EEET 5101 Information TheoryChapter 1

BY Wai (W2-4)

siuwai.ho@unisa.edu.au

Introduction

Probability Theory

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Basic Course Information Lecturers:

Dr Siu Wai Ho, W2-4, Mawson Lakes Dr Badri Vellambi Ravisankar, W1-22, Mawson Lakes Dr Roy Timo, W1-7, Mawson Lakes

Office Hour: Tue 2:00-5:00pm (starting from 27/7/2010)

Class workload: Homework Assignment 25% Mid-term 25% Final 50%

Textbook: T. M. Cover and J. M. Thomas, Elements of Information Theory, 2nd, Wiley-Interscience, 2006.

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Basic Course Information References: OTHER RELEVANT TEXTS (Library):

1. Information Theory and Network Coding by Raymond Yeung 2. Information Theory: Coding Theorems for Discrete Memoryless Systems by

Imre Csiszar and Janos Korner.

OTHER RELEVANT TEXTS (Online): 3. Probability, Random Processes, and Ergodic Properties by Robert Gray 4. Introduction to Statistical Signal Processing Robert Gray and L. Davisson 5. Entropy and Information Theory by Robert Gray http://ee.stanford.edu/~gray/

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The Beginning of Information Theory

In 1948, Claude E. Shannon published his paper “A Mathematical Theory of Communication” in the Bell Systems Technical Journal.

He introduced two fundamental concepts about “information”: Information can be measured by entropy Information to be transmitted is digital

Information Source Transmitter

Message

Receiver Destination

Message

Signal ReceivedSignal

Noise Source

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The Beginning of Information Theory

In the same paper, he has answered two fundamental questions in communication theory:

What is the ultimate data compression ?

How to minimize the compression rate m/n with Pr{u v} = 0.

What is the ultimate transmission rate of communication?

How to maximize the transmission rate n/m with Pr{k k’} 0.

Source ReceiverEncoder Decoderu = u1 … un v = v1 … vnx1 … xm

Source ReceiverEncoder Decoderk’x1 … xm y1 … ym

Channelk {1,…,2n}

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The Idea of Channel Capacity Example [MacKay 2003]: Suppose we are now provided a noisy channel

We test it 10000 times and find the following statistics Pr{y=0|x=0} = Pr{y=1|x=1} = 0.9; Pr{y=0|x=1} = Pr{y=1|x=0} = 0.1 The occurrence of difference is independent of the previous use

Suppose we want to send a message: s = 0 0 1 0 1 1 0 The error probability = 1 – Pr{no error} = 1 – 0.97 0.5217 How can we get a smaller error probability?

Channelx y

0.9

0.9

0.10 0

11

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The Idea of Channel Capacity Method 1: Repetition codes [R3] To replace the source message by 0 000; 1 111

The original bit error probability Pb : 0.1. The new Pb : = 3 0.9 0.12 + 0.13 = 0.028

bit error probability 0 rate 0 ??

s 0 0 1 0 1 1 0

t 000 000 111 000 111 111 000

n 000 001 000 000 101 000 000

r=tn

000 001 111 000 010 111 000

s’ 0 0 1 0 0 1 0

Majority voteat the receiver

3

1

use channel ofnumber The

ed transmittbits ofnumber The code a of Rate

t: transmitted symbolsn: noiser: received symbols

The Idea of Channel Capacity Method 1: Repetition codes

8pb 0 rate 0pb 0 rate 0

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The Idea of Channel Capacity Method 2: Hamming codes [(7,4) Hamming Code] group 4 bits into s. E.g., s = 0 0 1 0

Here t = GTs = 0 0 1 0 1 1 1, where

1101000

1110100

0110010

1010001

G

The Idea of Channel Capacity Method 2: Hamming codes

Is the search of a good code an everlasting job? Where is the destination? 10

The Idea of Channel Capacity Information theory tells us the fundamental limits.

It is impossible to design a code with coding rate and error probability on the right side of the line. 11

Shannon’ s Channel CodingTheorem

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Intersections with other Fields

Information theory showsthe fundamental limits indifferent communicationsystems

It also provides insightson how to achieve these limits

It also intersects other fields[Cover and Thomas 2006]

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Content in this course 2) Information Measures and Divergence:

2a) Entropy, Mutual Information and Kullback-Leibler Divergence

-Definitions, chain rules, relations 2b) Basic Lemmas & Inequalities:

-Data Processing Inequality, Fano’s Inequality.

3) Asymptotic Equipartition Property (AEP) for iid Random Processes:

3a) Weak Law of Large Numbers 3b) AEP as a consequence of the Weak Law of Large Numbers 3c) Tail event bounding:

-Markov, Chebychev and Chernoff bounds 3d) Types and Typicality

-Strong and weak typicality 3e) The lossless source coding theorem

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Content in this course 4) The AEP for Non-iid Random Processes:

4a) Random Processes with memory

-Markov processes, stationarity and ergodicity 4b) Entropy Rate 4c) The lossless source coding theorem

5) Lossy Compression: 5a) Motivation 5b) Rate-distortion (RD) theory for DMSs (Coding and Converse theorems). 5c) Computation of the RD function (numerical and analytical)

How to minimize the compression rate m/n with u and v satisfying certain distortion criteria.

