurrence of
a specific event.
Note that in this version of {\em probability}, it is possible that two different
observers may assign different probabilities to the same event.
Furthermore, the {\em probability} of an event, for me, is likely to change as
I learn more about the event, or the context of the event.
\pagedone
\item In some (possibly many) cases, we may be able to find a reasonable correspondence
between these two views of probability. In particular, we may sometimes be able
to understand the {\em observer relative} version of the probability of an event
to be an approximation to the {\em frequentist} version, and to view new knowledge
as providing us a better estimate of the relative frequencies.
\end{enumerate}
\end{itemize}
\pagedone
%Slide 4
\begin{itemize}
\item I won't go through much, but some probability basics, where $a$ and $b$ are events: \newline
$ P(not\ a) = 1 - P(a).$\newline
$ P(a\ or\ b) = P(a) + P(b) - P(a\ \mathrm{and}\ b).$ \newline
We will often denote $P(a\ \mathrm{and}\ b)$ by $P(a, b)$.
If $P(a, b) = 0$, we say $a$ and $b$ are mutually exclusive.\newline
\item Conditional probability: \newline\newline
$ P(a \vert b) $ is the probability of $a$, given that we know $b$.
The joint probability of both $a$ and $b$ is given by:
$$P(a, b) = P(a \vert b) P(b).$$
Since $P(a, b) = P(b, a)$, we have Bayes' Theorem:
$$P(a \vert b)P(b) = P(b \vert a) P(a),$$
or
$$P(a \vert b) = \frac{P(b \vert a) P(a)}{P(b)}.$$
\end{itemize}
\pagedone
%Slide 5
\begin{itemize}
\item If two events $a$ and $b$ are such that
$$P(a \vert b) = P(a),$$
we say that the events $a$ and $b$ are {\em independent}. Note that from
Bayes' Theorem, we will also have that
$$P(b \vert a) = P(b),$$
and furthermore,
$$P(a, b) = P(a \vert b)P(b) = P(a)P(b).$$
This last equation is often taken as the definition of {\em independence}.
\item We have in essence begun here the development of a mathematized methodology for drawing
inferences about the world from uncertain knowledge. We could say that our observation
of the coin showing heads gives us {\em information} about the world. We will develop a
formal mathematical definition of the {\em information} content of an event which occurs
with a certain probability.
\end{itemize}
\pagedone
\quotesection{Surprise, information, and miracles}
%%tth:\begin{quote}
``The opposite of a correct statement is a false statement. The opposite of a profound truth may well be another profound truth.''
- Niels Bohr (1885-1962)
``I heard someone tried the monkeys-on-typewriters bit trying for the plays of W. Shakespeare, but all they got was the collected works of Francis Bacon.''
- Bill Hirst
``There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.''
- Albert Einstein (1879-1955)
%%tth:\end{quote}
\pagedone
\sectionhead{Basics of information theory}
\begin{itemize}
\item We would like to develop a usable measure of the {\em information} we get
from observing the occurrence of an event having probability $p$ . Our first reduction
will be to ignore any particular features of the event, and only observe whether or
not it happened. In essence this means that we can think of the event as the
observance of a symbol whose probability of occurring is $p$.
We will thus be defining the {\em information} in terms of the probability $p$.
\pagedone
\item We will want our {\em information} measure $I(p)$ to have several properties:
\begin{enumerate}
\item Information is a non-negative quantity: $I(p) \ge 0$.
\item If an event has probability 1, we get no information from the occurrence
of the event: $I(1) = 0$.
\item If two independent events occur (whose joint probability is the product of
their individual probabilities), then the information we get from observing
the events is the sum of the two informations:
$I(p_1*p_2) = I(p_1) + I(p_2).$
(This is the critical property \ldots)
\item We will want our {\em information} measure to be a continuous (and, in fact,
monotonic) function of the probability (slight changes in probability should result
in slight changes in {\em information}).
\end{enumerate}
\pagedone
\item We can therefore derive the following:
\begin{enumerate}
\item $I(p^2) = I(p*p) = I(p) + I(p) = 2*I(p)$
\item Thus, further, $I(p^n) = n*I(p)$ \newline
(by induction \ldots)
\item $I(p) = I((p^{1/m})^m) = m * I(p^{1/m})$, so $I(p^{1/m}) = \frac{1}{m}*I(P)$
and thus in general
$$I(p^{n/m}) = \frac{n}{m}*I(p)$$
\item And thus, by continuity, we get, for $0 < p \le 1$, and $a>0$ a real number:
$$I(p^a) = a*I(p)$$
\end{enumerate}
\item From this, we can derive the nice property:
$$I(p) = -\log_b(p) = \log_b(1/p)$$
for some base $b$.
\end{itemize}
\pagedone
%Slide 9
\begin{itemize}
\item Summarizing: from the four properties,
\begin{enumerate}
\item $I(p) \ge 0$
\item $I(p_1*p_2) = I(p_1) + I(p_2)$
\item $I(p)$ is monotonic and continuous in $p$
\item $I(1) = 0$
\end{enumerate}
we can derive that
$$I(p) = \log_b(1/p) = - \log_b(p),$$
for some positive constant $b$. The base $b$ determines the units we
are using.
We can change the units by changing the base, using the formulas, for $b_1, b_2, x > 0$,
$$x = b_1^{\log_{b_1}(x)}$$
and therefore
$$\log_{b_2}(x) = \log_{b_2}(b_1^{\log_{b_1}(x)}) = (\log_{b_2}(b_1))(\log_{b_1}(x)).$$
\end{itemize}
\pagedone
\begin{itemize}
\item Thus, using different bases for the logarithm results in {\em information} measures
which are just constant multiples of each other, corresponding with measurements
in different units:
\begin{enumerate}
\item $\log_2$ units are {\em bits} (from 'binary')
\item $\log_3$ units are {\em trits}(from 'trinary')
\item $\log_e$ units are {\em nats} (from 'natural logarithm') (We'll use $\ln(x)$
for $\log_e(x)$)
\item $\log_{10}$ units are {\em Hartleys}, after an early worker in the field.
\end{enumerate}
\item Unless we want to emphasize the units, we need not bother to specifiy
the base for the logarithm, and will write $\log(p)$. Typically, we will think
in terms of $\log_2(p)$.
\end{itemize}
\pagedone
\begin{itemize}
\item For example, flipping a fair coin once will give us events $h$ and $t$ each
with probability $1/2$, and thus a single flip of a coin gives us $-\log_2(1/2) = 1$ bit
of information (whether it comes up $h$ or $t$).
Flipping a fair coin $n$ times (or, equivalently, flipping $n$ fair coins) gives us
$-\log_2((1/2)^n) = \log_2(2^n) = n*\log_2(2) = n$ bits of information.
We could enumerate a sequence of 25 flips as, for example:
$$hthhtththhhthttththhhthtt$$
or, using $1$ for $h$ and $0$ for $t$, the 25 bits
$$1011001011101000101110100.$$
We thus get the nice fact that $n$ flips of a fair coin gives us $n$ bits of information,
and takes $n$ binary digits to specify. That these two are the same reassures us that
we have done a good job in our definition of our {\em information} measure \ldots
\end{itemize}
\pagedone
\quotesection{Information (and hope)}
%%tth:\begin{quote}
``In Cyberspace, the First Amendment is a local ordinance.''
- John Perry Barlow
``Groundless hope, like unconditional love, is the only kind worth having.''
- John Perry Barlow
``The most interesting facts are those which can be used several times, those which have a chance of recurring. \ldots Which, then, are the facts that have a chance of recurring? In the first place, simple facts.''
H. Poincare, 1908
%%tth:\end{quote}
\pagedone
%Slide 10
\sectionhead{Some entropy theory}
\begin{itemize}
\item Suppose now that we have $n$ symbols $\{a_1, a_2, \ldots, a_n\}$, and some
source is providing us with a stream of these symbols. Suppose further that
the source emits the symbols with probabilities $\{p_1, p_2, \ldots, p_n\}$, respectively.
For now, we also assume that the symbols are emitted independently
(successive symbols do not depend in any way on past symbols).
What is the average amount of {\em information} we get from each symbol we see
in the stream?
\end{itemize}
\pagedone
\begin{itemize}
\item What we really want here is a weighted average. If we observe the symbol $a_i$,
we will get be getting $\log(1/p_i)$ {\em information} from that particular observation.
In a long run (say $N$) of observations, we will see (approximately)
$N*p_i$ occurrences of symbol $a_i$ (in the frequentist sense, that's what it means to say
that the probability of seeing $a_i$ is $p_i$). Thus, in the $N$ (independent) observations, we
will get total information $I$ of
$$I = \sum_{i=1}^{n}(N*p_i)*\log(1/p_i).$$
But then, the average information we get per symbol observed will be
\begin{eqnarray*}
I/N & = & (1/N)\sum_{i=1}^{n}(N*p_i)*\log(1/p_i)\\
& = & \sum_{i=1}^{n}p_i*\log(1/p_i)
\end{eqnarray*}
Note that $\lim_{x\rightarrow0} x*\log(1/x) = 0$, so we can, for our purposes,
define $p_i*\log(1/p_i)$ to be $0$ when $p_i = 0$.
\end{itemize}
\pagedone
%Slide 11
\begin{itemize}
\item This brings us to a fundamental definition. This definition is essentially due to
Shannon in 1948, in the seminal papers in the field of information theory.
As we have observed, we have defined {\em information} strictly in terms of
the probabilities of events. Therefore, let us suppose that we have a set of
probabilities (a probability distribution) $P = \{p_1, p_2, \ldots, p_n\}$.
We define the {\em entropy} of the distribution $P$ by:
$$H(P) = \sum_{i=1}^{n}p_i*\log(1/p_i).$$
I'll mention here the obvious generalization, if we have a continuous rather
than discrete probability distribution $P(x)$:
$$H(P) = \int P(x)*\log(1/P(x))dx.$$
\end{itemize}
\pagedone
%Slide 12
\begin{itemize}
\item Another worthwhile way to think about this is in terms of
{\em expected value}. Given a discrete probability distribution
$P = \{p_1, p_2, \ldots, p_n\}$, with $p_i \ge 0$ and $\sum_{i=1}^n p_i = 1$,
or a continuous distribution $P(x)$ with $P(x) \ge 0$ and $\int P(x)dx = 1$,
we can define the {\em expected value} of an associated discrete set
$F = \{f_1, f_2, \ldots, f_n\}$ or function $F(x)$ by:
$$ = \sum_{i=1}^n f_i p_i$$
or
$$ = \int F(x) P(x) dx.$$
With these definitions, we have that:
$$H(P) = *.$$
In other words, the {\em entropy} of a probability distribution is just the
{\em expected value} of the {\em information} of the distribution.
