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Course by: Behnaam Aazhang. E-mail the author

# Typical Sequences

Module by: Behnaam Aazhang. E-mail the author

Summary: A description of typical sequences.

If the binary symmetric channel has crossover probability εε then if xx is transmitted then by the Law of Large Numbers the output yy is different from xx in nε n ε places if nn is very large.

d H xynε d H x y n ε
(1)
The number of sequences of length nn that are different from xx of length nn at nε n ε is
nnε=n!(nε)!(nnε)! n n ε n n ε n n ε
(2)

## Example 1

x=000T x 0 0 0 and ε=13 ε 1 3 and nε=3×13 n ε 3 1 3 . The number of output sequences different from xx by one element: 3!1!2!=3×2×11×2=3 3 1 2 3 2 1 1 2 3 given by 101T 1 0 1 , 011T 0 1 1 , and 000T 0 0 0 .

Using Stirling's approximation

n!nnen2πn n n n n 2 n
(3)
we can approximate
nnε2n(((εlog 2 ε))(1ε)log 2 (1ε))=2n H b ε n n ε 2 n ε 2 ε 1 ε 2 1 ε 2 n H b ε
(4)
where H b ε((εlog 2 ε))(1ε)log 2 (1ε) H b ε ε 2 ε 1 ε 2 1 ε is the entropy of a binary memoryless source. For any xx there are 2n H b ε 2 n H b ε highly probable outputs that correspond to this input.

Consider the output vector YY as a very long random vector with entropy nHY n H Y . As discussed earlier, the number of typical sequences (or highly probably) is roughly 2nHY 2 n H Y . Therefore, 2n 2 n is the total number of binary sequences, 2nHY 2 n H Y is the number of typical sequences, and 2n H b ε 2 n H b ε is the number of elements in a group of possible outputs for one input vector. The maximum number of input sequences that produce nonoverlapping output sequences

M=2nHY2n H b ε=2n(HY H b ε) M 2 n H Y 2 n H b ε 2 n H Y H b ε
(5)

The number of distinguishable input sequences of length nn is

2n(HY H b ε) 2 n H Y H b ε
(6)
The number of information bits that can be sent across the channel reliably per nn channel uses n(HY H b ε) n H Y H b ε The maximum reliable transmission rate per channel use
R=log 2 Mn=n(HY H b ε)n=HY H b ε R 2 M n n H Y H b ε n H Y H b ε
(7)
The maximum rate can be increased by increasing HY H Y . Note that H b ε H b ε is only a function of the crossover probability and can not be minimized any further.

The entropy of the channel output is the entropy of a binary random variable. If the input is chosen to be uniformly distributed with p X 0= p X 1=12 p X 0 p X 1 1 2 .

Then

p Y 0=1 p X 0+ε p X 1=12 p Y 0 1 ε p X 0 ε p X 1 1 2
(8)
and
p Y 1=1 p X 1+ε p X 0=12 p Y 1 1 ε p X 1 ε p X 0 1 2
(9)
Then, HY H Y takes its maximum value of 1. Resulting in a maximum rate R=1 H b ε R 1 H b ε when p X 0= p X 1=12 p X 0 p X 1 1 2 . This result says that ordinarily one bit is transmitted across a BSC with reliability 1ε 1 ε . If one needs to have probability of error to reach zero then one should reduce transmission of information to 1 H b ε 1 H b ε and add redundancy.

Recall that for Binary Symmetric Channels (BSC)

H Y | X = p x 0H Y | X = 0 + p x 1H Y | X = 1 = p x 0(((1ε)log 2 (1ε)εlog 2 ε))+ p x 1(((1ε)log 2 (1ε)εlog 2 ε))=(((1ε)log 2 (1ε)))εlog 2 ε= H b ε H Y | X p x 0 H Y | X = 0 p x 1 H Y | X = 1 p x 0 1 ε 2 1 ε ε 2 ε p x 1 1 ε 2 1 ε ε 2 ε 1 ε 2 1 ε ε 2 ε H b ε
(10)
Therefore, the maximum rate indeed was
R=HYH Y | X =X;Y R H Y H Y | X X Y
(11)

## Example 2

The maximum reliable rate for a BSC is 1 H b ε 1 H b ε . The rate is 1 when ε=0 ε 0 or ε=1 ε 1 . The rate is 0 when ε=12 ε 1 2

This module provides background information necessary for an understanding of Shannon's Noisy Channel Coding Theorem. It is also closely related to material presented in Mutual Information.

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