Difference between revisions of "Information Theory/Discrete Sources with Memory"

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{{Header
|Untermenü=Entropie wertdiskreter Nachrichtenquellen
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|Untermenü=Entropy of Discrete Sources
|Vorherige Seite=Gedächtnislose Nachrichtenquellen
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|Vorherige Seite=Discrete Memoryless Sources
|Nächste Seite=Natürliche wertdiskrete Nachrichtenquellen
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|Nächste Seite=Natural Discrete Sources
 
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  \hspace{0.01cm}.$$
 
  \hspace{0.01cm}.$$
  
Due to the unequal symbol probabilities the entropy is smaller than the decision content  $H_0 = \log_2 M = 2 \hspace{0.05cm}. \rm bit/symbol$.
+
Due to the unequal symbol probabilities the entropy is smaller than  $H_0 = \log_2 M = 2 \hspace{0.05cm} \rm bit/symbol$.
  
 
[[File:EN_Inf_T_1_2_S1a.png|right|frame|Quaternary source without/with memory]]
 
[[File:EN_Inf_T_1_2_S1a.png|right|frame|Quaternary source without/with memory]]
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  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
*The index "1" is supposed to indicate that  $H_1$  considers only the symbol probabilities and not statistical bindings between symbols within the sequence.  With the decision content  $H_0 = \log_2 \ M$  the following size relation results:
+
*The index "1" is supposed to indicate that  $H_1$  considers only the symbol probabilities and not statistical bindings between symbols within the sequence.  With  $H_0 = \log_2 \ M$  the following size relation results:
 
   
 
   
$$H_0 \ge H_1 \ge H_2
+
:$$H_0 \ge H_1 \ge H_2
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
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Let us now illustrate the calculation of entropy approximations  $H_1$  and  $H_2$  with three examples.
 
Let us now illustrate the calculation of entropy approximations  $H_1$  and  $H_2$  with three examples.
  
 +
[[File:Inf_T_1_2_S2_vers2.png|right|frame|Ternary symbol sequence and formation of two-tuples]]
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
 
$\text{Example 2:}$ 
 
$\text{Example 2:}$ 
We will first look at the sequence  $〈 q_1$, ... , $q_{50} \rangle $  according to the graphic, where the sequence elements  $q_ν$  originate from the alphabet $\rm \{A, \ B, \ C \}$   ⇒   the symbol range is  $M = 3$.
+
We will first look at the sequence  $〈 q_1$, ... , $q_{50} \rangle $  according to the graphic, where the sequence elements  $q_ν$  originate from the alphabet $\rm \{A, \ B, \ C \}$   ⇒   the symbol set size is  $M = 3$.
 
 
[[File:Inf_T_1_2_S2_vers2.png|center|frame|Ternary symbol sequence and formation of two-tuples]]
 
  
 
By time averaging over the  $50$  symbols one gets the symbol probabilities  $p_{\rm A} ≈ 0.5$,   $p_{\rm B} ≈ 0.3$  and  $p_{\rm C} ≈ 0.2$, with which one can calculate the first order entropy approximation:
 
By time averaging over the  $50$  symbols one gets the symbol probabilities  $p_{\rm A} ≈ 0.5$,   $p_{\rm B} ≈ 0.3$  and  $p_{\rm C} ≈ 0.2$, with which one can calculate the first order entropy approximation:
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{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
 
$\text{Example 3:}$ 
 
$\text{Example 3:}$ 
Now let us consider a  ''memoryless  binary source''  with equally probable symbols, i.e. there would be  $p_{\rm A} = p_{\rm B} = 1/2$.  The first twenty subsequent elements are   $〈 q_ν 〉 =\rm BBAAABAABBBBBAAAABABAB$ ...
+
Now let us consider a  '''memoryless  binary source'''  with equally probable symbols, i.e. there would be  $p_{\rm A} = p_{\rm B} = 1/2$.  The first twenty subsequent elements are   $〈 q_ν 〉 =\rm BBAAABAABBBBBAAAABABAB$ ...
 
*Because of the equally probable binary symbols   $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$.
 
*Because of the equally probable binary symbols   $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$.
*The compound probability  $p_{\rm AB}$  of the combination  $\rm AB$  is equal to  $p_{\rm A} - p_{\rm B} = 1/4$.  Also $p_{\rm AA} = p_{\rm BB} = p_{\rm BA} = 1/4$.  
+
*The compound probability  $p_{\rm AB}$  of the combination  $\rm AB$  is equal to  $p_{\rm A} \cdot p_{\rm B} = 1/4$.  Also $p_{\rm AA} = p_{\rm BB} = p_{\rm BA} = 1/4$.  
 
*This gives for the second entropy approximation
 
*This gives for the second entropy approximation
 
   
 
   
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The third sequence considered here results from the binary sequence of  $\text{Example 3}$  by using a simple repeat code:  
 
The third sequence considered here results from the binary sequence of  $\text{Example 3}$  by using a simple repeat code:  
 
:$$〈 q_ν 〉 =\rm BbBbAaAaAaBbAaAaBbBb \text{...} $$
 
:$$〈 q_ν 〉 =\rm BbBbAaAaAaBbAaAaBbBb \text{...} $$
*The repeated symbols are marked by corresponding lower case letters.  It still applies  $M=2$.
+
*The repeated symbols are marked by corresponding lower case letters.  But it still applies  $M=2$.
 
*Because of the equally probable binary symbols, this also results in  $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$.
 
*Because of the equally probable binary symbols, this also results in  $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$.
 
*As shown in  [[Aufgaben:1.3_Entropienäherungen|Exercise 1.3]]  for the compound probabilities we obtain  $p_{\rm AA}=p_{\rm BB} = 3/8$  and  $p_{\rm AB}=p_{\rm BA} = 1/8$.  Hence
 
*As shown in  [[Aufgaben:1.3_Entropienäherungen|Exercise 1.3]]  for the compound probabilities we obtain  $p_{\rm AA}=p_{\rm BB} = 3/8$  and  $p_{\rm AB}=p_{\rm BA} = 1/8$.  Hence
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A closer look at the task at hand leads to the following conclusion:  
 
A closer look at the task at hand leads to the following conclusion:  
*The entropy should actually be  $H = 0.5 \hspace{0.05cm} \rm bit/symbol$  its (every second symbol does not provide new information)  
+
*The entropy should actually be  $H = 0.5 \hspace{0.05cm} \rm bit/symbol$  (every second symbol does not provide new information)  
*The second entropy approximation  $H_2 = 0.906 \hspace{0.05cm} \rm bit/symbol$  but is much larger than the entropy  $H$.
+
*The second entropy approximation  $H_2 = 0.906 \hspace{0.05cm} \rm bit/symbol$  is much larger than the entropy  $H$.
*For the determination of entropy the second order approximation is not sufficient.  rather, larger continuous blocks must be considered with  $k > 2$  symbols.  
+
*For the entropy  determinationthe second order approximation is not sufficient.  Rather, one must consider larger contiguous blocks of  $k > 2$  symbols.
*In the following, such a block is referred to as  $k$-tuple.}}
+
*In the following, such a block is referred to as  "$k$–tuple".}}
  
 
 
 
 
 
==Generalization to $k$-tuple and boundary crossing ==
 
==Generalization to $k$-tuple and boundary crossing ==
 
<br>
 
<br>
For abbreviation we write using the compound probability&nbsp; $p_i^{(k)}$&nbsp; a&nbsp; $k$-tuple in general:
+
For abbreviation we write using the compound probability&nbsp; $p_i^{(k)}$&nbsp; for a&nbsp; $k$&ndash;tuple in general:
 
   
 
   
:$$H_k = \frac{1}{k} \cdot \sum_{i=1}^{M^k} p_i^{(k)} \cdot {\rm log}_2\hspace{0.1cm} \frac{1}{p_i^{(k)}} \hspace{0.5cm}({\rm Einheit\hspace{-0.1cm}: \hspace{0.1cm}bit/Symbol})
+
:$$H_k = \frac{1}{k} \cdot \sum_{i=1}^{M^k} p_i^{(k)} \cdot {\rm log}_2\hspace{0.1cm} \frac{1}{p_i^{(k)}} \hspace{0.5cm}({\rm unit\hspace{-0.1cm}: \hspace{0.1cm}bit/symbol})
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
The control variable&nbsp; $i$&nbsp; stands for one of the&nbsp; $M^k$ Tuple.&nbsp; The previously calculated approximation&nbsp; $H_2$&nbsp; results with&nbsp; $k = 2$.
+
*The control variable&nbsp; $i$&nbsp; stands for one of the&nbsp; $M^k$&nbsp; tuples.&nbsp;  
 +
*The previously calculated approximation&nbsp; $H_2$&nbsp; results with&nbsp; $k = 2$.
 +
 
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Definition:}$&nbsp;
 
$\text{Definition:}$&nbsp;
The&nbsp; '''Entropy of a message source with memory'''&nbsp; has the following limit  
+
The&nbsp; '''entropy of a discrete source with memory'''&nbsp; has the following limit:
 
:$$H = \lim_{k \rightarrow \infty }H_k \hspace{0.05cm}.$$
 
:$$H = \lim_{k \rightarrow \infty }H_k \hspace{0.05cm}.$$
  
For the&nbsp; ''entropy approximations''&nbsp; $H_k$&nbsp; the following relations apply&nbsp; $(H_0$ is the decision content$)$:
+
For the&nbsp; '''entropy approximations'''&nbsp; $H_k$&nbsp; the following relations apply&nbsp; $(H_0=\log_2  M)$:
 
:$$H \le \text{...} \le H_k \le \text{...} \le H_2 \le H_1 \le H_0 \hspace{0.05cm}.$$}}
 
:$$H \le \text{...} \le H_k \le \text{...} \le H_2 \le H_1 \le H_0 \hspace{0.05cm}.$$}}
  
  
The computational effort will increase with increasing&nbsp; $k$&nbsp; except for a few special cases (see the following example) and depends on the symbol size&nbsp; $M$&nbsp; of course:
+
The computational effort will increase with increasing&nbsp; $k$&nbsp; except for a few special cases (see the following example) and depends on the symbol set size&nbsp; $M$&nbsp; of course:
*For the calculation of&nbsp; $H_{10}$&nbsp; a binary source&nbsp; $(M = 2)$&nbsp; is to be averaged over&nbsp; $2^{10} = 1024$&nbsp; terms.  
+
*For the calculation of&nbsp; $H_{10}$&nbsp; for a binary source&nbsp; $(M = 2)$&nbsp; one must average over&nbsp; $2^{10} = 1024$&nbsp; terms.  
 
*With each further increase of&nbsp; $k$&nbsp; by&nbsp; $1$&nbsp; the number of sum terms doubles.
 