\end{itemize}
\pagedone
%Slide 12
Several questions probably come to mind at this point:
\begin{itemize}
\item What properties does the function $H(P)$ have? For example,
does it have a maximum, and if so where?
\item Is {\em entropy} a reasonable name for this? In particular, the
name {\em entropy} is already in use in thermodynamics. How are
these uses of the term related to each other?
\item What can we do with this new tool?
\item Let me start with an easy one. Why use the letter $H$ for entropy?
What follows is a slight variation of a footnote, p. 105, in the book
{\em Spikes} by Rieke, et al. :-)
\end{itemize}
\pagedone
\quotesection{H (or S) for Entropy}
%%tth:\begin{quote}
``The enthalpy is [often] written U. V is the volume, and Z is the partition function. P and
Q are the position and momentum of a particle. R is the gas constant, and of course T is
temperature. W is the number of ways of configuring our system (the number of states),
and we have to keep X and Y in case we need more variables. Going back to the first half
of the alphabet, A, F, and G are all different kinds of free energies (the last named for
Gibbs). B is a virial coefficient or a magnetic field. I will be used as a symbol for
information; J and L are angular momenta. K is Kelvin, which is the proper unit of T.
M is magnetization, and N is a number, possibly Avogadro's, and O is too easily
confused with 0. This leaves S . . .'' and H. In {\em Spikes} they also eliminate H
(e.g., as the Hamiltonian). I, on the other hand, along with Shannon and others, prefer
to honor Hartley. Thus, H for entropy . . .
%%tth:\end{quote}
\pagedone
%Slide 14
\sectionhead{The Gibbs inequality}
\begin{itemize}
\item First, note that the function $\ln(x)$ has derivative $1/x$. From this, we find that
the tangent to $\ln(x)$ at $x = 1$ is the line $y = x - 1$. Further, since $\ln(x)$
is concave down, we have, for $x > 0$, that
$$\ln(x) \le x - 1,$$
with equality only when $x = 1$.
Now, given two probability distributions, $ P = \{p_1, p_2, \ldots, p_n\}$
and $Q = \{q_1, q_2, \ldots, q_n\}$, where $p_i, q_i \ge 0$ and $\sum_i p_i = \sum_i q_i = 1$,
we have
\begin{eqnarray*}
\sum_{i=1}^n p_i \ln\left(\frac{q_i}{p_i}\right) & \le & \sum_{i=1}^n p_i
\left(\frac{q_i}{p_i} - 1\right)
= \sum_{i=1}^n(q_i - p_i)\\
& = & \sum_{i=1}^n q_i - \sum_{i=1}^n p_i = 1 - 1 = 0,
\end{eqnarray*}
with equality only when $p_i = q_i$ for all $i$. It is easy to see that the inequality
actually holds for any base, not just $e$.
\end{itemize}
\pagedone
% \sectionhead{Magic}
% \begin{quote}
% ``The Universe is full of magical things patiently waiting for our wits to
% grow sharper.''
%
% -Eden Phillpotts
% \end{quote}
%
% \begin{quote}
% ``Any sufficiently advanced technology is indistinguishable from magic.''
%
% -Arthur C. Clarke
% \end{quote}
%
% \pagedone
%Slide 15
\begin{itemize}
\item We can use the Gibbs inequality to find the probability distribution which
maximizes the entropy function. Suppose $ P = \{p_1, p_2, \ldots, p_n\} $ is a
probability distribution. We have
\begin{eqnarray*}
H(P) - \log(n) & = & \sum_{i=1}^n p_i \log(1/p_i) - \log(n)\\
& = & \sum_{i=1}^n p_i \log(1/p_i) - \log(n) \sum_{i=1}^n p_i \\
& = & \sum_{i=1}^n p_i \log(1/p_i) - \sum_{i=1}^n p_i \log(n)\\
& = & \sum_{i=1}^n p_i(\log(1/p_i) - \log(n))\\
& = & \sum_{i=1}^n p_i(\log(1/p_i) + \log(1/n))\\
& = & \sum_{i=1}^n p_i\log\left(\frac{1/n}{p_i}\right)\\
\\
& \le & 0,
\end{eqnarray*}
with equality only when $p_i = \frac{1}{n}$ for all $i$. The last step is the application
of the Gibbs inequality.
\end{itemize}
\pagedone
%Slide 16
\begin{itemize}
\item What this means is that
$$0 \le H(P) \le \log(n).$$
We have $H(P) = 0$ when exactly one of the $p_i$'s is one and all the rest
are zero. We have $H(P) = \log(n)$ only when all of the events have the same
probability $\frac{1}{n}$.
That is, the maximum of the entropy function is the $\log()$ of the number of possible
events, and occurs when all the events are equally likely.
\item An example illustrating this result:
How much information can a student get from a single grade? First, the
maximum information occurs if all grades have equal probability (e.g.,
in a pass/fail class, on average half should pass if we want to maximize
the information given by the grade).
\pagedone
The maximum information the
student gets from a grade will be:
Pass/Fail : 1 bit.
A, B, C, D, F : 2.3 bits.
A, A-, B+, . . ., D-, F : 3.6 bits.
Thus, using +/- grading gives the students about 1.3 more bits of information
per grade than without +/-, and about 2.6 bits per grade more than pass/fail.
\item If a source provides us with a sequence chosen from 4 symbols (say
A, C, G, T), then the maximum average information per symbol is 2 bits.
If the source provides blocks of 3 of these symbols, then the maximum
average information is 6 bits per block (or, to use different units, 4.159 nats
per block).
\end{itemize}
\pagedone
%Slide 17
We ought to note several things.
\begin{itemize}
\item First, these definitions of {\em information}
and {\em entropy} may not match with some other uses of the terms.
For example, if we know that a source will, with equal probability, transmit either the
complete text of Hamlet or the complete text of Macbeth (and nothing else), then receiving
the complete text of Hamlet provides us with precisely 1 bit of information.
Suppose a book contains ascii characters. If the book is to provide us with {\em information}
at the maximum rate, then each ascii character will occur with equal probability -- it will be
a random sequence of characters.
\end{itemize}
\pagedone
\begin{itemize}
\item Second, it is important to recognize that our definitions of {\em information} and
{\em entropy} depend only on the probability distribution. In general, it won't make
sense for us to talk about the {\em information} or the {\em entropy} of a source
without specifying the probability distribution.
Beyond that, it can certainly happen that two different observers of the same
data stream have different models of the source, and thus associate different
probability distributions to the source. The two observers will then assign
different values to the {\em information} and {\em entropy} associated with
the source.
This observation (almost :-) accords with our intuition: two people listening to the same
lecture can get very different information from the lecture. For example, without
appropriate background, one person might not understand anything at all, and therefore
have as probability model a completely random source, and therefore get much
more information than the listener who understands quite a bit, and can therefore
anticipate much of what goes on, and
therefore assigns non-equal probabilities to successive words . . .
\end{itemize}
\pagedone
\quotesection{Thermodynamics}
%%tth:\begin{quote}
``A theory is the more impressive the greater the simplicity of its premises is, the more different kinds of things it relates, and the more extended its area of applicability. Therefore the deep impression which classical thermodynamics made upon me. It is the only physical theory of universal content which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown (for the special attention of those who are skeptics on principle).''
- A. Einstein, 1946
``Thermodynamics would hardly exist as a profitable discipline if it were not that the natural limit to the size of so many types of instruments which we now make in the laboratory falls in the region in which the measurements are still smooth.''
- P. W. Bridgman, 1941
%%tth:\end{quote}
\pagedone
\sectionhead{A simple physical example (gases)}
\begin{itemize}
\item Let us work briefly with a simple model for an idealized gas. Let us assume
that the gas is made up of $N$ point particles, and that at some time $t_0$ all
the particles are contained within a (cubical) volume $V$. Assume that through
some mechanism, we can determine the location of each particle sufficiently
well as to be able to locate it within a box with sides $1/100$ of the sides
of the containing volume $V$. There are $10^6$ of these small boxes within
$V$.
\item We can now develop a (frequentist) probability model for this system. For
each of the $10^6$ small boxes, we can assign a probability $p_i$ of finding a gas
particle in the small box by counting the number of particles $n_i$ in the box,
and dividing by $N$. That is, $p_i = \frac{n_i}{N}$. From this probability
distribution, we can calculate an entropy:
\begin{eqnarray*}
H(P) & = & \sum_{i=1}^{10^6}p_i*\log(1/p_i)\\
& = & \sum_{i=1}^{10^6}\frac{n_i}{N} * \log(N/n_i)
\end{eqnarray*}
If the particles are evenly distributed among the $10^6$ boxes, then we will
have that each $n_i = N/10^6$, and in this case the entropy will be:
\begin{eqnarray*}
H(evenly) & = & \sum_{i=1}^{10^6}\frac{N/10^6}{N} * \log\left(\frac{N}{N/10^6}\right)\\
& = & \sum_{i=1}^{10^6}\frac{1}{10^6} * \log(10^6)\\
& = & \log(10^6).
\end{eqnarray*}
\end{itemize}
\pagedone
There are several ways to think about this example.
\begin{itemize}
\item First, notice that the
calculated entropy of the system depends in a strong way on the relative
scale of measurement. For example, if the particles are evenly distributed, and
we increase our accuracy of measurement by a factor of $10$ (i.e., if each
small box is $1/1000$ of the side of $V$), then the calculated maximum entropy will be
$\log(10^9)$ instead of $\log(10^6)$.
For physical systems, we know that quantum limits (e.g., Heisenberg uncertainty
relations) will give us a bound on the accuracy of our measurements, and
thus a more or less natural scale for doing entropy calculations. On the other
hand, for macroscopic systems, we are likely to find that we can only make
relative rather than absolute entropy calculations.
\item Second, we have simplified our model of the gas particles to the extent that
they have only one property, their position. If we want to talk about the {\em state}
of a particle, all we can do is specify the small box the particle is in at time
$t_0$. There are thus $Q=10^6$ possible {\em states} for a particle, and the
maximum entropy for the system is $\log(Q)$. This may look familiar for
equilibrium statistical mechanics \ldots
\item Third, suppose we generalize our model slightly, and allow the particles to
move about within $V$. A {\em configuration} of the system is then simply a list of
$10^6$ numbers $b_i$ with $1 \le b_i \le N$ (i.e., a list of the numbers of particles
in each of the boxes). Suppose that the motions of the
particles are such that for each particle, there is an equal probability that it
will move into any given new small box during one (macroscopic) time step.
