Any message that brings a specification in a problem which involves a certain degree of uncertainty is called information and it was Shannon [1] who named this measure of information as entropy. In coding theory, the operational role of entropy comes from the source coding theorem which states that if is the entropy of the source letters for a discrete memoryless source, then the sequence of source outputs cannot be represented by a binary sequence using fewer than binary digits per source digit on the average, but it can be represented by a binary sequence using as close to binary digits per source digit on the average as desired. To be clearer, let us consider the discrete source that emits symbols with probability distribution

where. The aim of source coding is to encode the source using an alphabet of size, that is, to map each symbol to a codeword of length expressed using the letters of the alphabet. It is known that if the set of lengths satisfies Kraft’s [2] inequality

then there exists a uniquely decodable code with these lengths, which means that any sequence can be decoded unambiguously into a sequence of symbols. In this respect, Shannon [1] proved the first noiseless coding theorem for the uniquely decipherable code in the form of following inequality

where is a Shannon’s entropy and is the mean codeword length.

Later, Campbell [3] and Kapur [4] proved the source coding theorems for their own exponentiated mean codeword length in the form of following inequalities

and

respectively, where

is Campbell’s [3] mean codeword length,

is Kapur’s [4] mean codeword length and

is Renyi’s [5] measure of entropy.

Recently, Parkash and Kakkar [6] introduced two mean codeword lengths given by

and

Further, the authors provided two source coding theorems which show that for all uniquely decipherable codes, the mean codeword lengths and satisfy the relation:

and

respectively where

is a Kapur’s [4] two parameter additive measure of entropy and

is measure of entropy developed by Parkash and Kakkar [6].

This is to emphasize that in the entire literature of source coding theorems, one can observe that the mean codeword length is lower bounded by the entropy of the source and it can never be less than the entropy of the source but can be made closer to it. This phenomenon provides the idea of absolute redundancy which is the number of bits used to transmit a message minus the number of bits of actual information in the message, that is, the mean codeword length minus the entropy of the source. The objective of the present communication is to minimize this redundancy in order to increase the efficiency of the source encoding. For this purpose we have made use of the concept of escort distribution as follows:

If is the original distribution, then its escort distribution is given by

where for some parameter. Many researchers including Harte [7], Bercher [8,9], Beck and Schloegl [10] etc. used this distribution in their respective findings.

The aim of the present paper is to obtain the optimum probability distribution with which the source should deliver messages in order to minimize the absolute redundancy. To obtain our goal, we have taken into consideration the above mentioned generalized mean codeword lengths. Moreover, the upper bound to these codeword lengths has been found for Huffman [4] encoding.

Let us assume that for discrete source that emits symbols with probability distribution, the codewords having lengths, have been obtained using some encoding procedure on noiseless channel. Further, we assume that entropy of the source is and average codeword length is. Since from (1.7), we have, therefore, the average redundancy of the source code is given by

where and.

In order to minimize the average redundancy, we resort to the following theorem:

Theorem 1: The optimum probability distribution that minimizes the absolute redundancy of the source with entropy and the mean codeword length is the escort distribution, given by

Proof: To minimize the redundancy, we need to minimize

subject to the constraint

To prove this, we first of all, find the extremum of which is equivalent to extremizing and then use the fact that is minimum or maximum will depend upon the value of parameter.

So, in order to extremize, we consider the Lagrangian given by

where is Lagrange’s multiplier.

Now

Letting, we get

Substituting (2.6) in (2.4), we get

Substituting (2.7) in (2.6), we get the result (2.2).

Now,

We see that for and for.

Also,

So, has minimum value for and maximum for.

Thus, has minimum value for and maximum for and consequently, observing the function, we see that it has minimum value for,.

Thus, the minimum value is given by

Again, the necessary condition for the construction of uniquely decipherable codes is given by

Therefore, from (2.9), we have.

NOTE: It is to be noted that if the source is Huffman [11] encoded since for the Huffman encoding, we have

Therefore, for this case, (2.2) becomes

Similarly, if we consider the codeword length which satisfies the relation, then the absolute redundancy of the source code in this case is given by

where and.

Theorem 2. The optimum probability distribution that minimizes the absolute redundancy of the source with entropy and mean codeword length is the escort distribution, given by

Proof: We will find the extremum of which is equivalent to extremizing subject to constraint

Let us consider the Lagrangian given by

where is a Lagrange’s multiplier.

For an extremum, let, that is,

Using (2.14), we get

Substituting (2.17) in (2.16), we get (2.13).

Also,

and

So, reaches its minimum value when and is given by

that is,

Note: Again in this case also, if the source is Huffman [11] encoded, then the probabilities are given by

Next, we will find the upper bound on the codeword lengths and when the source is Huffman encoded.

Theorem 3. The exponentiated codeword length satisfies the following inequality

if the source is encoded using Huffman procedure.

Proof: The exponentiated codeword length can be written in the following form

where.

Considering (2.12), (2.19) becomes

where.

We need to find the extremum of subject to constraint (as the source is encoded using Huffman Procedure).

For this purpose, we first of all, find the extremum of which is equivalent to extremizing and then use the fact that is minimum or maximum depending upon the value of parameter.

So, we consider the Lagrangian given by

where is a Lagrange’s multiplier Put, (2.21) becomes

Letting, we get

Now, gives

Using (2.23) in (2.22), we get

that is,

Now,

We see that for and for.

Also,

So, has minimum value for and maximum for.

Therefore, has minimum value for and maximum for and consequently, observing the exponentiated mean codeword length, we see that it has maximum value for,.

Thus, the maximum value is given by

.

Theorem 4. The mean codeword length is upper bounded by , that is,

if the source is encoded using Huffman procedure.

Proof: The exponentiated codeword length can be written in the following form

We need to find the extremum of subject to constraint (as the source is encoded using Huffman Procedure).

So, we consider the Lagrangian given by

where is a Lagrange’s multiplier .

Letting, we get

Since , we have

Substitute (2.28) in (2.27), we get

Now,

Also,

So, the mean codeword length has maximum value when , and is given by

.

Note-I: For the case of Campbell’s codeword length, we have from (1.3),. So, the average redundancy of the source code in this case is given by

where and

The absolute redundancy in the case of Campbell’s [3] mean codeword length is the same as in case of exponentiated mean codeword length developed by Parkash and Kakkar [6] as given in (2.1). Thus, we see that similar results as proved in theorem (2.1) and theorem (2.3) hold for Campbell’s case also.

Note-II: Absolute redundancy when we use Kapur’s[4] mean codeword length is given by

where

Theorem 5: The optimum probability distribution that minimizes the absolute redundancy of the source with entropy and mean codeword length is given by

Proof: To minimize the redundancy, we need to minimize

subject to the constraint

To prove this, we first of all find the extremum of which is equivalent to extremizing and then using the fact that is minimum or maximum depending upon the value of parameter.

So, in order to extremize, we consider the Lagrangian given by

where is Lagrange’s multiplier.

Letting, we get

Substituting (2.33) in (2.31), we get

Substituting (2.34) in (2.33), we get the result (2.29).

Now,

We see that for and for.

Also,

So, has maximum value for and minimum value for.

Therefore, has maximum value for and minimum value for and consequently observing the function, we see that it has minimum value for,.

The minimum value is given by

.

Theorem 6. The Kapur’s [8] mean codeword length satisfies the following inequality

if the source is encoded using Huffman procedure.

Proof: Proceeding as in Theorem 2.3, we can prove the Theorem 6.

The authors are thankful to Council of Scientific and Industrial Research, New Delhi, for providing the financial assistance for the preparation of the manuscript.