*With each further increase of&nbsp; $k$&nbsp; by&nbsp; $1$&nbsp; the number of sum terms doubles.
*In case of a quaternary source&nbsp; $(M = 4)$&nbsp; it must already be averaged over&nbsp; $4^{10} = 1\hspace{0.08cm}048\hspace{0.08cm}576$&nbsp; summation terms must be averaged for the determination of&nbsp; $H_{10}$.
+
*In case of a quaternary source&nbsp; $(M = 4)$&nbsp; it must already be averaged over&nbsp; $4^{10} = 1\hspace{0.08cm}048\hspace{0.08cm}576$&nbsp; summation terms for the determination of&nbsp; $H_{10}$.
* Considering that each of these&nbsp; $4^{10} =2^{20} >10^6$&nbsp; $k$-tuple should occur in simulation/time averaging about&nbsp; $100$&nbsp; times (statistical guideline) to ensure sufficient simulation accuracy, it follows that the sequence length should be greater than&nbsp; $N = 10^8$&nbsp;.
+
* Considering that each of these&nbsp; $4^{10} =2^{20} >10^6$&nbsp; $k$&ndash;tuple should occur in simulation/time averaging about&nbsp; $100$&nbsp; times (statistical guideline) to ensure sufficient simulation accuracy, it follows that the sequence length should be in this case greater than&nbsp; $N = 10^8$.
  
  
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The (actual) entropy of this alternating binary sequence is therefore
+
The entropy of this alternating binary sequence is therefore
 
:$$H = \lim_{k \rightarrow \infty }{1}/{k} = 0 \hspace{0.05cm}.$$
 
:$$H = \lim_{k \rightarrow \infty }{1}/{k} = 0 \hspace{0.05cm}.$$
  
The result was to be expected, since the considered sequence has only minimal information, which does not affect the entropy end value&nbsp; $H$&nbsp; namely: <br> &nbsp; &nbsp; "Does&nbsp; $\rm A$&nbsp; occur at the even or the odd times?"
+
The result was to be expected, since the considered sequence has only minimal information, which does not affect the entropy end value&nbsp; $H$,&nbsp; namely: <br> &nbsp; &nbsp; "Does&nbsp; $\rm A$&nbsp; occur at the even or the odd times?"
  
 
You can see that&nbsp; $H_k$&nbsp; comes closer to this final value&nbsp; $H = 0$&nbsp; very slowly:&nbsp; The twentieth entropy approximation still returns&nbsp; $H_{20} = 0.05 \hspace{0.05cm} \rm bit/symbol$. }}
 
You can see that&nbsp; $H_k$&nbsp; comes closer to this final value&nbsp; $H = 0$&nbsp; very slowly:&nbsp; The twentieth entropy approximation still returns&nbsp; $H_{20} = 0.05 \hspace{0.05cm} \rm bit/symbol$. }}
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$\text{Summary of the results of the last pages:}$&nbsp;
 
$\text{Summary of the results of the last pages:}$&nbsp;
  
*Generally it applies to the&nbsp; '''Entropy of a message source''':
+
*Generally it applies to the&nbsp; '''entropy of a message source''':
 
:$$H \le \text{...} \le H_3 \le H_2 \le H_1 \le H_0  
 
:$$H \le \text{...} \le H_3 \le H_2 \le H_1 \le H_0  
 
  \hspace{0.05cm}.$$  
 
  \hspace{0.05cm}.$$  
*A&nbsp; '''redundancy-free source'' &nbsp; exists if all&nbsp; $M$&nbsp; symbols are equally probable and there are no statistical bonds within the sequence. <br> For these,&nbsp; $(r$&nbsp; denotes the ''relative redundancy'' $)$:
+
*A&nbsp; '''redundancy-free source''' &nbsp; exists if all&nbsp; $M$&nbsp; symbols are equally probable and there are no statistical bindings within the sequence. <br> For these hold&nbsp; $(r$&nbsp; denotes the ''relative redundancy'' $)$:
 
:$$H = H_0 = H_1 = H_2 = H_3 = \text{...}\hspace{0.5cm}
 
:$$H = H_0 = H_1 = H_2 = H_3 = \text{...}\hspace{0.5cm}
 
\Rightarrow \hspace{0.5cm} r = \frac{H - H_0}{H_0}= 0 \hspace{0.05cm}.$$  
 
\Rightarrow \hspace{0.5cm} r = \frac{H - H_0}{H_0}= 0 \hspace{0.05cm}.$$  
*A&nbsp; '''memoryless source''' &nbsp; can be quite redundant&nbsp; $(r> 0)$.&nbsp; This redundancy then is solely due to the deviation of the symbol probabilities from the uniform distribution.&nbsp; Here the following relations are valid
+
*A&nbsp; '''memoryless source''' &nbsp; can be quite redundant&nbsp; $(r> 0)$.&nbsp; This redundancy then is solely due to the deviation of the symbol probabilities from the uniform distribution.&nbsp; Here the following relations are valid:
 
:$$H = H_1 = H_2 = H_3 = \text{...} \le H_0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}0 \le r = \frac{H_1 - H_0}{H_0}< 1 \hspace{0.05cm}.$$  
 
:$$H = H_1 = H_2 = H_3 = \text{...} \le H_0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}0 \le r = \frac{H_1 - H_0}{H_0}< 1 \hspace{0.05cm}.$$  
 
*The corresponding condition for a&nbsp; '''source with memory'''&nbsp; is
 
*The corresponding condition for a&nbsp; '''source with memory'''&nbsp; is
 
:$$ H <\text{...} < H_3 < H_2 < H_1 \le H_0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} 0 < r = \frac{H_1 - H_0}{H_0}\le1 \hspace{0.05cm}.$$
 
:$$ H <\text{...} < H_3 < H_2 < H_1 \le H_0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} 0 < r = \frac{H_1 - H_0}{H_0}\le1 \hspace{0.05cm}.$$
 
*If&nbsp; $H_2 < H_1$, then&nbsp; $H_3 < H_2$, &nbsp; $H_4 < H_3$, etc. &nbsp; &rArr; &nbsp; In the general equation, the&nbsp; "≤" character must be replaced by the&nbsp; "<" character.  
 
*If&nbsp; $H_2 < H_1$, then&nbsp; $H_3 < H_2$, &nbsp; $H_4 < H_3$, etc. &nbsp; &rArr; &nbsp; In the general equation, the&nbsp; "≤" character must be replaced by the&nbsp; "<" character.  
*If the symbols are equally probable, then again&nbsp; $H_1 = H_0$, while&nbsp; $H_1 < H_0$&nbsp; applies to symbols which are not equally probable.}}
+
*If the symbols are equally probable, then again&nbsp; $H_1 = H_0$,&nbsp; while&nbsp; $H_1 < H_0$&nbsp; applies to symbols which are not equally probable.}}
 
 
 
 
  
==The Entropy of the AMI code ==
+
==The entropy of the AMI code ==
 
<br>
 
<br>
In chapter&nbsp; [[Digital_Signal_Transmission/Symbol-Wise Coding with Pseudo Ternary Codes#Properties_of_AMI Code|Symbol-wise Coding with Pseudo-Ternary Codes]]&nbsp; of the book "Digital Signal Transmission", among other things, the AMI pseudo-ternary code is discussed.  
+
In chapter&nbsp; [[Digital_Signal_Transmission/Symbol-Wise Coding with Pseudo Ternary Codes#Properties_of_AMI Code|Symbol-wise Coding with Pseudo-Ternary Codes]]&nbsp; of the book&nbsp; "Digital Signal Transmission", among other things, the AMI pseudo-ternary code is discussed.
 +
[[File:EN_Inf_T_1_2_S4.png|right|frame|Signals and symbol sequences for AMI code]]
 +
 
*This converts the binary sequence&nbsp; $〈 q_ν 〉$&nbsp; with&nbsp; $q_ν ∈ \{ \rm L, \ H \}$&nbsp; into the ternary sequence&nbsp; $〈 c_ν 〉$&nbsp; with&nbsp; $q_ν ∈ \{ \rm M, \ N, \ P \}$.
 
*This converts the binary sequence&nbsp; $〈 q_ν 〉$&nbsp; with&nbsp; $q_ν ∈ \{ \rm L, \ H \}$&nbsp; into the ternary sequence&nbsp; $〈 c_ν 〉$&nbsp; with&nbsp; $q_ν ∈ \{ \rm M, \ N, \ P \}$.
*The names of the source symbols stand for&nbsp; $\rm L$ow&nbsp; and&nbsp; $\rm H$igh&nbsp; and those of the code symbols for&nbsp; $\rm M$inus,&nbsp; $\rm N$ull&nbsp; and&nbsp; $\rm P$lus".  
+
*The names of the source symbols stand for&nbsp; $\rm L$(ow)&nbsp; and&nbsp; $\rm H$(igh)&nbsp; and those of the code symbols for&nbsp; $\rm M$(inus),&nbsp; $\rm P$(lus),&nbsp; and&nbsp; $\rm N$(ull) &nbsp; &rArr; &nbsp; "Zero".  
  
  
The coding rule of the AMI code ("Alternate Mark Inversion") is
+
The coding rule of the AMI code ("Alternate Mark Inversion") is:
[[File:EN_Inf_T_1_2_S4.png|right|frame|Signals and symbol sequences for AMI code]]
 
  
 
*Each binary symbol&nbsp; $q_ν =\rm L$&nbsp; is represented by the code symbol&nbsp; $c_ν =\rm N$&nbsp;.
 
*Each binary symbol&nbsp; $q_ν =\rm L$&nbsp; is represented by the code symbol&nbsp; $c_ν =\rm N$&nbsp;.
*In contrast,&nbsp; $q_ν =\rm H$&nbsp; alternates with&nbsp; $c_ν =\rm P$&nbsp; and&nbsp; $c_ν =\rm M$&nbsp; coded &nbsp; ⇒ &nbsp; name "AMI".
+
*In contrast,&nbsp; $q_ν =\rm H$&nbsp; alternates with&nbsp; $c_ν =\rm P$&nbsp; and&nbsp; $c_ν =\rm M$&nbsp; coded &nbsp; ⇒ &nbsp; name "Alternate Mark Inversion".
 
+
*This special encoding adds redundancy with the sole purpose of ensuring that the code sequence does not contain a DC component.  
 
 
This special encoding adds redundancy with the sole purpose of ensuring that the code sequence does not contain a DC component.  
 