How likely is it that at some later time we will find the system
in a ``high'' entropy configuration? How likely is it that if we start the system
in a ``low'' entropy configuration, it will stay in a ``low'' entropy configuration
for an appreciable length of time? If the system is not currently in a ``maximum''
entropy configuration, how likely is it that the entropy will increase in succeeding
time steps (rather than stay the same or decrease)?
Let's do a few computations using combinations:
$$\binom{n}{m} = \frac{n!}{m! * (n - m)!},$$
and Stirling's approximation:
$$n! \approx \sqrt{2\pi}\ n^ne^{-n}\sqrt{n}.$$
\pagedone
Let us start here:
There are $10^6$ configurations with all the particles sitting in exactly
one small box, and the entropy of each of those configurations is:
$$H(all\ in\ one) = \sum_{i=1}^{10^6}p_i * \log(1/p_i) = 0,$$
since exactly one $p_i$ is $1$ and the rest are $0$.
These are obviously minimum entropy configurations.
Now consider pairs of small boxes. The number of configurations with all the particles
evenly distributed between two boxes is:
\begin{eqnarray*}
\binom{10^6}{2} & = & \frac{10^6!}{(2)!(10^6 - 2)!}\\
& = & \frac{10^6 * (10^6 - 1)}{2}\\
& = & 5 * 10^{11},
\end{eqnarray*}
which is a (comparatively :-) large number. The entropy of each of these configurations is:
$$H(two\ boxes) = 1/2 * \log(2) + 1/2 * \log(2) = \log(2).$$
We thus know that there are at least $5 * 10^{11} + 10^6$ configurations.
If we start the system in a configuration with entropy $0$,
then the probability that at some later time it will be in a configuration with
entropy $\ge \log(2)$ will be
\begin{eqnarray*}
& \ge & \frac{5*10^{11}}{5*10^{11} + 10^6} = (1 - \frac{10^6}{5*10^{11} + 10^6})\\
& \ge &(1 - 10^{-5}).
\end{eqnarray*}
As an example at the other end, consider the number of configurations with
the particles distributed almost equally, except that half the boxes are short
by one particle, and the rest have an extra. The number of such configurations
is:
\begin{eqnarray*}
\binom{10^6}{10^6/2} & = & \frac{10^6!}{(10^6/2)!(10^6 - 10^6/2)!}\\
& = & \frac{10^6!}{((10^6/2)!)^2}
\end{eqnarray*}
\begin{eqnarray*}
& \approx & \frac{\sqrt{2\pi}{(10^6)}^{10^6}e^{-10^6}\sqrt{10^6}}
{(\sqrt{2\pi}(10^6/2)^{10^6/2}e^{-(10^6/2)}\sqrt{10^6/2})^2}\\
& = & \frac{\sqrt{2\pi}{(10^6)}^{10^6}e^{-10^6}\sqrt{10^6}}
{{2\pi}(10^6/2)^{10^6}e^{-(10^6)}{10^6/2}}\\
& = & \frac{2^{10^6+1}\sqrt{10^6}}
{\sqrt{2\pi}\sqrt{10^6}}\\
& \approx & 2^{10^6}\\
& = & (2^{10})^{10^5}\\
& \approx & 10^{3*10^5}.
\end{eqnarray*}
Each of these configurations has entropy essentially equal to $\log(10^6)$.
From this, we can conclude that if we start the system in a configuration with
entropy $0$ (i.e., all particles in one box), the probability that later it will be
in a higher entropy configuration will be $> (1 - 10^{-3*10^5})$.
% \begin{eqnarray*}
% & \ge & \frac{10^{3*10^5}}{10^3*10^5 + 10^6} = (1 - \frac{10^6}{10^{3*10^5} + 10^6})\\
% & > &(1 - 10^{-3*10^5 + 6}).
% \end{eqnarray*}
Similar arguments (with similar results in terms of probabilities) can be made
for starting in any configuration with entropy appreciably less than $\log(10^6)$
(the maximum). In other words, it is overwhelmingly probable that as time passes,
macroscopically, the system will increase in entropy until it reaches the maximum.
In many respects, these general arguments can be thought of as a ``proof''
(or at least an explanation) of
a version of the second law of thermodynamics:
Given any macroscopic system which is free to change configurations, and given any
configuration with entropy less than the maximum, there will be overwhelmingly many
more accessible configurations with higher entropy than lower entropy, and thus, with
probability indistinguishable from $1$, the system will (in macroscopic time steps)
successively change to configurations with higher entropy until it reaches the maximum.
\end{itemize}
\pagedone
\pagedone
\quotesection{Language, and putting things together}
%%tth:\begin{quote}
``An essential distinction between language and experience is that language separates out from the living matrix little bundles and freezes them; in doing this it produces something totally unlike experience, but nevertheless useful.''
- P. W. Bridgman, 1936
``One is led to a new notion of unbroken wholeness which denies the classical
analyzability of the world into separately and independently existing parts.
The inseparable quantum interconnectedness of the whole universe is the
fundamental reality.''
- David Bohm
%%tth:\end{quote}
\pagedone
% \sectionhead{Shocking}
% \begin{quote}
% ``Anyone who is not shocked by quantum theory has not understood it.'' --Neils Bohr
% \end{quote}
%
% \begin{quote}
% ``One is led to a new notion of unbroken wholeness which denies the classical
% analyzability of the world into separately and independently existing parts.
% The inseparable quantum interconnectedness of the whole universe is the
% fundamental reality.'' --David Bohm
% \end{quote}
%
% \begin{quote}
% ``I don't like it, and I'm sorry I ever had anything to do with it.'' --Erwin Schrodinger
% \end{quote}
%
% \pagedone
% %Slide 19
\sectionhead{Shannon's communication theory}
\begin{itemize}
\item In his classic 1948 papers, Claude Shannon laid the foundations
for contemporary {\em information}, {\em coding}, and {\em communication}
theory. He developed a general model for communication systems, and a set
of theoretical tools for analyzing such systems.
His basic model consists of three parts: a sender (or source), a
channel, and a receiver (or sink). His general model also includes
encoding and decoding elements, and noise within the channel.
\end{itemize}
\vfil
%\centerline{\epsfxsize=5in \epsfbox{Communication_model.epsf}}
\centerline{\includegraphics[width = 5in]{Communication_Model}}
\centerline{Shannon's communication model}
\pagedone
\begin{itemize}
\item In Shannon's discrete model, it is assumed that the source provides a stream
of symbols selected from a finite alphabet $A = \{a_1, a_2, \ldots, a_n\}$, which
are then encoded. The code is sent through the channel (and possibly disturbed
by noise). At the other end of the channel, the receiver will decode, and derive
information from the sequence of symbols.
Let me mention at this point that sending information from {\em now} to
{\em then} is equivalent to sending information from {\em here} to {\em there},
and thus Shannon's theory applies equally as well to information storage questions
as to information transmission questions.
\end{itemize}
\pagedone
\begin{itemize}
\item One important question we can ask is, how efficiently can we encode
information that we wish to send through the channel? For the moment,
let's assume that the channel is noise-free, and, of course, that
the receiver can accurately recover everything that is transmitted through
the channel. What we need, then, is a way to encode the stream of source
symbols for transmission through the channel in such a way that the encoded
stream can be uniquely decoded at the receiving end.
If the alphabet of the channel (i.e., the symbols that can actually be
carried by the channel) is $C = \{c_1, c_2, \ldots, c_r\}$, then an encoding
of the source alphabet $A$ is just a function $f: A \to C^*$ (where $C^*$ is
the set of all possible finite strings of symbols from $C$). For future
calculations, let $l_i = |f(a_i)|, i = 1, 2, \ldots, n$ (i.e., $l_i$ is the
length of the string encoding the symbol $a_i \in A$).
\pagedone
\item There is a nice inequality concerning the lengths of code strings for
uniquely decodable (and/or instantaneous) codes, called the
McMillan/Kraft inequality. There is a uniquely decodable code with lengths
$l_1, l_2, \ldots, l_n$ if and only if
$$\sum_{i=1}^n \frac{1}{r^{l_i}} \le 1.$$
The necessity of this inequality can be seen from looking at
$$K^n = \left[\sum_{i=1}^n \frac{1}{r^{l_i}}\right]^n.$$
We can rewrite this as
$$K^n = \sum_{k=n}^{nl} \frac{N_k}{r^k}$$
where $l$ is the length of the longest code and $N_k$ is the number of
encodings of strings having encoded length $k$.
\pagedone
Note that $N_k$ cannot be greater than
$r^k$ (the total number of strings of length $k$, whether they encode
anything or not). From this we can see that
$$K^n \le \sum_{k=n}^{nl} \frac{r^k}{r^k} = nl - n + 1 \le nl.$$
From this we can conclude that $K \le 1$ (as desired), since otherwise
$K^n$ would exceed $nl$ for some (possibly large) $n$.
We can now prove a very important property of the entropy -- the entropy
gives a lower bound for the efficiency of an encoding scheme (in other
words, a lower bound on the possible compression of a data stream).
With $K$ defined as above, we can define a set of numbers $Q_i$
(pseudo-probabilities) by
$$Q_i = \frac{r^{-l_i}}{K}.$$
We call these pseudo-probabilities because we have $0 < Q_i \le 1$ for
all $i$, and
$$\sum_{i = 1}^n Q_i = 1.$$
If $p_i$ is the probabilities of observing $a_i$ in the data stream,
then we can apply the Gibbs inequality to get
$$\sum_{i = 1}^n p_i \log \left(\frac{Q_i}{p_i}\right) \le 0,$$
or
$$\sum_{i = 1}^n p_i \log \left(\frac{1}{p_i}\right) \le
\sum_{i = 1}^n p_i \log \left(\frac{1}{Q_i}\right).$$
The left hand side is the entropy of the source, say $H(S)$.
Recalling the definition of $Q_i$, and that $K \le 1$, we find
$$H(S) \le \sum_{i = 1}^n p_i \left( \log(K) - log\left(r^{-l_i}\right)\right)$$
$$ = \log(K) + \sum_{i = 1}^n p_i l_i \log(r) \le \log(r)
\sum_{i = 1}^n p_i l_i.$$
\pagedone
\item From this, we can draw an important conclusion. If we let
$L = \sum_{i = 1}^n p_i l_i$, then $L$ is just the average length of code words
in the encoding. What we have, then, is that
$$H(S) \le L \log(r).$$
In other words, the entropy gives us a lower bound on average code length
for any uniquely decodable symbol-by-symbol encoding of our data stream.