Source ReceiverEncoder Decoderu = u1 … un v = v1 … vnx1 … xm

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Content in this course 6) Reliable Communication over Noisy Channels:

6a) Discrete memoryless channels

-Codes, rates, redundancy and reliable communication 6b) Shannon’s channel coding theorem and its converse 6c) Computation of channel capacity (numerical and analytical) 6d) Joint source-channel coding and the principle of separation 6e) Dualities between channel capacity and rate-distortion theory 6f) Extensions of Shannon’s capacity to channels with memory (if time

permits)

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Content in this course 7) Lossy Source Coding and Channel Coding with Side-

Information: 7a) Rate Distortion with Side Information

-Joint and conditional rate-distortion theory, Wyner-Ziv coding, extended Shannon lower bound, numerical computation

7b) Channel Capacity with Side Information 7c) Dualities

8) Introduction to Multi-User Information Theory (If time permits):

Possible topics: lossless and lossy distributed source coding, multiple access channels, broadcast channels, interference channels, multiple descriptions, successive refinement of information, and the failure of source-channel separation.

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Prerequisites – Probability Theory

Let X be a discrete random variable taking values from the alphabet The probability distribution of X is denoted by pX = {pX(x), x X}, where

pX(x) means the probability that X = x.

pX(x) 0

x pX(x) = 1

Let SX be the support of X, i.e. SX = {x X: p(x) > 0}.

Example : Let X be the outcome of a dice

Let = {1, 2, 3, 4, 5, 6, 7, 8, 9, …} equal to all positive integers.In this case, is a countably infinite alphabet

SX = {1, 2, 3, 4, 5, 6} which is a finite alphabet

If the dice is fair, then pX(1) = pX(2) = = pX(6) = 1/6.

If is a subset of real numbers, e.g., = [0, 1], is a continuous alphabet and X is a continuous random variable

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Prerequisites – Probability Theory

Let X and Y be random variable taking values from the alphabet X and Y, respectively

The joint probability distribution of X and Y is denoted by pXY and

pXY(xy) means the probability that X = x and Y = y

pX(x), pY(y), pXY(xy) p(x), p(y), p(xy) when there is no ambiguity.

pXY(x) 0

xy pXY(x) = 1

Marginal distributions: pX(x) = y pXY(xy) and pY(y) = x pXY(xy)

Conditional probability: for pX(x) > 0, pY|X(y|x) = pXY(xy)/ pX(x) which denotes the probability that Y = y given the conditional that X = x

Consider a function f: X Y If X is a random variable, f(X) is also random. Let Y = f(X). E.g., X is the outcome of a fair dice and f(X) = (X – 3.5)2

What is pXY?

PY|XX Y

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Expectation and Variance The expectation of X is given by E[X] = x pX(x) x The variance of X is given by E[(X – E[X])2] = E[X2] – (E[X])2

The expected value of f(X) is E[f(X)] = x pX(x) f(x)

The expected value of k(X, Y) is E[k(X, Y)] = xy pXY(xy) k(x,y) We can take the expectation on only Y, i.e.,

EY[k(X, Y)] = y pY(y) k(X,y) which is still a random variable

E.g., Suppose some real-valued functions f, g, k and l are given. What is E[f(X, g(Y), k(X,Y))l(Y)]?

xy pXY(xy) f(x, g(y), k(x,y))l(y) which gives a real value

What is EY[f(X, g(Y), k(X,Y)]l(Y)? y pY(y) f(X, g(y), k(X,y))l(y) which is still a random variable. Usually, this can be done only if X and Y are independent.

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Conditional Independent Two r.v. X and Y are independent if p(xy) = p(x)p(y) x, y For r.v. X, Y and Z, X and Z are independent conditioning on

Y, denoted by X Z | Y if p(xyz)p(y) = p(xy)p(yz) x, y, z ----- (1)

Assume p(y) > 0, p(x, z|y) = p(x|y)p(z|y) x, y, z ----- (2)

If (1) is true, then (2) is also true given p(y) > 0 If p(y) = 0, p(x, z|y) may be undefined for a given p(x, y, z). Regardless whether p(y) = 0 for some y, (1) is a sufficient

condition to test X Z | Y p(xy) = p(x)p(y) is also called pairwise independent

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Mutual and Pairwise Independent

Mutual Indep. : p(x1, x2, …, xn) = p(x1)p(x2) p(xn) Mutual Independent Pairwise Independent Suppose we have i, j s.t. i, j [1, n] and i j Let = [1, n] \ {i, j}

iix

nxpxpxpiix

nxxxpnXXXnXXX

:)()2()1(

:),...,2,1(

2121

)()(

)()()(

2

)2(2

1

)1(1

),(

jxpixpnx

nxnXpjx

jXpixiXp

xxXp

xxXpjxix

jXiXp

ji XX

Pairwise Independent Mutual Independent

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Mutual and Pairwise Independent

Example : Z = X Y and Pr{X=0} = Pr{X=1} = Pr{Y=0} = Pr{Y=1} = 0.5 Pr{Z=0} = Pr{X=0}Pr{Y=0} + Pr{X=1}Pr{Y=1} = 0.5 Pr{Z=1} = 0.5 Pr{X=0, Y=0} = 0.25 = Pr{X=0}Pr{Y=0} Pr{X=0, Z=1} = 0.25 = Pr{X=0}Pr{Z=1} Pr{Y=1, Z=1} = 0.25 = Pr{Y=1}Pr{Z=1} …….. So X, Y and Z are pairwise Independent However, Pr{X=0, Y=0, Z=0} = Pr{X=0}Pr{Y=0} = 0.25 Pr{X=0}Pr{Y=0}Pr{Z=0} = 0.125 X, Y and Z are not mutually Independent but pairwise Independent