 
<br clear=all>
 
<br clear=all>
 
However, we do not consider the spectral properties of the AMI code here, but interpret this code information-theoretically:
 
However, we do not consider the spectral properties of the AMI code here, but interpret this code information-theoretically:
*Based on the number of steps&nbsp; $M = 3$&nbsp; the decision content of the (ternary) code sequence is equal to&nbsp; $H_0 = \log_2 \ 3 ≈ 1.585 \hspace{0.05cm} \rm bit/symbol$.&nbsp; The first entropy approximation returns&nbsp; $H_1 = 1.5 \hspace{0.05cm} \rm bit/Symbol$, as shown in the following calculation:
+
*Based on the symbol set size&nbsp; $M = 3$&nbsp; the decision content of the (ternary) code sequence is equal to&nbsp; $H_0 = \log_2 \ 3 ≈ 1.585 \hspace{0.05cm} \rm bit/symbol$.&nbsp; The first entropy approximation returns&nbsp; $H_1 = 1.5 \hspace{0.05cm} \rm bit/Symbol$, as shown in the following calculation:
 
    
 
    
 
:$$p_{\rm H} = p_{\rm L} = 1/2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 
:$$p_{\rm H} = p_{\rm L} = 1/2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
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  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
*Let's now look at two-tuples.&nbsp; With AMI code,&nbsp; $\rm P$&nbsp; cannot follow&nbsp; $\rm P$&nbsp; and&nbsp; $\rm M$&nbsp; cannot follow&nbsp; $\rm M$&nbsp; follow&nbsp; The probability for&nbsp; $\rm NN$&nbsp; is equal to&nbsp; $p_{\rm L} - p_{\rm L} = 1/4$.&nbsp; All other (six) two-tuples occur with the probability&nbsp; $1/8$&nbsp; on.&nbsp; From this follows for the second entropy approximation:
+
*Let's now look at two-tuples.&nbsp; With the AMI code,&nbsp; $\rm P$&nbsp; cannot follow&nbsp; $\rm P$&nbsp; and&nbsp; $\rm M$&nbsp; cannot follow&nbsp; $\rm M$.&nbsp; The probability for&nbsp; $\rm NN$&nbsp; is equal to&nbsp; $p_{\rm L} \cdot p_{\rm L} = 1/4$.&nbsp; All other (six) two-tuples occur with the probability&nbsp; $1/8$.&nbsp; From this follows for the second entropy approximation:
 
:$$H_2 = 1/2 \cdot \big [ 1/4 \cdot {\rm log_2}\hspace{0.1cm}4 + 6 \cdot 1/8 \cdot {\rm log_2}\hspace{0.1cm}8 \big ] = 1,375 \,{\rm bit/symbol} \hspace{0.05cm}.$$
 
:$$H_2 = 1/2 \cdot \big [ 1/4 \cdot {\rm log_2}\hspace{0.1cm}4 + 6 \cdot 1/8 \cdot {\rm log_2}\hspace{0.1cm}8 \big ] = 1,375 \,{\rm bit/symbol} \hspace{0.05cm}.$$
  
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:$$ H < \hspace{0.05cm}\text{...}\hspace{0.05cm} < H_5 < H_4 < H_3 < H_2 = 1.375 \,{\rm bit/symbol} \hspace{0.05cm}.$$
 
:$$ H < \hspace{0.05cm}\text{...}\hspace{0.05cm} < H_5 < H_4 < H_3 < H_2 = 1.375 \,{\rm bit/symbol} \hspace{0.05cm}.$$
  
*Exceptionally in this example we know the actual entropy&nbsp; $H$&nbsp; of the code symbol sequence&nbsp; $〈 c_ν 〉$: &nbsp; since no new information is added by the coder and no information is lost, it is the same entropy&nbsp; $H = 1 \,{\rm bit/symbol} $&nbsp; as the one of the redundancy-free binary sequence $〈 q_ν 〉$.
+
*Exceptionally in this example we know the actual entropy&nbsp; $H$&nbsp; of the code symbol sequence&nbsp; $〈 c_ν 〉$: &nbsp; Since no new information is added by the coder and no information is lost, it is the same entropy&nbsp; $H = 1 \,{\rm bit/symbol} $&nbsp; as the one of the redundancy-free binary sequence $〈 q_ν 〉$.
  
  
The&nbsp; [[Aufgaben:1.4_Entropienäherungen_für_den_AMI-Code|Exercise 1.4]]&nbsp; shows the considerable effort required to calculate the entropy approximation&nbsp; $H_3$. &nbsp; Moreover,&nbsp; $H_3$&nbsp; still deviates significantly from the final value&nbsp; $H = 1 \,{\rm bit/symbol} $.&nbsp; A faster result is achieved if the AMI code is described by a markov chain as explained in the next section.
+
The&nbsp; [[Aufgaben:1.4_Entropienäherungen_für_den_AMI-Code|Exercise 1.4]]&nbsp; shows the considerable effort required to calculate the entropy approximation&nbsp; $H_3$. &nbsp; Moreover,&nbsp; $H_3$&nbsp; still deviates significantly from the final value&nbsp; $H = 1 \,{\rm bit/symbol} $.&nbsp; A faster result is achieved if the AMI code is described by a Markov chain as explained in the next section.
  
  
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[[File:Inf_T_1_2_S5_vers2.png|right|frame|Markov processes with&nbsp; $M = 2$&nbsp; states]]
 
[[File:Inf_T_1_2_S5_vers2.png|right|frame|Markov processes with&nbsp; $M = 2$&nbsp; states]]
  
Sequences with statistical bonds between the sequence elements (symbols) are often modeled by&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov processes]]&nbsp; where we limit ourselves here to first-order Markov processes.&nbsp; First we consider a binary Markov process&nbsp; $(M = 2)$&nbsp; with the states (symbols)&nbsp; $\rm A$&nbsp; and&nbsp; $\rm B$.
+
Sequences with statistical bindings between the sequence elements (symbols) are often modeled by&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov processes]]&nbsp; where we limit ourselves here to first-order Markov processes.&nbsp; First we consider a binary Markov process&nbsp; $(M = 2)$&nbsp; with the states (symbols)&nbsp; $\rm A$&nbsp; and&nbsp; $\rm B$.
  
 
On the right, you can see the transition diagram for a first-order binary Markov process&nbsp; however, only two of the four transfer probabilities given are freely selectable, for example
 
On the right, you can see the transition diagram for a first-order binary Markov process&nbsp; however, only two of the four transfer probabilities given are freely selectable, for example
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For the other two transition probabilities&nbsp; $p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} = 1- p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}$ &nbsp;and &nbsp; $p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B} = 1- p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}
+
For the other two transition probabilities:&nbsp; $p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} = 1- p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}$ &nbsp;and &nbsp; $p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B} = 1- p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}
 
  \hspace{0.05cm}.$
 
  \hspace{0.05cm}.$
  
Due to the presupposed properties&nbsp; [[Theory_of_Stochastic_Signals/Auto Correlation Function (ACF)#Stationary_Random Processes|Stationarity]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Auto Correlation Function (ACF)#Ergodic_Random Processes|Ergodicity]]&nbsp; the following applies to the state or symbol probabilities:
+
Due to the presupposed properties&nbsp; [[Theory_of_Stochastic_Signals/Auto-Correlation_Function_(ACF)#Stationary_random_processes|"Stationarity"]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Auto-Correlation_Function_(ACF)#Ergodic_random_processes|"Ergodicity"]]&nbsp; the following applies to the state or symbol probabilities:
 
   
 
   
 
:$$p_{\rm A} = {\rm Pr}({\rm A}) = \frac{p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}}{p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} + p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}}
 
:$$p_{\rm A} = {\rm Pr}({\rm A}) = \frac{p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}}{p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} + p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}}
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  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
These equations allow first information theoretical statements about the Markov processes:
+
These equations allow first information&ndash;theoretical statements about the Markov processes:
* For&nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}} = p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$&nbsp; the symbols are equally likely &nbsp; ⇒ &nbsp; $p_{\text{A}} = p_{\text{B}}= 0.5$.&nbsp; The first entropy approximation returns&nbsp; $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$, independent of the actual values of the (conditional) transition probabilities&nbsp; $p_{\text{A|B}}$&nbsp; &nbsp;or &nbsp; $p_{\text{B|A}}$.
+
* For&nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}} = p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$&nbsp; the symbols are equally likely &nbsp; ⇒ &nbsp; $p_{\text{A}} = p_{\text{B}}= 0.5$.&nbsp; The first entropy approximation returns&nbsp; $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$, independent of the actual values of the (conditional) transition probabilities&nbsp; $p_{\text{A|B}}$&nbsp; and&nbsp; $p_{\text{B|A}}$.
*The source entropy&nbsp; $H$&nbsp; as the threshold value&nbsp; $($for&nbsp; $k \to \infty)$&nbsp; of the&nbsp; [[Information_Theory/Discrete Memoryless Sources#Generalization_to_.7F.27 .22.60UNIQ-MathJax109-QINU.60.22.27.7F.E2.80.93Tuple_and_boundary.C3.BCtransition|Entropy approximation&nbsp; $k$th order]] &nbsp; &rArr; &nbsp; $H_k$&nbsp; but depends very much on the actual values of&nbsp; $p_{\text{A|B}}$ &nbsp;and&nbsp; $p_{\text{B|A}}$&nbsp; and not only on their quotients.&nbsp; This is shown by the following example.
+
*But the source entropy&nbsp; $H$&nbsp; as the limit value&nbsp; $($for&nbsp; $k \to \infty)$&nbsp; of the&nbsp; [[Information_Theory/Discrete Memoryless Sources#Generalization_to_.7F.27 .22.60UNIQ-MathJax109-QINU.60.22.27.7F.E2.80.93Tuple_and_boundary.C3.BCtransition|Entropy approximation of order&nbsp; $k$]] &nbsp; &rArr; &nbsp; $H_k$&nbsp; depends very much on the actual values of&nbsp; $p_{\text{A|B}}$ &nbsp;and&nbsp; $p_{\text{B|A}}$&nbsp; and not only on their quotients.&nbsp; This is shown by the following example.
  
  
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
 
$\text{Example 6:}$&nbsp;
 
$\text{Example 6:}$&nbsp;
We consider three binary symmetric Markov sources that are represented by the numerical values of the symmetric transition probabilities&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} }$&nbsp;.&nbsp; For the symbol probabilities the following applies:&nbsp; $p_{\rm A} = p_{\rm B}= 0.5$, and the other transition probabilities have the values
+
We consider three binary symmetric Markov sources that are represented by the numerical values of the symmetric transition probabilities&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} }$.&nbsp; For the symbol probabilities the following applies:&nbsp; $p_{\rm A} = p_{\rm B}= 0.5$, and the other transition probabilities have the values
  
[[File:EN_Inf_T_1_2_S5b.png|right|frame|Three examples of binary Markov sources]]  
+
[[File:EN_Inf_T_1_2_S5b.png|right|frame|Three examples of binary Markov sources with&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} }$&nbsp;]]  
 
:$$p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 1 - p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } =
 
:$$p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 1 - p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } =
 
p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}B} }.$$
 
p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}B} }.$$
  
*The middle (blue) symbol sequence with&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.5$&nbsp; has the entropy&nbsp; $H ≈ 1 \hspace{0.1cm}  \rm bit/symbol$.&nbsp; That means: &nbsp; In this special case there are no statistical bonds within the sequence.
+
*The middle (blue) sequence with&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.5$&nbsp; has the entropy&nbsp; $H ≈ 1 \hspace{0.1cm}  \rm bit/symbol$.&nbsp; That means: &nbsp; In this special case there are no statistical bindings within the sequence.
  