Note that, for example, if we calculate entropy in bits and use binary
($r = 2$) encoding, then we have simply
$$H(S) \le L.$$
Shannon went beyond this, and showed that the bound (appropriately recast)
holds even if we use extended coding systems where we group symbols together
(into ``words'') before doing our encoding. The generalized form of this
inequality is called {\em Shannon's noiseless coding theorem}.
\pagedone
\item In building encoding schemes for data streams (or, alternatively, in
building data compression schemes), we will want to use our best
understandings of the structure of the data stream -- in other words,
we will want to use our best probability model of the data stream.
Shannon's theorem tells us that, since the entropy gives us a lower
bound on our encoding efficiency, if we want to improve our schemes,
we will have to develop successively better probability models.
One way to think about a scientific theory is that a theory is just
an efficient way of encoding (i.e., structuring) our knowledge about
(some aspect of) the world. A good theory is one which reduces the
(relative) entropy of our
(probabilistic) understanding of the system (i.e., that decreases
our average lack of knowledge about the system) \ldots
\end{itemize}
\pagedone
\begin{itemize}
\item Shannon went on to generalize to the (more realistic) situation in
which the channel itself is noisy. In other words, not only are we
unsure about the data stream we will be transmitting through the
channel, but the channel itself adds an additional layer of probability
to our transmissions.
Given a source of symbols and a channel with noise (in particular, given
probability models for the source and the channel noise), we can talk about
the {\em capacity} of the channel.
The general model Shannon worked with involved two sets of symbols, the input
symbols and the output symbols. Let us say the two sets of symbols are
$A = \{a_1, a_2, \ldots, a_n\}$ and $B = \{b_1, b_2, \ldots, b_m\}$. Note
that we do not necessarily assume the same number of symbols in the two sets.
Given the noise in the channel, when symbol $b_j$ comes out of the channel,
we can not be sure which $a_i$ was put in. The channel is characterized by
the set of probabilities $\{P(a_i|b_j)\}$.
\item We can then consider various
related information and entropy measures. First, we can consider the
information we get from observing a symbol $b_j$. Given a probability model
of the source, we have an {\em a priori} estimate $P(a_i)$ that symbol $a_i$
will be sent next. Upon observing $b_j$, we can revise our estimate to $P(a_i|b_j)$.
The change in our information (the {\em mutual information}) will be given by:
\begin{eqnarray*}
I(a_i; b_j) & = & \log\left(\frac{1}{P(a_i)}\right) - \log\left(\frac{1}{P(a_i|b_j)}\right)\\
& = & \log\left(\frac{P(a_i|b_j)}{P(a_i)}\right)
\end{eqnarray*}
We have the properties:
\begin{eqnarray*}
I(a_i; b_j) & = & I(b_j; a_i)\\
I(a_i; b_j) & = & \log(P(a_i|b_j)) + I(a_i)\\
I(a_i; b_j) & \le & I(a_i)
\end{eqnarray*}
If $a_i$ and $b_j$ are independent (i.e., if $P(a_i, b_j) = P(a_i) * P(b_j)$), then
$I(a_i; b_j) = 0.$
\item What we actually want is to average the {\em mutual information} over all the
symbols:
\begin{eqnarray*}
I(A; b_j) & = & \sum_iP(a_i|b_j) * I(a_i;b_j)\\
& = & \sum_iP(a_i|b_j) * \log\left(\frac{P(a_i|b_j)}{P(a_i)}\right)\\
I(a_i;B) & = & \sum_jP(a_i|b_j) * \log\left(\frac{P(b_j|a_i)}{P(b_j)}\right),
\end{eqnarray*}
and from these,
\begin{eqnarray*}
I(A; B) & = & \sum_iP(a_i) * I(a_i;B)\\
& = & \sum_i\sum_jP(a_i, b_j) * \log\left(\frac{P(a_i, b_j)}{P(a_i)P(b_j)}\right)\\
& = & I(B; A).
\end{eqnarray*}
We have the properties:
$ I(A; B) \ge 0,$
and
$ I(A; B) = 0$
if and only if $A$ and $B$ are independent.
\pagedone
\item We then have the definitions and properties:
\begin{eqnarray*}
H(A) & = & \sum_{i = 1}^{n} P(a_i) * \log (1/P(a_i)) \\
H(B) & = & \sum_{j = 1}^{m} P(b_j) * \log (1/P(b_j)) \\
H(A|B) & = & \sum_{i = 1}^{n}\sum_{j = 1}^mP(a_i|b_j) * \log (1/P(a_i|b_j)) \\
H(A, B) & = & \sum_{i = 1}^{n}\sum_{j = 1}^mP(a_i, b_j) * \log (1/P(a_i, b_j)) \\
H(A, B) & = & H(A) + H(B|A)\\
&= & H(B) + H(A|B),
\end{eqnarray*}
and furthermore:
\begin{eqnarray*}
I(A; B) & = & H(A) + H(B) - H(A, B)\\
& = & H(A) - H(A|B)\\
& = & H(B) - H(B|A)\\
& \ge & 0
\end{eqnarray*}
\pagedone
\item If we are given a channel, we could ask what is the maximum possible information
that can be transmitted through the channel. We could also ask what mix of the symbols
$\{a_i\}$ we should use to achieve the maximum. In particular, using the definitions
above, we can define the {\em Channel Capacity} $C$ to be:
$$C = \max_{P(a)}\ I(A; B).$$
\item We have the nice property that if we are using the channel at its capacity, then
for each of the $a_i$,
$$I(a_i;B) = C,$$
and thus, we can maximize channel use by maximizing the use for each symbol
independently.
\pagedone
\item We also have Shannon's main theorem:
For any channel, there exist ways of encoding input symbols such that we can
simultaneously utilize the channel as closely as we wish to the capacity, and
at the same time have an error rate as close to zero as we wish.
\item This is actually quite a remarkable theorem. We might naively guess that
in order to minimize the error rate, we would have to use more of the channel
capacity for error detection/correction, and less for actual transmission of
information. Shannon showed that it is possible to keep error rates low and
still use the channel for information transmission at (or near) its capacity.
\pagedone
\item Unfortunately, Shannon's proof has a a couple of downsides. The first is that
the proof is non-constructive. It doesn't tell us how to construct the coding system
to optimize channel use, but only tells us that such a code exists. The second is that
in order to use the capacity with a low error rate, we may have to encode very large
blocks of data. This means that if we are attempting to use the channel in real-time,
there may be time lags while we are filling buffers. There is thus still much work
possible in the search for efficient coding schemes.
Among the things we can do is look at natural coding systems (such as, for example,
the DNA coding system, or neural systems) and see how they use the capacity of their
channel. It is not unreasonable to assume that evolution will have done a pretty good
job of optimizing channel use \ldots
\end{itemize}
\pagedone
\quotesection{Tools}
%%tth:\begin{quote}
``It is a recurring experience of scientific progress that what was yesterday an object of study, of interest in its own right, becomes today something to be taken for granted, something understood and reliable, something known and familiar -- a tool for further research and discovery.''
-J. R. Oppenheimer, 1953
``Nature uses only the longest threads to weave her patterns, so that each
small piece of her fabric reveals the organization of the entire tapestry.''
- Richard Feynman
%%tth:\end{quote}
\pagedone
\sectionhead{Application to Biology (analyzing genomes)}
\begin{itemize}
\item Let us apply some of these ideas to the (general) problem of analyzing genomes.
We can start with an example such as the comparatively small genome of Escherichia coli,
strain K-12, substrain MG1655, version M52. This example has the convenient features:
\begin{enumerate}
\item It has been completely sequenced.
\item The sequence is available for downloading
\href{http://www.genome.wisc.edu/}{(http://www.genome.wisc.edu/)}.
\item Annotated versions are available for further work.
\item It is large enough to be interesting (somewhat over 4 mega-bases, or 4 million
nucleotides), but not so huge as to be completely unwieldy.
\item The labels on the printouts tend to make other people using the printer
a little nervous :-)
\item Here's the beginning of the file:
\begin{verbatim}
>gb|U00096|U00096 Escherichia coli
K-12 MG1655 complete genome
AGCTTTTCATTCTGACTGCAACGGGCAATATGTCT
CTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC
TTCTGAACTGGTTACCTGCCGTGAGTAAATTAAAA
TTTTATTGACTTAGGTCACTAAATACTTTAACCAA
TATAGGCATAGCGCACAGACAGATAAAAATTACAG
AGTACACAACATCCATGAAACGCATTAGCACCACC
ATTACCACCACCATCACCATTACCACAGGTAACGG
TGCGGGCTGACGCGTACAGGAAACACAGAAAAAAG
CCCGCACCTGACAGTGCGGGCTTTTTTTTTCGACC
AAAGGTAACGAGGTAACAACCATGCGAGTGTTGAA
\end{verbatim}
\end{enumerate}
\end{itemize}
\pagedone
\begin{itemize}
\item In this exploratory project, my goal has been to apply the {\em information}
and {\em entropy} ideas outlined above to genome analysis. Some of the results I have
so far are tantalizing. For a while, I'll just walk you through some preliminary work.
While I am not an expert in genomes/DNA, I am hoping that
some of what I am doing can bring fresh eyes to the problems
of analyzing genome sequences, without too many preconceptions. It is
at least conceivable that my naivet\'e will be an advantage \ldots
\pagedone
\item My first step was to generate for myself a ``random genome'' of comparable
size to compare things with. In this case, I simply used the Unix `random'
function to generate a file containing a random sequence of about 4 million
A, C, G, T. In the actual genome, these letters stand for the nucleotides
adenine, cytosine, guanine, and thymine.
Other people working in this area have taken some other approaches to this
process, such as randomly shuffling an actual genome (thus maintaining the
relative proportions of A, C, G, and T). Part of the justification for this
methodology is that actual (identified) coding sections of DNA tend to have
a ratio of C+G to A+T different from one. I didn't worry about this issue
(for various reasons).
\pagedone
\item My next step was to start developing a (variety of) probability model(s) for
the genome. The general idea that I am working on is to build
some automated tools to locate ``interesting'' sections of a genome. Thinking
of DNA as a coding system, we can hope that ``important'' stretches
of DNA will have entropy different from other stretches. Of course,
as noted above, the entropy measure depends in an essential way on the
probability model attributed to the source. We will want to try to build a
model that catches important aspects of what we find interesting or significant.
We will want to use our knowledge of the systems in which DNA is embedded
to guide the development of our models. On the other hand, we probably
don't want to constrain the model too much. Remember that information
and entropy are measures of {\em unexpectedness}. If we constrain our model
too much, we won't leave any room for the unexpected!
\pagedone
\item We know, for example, that simple repetitions have low entropy. But if
the code being used is redundant (sometimes called {\em degenerate}), with multiple
encodings for the same symbol (as is the case for DNA codons), what looks
to one observer to be a random stream may be recognized by another observer
(who knows the code) to be a simple repetition.
\item The first element of my probability model(s) involves the observation that
coding sequences for peptides and proteins are encoded via {\em codons},
that is, by sequences of blocks of triples of nucleotides. Thus, for example,
the codon AGC on mRNA (messenger RNA) codes for the amino acid serine
(or, if we happen to be reading in the reverse direction, it might code for alanine).
On DNA, AGC codes for UCG or CGA on the mRNA, and thus could code for
cysteine or arginine.
\pagedone
% \centerline{\includegraphics[bb = 0 0 450 562, width = 5in]{genetic_code}}
\centerline{\includegraphics[width = 5in]{genetic_code}}
% \includegraphics[bb = 0 0 450 562, width = 5in]{genetic_code}
Amino acids specified by each codon sequence on mRNA. \\
A = adenine G = guanine C = cytosine T = thymine U = uracil\\
Table from \href{http://www.accessexcellence.org/AB/GG/}
{http://www.accessexcellence.org}
\pagedone
Key for the above table:
Ala: Alanine \\ Arg: Arginine \\ Asn: Asparagine \\ Asp: Aspartic acid \\
Cys: Cysteine \\ Gln: Glutamine \\ Glu: Glutamic acid \\ Gly: Glycine \\
His: Histidine \\ Ile: Isoleucine \\ Leu: Leucine \\ Lys: Lysine \\
Met: Methionine \\ Phe: Phenylalanine \\ Pro: Proline \\ Ser: Serine \\
Thr: Threonine \\ Trp: Tryptophane \\ Tyr: Tyrosine \\ Val: Valine \\
\end{itemize}
\pagedone
\begin{itemize}
\item For our first model, we will consider each three-nucleotide codon to be a
distinct symbol. We can then take a chunk of genome and estimate the probability
of occurence of each codon by simply counting and dividing by the length.
At this level, we are assuming we have no knowledge of where codons start,
and so in this model, we assume that ``readout'' could begin at any nucleotide.
We thus use each three adjacent nucleotides.
For example, given the DNA chunk:
\begin{verbatim}
AGCTTTTCATTCTGACTGCAACGGGCAATATGTC
\end{verbatim}
we would count:
\begin{verbatim}
AAT 1 AAC 1 ACG 1 ACT 1 AGC 1
ATA 1 ATG 1 ATT 1 CAA 2 CAT 1
CGG 1 CTG 2 CTT 1 GAC 1 GCA 2
GCT 1 GGC 1 GGG 1 GTC 1 TAT 1
TCA 1 TCT 1 TGA 1 TGC 1 TGT 1
TTC 2 TTT 2
\end{verbatim}
\pagedone
\item We can then estimate the entropy of the chunk as:
$$\sum p_i * \log_2(1/p_i) = 4.7\ bits.$$
The maximum possible entropy for this chunk would be:
$$\log_2(27) = 4.755\ bits.$$
\item We want to find ``interesting'' sections (and features) of
a genome. As a starting place, we can slide a ``window'' over the genome,
and estimate the entropy within the window. The plot below shows the
entropy estimates for the E. coli genome, within a window of size 6561
($= 3^8$). The window
is slid in steps of size 81 ($= 3^4$). This results in 57,194 values, one for
each placement of the window. For comparison, the values for a ``random''
genome are also shown.
\end{itemize}
%\pagedone
\vfill
\centerline{\includegraphics[width = 5.5in]{ecoli+random_6561_81}}
\centerline{Entropy of E. coli and random}
%\centerline{and random,}
\centerline{window 6561, slide-step 81}
\pagedone
\begin{itemize}
\item At this level, we can make the simple observation that the actual genome
values are quite different from the comparative random string. The values
for E. coli range from about 5.8 to about 5.96, while the random values are clustered
quite closely above 5.99 (the maximum possible is $\log_2(64) = 6$).
\item From here, there are various directions we could go. With a given window
size and step size (e.g., 6561:81, as in the given plot),
we can look at interesting features of the {\em entropy} estimates.
For example, we could look at regions with high entropy, or low entropy.
We could look at regions where there are abrupt changes in entropy, or
regions where entropy stays relatively stable.
\pagedone
\item We could change the window size, and/or step size. We could work
to develop adaptive algorithms which zoom in on interesting regions, where
``interesting'' is determined by criteria such as the ones listed above.
\item We could take known coding regions of genomes, and develop entropy
``fingerprints'' which we could then try to match.
\item There are various ``data massage'' techniques we could use. For example,
we could take the fourier transform of the entropy estimates, and explore that.
Below is an example of such a fourier transform. Notice that it has some
interesting ``periodic'' features which might be worth exploring. It is also
interesting to note that the fourier transform of the entropy of a ``random''
genome has the shape of approximately $1/f = 1/f^1$ (not unexpected \ldots), whereas
the E. coli data are closer to $1/f^{1.5}$.
\item The discrete Fourier transform of a sequence $(a_j)_{j=0}^{q-1}$
is the sequence
$(A_k)_{k=0}^{q-1}$
where
$$A_k=\frac{1}{\sqrt{q}}\sum_{j=0}^{q-1}a_je^{\frac{2\pi ijk}{q}}$$
One way to think about this is that
$(A_k) = F ((a_j))$
where the linear transformation $F$ is given by:
$$[F]_{j,k}=\frac{1}{\sqrt{q}}e^{\frac{2\pi ijk}{q}}$$
Note that the inverse of $F$ is its conjugate transpose $F^\dag$ -- that is,
$$[F^{-1}]_{k,j}=\frac{1}{\sqrt{q}}e^{-\frac{2\pi ijk}{q}}.$$
The plots that follow are log-log plots of the norms
$|A_k| = (A_k\bar{A_k})^{1/2}$ (power spectra).
\end{itemize}
\pagedone
%\vfill
%\centerline{\includegraphics[width = 6in]{ecoli_fft}}
\centerline{\includegraphics[width = 5.5in]{ecoli_6561_81_fft}}
\centerline{Fourier transform of E. coli}
\centerline{window 6561, slide-step 81}
\pagedone
%\vfill
\centerline{\includegraphics[width = 5.5in]{random_6561_81_fft}}
\centerline{Fourier transform of random}
\centerline{window 6561, slide-step 81}
\pagedone
\pagedone
\sectionhead{Some other measures}
\begin{itemize}
\item There have been various approaches to expanding on the idea of entropy
as a measure of complexity. One useful generalization of entropy was developed
by the Hungarian mathematician A. R\'enyi. His method involves looking
at the moments of order $q$ of a probability distribution $\{p_i\}$:
$$S_q = \frac{1}{q-1}\log\sum_i p_i^q$$
If we take the limit as $q\rightarrow 1$, we get:
$$S_1 = \sum_i p_i\log(1/p_i),$$
the entropy we have previously defined. We can then think of $S_q$ as
a generalized entropy for any real number $q$.
\pagedone
\item Expanding on these generalized entropies, we can then define a
generalized {\em dimension} associated with a data set. If we imagine
the data set to be distributed among bins of diameter $r$, we can let $p_i$ be the
probability that a data item falls in the $i$'th bin (estimated by counting the
data elements in the bin, and dividing by the total number of items). We can
then, for each $q$, define a dimension:
$$D_q = \lim_{r\rightarrow 0}\frac{1}{q-1}\frac{\log\sum_ip_i^q}
{ \log(r)}.$$
\item Why do we call this a generalized {\em dimension}?
Consider $D_0$.
First, we will adopt the (analyst's?) convention that $p_i^0 = 0$ when $p_i = 0$. Also,
let $N_r$ be the number of non-empty bins (i.e., the number of bins of diameter
$r$ it takes to cover the data set).
Then we have:
$$D_0 = \lim_{r\rightarrow 0}\frac{\log\sum_ip_i^0}{\log(1/r)}
= \lim_{r\rightarrow 0}\frac{\log(N_r)}{\log(1/r)}$$
Thus, $D_0$ is the Hausdorff dimension $D$, which is frequently in the
literature called the {\em fractal dimension} of the set.
Three examples:
\begin{enumerate}
\item Consider the unit interval $[0,1]$. Let $r_k = 1/2^k$. Then
$N_{r_k} = 2^k$, and
$$D_0 = \lim_{k\rightarrow\infty}\frac{\log(2^k)}{\log(2^k)} = 1.$$
\item Consider the unit square $[0,1]X[0,1]$. Again, let $r_k = 1/2^k$. Then
$N_{r_k} = 2^{2k}$, and
$$D_0 = \lim_{k\rightarrow\infty}\frac{\log(2^{2k})}{\log(2^k)} = 2.$$
\pagedone
\item Consider the {\em Cantor set}:
\vspace{1cm}
%\centerline{
\includegraphics[width = 5.5in]{Cantor_set}
%}
The construction of the Cantor set is suggested by the diagram. The
Cantor set is what remains from the interval after we have removed
middle thirds countably many times. It is an
uncountable set, with measure (``length'') 0. For this set we will
let $r_k = 1/3^k$. Then $N_{r_k} = 2^k$, and
$$D_0 = \lim_{k\rightarrow\infty}\frac{\log(2^k)}{\log(3^k)}
= \frac{\log(2)}{\log(3)} \approx 0.631.$$
The Cantor set is a traditional example of a {\em fractal}. It is
self similar, and has $D_0 \approx 0.631$, which is strictly greater
than its topological dimension $(=\ 0)$.
\pagedone
It is an important example since many nonlinear dynamical systems have
trajectories which are locally the product of a Cantor set with a manifold
(i.e., Poincar\'e sections are generalized Cantor sets).
An interesting example of this phenomenon occurs with the {\em logistics}
equation:
$$x_{i+1} = k * x_i * (1 - x_i)$$
with $k > 4$. In this case (of which you rarely see pictures \ldots), most
starting points run off rapidly to $- \infty$, but there is a {\em strange
repellor}(!) which is a Cantor set. It is a {\em repellor} since arbitrarily
close to any point on the trajectory are points which run off to $- \infty$.
One thing this means is that any finite precision simulation will not capture
the {\em repellor} \ldots
\end{enumerate}
\pagedone
\item We can make several observations about $D_q$:
\begin{enumerate}
\item If $q_1 \le q_2$, then $D_{q_1} \le D_{q_2}$.
\item If the set is strictly self-similar with equal probabilities $p_i = 1/N$, then
we do not need to take the limit as $r\rightarrow 0$, and
\begin{eqnarray*}
D_q & = & \frac{1}{q-1}\frac{\log(N * (1/N)^q)}{\log(r)}\\
& = & \frac{\log(N)}{\log(1/r)}\\
& = & D_0
\end{eqnarray*}
for all $q$. This is the case, for example, for the Cantor set.