*The left (red) sequence with&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.2$&nbsp; has less changes between&nbsp; $\rm A$&nbsp; and&nbsp; $\rm B$&nbsp; and&nbsp; $\rm B$&nbsp; Due to statistical dependencies between neighboring symbols the entropy is now smaller&nbsp; $H ≈ 0.72 \hspace{0.1cm}  \rm bit/symbol$.
+
*The left (red) sequence with&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.2$&nbsp; has less changes between&nbsp; $\rm A$&nbsp; and&nbsp; $\rm B$.&nbsp; Due to statistical dependencies between neighboring symbols the entropy is now smaller:&nbsp; $H ≈ 0.72 \hspace{0.1cm}  \rm bit/symbol$.
  
*The right (green) symbol sequence with&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.8$&nbsp; has the exact same entropy&nbsp; $H ≈ 0.72 \hspace{0.1cm}  \rm bit/symbol$&nbsp; as the red sequence.&nbsp; Here you can see many areas with alternating symbols&nbsp; $($... $\rm ABABAB$ ... $)$.}}
+
*The right (green) symbol sequence with&nbsp; $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.8$&nbsp; has the exact same entropy&nbsp; $H ≈ 0.72 \hspace{0.1cm}  \rm bit/symbol$&nbsp; as the red sequence.&nbsp; Here you can see many regions with alternating symbols&nbsp; $($... $\rm ABABAB$ ... $)$.}}
  
  
 
This example is worth noting:
 
This example is worth noting:
*If you had not used the markup properties of the red and green sequences, you would have reached the respective result&nbsp; $H ≈ 0.72 \hspace{0.1cm}  \rm bit/symbol$&nbsp; only after lengthy calculations.
+
*If you had not used the Markov properties of the red and green sequences, you would have reached the respective result&nbsp; $H ≈ 0.72 \hspace{0.1cm}  \rm bit/symbol$&nbsp; only after lengthy calculations.
*The following pages show that for a source with mark properties the final value&nbsp; $H$&nbsp; can be determined from the entropy approximations&nbsp; $H_1$&nbsp; and&nbsp; $H_2$&nbsp; alone. &nbsp; Likewise, all entropy approximations&nbsp; $H_1$&nbsp; and&nbsp; $H_2$&nbsp; can also be calculated from&nbsp; $H_k$&nbsp; for&nbsp; $k$-tuples in a simple manner &nbsp; ⇒ &nbsp; $H_3$,&nbsp; $H_4$,&nbsp; $H_5$, ... &nbsp; $H_{100}$, ...
+
*The following pages show that for a source with Markov properties the final value&nbsp; $H$&nbsp; can be determined from the entropy approximations&nbsp; $H_1$&nbsp; and&nbsp; $H_2$&nbsp; alone. &nbsp; Likewise, all entropy approximations&nbsp; $H_1$&nbsp; and&nbsp; $H_2$&nbsp; can also be calculated from&nbsp; $H_k$&nbsp; for&nbsp; $k$&ndash;tuples in a simple manner &nbsp; ⇒ &nbsp; $H_3$,&nbsp; $H_4$,&nbsp; $H_5$, ... &nbsp; $H_{100}$, ...
 
 
 
   
 
   
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[[File:Inf_T_1_2_S5_vers2.png|right|frame|Markov processes with&nbsp; $M = 2$&nbsp; states]]
 
[[File:Inf_T_1_2_S5_vers2.png|right|frame|Markov processes with&nbsp; $M = 2$&nbsp; states]]
 
We continue to assume the first-order symmetric binary Markov source.&nbsp; As on the previous page, we use the following nomenclature for
 
We continue to assume the first-order symmetric binary Markov source.&nbsp; As on the previous page, we use the following nomenclature for
*the transition probabilities&nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}}$, &nbsp; $p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$,&nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}A}}= 1- p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$, &nbsp; $p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}B}} = 1 - p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}}$, &nbsp;  
+
*the transition probabilities &nbsp; &rArr; &nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}}$, &nbsp; $p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$,&nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}A}}= 1- p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$, &nbsp; $p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}B}} = 1 - p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}}$, &nbsp;  
*the ergodic probabilities&nbsp; $p_{\text{A}}$&nbsp; and&nbsp; $p_{\text{B}}$,
+
*the ergodic probabilities &nbsp; &rArr; &nbsp; $p_{\text{A}}$&nbsp; and&nbsp; $p_{\text{B}}$,
*the compound probabilities, for example&nbsp; $p_{\text{AB}} = p_{\text{A}} - p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$.
+
*the compound probabilities, for example &nbsp; &rArr; &nbsp; $p_{\text{AB}} = p_{\text{A}} - p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$.
  
  
  
We now compute the&nbsp; [[Information_Theory/Sources with Memory#Entropy_in_Two_Tuple|Entropy of a two-tuple]]&nbsp; (with the unit "bit/two-tuple"):
+
We now compute the&nbsp; [[Information_Theory/Discrete_Sources_with_Memory#Entropy_with_respect_to_two-tuples|Entropy of a two-tuple]]&nbsp; (with the unit "bit/two-tuple"):
 
   
 
   
 
:$$H_2\hspace{0.05cm}' = p_{\rm A}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm A}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm B}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}} + p_{\rm B}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B}}
 
:$$H_2\hspace{0.05cm}' = p_{\rm A}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm A}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm B}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B}  \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}} + p_{\rm B}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B}  \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B}}
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{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
$\text{Conclusion:}$&nbsp; Thus the&nbsp; '''second entropy approximation'''&nbsp; (with the unit "bit/symbol"):
+
$\text{Conclusion:}$&nbsp; Thus the&nbsp; '''second entropy approximation'''&nbsp; (with the unit "bit/symbol")&nbsp; is:
 
:$$H_2 = {1}/{2} \cdot {H_2\hspace{0.05cm}'} = {1}/{2} \cdot \big [ H_{\rm 1} + H_{\rm M} \big]  
 
:$$H_2 = {1}/{2} \cdot {H_2\hspace{0.05cm}'} = {1}/{2} \cdot \big [ H_{\rm 1} + H_{\rm M} \big]  
 
  \hspace{0.05cm}.$$}}
 
  \hspace{0.05cm}.$$}}
  
  
is to be noted:
+
It is to be noted:
 
*The first summand&nbsp; $H_1$ &nbsp; ⇒ &nbsp; first entropy approximation depends only on the symbol probabilities.
 
*The first summand&nbsp; $H_1$ &nbsp; ⇒ &nbsp; first entropy approximation depends only on the symbol probabilities.
*For a symmetrical markov process &nbsp; ⇒ &nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}} = p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}} $ &nbsp; ⇒ &nbsp; $p_{\text{A}} = p_{\text{B}} = 1/2$ &nbsp; this is the result for this first summand&nbsp; $H_1 = 1 \hspace{0.1cm} \rm bit/symbol$.
+
*For a symmetrical Markov process &nbsp; ⇒ &nbsp; $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}} = p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}} $ &nbsp; ⇒ &nbsp; $p_{\text{A}} = p_{\text{B}} = 1/2$ &nbsp; the result for this first summand is&nbsp; $H_1 = 1 \hspace{0.1cm} \rm bit/symbol$.
 
*The second summand&nbsp; $H_{\text{M}}$&nbsp; must be calculated according to the second of the two upper equations.  
 
*The second summand&nbsp; $H_{\text{M}}$&nbsp; must be calculated according to the second of the two upper equations.  
*For a symmetrical Markov process you get&nbsp; $H_{\text{M}} = H_{\text{bin}}(p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B})$.
+
*For a symmetrical Markov process you get&nbsp; $H_{\text{M}} = H_{\text{bin}}(p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}})$.
  
  
Now, this result is extended to the&nbsp; $k$-th entropy approximation.&nbsp; Here, the advantage of Markov sources over other sources is taken advantage of, that the entropy calculation for&nbsp; $k$-tuples is very simple.&nbsp; For each Markov source, the following applies
+
Now, this result is extended to the&nbsp; $k$&ndash;th entropy approximation.&nbsp; Here, the advantage of Markov sources over other sources is taken advantage of, that the entropy calculation for&nbsp; $k$&ndash;tuples is very simple.&nbsp; For each Markov source, the following applies
 
   
 
   
 
:$$H_k = {1}/{k} \cdot \big [ H_{\rm 1} + (k-1) \cdot H_{\rm M}\big ] \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
 
:$$H_k = {1}/{k} \cdot \big [ H_{\rm 1} + (k-1) \cdot H_{\rm M}\big ] \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
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However, these approximations are not very important.&nbsp; Mostly only the limit value&nbsp; $H$.&nbsp; For sources without markov properties the approximations&nbsp; $H_k$&nbsp; are calculated only to be able to estimate this limit value, i.e. the actual entropy.
+
However, these approximations are not very important.&nbsp; Mostly only the limit value&nbsp; $H$&nbsp; is of interest.&nbsp; For sources without Markov properties the approximations&nbsp; $H_k$&nbsp; are calculated only to be able to estimate this limit value, i.e. the actual entropy.
  
  
 
''Notes'':  
 
''Notes'':  
*In the&nbsp; [[Aufgaben:1.5_Binäre_Markovquelle|Exercise 1.5]]&nbsp; the above equations are applied to the more general case of an asymmetric binary source.
+
*In the&nbsp; [[Aufgaben:Exercise_1.6:_Non-Binary_Markov_Sources|Exercise 1.5]]&nbsp; the above equations are applied to the more general case of an asymmetric binary source.
 
*All equations on this page also apply to non-binary Markov sources&nbsp; $(M > 2)$ as shown on the next page.
 
*All equations on this page also apply to non-binary Markov sources&nbsp; $(M > 2)$ as shown on the next page.
  
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==Non-binary Markov sources ==
 
==Non-binary Markov sources ==
 
<br>
 
<br>
[[File:EN_Inf_T_1_2_S6.png|right|frame|Ternary and Quaternary First Order Markov Source]]
+
[[File:EN_Inf_T_1_2_S6.png|right|frame|Ternary and quaternary first order Markov source]]
  
The following equations apply to each Markov source regardless of the symbol range:
+
The following equations apply to each Markov source regardless of the symbol set size&nbsp; $M$:
 
   
 
   
 
:$$H = 2 \cdot H_2 - H_{\rm 1}  \hspace{0.05cm},$$
 
:$$H = 2 \cdot H_2 - H_{\rm 1}  \hspace{0.05cm},$$
Line 348: Line 348:
 
These equations allow the simple calculation of the entropy&nbsp; $H$&nbsp; from the approximations&nbsp; $H_1$&nbsp; and&nbsp; $H_2$.
 
These equations allow the simple calculation of the entropy&nbsp; $H$&nbsp; from the approximations&nbsp; $H_1$&nbsp; and&nbsp; $H_2$.
  
We now look at the transition diagrams sketched on the right
+
We now look at the transition diagrams sketched on the right:
*a ternary Markov source&nbsp; $\rm MQ3$&nbsp; $($&nbsp; $M = 3$,&nbsp; blue coloring$)$ and  
+
*A ternary Markov source&nbsp; $\rm MQ3$&nbsp; $(M = 3$,&nbsp; blue coloring$)$,&nbsp; and  
 
*a quaternary Markov source&nbsp; $\rm MQ4$&nbsp; $(M = 4$,&nbsp; red color$)$.  
 