\item $D_1$ is usually called the {\em information dimension}:
$$D_1 = \lim_{r\rightarrow 0}\frac{\sum_ip_i * \log(1/p_i)}{\log(r)}$$
The numerator is just the entropy of the probability distribution.
\item $D_2$ is usually called the {\em correlation dimension}:
$$D_2 = \lim_{r\rightarrow 0}\frac{\log \sum_ip_i^2}{\log(r)}$$
This dimension is related to the probability of finding two elements of
the set within a distance $r$ of each other.
\end{enumerate}
\end{itemize}
\pagedone
\sectionhead{Some additional material}
What follows are some additional examples, and expanded discussion of
some topics \ldots
\pagedone
\sectionhead{Examples using Bayes' Theorem}
\begin{itemize}
\item A quick example: \newline
Suppose that you are asked by a friend to help them understand the results of
a genetic screening test they have taken. They have been told that they
have tested positive, and that the test is 99\% accurate. What is the
probability that they actually have the anomaly?
%% (Hint: We don't yet have enough information \ldots)
%% \pagedone
You do some research, and find out that the test screens for a genetic anomaly
that is believed to occur in one person out of 100,000 on average. The lab that
does the tests guarantees that the test is 99\% accurate. You push the question,
and find that the lab says that one percent of the time, the test falsely reports
the absence of the anomaly when it is there, and one percent of the time the test
falsely reports the presence of the anomaly when it is not there.
The test has come
back positive for your friend. How worried should they be? Given this much
information, what can you calculate as the probability they actually have the anomaly?
In general, there are four possible situations for an individual being tested:
\begin{enumerate}
\item Test positive (Tp), and have the anomaly (Ha).
\item Test negative (Tn), and don't have the anomaly (Na).
\item Test positive (Tp), and don't have the anomaly (Na).
\item Test negative (Tn), and have the anomaly (Ha).
\end{enumerate}
\pagedone
We would like to calculate for our friend the probability they actually have
the anomaly (Ha), given that they have tested positive (Tp):
$$P(Ha \vert Tp).$$
We can do this using Bayes' Theorem.
We can calculate:
\begin{eqnarray*}
P(Ha \vert Tp) & = & \frac{P(Tp \vert Ha) * P(Ha)}{P(Tp)}.
\end{eqnarray*}
We need to figure out the three items on the right side of the equation. We can
do this by using the information given.
\pagedone
Suppose the screening test was done on
10,000,000 people. Out of these $10^7$ people, we expect there to be
$10^7/10^5 = 100$ people with the anomaly, and 9,999,900 people without the
anomaly. According to the lab, we would expect the test results to be:
\begin{itemize}
\item Test positive (Tp), and have the anomaly (Ha):
$$ 0.99 * 100 = 99\ \mathrm{people}.$$
\item Test negative (Tn), and don't have the anomaly (Na):
$$ 0.99 * 9,999,900 = 9,899,901\ \mathrm{people}.$$
\item Test positive (Tp), and don't have the anomaly (Na):
$$ 0.01 * 9,999,900 = 99,999\ \mathrm{people}.$$
\item Test negative (Tn), and have the anomaly (Ha):
$$ 0.01 * 100 = 1\ \mathrm{person}.$$
\end{itemize}
\pagedone
Now let's put the the pieces together:
\begin{eqnarray*}
P(Ha) & = & \frac{1}{100,000}\\
\\
& = & 10^{-5}\\
\\
P(Tp) & = & \frac{99 + 99,999}{10^7}\\
\\
& = & \frac{100,098}{10^7}\\
\\
& = & 0.0100098\\
\\
P(Tp \vert Ha) & = & 0.99
\end{eqnarray*}
\pagedone
Thus, our calculated probability that our friend actually has the anomaly is:
\begin{eqnarray*}
P(Ha \vert Tp) & = & \frac{P(Tp \vert Ha) * P(Ha)}{P(Tp)}\\
\\
& = & \frac{0.99 * 10^{-5}}{0.0100098}\\
\\
& = & \frac{9.9 * 10^{-6}}{1.00098 * 10^{-2}}\\
\\
& = & 9.890307 * 10^{-4}\\
\\
& < & 10^{-3}
\end{eqnarray*}
In other words, our friend, who has tested {\em positive}, with a test that
is 99\% correct, has less that one chance in 1000 of actually having the anomaly!
\pagedone
\item There are a variety of questions we could ask now, such as, ``For this anomaly,
how accurate would the test have to be for there to be a greater than 50\%
probability that someone who tests positive actually has the anomaly?''
For this, we need fewer false positives than true positives. Thus, in
the example, we would need fewer than 100 false positives out of the
9,999,900 people who do not have the anomaly. In other words, the proportion
of those without the anomaly for whom the test would have to be correct
would need to be greater than:
$$\frac{9,999,800}{9,999,900} = 99.999\%$$
\pagedone
\item Another question we could ask is, ``How prevalent would an anomaly have to be
in order for a 99\% accurate test (1\% false positive and 1\% false negative)
to give a greater than 50\% probability of actually having the anomaly when
testing positive?''
Again, we need fewer false positives than true positives. We would therefore
need the actual occurrence to be greater than 1 in 100 (each false positive
would be matched by at least one true positive, on average).
\pagedone
\item Note that the current population of the US is about 280,000,000 and the
current population of the world is about 6,200,000,000. Thus, we could
expect an anomaly that affects 1 person in 100,000 to affect about 2,800
people in the US, and about 62,000 people worldwide, and one affecting
one person in 100 would affect 2,800,000 people in the US, and 62,000,000
people worldwide \ldots
%%\pagedone
\item Another example: suppose the test were not so accurate? Suppose the test
were 80\% accurate (20\% false positive and 20\% false negative). Suppose
that we are testing for a condition expected to affect 1 person in 100.
What would be the probability that a person testing positive actually has
the condition?
\pagedone
We can do the same sort of calculations.
Let's use 1000
people this time. Out of this sample, we would expect 10 to have the
condition.
\begin{itemize}
\item Test positive (Tp), and have the condition (Ha):
$$ 0.80 * 10 = 8\ \mathrm{people}.$$
\item Test negative (Tn), and don't have the condition (Na):
$$ 0.80 * 990 = 792\ \mathrm{people}.$$
\item Test positive (Tp), and don't have the condition (Na):
$$ 0.20 * 990 = 198\ \mathrm{people}.$$
\item Test negative (Tn), and have the condition (Ha):
$$ 0.20 * 10 = 2\ \mathrm{people}.$$
\end{itemize}
\pagedone
Now let's put the the pieces together:
\begin{eqnarray*}
P(Ha) & = & \frac{1}{100}\\
\\
& = & 10^{-2}\\
\\
P(Tp) & = & \frac{8 + 198}{10^3}\\
\\
& = & \frac{206}{10^3}\\
\\
& = & 0.206\\
\\
P(Tp \vert Ha) & = & 0.80
\end{eqnarray*}
\pagedone
Thus, our calculated probability that our friend actually has the anomaly is:
\begin{eqnarray*}
P(Ha \vert Tp) & = & \frac{P(Tp \vert Ha) * P(Ha)}{P(Tp)}\\
\\
& = & \frac{0.80 * 10^{-2}}{0.206}\\
\\
& = & \frac{8 * 10^{-3}}{2.06 * 10^{-1}}\\
\\
& = & 3.883495 * 10^{-2}\\
\\
& < & .04
\end{eqnarray*}
In other words, one who has tested {\em positive}, with a test that
is 80\% correct, has less that one chance in 25 of actually having this
condition. (Imagine for a moment, for example, that this is a drug test
being used on employees of some corporation \ldots)
\pagedone
\item We could ask the same kinds of questions we asked before:
\begin{enumerate}
\item How accurate would the test have to be to get a better than 50\%
chance of actually having the condition when testing positive?
(99\%)
\item For an 80\% accurate test, how frequent would the condition
have to be to get a better than 50\% chance?
(1 in 5)
\end{enumerate}
\pagedone
\item Some questions:
\begin{enumerate}
\item Are these examples realistic? If not, why not?
\item What sorts of things could we do to improve our results?
\item Would it help to repeat the test? For example, if the
probability of a false positive is 1 in 100, would that mean
that the probability of two false positives on the same
person would be 1 in 10,000 ($\frac{1}{100} * \frac{1}{100}$)?
If not, why not?
\item In the case of a medical condition such as a genetic anomaly,
it is likely that the test would not be applied randomly, but would
only be ordered if there were other symptoms suggesting the anomaly.
How would this affect the results?
\end{enumerate}
\item Another example: \newline
Suppose that Tom, having had too much time on his hands while an undergraduate
Philosophy major, through much practice at prestidigitation, got to the point where if
he flipped a coin, his flips would have the probabilities:
$$P(h) = 0.7,\ P(t) = 0.3.$$
Now suppose further that you are brought into a room with 10 people in it, including
Tom, and on a table is a coin showing heads. You are told further that one of the 10
people was chosen at random, that the chosen person flipped the coin and put it on the
table, and that research shows that the overall average for the 10 people each flipping
coins many times is:
$$P(h) = 0.52,\ P(t) = 0.48.$$
What is the probability that it was Tom who flipped the coin?
By Bayes' Theorem, we can calculate:
\begin{eqnarray*}
P(\mathrm{Tom} \vert h) & = & \frac{P(h \vert \mathrm{Tom}) P(\mathrm{Tom})}{P(h)}
= \frac{0.7 * 0.1}{0.52}\\
& = & 0.1346.
\end{eqnarray*}
Note that this estimate revises our {\em a priori} estimate of the probability of Tom being
the flipper up from 0.10.
This process (revising estimated probability) of course depends in a critical way on
having {\em a priori} estimates in the first place \ldots
\end{itemize}
\pagedone
\sectionhead{Analog channels}
\begin{itemize}
\item The part of Shannon's work we have looked at so far deals with discrete
(or digital) signalling systems. There are related ideas for continuous
(or analog) systems. What follows gives a brief hint of some of
the ideas, without much detail.