*a quaternary Markov source&nbsp; $\rm MQ4$&nbsp; $(M = 4$,&nbsp; red color$)$.  
 
<br clear=all>
 
<br clear=all>
In&nbsp; [[Aufgaben:1.6_Nichtbinäre_Markovquellen|Exercise 1.6]]&nbsp; the entropy approximations&nbsp; $H_k$&nbsp; and the source entropy&nbsp; $H$&nbsp; are calculated as the limit of&nbsp; $H_k$&nbsp; for&nbsp; $k \to \infty$&nbsp;. &nbsp; The results are shown in the following figure.&nbsp; All entropies specified there have the unit "bit/symbol".
+
In&nbsp; [[Aufgaben:1.6_Nichtbinäre_Markovquellen|Exercise 1.6]]&nbsp; the entropy approximations&nbsp; $H_k$&nbsp; and the total entropy&nbsp; $H$&nbsp; are calculated as the limit of&nbsp; $H_k$&nbsp; for&nbsp; $k \to \infty$&nbsp;. &nbsp; The results are shown in the following figure.&nbsp; All entropies specified there have the unit "bit/symbol".
  
[[File:EN_Inf_T_1_2_S6b.png|center|frame|Entropies for&nbsp; $\rm MQ3$,&nbsp; $\rm MQ4$&nbsp; and the&nbsp; $\rm AMI-Code$]]
+
[[File:EN_Inf_T_1_2_S6b.png|right|frame|Entropies for&nbsp; $\rm MQ3$,&nbsp; $\rm MQ4$&nbsp; and the&nbsp; $\rm AMI$&nbsp; code]]
  
 
These results can be interpreted as follows:
 
These results can be interpreted as follows:
*For the ternary markov source&nbsp; $\rm MQ3$&nbsp; the entropy approximations of&nbsp; $H_1 = 1,500$&nbsp; above&nbsp; $H_2 = 1,375$&nbsp; up to the limit&nbsp; $H = 1,250$&nbsp; continuously decreasing&nbsp; paths&nbsp; $M = 3$&nbsp; the decision content is&nbsp; $H_0 = 1,585$.
+
*For the ternary markov source&nbsp; $\rm MQ3$&nbsp; the entropy approximations of&nbsp; $H_1 = 1.500$&nbsp; above&nbsp; $H_2 = 1.375$&nbsp; up to the limit&nbsp; $H = 1.250$&nbsp; continuously decreasing.&nbsp; Because of&nbsp; $M = 3$&nbsp; &rArr; &nbsp; $H_0 = 1.585$.
*For the quaternary Markov source&nbsp; $\rm MQ4$&nbsp; one receives&nbsp; $H_0 = H_1 = 2,000$&nbsp; (since four equally probable states) and&nbsp; $H_2 = 1.5$. &nbsp; From the&nbsp; $H_1$-&nbsp; and&nbsp; $H_2$-value all entropy approximations&nbsp; $H_k$&nbsp; and the final value&nbsp; $H = 1.000$&nbsp; can be calculated.
+
*For the quaternary Markov source&nbsp; $\rm MQ4$&nbsp; one receives&nbsp; $H_0 = H_1 = 2.000$&nbsp; (since four equally probable states) and&nbsp; $H_2 = 1.5$. &nbsp; From the&nbsp; $H_1$-&nbsp; and&nbsp; $H_2$-values all entropy approximations&nbsp; $H_k$&nbsp; and the final value&nbsp; $H = 1.000$&nbsp; can be calculated.
*The two models&nbsp; $\rm MQ3$&nbsp; and&nbsp; $\rm MQ4$&nbsp; were created during the attempt to calculate the&nbsp; [[Information_Theory/Sources_with_Memory#The_Entropy_of_AMI.E2.80. 93Codes|AMI-Code]]&nbsp; to be described information theoretically by Markov sources.&nbsp; The symbols&nbsp; $\rm M$,&nbsp; $\rm N$&nbsp; and&nbsp; $\rm P$&nbsp; stand for "minus", "zero" and "plus".
+
*The two models&nbsp; $\rm MQ3$&nbsp; and&nbsp; $\rm MQ4$&nbsp; were created during the attempt to calculate the&nbsp; [[Information_Theory/Sources_with_Memory#The_Entropy_of_AMI.E2.80. 93Codes|AMI code]]&nbsp; to be described information&ndash;theoretically by Markov sources.&nbsp; The symbols&nbsp; $\rm M$,&nbsp; $\rm N$&nbsp; and&nbsp; $\rm P$&nbsp; stand for&nbsp; "minus",&nbsp; "zero"&nbsp; and&nbsp; "plus".
*The entropy approximations&nbsp; $H_1$,&nbsp; $H_2$&nbsp; and&nbsp; $H_3$&nbsp; of the AMI code (green markers) were calculated in the&nbsp; [[Aufgaben:Aufgabe 1.4: Entropienäherungen für den AMI-Code|Exercise 1.4]]&nbsp; on the calculation of&nbsp; $H_4$,&nbsp; $H_5$, ... had to be omitted for reasons of effort.&nbsp; But the final value of&nbsp; $H_k$&nbsp; for&nbsp; $k \to \infty$ &nbsp; ⇒ &nbsp; $H = 1,000$ is known.
+
*The entropy approximations&nbsp; $H_1$,&nbsp; $H_2$&nbsp; and&nbsp; $H_3$&nbsp; of the AMI code (green markers) were calculated in the&nbsp; [[Aufgaben:Aufgabe 1.4: Entropienäherungen für den AMI-Code|Exercise 1.4]].&nbsp; On the calculation of&nbsp; $H_4$,&nbsp; $H_5$, ... had to be omitted for reasons of effort.&nbsp; But the final value of&nbsp; $H_k$&nbsp; for&nbsp; $k \to \infty$ &nbsp; ⇒ &nbsp; $H = 1.000$&nbsp; is known.
*You can see that the Markov model&nbsp; $\rm MQ3$&nbsp; for&nbsp; $H_0 = 1,585$,&nbsp; $H_1 = 1,500$&nbsp; and&nbsp; $H_2 = 1,375$&nbsp; yields exactly the same numerical values as the AMI code. &nbsp; On the other hand&nbsp; $H_3$&nbsp; &rArr; &nbsp; $(1,333$&nbsp; instead of&nbsp; $1,292)$&nbsp; and especially the final value&nbsp; $H$&nbsp; &rArr; &nbsp; $(1,250$&nbsp; compared to&nbsp; $1,000)$.
+
*You can see that the Markov model&nbsp; $\rm MQ3$&nbsp; for&nbsp; $H_0 = 1.585$,&nbsp; $H_1 = 1.500$&nbsp; and&nbsp; $H_2 = 1.375$&nbsp; yields exactly the same numerical values as the AMI code. &nbsp; On the other hand&nbsp; $H_3$&nbsp; &rArr; &nbsp; $(1.333$&nbsp; instead of&nbsp; $1.292)$&nbsp; and especially the final value&nbsp; $H$&nbsp; &rArr; &nbsp; $(1.250$&nbsp; compared to&nbsp; $1.000)$.
 
*The model&nbsp; $\rm MQ4$&nbsp; $(M = 4)$&nbsp; differs from the AMI code&nbsp; $(M = 3)$&nbsp; with respect to the decision content&nbsp; $H_0$&nbsp; and also with respect to all entropy approximations&nbsp; $H_k$. &nbsp; Nevertheless, $\rm MQ4$&nbsp; is the more suitable model for the AMI code, since the final value&nbsp; $H = 1,000$&nbsp; is the same.
 
*The model&nbsp; $\rm MQ4$&nbsp; $(M = 4)$&nbsp; differs from the AMI code&nbsp; $(M = 3)$&nbsp; with respect to the decision content&nbsp; $H_0$&nbsp; and also with respect to all entropy approximations&nbsp; $H_k$. &nbsp; Nevertheless, $\rm MQ4$&nbsp; is the more suitable model for the AMI code, since the final value&nbsp; $H = 1,000$&nbsp; is the same.
 
*The model&nbsp; $\rm MQ3$&nbsp; yields entropy values that are too large, since the sequences&nbsp; $\rm PNP$&nbsp; and&nbsp; $\rm MNM$&nbsp; are possible here, which cannot occur in the AMI code. &nbsp; Already with the approximation&nbsp; $H_3$&nbsp; the difference is slightly noticeable, in the final value&nbsp; $H$&nbsp; even clearly&nbsp; $(1.25$&nbsp; instead of&nbsp; $1)$.
 
*The model&nbsp; $\rm MQ3$&nbsp; yields entropy values that are too large, since the sequences&nbsp; $\rm PNP$&nbsp; and&nbsp; $\rm MNM$&nbsp; are possible here, which cannot occur in the AMI code. &nbsp; Already with the approximation&nbsp; $H_3$&nbsp; the difference is slightly noticeable, in the final value&nbsp; $H$&nbsp; even clearly&nbsp; $(1.25$&nbsp; instead of&nbsp; $1)$.
  
  
In the model&nbsp; $\rm MQ4$&nbsp; the state&nbsp; "Null"&nbsp; was split into two states&nbsp; $\rm N$&nbsp; and&nbsp; $\rm O$&nbsp; (see upper right figure on this page):
+
In the model&nbsp; $\rm MQ4$&nbsp; the state&nbsp; "Zero"&nbsp; was split into two states&nbsp; $\rm N$&nbsp; and&nbsp; $\rm O$&nbsp; (see upper right figure on this page):
*Here applies to the state&nbsp; $\rm N$: &nbsp; The current binary symbol&nbsp; $\rm L$&nbsp; is displayed with the amplitude value&nbsp; $0$&nbsp; (zero), as per the AMI rule.&nbsp; The next occurring&nbsp; $\rm H$ symbol, on the other hand, is displayed as&nbsp; $-1$&nbsp; (minus), because the last&nbsp; $\rm H$ symbol was encoded as&nbsp; $+1$&nbsp; (plus).
+
*Here applies to the state&nbsp; $\rm N$: &nbsp; The current binary symbol&nbsp; $\rm L$&nbsp; is encoded with the amplitude value&nbsp; $0$&nbsp; (zero), as per the AMI rule.&nbsp; The next occurring&nbsp; $\rm H$ symbol, on the other hand, is displayed as&nbsp; $-1$&nbsp; (minus), because the last symol&nbsp; $\rm H$&nbsp; was encoded as&nbsp; $+1$&nbsp; (plus).
*The current binary symbol&nbsp; $\rm O$&nbsp; is also displayed with the state&nbsp; $\rm L$&nbsp; with the ternary value&nbsp; $0$&nbsp;. &nbsp; In contrast to the state&nbsp; $\rm N$&nbsp; however, the next occurring&nbsp; $\rm H$ symbol is now displayed as&nbsp; $+1$&nbsp; (Plus) since the last&nbsp; $\rm H$ symbol was encoded as&nbsp; $-1$&nbsp; (Minus).
+
*Also with the state&nbsp; $\rm O$&nbsp; the current binary symbol&nbsp; $\rm L$&nbsp; is represented with the ternary value&nbsp; $0$.&nbsp; In contrast to the state&nbsp; $\rm N$&nbsp; however, the next occurring&nbsp; $\rm H$ symbol is now encoded as&nbsp; $+1$&nbsp; (plus) since the last&nbsp; $\rm H$ symbol was encoded as&nbsp; $-1$&nbsp; (minus).
  