\item Suppose we have a signalling system using band-limited signals (i.e.,
the frequencies of the transmissions are restricted to lie within some
specified range). Let us call the bandwidth $W$. Let us further assume
we are transmitting signals of duration $T$. In order to reconstruct a given
signal, we will need $2WT$ samples of the signal. Thus, if we are sending
continuous signals, each signal can be represented by $2WT$ numbers $x_i$,
taken at equal intervals. We can associate with each signal an {\em energy},
given by:
$$E = \frac{1}{2W}\sum_{i = 1}^{2WT}x_i^2.$$
The distance of the signal (from the origin) will be
$$r = \left(\sum x_i^2\right)^{1/2} = (2WE)^{1/2}$$
We can define the {\em signal power} to be the average energy:
$$S = \frac{E}{T}.$$
Then the radius of the sphere of transmitted signals will be:
$$r = (2WST)^{1/2}.$$
Each signal will be disturbed by the noise in the channel. If we measure the power
of the noise $N$ added by the channel, the disturbed signal will lie in a sphere
around the original signal of radius $(2WNT)^{1/2}$. Thus the original
sphere must be enlarged to a larger radius to enclose the disturbed signals.
The new radius will be:
$$r = \left(2WT(S + N)\right)^{1/2}.$$
In order to use the channel effectively and minimize error (misreading of
signals), we will want to put the signals in the sphere, and separate them
as much as possible (and have the distance between the signals at least twice
what the noise contributes \ldots). We thus want to divide the sphere up
into sub-spheres of radius $= (2WNT)^{1/2}.$ From this, we can get an
upper bound on the number $M$ of possible messages that we can reliably
distinguish. We can use the formula for the volume of an $n$-dimensional
sphere:
$$V(r, n) = \frac{\pi^{n/2}r^n}{\Gamma(n/2 + 1)}.$$
We have the bound:
\begin{eqnarray*}
M & \le & \frac{\pi^{WT}\left(2WT(S + N)\right)^{WT}}{\Gamma(WT + 1)}
\frac{\Gamma(WT + 1)}{\pi^{WT}(2WTN)^{WT}}\\
& = & \left(1 + \frac{S}{N}\right)^{WT}
\end{eqnarray*}
The information sent is the $\log$ of the number of messages sent (assuming
they are equally likely), and hence:
$$I = \log(M) = WT * \log\left(1 + \frac{S}{N}\right),$$
and the rate at which information is sent will be:
$$W * \log\left(1 + \frac{S}{N}\right).$$
We thus have the usual {\em signal/noise} formula for channel capacity \ldots
\pagedone
\item An amusing little side light: ``Random'' band-limited natural phenoma
typically display a power spectrum that obeys a power law of the general form $\frac{1}{f^\alpha}$. On the other
hand, from what we have seen, if we want to use a channel optimally,
we should have essentially equal power at all frequencies in the band.
This means that a possible way to engage in SETI (the search for
extra-terrestrial intelligence) will be to look for bands in which there
is white noise! White noise is likely to be the signature of (intelligent)
optimal use of a channel \ldots
\end{itemize}
\pagedone
\sectionhead{A Maximum Entropy Principle}
\begin{itemize}
\item Suppose we have a system for which we can measure certain macroscopic
characteristics. Suppose further that the system is made up of many microscopic
elements, and that the system is free to vary among various states. Given
the discussion above, let us assume that with probability essentially equal to
1, the system will be observed in states with maximum entropy.
We will then sometimes be able to gain understanding of the system by
applying a {\em maximum information entropy} principle (MEP), and, using
Lagrange multipliers, derive formulae for aspects of the system.
\pagedone
\item Suppose we have a set of macroscopic measurable characteristics $f_k$,
$k = 1, 2, \ldots, M$
(which we can think of as constraints on the system), which we assume are
related to microscopic characteristics via:
$$\sum_ip_i * f_i^{(k)} = f_k.$$
Of course, we also have the constraints:
$$p_i \ge 0,\ \mathrm{and}$$
$$\sum_ip_i = 1.$$
We want to maximize the entropy, $\sum_ip_i\log(1/p_i)$, subject to these
constraints. Using Lagrange multipliers $\lambda_k$ (one for each constraint),
we have the general solution:
$$p_i = \exp\left(- \lambda - \sum_k\lambda_kf_i^{(k)}\right).$$
If we define $Z$, called the partition function, by
$$Z(\lambda_1, \ldots, \lambda_M) = \sum_i\exp\left(-\sum_k\lambda_kf_i^{(k)}\right),$$
then we have $e^\lambda = Z$, or $\lambda = \ln(Z)$.
\end{itemize}
\pagedone
\sectionhead{Application: Economics I (a Boltzmann Economy)}
\begin{itemize}
\item Our first example here is a very simple economy. Suppose there is
a fixed amount of money ($M$ dollars), and a fixed number of agents ($N$)
in the economy.
Suppose that during each time step, each agent randomly selects another
agent and transfers one dollar to the selected agent. An agent having
no money doesn't go in debt. What
will the long term (stable) distribution of money be?
This is not a very realistic economy -- there is no growth, only a redistribution
of money (by a random process). For the sake of argument, we can imagine
that every agent starts with approximately the same amount of money, although
in the long run, the starting distribution shouldn't matter.
\pagedone
\item For this example, we are interested in looking at the distribution of money
in the economy, so we are looking at the
probabilities $\{p_i\}$ that an agent has the amount of money $i$. We are
hoping to develop a model for the collection $\{p_i\}$.
If we let $n_i$ be the number of agents who have $i$ dollars, we have two
constraints:
$$\sum_in_i * i = M$$
and
$$\sum_in_i = N.$$
Phrased differently (using $p_i = \frac{n_i}{N}$), this says
$$\sum_ip_i * i = \frac{M}{N}$$
and
$$\sum_ip_i = 1.$$
\pagedone
\item We now apply Lagrange multipliers:
\begin{eqnarray*}
L = \sum_ip_i\ln(1/p_i)
& - &\lambda\left[\sum_ip_i*i - \frac{M}{N}\right] \\
& - & \mu\left[\sum_ip_i - 1\right],
\end{eqnarray*}
from which we get
$$\frac{\partial L}{\partial p_i} = -[1 + \ln(p_i)] - \lambda i - \mu = 0.$$
We can solve this for $p_i$:
$$\ln(p_i) = - \lambda i - (1 + \mu)$$
and so
$$p_i = e^{-\lambda_0}e^{-\lambda i}$$
(where we have set $1 + \mu \equiv \lambda_0)$.
\pagedone
\item Putting in constraints, we have
\begin{eqnarray*}
1 & = & \sum_i p_i \\
& = & \sum_i e^{-\lambda_0}e^{-\lambda i}\\
& = & e^{-\lambda_0} \sum_{i = 0}^M e^{-\lambda i},
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{M}{N} & = & \sum_i p_i * i \\
& = & \sum_i e^{-\lambda_0}e^{-\lambda i} * i \\
& = & e^{-\lambda_0} \sum_{i = 0}^M e^{-\lambda i} * i.
\end{eqnarray*}
We can approximate (for large $M$)
$$ \sum_{i = 0}^M e^{-\lambda i} \approx \int_0^Me^{-\lambda x}dx
\approx \frac{1}{\lambda},$$
and
$$\sum_{i = 0}^M e^{-\lambda i} * i \approx \int_0^Mxe^{-\lambda x}dx
\approx \frac{1}{\lambda^2}.$$
\pagedone
From these we have (approximately)
$$e^{\lambda_0} = \frac{1}{\lambda}$$
and
$$e^{\lambda_0}\frac{M}{N} = \frac{1}{\lambda^2}.$$
From this, we get
$$\lambda = \frac{N}{M} = e^{-\lambda_0},$$
and thus (letting $T = \frac{M}{N}$) we have:
\begin{eqnarray*}
p_i & = & e^{-\lambda_0}e^{-\lambda i} \\
& = & \frac{1}{T}e^{-\frac{i}{T}}.
\end{eqnarray*}
This is a Boltzmann-Gibbs distribution, where we can think of $T$ (the
average amount of money per agent) as the
``temperature,'' and thus we have a ``Boltzmann economy'' \ldots
Note: this distribution also solves the functional equation
$$p(m_1)p(m_2) = p(m_1 + m_2).$$
\pagedone
\item This example, and related topics, are discussed in
{\em Statistical mechanics of money}
by Adrian Dragulescu and Victor M. Yakovenko,
\tthdump{\href{http://arxiv.org/abs/cond-mat/0001432}
{http://arxiv.org/abs/cond-mat/0001432}}
%%tth:\href{http://arxiv.org/abs/cond-mat/0001432}
%%tth:{http://arxiv.org/abs/cond-mat/0001432}
and
{\em Statistical mechanics of money: How saving propensity affects its distribution}
by Anirban Chakraborti and Bikas K. Chakrabarti
\tthdump{\href{http://arxiv.org/abs/cond-mat/0004256}
{http://arxiv.org/abs/cond-mat/0004256}}
%%tth:\href{http://arxiv.org/abs/cond-mat/0004256}
%%tth:{http://arxiv.org/abs/cond-mat/0004256}
\end{itemize}
\pagedone
\sectionhead{Application: Economics II (a power law)}
\begin{itemize}
\item Suppose that a (simple) economy is made up of many agents $a$, each with
wealth at time $t$ in the amount of $w(a, t)$. (I'll leave it to you to
come up with a reasonable definition of ``wealth'' -- of course we will
want to make sure that the definition of ``wealth'' is applied consistently
across all the agents.) We can also look at the total wealth in the economy
$W(t) = \sum_aw(a,t)$.
For this example, we are interested in looking at the distribution of wealth
in the economy, so we will assume there is some collection $\{w_i\}$ of
possible values for the wealth an agent can have, and associated probabilities
$\{p_i\}$ that an agent has wealth $w_i$. We are hoping to develop a
model for the collection $\{p_i\}$.
\pagedone
\item In order to apply the maximum entropy principle, we want to look at
global (aggregate/macro) observables of the system that reflect (or are made
up of) characteristics of (micro) elements of the system.
For this example, we can look at the growth rate of the economy. A reasonable
way to think about this is to let $R_i = w_i(t_1) / w_i(t_0)$ and
$R = W(t_1)/W(t_0)$ (where $t_0$ and $t_1$ represent time steps of the economy).
The growth rate will then be $\ln(R)$. We then have the two constraints on the
$p_i$:
$$\sum_ip_i * \ln(R_i) = \ln(R)$$
and
$$\sum_ip_i = 1.$$
\pagedone
\item We now apply Lagrange multipliers:
\begin{eqnarray*}
L = \sum_ip_i\ln(1/p_i)
& - &\lambda\left[\sum_ip_i\ln(R_i) - \ln(R)\right] \\
& - & \mu\left[\sum_ip_i - 1\right],
\end{eqnarray*}
from which we get
$$\frac{\partial L}{\partial p_i} = -[1 + \ln(p_i)] - \lambda \ln(R_i) - \mu = 0.$$
We can solve this for $p_i$:
$$p_i = e^{-\lambda_0}e^{-\lambda\ln(R_i)} = e^{-\lambda_0}R_i^{-\lambda}$$
(where we have set $1 + \mu \equiv \lambda_0)$.