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Conclusion:}$&nbsp;  
 
$\text{Conclusion:}$&nbsp;  
*The&nbsp; $\rm MQ4$&ndash;Output sequence actually follows the rules of the AMI code and assigns the entropy&nbsp; $H = 1.000 \hspace{0.15cm} \rm bit/symbol$&nbsp; up.  
+
*The&nbsp; $\rm MQ4$ output sequence actually follows the rules of the AMI code and has entropy&nbsp; $H = 1.000 \hspace{0.15cm} \rm bit/symbol$.  
*Because of the new state&nbsp; $\rm O$&nbsp; but is now&nbsp; $H_0 = 2.000 \hspace{0.15cm} \rm bit/symbol$&nbsp; $($against&nbsp; $1.585 \hspace{0.15cm} \rm bit/symbol)$&nbsp; clearly too large.  
+
*But because of the new state&nbsp; $\rm O$&nbsp; now&nbsp; $H_0 = 2.000 \hspace{0.15cm} \rm bit/symbol$&nbsp; $($against&nbsp; $1.585 \hspace{0.15cm} \rm bit/symbol)$&nbsp; is clearly too large.  
*Also all&nbsp; $H_k$ approximations are larger than in AMI code.  
+
*Also all&nbsp; $H_k$ approximations are larger than with the AMI code.&nbsp; First for &nbsp;$k \to \infty$&nbsp; the model&nbsp; $\rm MQ4$&nbsp; and the AMI code match exactly: &nbsp; $H = 1.000 \hspace{0.15cm} \rm bit/symbol$.}}
*First for &nbsp;$k \to \infty$&nbsp; the model&nbsp; $\rm MQ4$&nbsp; and the AMI code match exactly: &nbsp; $H = 1,000 \hspace{0.15cm} \rm bit/symbol$.}}
 
  
 
   
 
   
 
== Exercises for the chapter==
 
== Exercises for the chapter==
 
<br>
 
<br>
[[Aufgaben:1.3 Entropienäherungen|Aufgabe 1.3: Entropienäherungen]]
+
[[Aufgaben:Exercise_1.3:_Entropy_Approximations|Exercise 1.3: Entropy Approximations]]
  
[[Aufgaben:1.4 Entropienäherungen für den AMI-Code|Aufgabe 1.4: Entropienäherungen für den AMI-Code]]
+
[[Aufgaben:Exercise_1.4:_Entropy_Approximations_for_the_AMI_Code|Exercise 1.4: Entropy Approximations for the AMI Code]]
  
[[Aufgaben:1.4Z Entropie der AMI-Codierung|Zusatzaufgabe 1.4Z: Entropie der AMI-Codierung]]
+
[[Aufgaben:Exercise_1.4Z:_Entropy_of_the_AMI_Code|Exercise 1.4Z: Entropy of the AMI Code]]
  
[[Aufgaben:1.5 Binäre Markovquelle|Aufgabe 1.5: Binäre Markovquelle]]
+
[[Aufgaben:Exercise_1.5:_Binary_Markov_Source|Exercise 1.5: Binary Markov Source]]
  
[[Aufgaben:1.5Z Symmetrische Markovquelle|Aufgabe 1.5Z: Symmetrische Markovquelle]]
+
[[Aufgaben:Exercise_1.5Z:_Symmetrical_Markov_Source|Exercise 1.5Z: Symmetrical Markov Source]]
  
[[Aufgaben:1.6 Nichtbinäre Markovquellen|Aufgabe 1.6: Nichtbinäre Markovquellen]]
+
[[Aufgaben:Exercise_1.6:_Non-Binary_Markov_Sources|Exercise 1.6: Non-Binary Markov Sources]]
  
[[Aufgaben:1.6Z Ternäre Markovquelle|Aufgabe 1.6Z:Ternäre Markovquelle]]
+
[[Aufgaben:Exercise_1.6Z:_Ternary_Markov_Source|Exercise 1.6Z:Ternary Markov Source]]
  
  
 
{{Display}}
 
{{Display}}

Latest revision as of 15:27, 11 July 2021

A simple introductory example


$\text{Example 1:}$  At the  beginning of the first chapter  we have considered a memoryless message source with the symbol set  $\rm \{A, \ B, \ C, \ D\}$   ⇒   $M = 4$  .   An exemplary symbol sequence is shown again in the following figure as source  $\rm Q1$ .

With the symbol probabilities  $p_{\rm A} = 0.4 \hspace{0.05cm},\hspace{0.2cm}p_{\rm B} = 0.3 \hspace{0.05cm},\hspace{0.2cm}p_{\rm C} = 0.2 \hspace{0.05cm},\hspace{0.2cm} p_{\rm D} = 0.1\hspace{0.05cm}$  the entropy is

$$H \hspace{-0.05cm}= 0.4 \cdot {\rm log}_2\hspace{0.05cm}\frac {1}{0.4} + 0.3 \cdot {\rm log}_2\hspace{0.05cm}\frac {1}{0.3} + 0.2 \cdot {\rm log}_2\hspace{0.05cm}\frac {1}{0.2} + 0.1 \cdot {\rm log}_2\hspace{0.05cm}\frac {1}{0.1} \approx 1.84 \hspace{0.05cm}{\rm bit/symbol} \hspace{0.01cm}.$$

Due to the unequal symbol probabilities the entropy is smaller than  $H_0 = \log_2 M = 2 \hspace{0.05cm} \rm bit/symbol$.

Quaternary source without/with memory



The source  $\rm Q2$  is almost identical to the source  $\rm Q1$, except that each individual symbol is output not only once, but twice in a row:  $\rm A ⇒ AA$,  $\rm B ⇒ BB$,  and so on.

  • It is obvious that  $\rm Q2$  has a smaller entropy (uncertainty) than  $\rm Q1$.
  • Because of the simple repetition code,  the entropy
$$H = 1.84/2 = 0.92 \hspace{0.05cm} \rm bit/symbol$$
has only half the size, although the occurrence probabilities have not changed.


$\text{Conclusion:}$  This example shows:

  • The entropy of a source with memory is smaller than the entropy of a memoryless source with equal symbol probabilities.
  • The statistical bindings within the sequence  $〈 q_ν 〉$  have to be considered now,
  • namely the dependency of the symbol  $q_ν$  from the predecessor symbols  $q_{ν-1}$,  $q_{ν-2}$, ...


Entropy with respect to two-tuples


We continue to look at the source symbol sequence  $〈 q_1, \hspace{0.05cm} q_2,\hspace{0.05cm}\text{ ...} \hspace{0.05cm}, q_{ν-1}, \hspace{0.05cm}q_ν, \hspace{0.05cm}\hspace{0.05cm}q_{ν+1} .\hspace{0.05cm}\text{...} \hspace{0.05cm}〉$  and now consider the entropy of two successive source symbols.

  • All source symbols  $q_ν$  are taken from an alphabet with the symbol set size  $M$, so that for the combination  $(q_ν, \hspace{0.05cm}q_{ν+1})$  there are exactly  $M^2$  possible symbol pairs with the following combined probabilities:
$${\rm Pr}(q_{\nu}\cap q_{\nu+1})\le {\rm Pr}(q_{\nu}) \cdot {\rm Pr}( q_{\nu+1}) \hspace{0.05cm}.$$
  • From this, the  compound entropy  of two consecutive symbols can be computed:
$$H_2\hspace{0.05cm}' = \sum_{q_{\nu}\hspace{0.05cm} \in \hspace{0.05cm}\{ \hspace{0.05cm}q_{\mu}\hspace{0.01cm} \}} \ \sum_{q_{\nu+1}\hspace{0.05cm} \in \hspace{0.05cm}\{ \hspace{0.05cm} q_{\mu}\hspace{0.01cm} \}}\hspace{-0.1cm}{\rm Pr}(q_{\nu}\cap q_{\nu+1}) \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{{\rm Pr}(q_{\nu}\cap q_{\nu+1})} \hspace{0.4cm}({\rm unit\hspace{-0.1cm}: \hspace{0.1cm}bit/two–tuple}) \hspace{0.05cm}.$$
The index "2" symbolizes that the entropy thus calculated refers to  two-tuples.
  • To get the average information content per symbol,  $H_2\hspace{0.05cm}'$  has to be divided in half:
$$H_2 = {H_2\hspace{0.05cm}'}/{2} \hspace{0.5cm}({\rm unit\hspace{-0.1cm}: \hspace{0.1cm}bit/symbol}) \hspace{0.05cm}.$$
  • In order to achieve a consistent nomenclature, we now label the entropy defined in chapter  Memoryless Message Sources  with  $H_1$:
$$H_1 = \sum_{q_{\nu}\hspace{0.05cm} \in \hspace{0.05cm}\{ \hspace{0.05cm}q_{\mu}\hspace{0.01cm} \}} {\rm Pr}(q_{\nu}) \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{{\rm Pr}(q_{\nu})} \hspace{0.5cm}({\rm unit\hspace{-0.1cm}: \hspace{0.1cm}bit/symbol}) \hspace{0.05cm}.$$
  • The index "1" is supposed to indicate that  $H_1$  considers only the symbol probabilities and not statistical bindings between symbols within the sequence.  With  $H_0 = \log_2 \ M$  the following size relation results:
$$H_0 \ge H_1 \ge H_2 \hspace{0.05cm}.$$
  • With statistical independence of the sequence elements   ⇒   $H_2 = H_1$.


The previous equations each indicate an  "ensemble mean value".   The probabilities required for the calculation of  $H_1$  and  $H_2$  can, however, also be calculated as time averages from a very long sequence or, more precisely, approximated by the corresponding  relative frequencies.

Let us now illustrate the calculation of entropy approximations  $H_1$  and  $H_2$  with three examples.

Ternary symbol sequence and formation of two-tuples

$\text{Example 2:}$  We will first look at the sequence  $〈 q_1$, ... , $q_{50} \rangle $  according to the graphic, where the sequence elements  $q_ν$  originate from the alphabet $\rm \{A, \ B, \ C \}$   ⇒   the symbol set size is  $M = 3$.