Solving, we get $\lambda_0 = \ln(Z(\lambda))$, where $Z(\lambda) \equiv
\sum_i R_i^{-\lambda}$ (the partition function) normalizes the probability
distribution to sum to 1. From this we see the power law (for $\lambda > 1$):
$$p_i = \frac{R_i^{-\lambda}}{Z(\lambda)}.$$
\pagedone
\item We might actually like to calculate specific values of $\lambda$, so we
will do the process again in a continuous version. In this version, we
will let $R = w(T)/w(0)$ be the relative wealth at time T. We want to
find the probability density function $f(R)$, that is:
$$\max_{\{f\}} H(f) = - \int_1^\infty f(R) \ln(f(R))dR,$$
subject to
\begin{eqnarray*}
\int_1^\infty f(R)dR & = & 1, \\
\int_1^\infty f(R)\ln(R)dR & = & C\ln(R),
\end{eqnarray*}
where $C$ is the average number of transactions per time step.
We need to apply the calculus of variations to maximize over a class
of functions.
\pagedone
When we are solving an extremal problem of the form
$$\int F[x, f(x), f'(x)]dx,$$
we work to solve
$$\frac{\partial F}{\partial f(x)} - \frac{d}{dx}
\left(\frac{\partial F}{\partial f'(x)}\right) = 0.$$
Our Lagrangian is of the form
\begin{eqnarray*}
L \equiv & - & \int_1^\infty f(R) \ln(f(R)) dr -
\mu \left(\int_1^\infty f(R) dR - 1\right) \\
& - & \lambda\left(\int_1^\infty f(R)\ln(R)dR - C*\ln(R)\right).
\end{eqnarray*}
Since this does not depend on $f'(x)$, we look at:
$$\frac{\partial[-f(R)\ln f(R) - \mu(f(R) - 1) - \lambda(f(R)\ln R - R)]}
{\partial f(R)}$$
$$= 0$$
from which we get
$$f(R) = e^{-(\lambda_0 - \lambda \ln(R))} = R^{-\lambda}e^{-\lambda_0},$$
where again $\lambda_0 \equiv 1 + \mu$.
\pagedone
We can use the first constraint to solve for $e^{\lambda_0}$:
$$e^{\lambda_0} = \int_1^\infty R^{-\lambda}dR
= \left[\frac{R^{-\lambda + 1}}{1 - \lambda}\right]_1^\infty
= \frac{1}{\lambda - 1},
$$
assuming $\lambda > 1$. We therefore have a power law distribution for
wealth of the form:
$$f(R) = (\lambda - 1)R^{-\lambda}.$$
To solve for $\lambda$, we use:
$$C * \ln(R) = (\lambda - 1)\int_1^\infty R^{-\lambda}\ln(R)dR.$$
Using integration by parts, we get
\begin{eqnarray*}
C*\ln(R)
& = & (\lambda - 1)\left[\ln(R)\frac{R^{1 - \lambda}}{1 - \lambda}
\right]_1^\infty \\
&\ \ & - (\lambda - 1)\int_1^\infty\frac{R^{-\lambda}}{1 - \lambda}dR \\
& = & (\lambda - 1)\left[\ln(R)\frac{R^{1 - \lambda}}{1 - \lambda}
\right]_1^\infty
+ \left[\frac{R^{1 - \lambda}}{1 - \lambda}\right]_1^\infty.
\end{eqnarray*}
\pagedone
By L'H\^opital's rule, the first term goes to zero as $R \to \infty$,
so we are left with
$$C * \ln(R) = \left[\frac{R^{1 - \lambda}}{1 - \lambda}\right]_1^\infty
= \frac{1}{\lambda - 1},$$
or, in other terms,
$$\lambda - 1 = C * \ln(R^{-1}).$$
For much more discussion of this example, see the paper {\em A Statistical
Equilibrium Model of Wealth Distribution} by Mishael Milakovic, February, 2001,
available on the web at:
% \tthdump{\href{http://www.econometricsociety.org/cgi-bin/conference/download.cgi?db\_name=SCE2001\&paper\_id=214}
% {http://www.econometricsociety.org/
% cgi-bin/conference/download.cgi
% ?db\_name=SCE2001\&paper\_id=214}}
% %%tth:\href{http://www.econometricsociety.org/cgi-bin/conference/download.cgi?db\_name=SCE2001\&pape\r_id=214}
% %%tth:{http://www.econometricsociety.org/cgi-bin/conference/download.cgi?db\_name=SCE2001\&paper\_id=214}
\tthdump{\href{http://astarte.csustan.edu/\~tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}
{http://astarte.csustan.edu/\~\ tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}}
%%tth:\href{http://astarte.csustan.edu/~tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}
%%tth:{http://astarte.csustan.edu/~tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}
\end{itemize}
\pagedone
\sectionhead{Application to Physics (lasers)}
\begin{itemize}
\item We can also apply this maximum entropy principle to physics examples. Here is
how it looks applied to a single mode laser. For a laser,
we will be interested in the intensity of the light emitted, and the coherence property
of the light will be observed in the second moment of the intensity. The
electric field strength of such a laser will have the form
$$E(x, t) = E(t)\sin(kx),$$
and E(t) can be decomposed in the form
$$E(t) = Be^{-i\omega t} + B^*e^{i\omega t}.$$
If we measure the intensity of the light over time intervals long compared to the
frequency, but small compared to fluctuations of $B(t)$, the output will be
proportional to $BB^*$ and to the loss rate, $2\kappa$, of the laser:
$$I = 2\kappa B B^*.$$
The intensity squared will be
$$I^2 = 4\kappa^2B^2B^{*2}.$$
\pagedone
\item If we assume that $B$ and $B^*$ are continuous random variables associated with
a stationary process, then the information entropy of the system will be:
$$H = \int p(B,B^*)\log\left(\frac{1}{p(B,B^*)}\right)d^2B.$$
The two constraints on the system will be the averages of the intensity and the
square of the intensity:
\begin{eqnarray*}
f_1 & = & <2\kappa B B^*>,\\
f_2 & = & <4\kappa^2B^2B^{*2}>.
\end{eqnarray*}
Then, of course, we will let
\begin{eqnarray*}
f_{B,B^*}^{(1)} & = & 2\kappa B B^*,\\
f_{B,B^*}^{(2)} & = & 4\kappa^2B^2B^{*2}.
\end{eqnarray*}
We can now use the method outlined above, finding the maximum entropy
general solution derived via Lagrange multipliers for this system.
\pagedone
\item Applying the general solution, we get:
$$p(B, B^*) = \exp\left[- \lambda - \lambda_12\kappa BB^*
- \lambda_24\kappa^2(BB^*)^2\right],$$
or, in other notation:
$$p(B, B^*) = N * \exp(- \alpha|B|^2 - \beta|B|^4).$$
This function in laser physics is typically derived by solving the
Fokker-Planck equation belonging to the Langevin equation for the system.
\item For quick reference, the typical generic Langevin equation looks like:
$$\dot{\mathbf{q}} = K(\mathbf{q}) + \mathbf{F}(t)$$
where $\mathbf{q}$ is a state vector, and the fluctuating forces $F_j(t)$ are typically
assumed to have
\begin{eqnarray*}
& = & 0\\
& = & Q_j\delta_{jj'}\delta(t - t').
\end{eqnarray*}
\pagedone
\item The associated generic Fokker-Planck equation for the distribution function
$f(q, t)$ then looks like:
$$\frac{\partial f}{\partial t} = - \sum_j\frac{\partial}{\partial q_j}(K_jf)
+ \frac{1}{2}\sum_{jk}Q_{jk}\frac{\partial^2}
{\partial q_j\partial q_k}f.$$
The first term is called the drift term, and the second the diffusion term.
This can typically be solved only for special cases \ldots
\item For much more discussion of these topics, I can recommend the
book {\em Information and Self-organization, A Macroscopic Approach
to Complex Systems} by Hermann Haken, Springer-Verlag Berlin, New
York, 1988.
\end{itemize}
\pagedone
\footnotesize
\bibliographystyle{plain}
\tthdump{\hypertarget{References}{}\hyperlink{Our general topics:}{\hfil \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ To top $\leftarrow$}}
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\vspace{-1.0 in}
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%
% \sectionhead{On-line references}
%
%
% Some of the references listed above are available on line. They are listed again here for easy access:
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% %
% %
%
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% Grover L K,
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% %in {\it LANL e-print} quant-ph/9711043, http://xxx.lanl.gov (1997)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9711043}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9711043}
% {http://xxx.lanl.gov/abs/quant-ph/9711043}
%
% %\bibitem{grover4}
% Grover L K,
% A fast quantum mechanical algorithm for estimating the median,
% %in {\it LANL e-print} quant-ph/9607024, http://xxx.lanl.gov (1997)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9607024}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9607024}
% {http://xxx.lanl.gov/abs/quant-ph/9607024}
%
%
% %\bibitem{knill4}
% Knill E, Laflamme R and Zurek W H 1997
% Resilient quantum computation: error models and thresholds
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% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9702058}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9702058}
% {http://xxx.lanl.gov/abs/quant-ph/9702058}
%
% \pagedone
%
%
% %\bibitem{preskill2}
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% %in {\it LANL e-print} quant-ph/9712048, http://xxx.lanl.gov (1997),
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% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9712048}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9712048}
% {http://xxx.lanl.gov/abs/quant-ph/9712048}
%
% %\bibitem{preskill3}
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% %\hyperURL{http}{www.theory.caltech.edu/people/preskill}{ph229}
% {http://www.theory.caltech.edu/people/preskill/ph229}
%
%
% %\bibitem{rieffel}
% Rieffel E, Polak W
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% %{\it LANL e-print} quant-ph/9809016, http://xxx.lanl.gov (1998),
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9809016}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9809016}
% {http://xxx.lanl.gov/abs/quant-ph/9809016}
%
% %\bibitem{Steane-97}
% Steane A,
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% %(preprint in {\it LANL e-print} quant-ph/9708022, http://xxx.lanl.gov)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9708022}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9708022}
% {http://xxx.lanl.gov/abs/quant-ph/9708022}
%
% %\bibitem{zalka2}
% Zalka C,
% Grover's quantum searching algorithm is optimal,
% %in {\it LANL e-print} quant-ph/9711070http://xxx.lanl.gov (1997)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9711070}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9711070}
% {http://xxx.lanl.gov/abs/quant-ph/9711070}
%
\end{document}
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