By time averaging over the  $50$  symbols one gets the symbol probabilities  $p_{\rm A} ≈ 0.5$,   $p_{\rm B} ≈ 0.3$  and  $p_{\rm C} ≈ 0.2$, with which one can calculate the first order entropy approximation:

$$H_1 = 0.5 \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{0.5} + 0.3 \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{0.3} + 0.2 \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{0.2} \approx \, 1.486 \,{\rm bit/symbol} \hspace{0.05cm}.$$

Due to the not equally probable symbols  $H_1 < H_0 = 1.585 \hspace{0.05cm}. \rm bit/symbol$.  As an approximation for the probabilities of two-tuples one gets from the above sequence:

$$\begin{align*}p_{\rm AA} \hspace{-0.1cm}& = \hspace{-0.1cm} 14/49\hspace{0.05cm}, \hspace{0.2cm}p_{\rm AB} = 8/49\hspace{0.05cm}, \hspace{0.2cm}p_{\rm AC} = 3/49\hspace{0.05cm}, \\\ p_{\rm BA} \hspace{-0.1cm}& = \hspace{0.07cm} 7/49\hspace{0.05cm}, \hspace{0.25cm}p_{\rm BB} = 2/49\hspace{0.05cm}, \hspace{0.2cm}p_{\rm BC} = 5/49\hspace{0.05cm}, \\\ p_{\rm CA} \hspace{-0.1cm}& = \hspace{0.07cm} 4/49\hspace{0.05cm}, \hspace{0.25cm}p_{\rm CB} = 5/49\hspace{0.05cm}, \hspace{0.2cm}p_{\rm CC} = 1/49\hspace{0.05cm}.\end{align*}$$

Please note that the  $50$  sequence elements can only be formed from  $49$  two-tuples  $(\rm AA$, ... , $\rm CC)$  which are marked in different colors in the graphic.

  • The entropy approximation  $H_2$  should actually be equal to  $H_1$  since the given symbol sequence comes from a memoryless source.
  • Because of the short sequence length  $N = 50$  and the resulting statistical inaccuracy, however, a smaller value results:  
$$H_2 ≈ 1.39\hspace{0.05cm} \rm bit/symbol.$$


$\text{Example 3:}$  Now let us consider a  memoryless  binary source  with equally probable symbols, i.e. there would be  $p_{\rm A} = p_{\rm B} = 1/2$.  The first twenty subsequent elements are   $〈 q_ν 〉 =\rm BBAAABAABBBBBAAAABABAB$ ...

  • Because of the equally probable binary symbols   $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$.
  • The compound probability  $p_{\rm AB}$  of the combination  $\rm AB$  is equal to  $p_{\rm A} \cdot p_{\rm B} = 1/4$.  Also $p_{\rm AA} = p_{\rm BB} = p_{\rm BA} = 1/4$.
  • This gives for the second entropy approximation
$$H_2 = {1}/{2} \cdot \big [ {1}/{4} \cdot {\rm log}_2\hspace{0.1cm}4 + {1}/{4} \cdot {\rm log}_2\hspace{0.1cm}4 +{1}/{4} \cdot {\rm log}_2\hspace{0.1cm}4 +{1}/{4} \cdot {\rm log}_2\hspace{0.1cm}4 \big ] = 1 \,{\rm bit/symbol} = H_1 = H_0 \hspace{0.05cm}.$$

Note:   Due to the short length of the given sequence the probabilities are slightly different:  $p_{\rm AA} = 6/19$,  $p_{\rm BB} = 5/19$,  $p_{\rm AB} = p_{\rm BA} = 4/19$.


$\text{Example 4:}$  The third sequence considered here results from the binary sequence of  $\text{Example 3}$  by using a simple repeat code:

$$〈 q_ν 〉 =\rm BbBbAaAaAaBbAaAaBbBb \text{...} $$
  • The repeated symbols are marked by corresponding lower case letters.  But it still applies  $M=2$.
  • Because of the equally probable binary symbols, this also results in  $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$.
  • As shown in  Exercise 1.3  for the compound probabilities we obtain  $p_{\rm AA}=p_{\rm BB} = 3/8$  and  $p_{\rm AB}=p_{\rm BA} = 1/8$.  Hence
$$\begin{align*}H_2 ={1}/{2} \cdot \big [ 2 \cdot {3}/{8} \cdot {\rm log}_2\hspace{0.1cm} {8}/{3} + 2 \cdot {1}/{8} \cdot {\rm log}_2\hspace{0.1cm}8\big ] = {3}/{8} \cdot {\rm log}_2\hspace{0.1cm}8 - {3}/{8} \cdot{\rm log}_2\hspace{0.1cm}3 + {1}/{8} \cdot {\rm log}_2\hspace{0.1cm}8 \approx 0.906 \,{\rm bit/symbol} < H_1 = H_0 \hspace{0.05cm}.\end{align*}$$

A closer look at the task at hand leads to the following conclusion:

  • The entropy should actually be  $H = 0.5 \hspace{0.05cm} \rm bit/symbol$  (every second symbol does not provide new information)
  • The second entropy approximation  $H_2 = 0.906 \hspace{0.05cm} \rm bit/symbol$  is much larger than the entropy  $H$.
  • For the entropy determination, the second order approximation is not sufficient.  Rather, one must consider larger contiguous blocks of  $k > 2$  symbols.
  • In the following, such a block is referred to as  "$k$–tuple".


Generalization to $k$-tuple and boundary crossing


For abbreviation we write using the compound probability  $p_i^{(k)}$  for a  $k$–tuple in general:

$$H_k = \frac{1}{k} \cdot \sum_{i=1}^{M^k} p_i^{(k)} \cdot {\rm log}_2\hspace{0.1cm} \frac{1}{p_i^{(k)}} \hspace{0.5cm}({\rm unit\hspace{-0.1cm}: \hspace{0.1cm}bit/symbol}) \hspace{0.05cm}.$$
  • The control variable  $i$  stands for one of the  $M^k$  tuples. 
  • The previously calculated approximation  $H_2$  results with  $k = 2$.


$\text{Definition:}$  The  entropy of a discrete source with memory  has the following limit:

$$H = \lim_{k \rightarrow \infty }H_k \hspace{0.05cm}.$$

For the  entropy approximations  $H_k$  the following relations apply  $(H_0=\log_2 M)$:

$$H \le \text{...} \le H_k \le \text{...} \le H_2 \le H_1 \le H_0 \hspace{0.05cm}.$$


The computational effort will increase with increasing  $k$  except for a few special cases (see the following example) and depends on the symbol set size  $M$  of course:

  • For the calculation of  $H_{10}$  for a binary source  $(M = 2)$  one must average over  $2^{10} = 1024$  terms.
  • With each further increase of  $k$  by  $1$  the number of sum terms doubles.
  • In case of a quaternary source  $(M = 4)$  it must already be averaged over  $4^{10} = 1\hspace{0.08cm}048\hspace{0.08cm}576$  summation terms for the determination of  $H_{10}$.
  • Considering that each of these  $4^{10} =2^{20} >10^6$  $k$–tuple should occur in simulation/time averaging about  $100$  times (statistical guideline) to ensure sufficient simulation accuracy, it follows that the sequence length should be in this case greater than  $N = 10^8$.


$\text{Example 5:}$  We consider an alternating binary sequence   ⇒   $〈 q_ν 〉 =\rm ABABABAB$ ...  .  Accordingly it holds that  $H_0 = H_1 = 1 \hspace{0.1cm} \rm bit/symbol$.

In this special case, the  $H_k$ approximation must be determined independently from  $k$  by averaging only two compound probabilities:

  • $k = 2$:    $p_{\rm AB} = p_{\rm BA} = 1/2$     ⇒     $H_2 = 1/2 \hspace{0.2cm} \rm bit/symbol$,
  • $k = 3$:    $p_{\rm ABA} = p_{\rm BAB} = 1/2$     ⇒     $H_3 = 1/3 \hspace{0.2cm} \rm bit/symbol$,
  • $k = 4$:    $p_{\rm ABAB} = p_{\rm BABA} = 1/2$     ⇒     $H_4 = 1/4 \hspace{0.2cm} \rm bit/symbol$.


The entropy of this alternating binary sequence is therefore

$$H = \lim_{k \rightarrow \infty }{1}/{k} = 0 \hspace{0.05cm}.$$

The result was to be expected, since the considered sequence has only minimal information, which does not affect the entropy end value  $H$,  namely:
    "Does  $\rm A$  occur at the even or the odd times?"

You can see that  $H_k$  comes closer to this final value  $H = 0$  very slowly:  The twentieth entropy approximation still returns  $H_{20} = 0.05 \hspace{0.05cm} \rm bit/symbol$.


$\text{Summary of the results of the last pages:}$ 

  • Generally it applies to the  entropy of a message source:
$$H \le \text{...} \le H_3 \le H_2 \le H_1 \le H_0 \hspace{0.05cm}.$$
  • redundancy-free source   exists if all  $M$  symbols are equally probable and there are no statistical bindings within the sequence.
    For these hold  $(r$  denotes the relative redundancy $)$:
$$H = H_0 = H_1 = H_2 = H_3 = \text{...}\hspace{0.5cm} \Rightarrow \hspace{0.5cm} r = \frac{H - H_0}{H_0}= 0 \hspace{0.05cm}.$$
  • memoryless source   can be quite redundant  $(r> 0)$.  This redundancy then is solely due to the deviation of the symbol probabilities from the uniform distribution.  Here the following relations are valid:
$$H = H_1 = H_2 = H_3 = \text{...} \le H_0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}0 \le r = \frac{H_1 - H_0}{H_0}< 1 \hspace{0.05cm}.$$
  • The corresponding condition for a  source with memory  is
$$ H <\text{...} < H_3 < H_2 < H_1 \le H_0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} 0 < r = \frac{H_1 - H_0}{H_0}\le1 \hspace{0.05cm}.$$
  • If  $H_2 < H_1$, then  $H_3 < H_2$,   $H_4 < H_3$, etc.   ⇒   In the general equation, the  "≤" character must be replaced by the  "<" character.
  • If the symbols are equally probable, then again  $H_1 = H_0$,  while  $H_1 < H_0$  applies to symbols which are not equally probable.


The entropy of the AMI code


In chapter  Symbol-wise Coding with Pseudo-Ternary Codes  of the book  "Digital Signal Transmission", among other things, the AMI pseudo-ternary code is discussed.

Signals and symbol sequences for AMI code
  • This converts the binary sequence  $〈 q_ν 〉$  with  $q_ν ∈ \{ \rm L, \ H \}$  into the ternary sequence  $〈 c_ν 〉$  with  $q_ν ∈ \{ \rm M, \ N, \ P \}$.
  • The names of the source symbols stand for  $\rm L$(ow)  and  $\rm H$(igh)  and those of the code symbols for  $\rm M$(inus),  $\rm P$(lus),  and  $\rm N$(ull)   ⇒   "Zero".


The coding rule of the AMI code ("Alternate Mark Inversion") is:

  • Each binary symbol  $q_ν =\rm L$  is represented by the code symbol  $c_ν =\rm N$ .
  • In contrast,  $q_ν =\rm H$  alternates with  $c_ν =\rm P$  and  $c_ν =\rm M$  coded   ⇒   name "Alternate Mark Inversion".
  • This special encoding adds redundancy with the sole purpose of ensuring that the code sequence does not contain a DC component.


However, we do not consider the spectral properties of the AMI code here, but interpret this code information-theoretically:

  • Based on the symbol set size  $M = 3$  the decision content of the (ternary) code sequence is equal to  $H_0 = \log_2 \ 3 ≈ 1.585 \hspace{0.05cm} \rm bit/symbol$.  The first entropy approximation returns  $H_1 = 1.5 \hspace{0.05cm} \rm bit/Symbol$, as shown in the following calculation:
$$p_{\rm H} = p_{\rm L} = 1/2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm N} = p_{\rm L} = 1/2\hspace{0.05cm},\hspace{0.2cm}p_{\rm M} = p_{\rm P}= p_{\rm H}/2 = 1/4\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_1 = 1/2 \cdot {\rm log}_2\hspace{0.1cm}2 + 2 \cdot 1/4 \cdot{\rm log}_2\hspace{0.1cm}4 = 1.5 \,{\rm bit/symbol} \hspace{0.05cm}.$$
  • Let's now look at two-tuples.  With the AMI code,  $\rm P$  cannot follow  $\rm P$  and  $\rm M$  cannot follow  $\rm M$.  The probability for  $\rm NN$  is equal to  $p_{\rm L} \cdot p_{\rm L} = 1/4$.  All other (six) two-tuples occur with the probability  $1/8$.  From this follows for the second entropy approximation:
$$H_2 = 1/2 \cdot \big [ 1/4 \cdot {\rm log_2}\hspace{0.1cm}4 + 6 \cdot 1/8 \cdot {\rm log_2}\hspace{0.1cm}8 \big ] = 1,375 \,{\rm bit/symbol} \hspace{0.05cm}.$$
  • For the further entropy approximations  $H_3$,  $H_4$, ...  and the actual entropy  $H$  will apply:
$$ H < \hspace{0.05cm}\text{...}\hspace{0.05cm} < H_5 < H_4 < H_3 < H_2 = 1.375 \,{\rm bit/symbol} \hspace{0.05cm}.$$
  • Exceptionally in this example we know the actual entropy  $H$  of the code symbol sequence  $〈 c_ν 〉$:   Since no new information is added by the coder and no information is lost, it is the same entropy  $H = 1 \,{\rm bit/symbol} $  as the one of the redundancy-free binary sequence $〈 q_ν 〉$.


The  Exercise 1.4  shows the considerable effort required to calculate the entropy approximation  $H_3$.   Moreover,  $H_3$  still deviates significantly from the final value  $H = 1 \,{\rm bit/symbol} $.  A faster result is achieved if the AMI code is described by a Markov chain as explained in the next section.


Binary sources with Markov properties


Markov processes with  $M = 2$  states

Sequences with statistical bindings between the sequence elements (symbols) are often modeled by  Markov processes  where we limit ourselves here to first-order Markov processes.  First we consider a binary Markov process  $(M = 2)$  with the states (symbols)  $\rm A$  and  $\rm B$.

On the right, you can see the transition diagram for a first-order binary Markov process  however, only two of the four transfer probabilities given are freely selectable, for example

  • $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}} = \rm Pr(A\hspace{0.01cm}|\hspace{0.01cm}B)$   ⇒   conditional probability that  $\rm A$  follows  $\rm B$ .
  • $p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}} = \rm Pr(B\hspace{0.01cm}|\hspace{0.01cm}A)$   ⇒   conditional probability that  $\rm B$  follows  $\rm A$ .


For the other two transition probabilities:  $p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} = 1- p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}$  and   $p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B} = 1- p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} \hspace{0.05cm}.$

Due to the presupposed properties  "Stationarity"  and  "Ergodicity"  the following applies to the state or symbol probabilities:

$$p_{\rm A} = {\rm Pr}({\rm A}) = \frac{p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}}{p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} + p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} \hspace{0.05cm}, \hspace{0.5cm}p_{\rm B} = {\rm Pr}({\rm B}) = \frac{p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}}{p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} + p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} \hspace{0.05cm}.$$

These equations allow first information–theoretical statements about the Markov processes:

  • For  $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}} = p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$  the symbols are equally likely   ⇒   $p_{\text{A}} = p_{\text{B}}= 0.5$.  The first entropy approximation returns  $H_1 = H_0 = 1 \hspace{0.05cm} \rm bit/symbol$, independent of the actual values of the (conditional) transition probabilities  $p_{\text{A|B}}$  and  $p_{\text{B|A}}$.
  • But the source entropy  $H$  as the limit value  $($for  $k \to \infty)$  of the  Entropy approximation of order  $k$   ⇒   $H_k$  depends very much on the actual values of  $p_{\text{A|B}}$  and  $p_{\text{B|A}}$  and not only on their quotients.  This is shown by the following example.


$\text{Example 6:}$  We consider three binary symmetric Markov sources that are represented by the numerical values of the symmetric transition probabilities  $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} }$.  For the symbol probabilities the following applies:  $p_{\rm A} = p_{\rm B}= 0.5$, and the other transition probabilities have the values

Three examples of binary Markov sources with  $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} }$ 
$$p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 1 - p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}B} }.$$
  • The middle (blue) sequence with  $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.5$  has the entropy  $H ≈ 1 \hspace{0.1cm} \rm bit/symbol$.  That means:   In this special case there are no statistical bindings within the sequence.
  • The left (red) sequence with  $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.2$  has less changes between  $\rm A$  and  $\rm B$.  Due to statistical dependencies between neighboring symbols the entropy is now smaller:  $H ≈ 0.72 \hspace{0.1cm} \rm bit/symbol$.
  • The right (green) symbol sequence with  $p_{\rm {A\hspace{0.01cm}\vert\hspace{0.01cm}B} } = p_{\rm {B\hspace{0.01cm}\vert\hspace{0.01cm}A} } = 0.8$  has the exact same entropy  $H ≈ 0.72 \hspace{0.1cm} \rm bit/symbol$  as the red sequence.  Here you can see many regions with alternating symbols  $($... $\rm ABABAB$ ... $)$.


This example is worth noting:

  • If you had not used the Markov properties of the red and green sequences, you would have reached the respective result  $H ≈ 0.72 \hspace{0.1cm} \rm bit/symbol$  only after lengthy calculations.
  • The following pages show that for a source with Markov properties the final value  $H$  can be determined from the entropy approximations  $H_1$  and  $H_2$  alone.   Likewise, all entropy approximations  $H_1$  and  $H_2$  can also be calculated from  $H_k$  for  $k$–tuples in a simple manner   ⇒   $H_3$,  $H_4$,  $H_5$, ...   $H_{100}$, ...


Simplified entropy calculation for Markov sources


Markov processes with  $M = 2$  states

We continue to assume the first-order symmetric binary Markov source.  As on the previous page, we use the following nomenclature for

  • the transition probabilities   ⇒   $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}}$,   $p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$,  $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}A}}= 1- p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$,   $p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}B}} = 1 - p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}}$,  
  • the ergodic probabilities   ⇒   $p_{\text{A}}$  and  $p_{\text{B}}$,
  • the compound probabilities, for example   ⇒   $p_{\text{AB}} = p_{\text{A}} - p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}}$.


We now compute the  Entropy of a two-tuple  (with the unit "bit/two-tuple"):

$$H_2\hspace{0.05cm}' = p_{\rm A} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm A} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm B} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}} + p_{\rm B} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B}} \hspace{0.05cm}.$$

If one now replaces the logarithms of the products by corresponding sums of logarithms, one gets the result  $H_2\hspace{0.05cm}' = H_1 + H_{\text{M}}$  with

$$H_1 = p_{\rm A} \cdot (p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} + p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A})\cdot {\rm log}_2\hspace{0.1cm}\frac {1}{p_{\rm A}} + p_{\rm B} \cdot (p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} + p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B})\cdot {\rm log}_2\hspace{0.1cm}\frac {1}{p_{\rm B}} = p_{\rm A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{p_{\rm A}} + p_{\rm B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{p_{\rm B}} = H_{\rm bin} (p_{\rm A})= H_{\rm bin} (p_{\rm B}) \hspace{0.05cm},$$
$$H_{\rm M}= p_{\rm A} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm A} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm B} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}} + p_{\rm B} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B}} \hspace{0.05cm}.$$

$\text{Conclusion:}$  Thus the  second entropy approximation  (with the unit "bit/symbol")  is:

$$H_2 = {1}/{2} \cdot {H_2\hspace{0.05cm}'} = {1}/{2} \cdot \big [ H_{\rm 1} + H_{\rm M} \big] \hspace{0.05cm}.$$


It is to be noted:

  • The first summand  $H_1$   ⇒   first entropy approximation depends only on the symbol probabilities.
  • For a symmetrical Markov process   ⇒   $p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}} = p_{\rm {B\hspace{0.01cm}|\hspace{0.01cm}A}} $   ⇒   $p_{\text{A}} = p_{\text{B}} = 1/2$   the result for this first summand is  $H_1 = 1 \hspace{0.1cm} \rm bit/symbol$.
  • The second summand  $H_{\text{M}}$  must be calculated according to the second of the two upper equations.
  • For a symmetrical Markov process you get  $H_{\text{M}} = H_{\text{bin}}(p_{\rm {A\hspace{0.01cm}|\hspace{0.01cm}B}})$.


Now, this result is extended to the  $k$–th entropy approximation.  Here, the advantage of Markov sources over other sources is taken advantage of, that the entropy calculation for  $k$–tuples is very simple.  For each Markov source, the following applies

$$H_k = {1}/{k} \cdot \big [ H_{\rm 1} + (k-1) \cdot H_{\rm M}\big ] \hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_2 = {1}/{2} \cdot \big [ H_{\rm 1} + H_{\rm M} \big ]\hspace{0.05cm}, \hspace{0.3cm} H_3 ={1}/{3} \cdot \big [ H_{\rm 1} + 2 \cdot H_{\rm M}\big ] \hspace{0.05cm},\hspace{0.3cm} H_4 = {1}/{4} \cdot \big [ H_{\rm 1} + 3 \cdot H_{\rm M}\big ] \hspace{0.05cm},\hspace{0.15cm}{\rm usw.}$$

$\text{Conclusion:}$  With the boundary condition for  $k \to \infty$, one obtains the actual source entropy

$$H = \lim_{k \rightarrow \infty} H_k = H_{\rm M} \hspace{0.05cm}.$$

From this simple result important insights for the entropy calculation follow:

  • For Markov sources it is sufficient to determine the entropy approximations  $H_1$  and  $H_2$.  Thus, the entropy of a Markov source is
$$H = 2 \cdot H_2 - H_{\rm 1} \hspace{0.05cm}.$$
  • Through  $H_1$  and  $H_2$  all further entropy approximations  $H_k$  are also fixed for  $k \ge 3$ :
$$H_k = \frac{2-k}{k} \cdot H_{\rm 1} + \frac{2\cdot (k-1)}{k} \cdot H_{\rm 2} \hspace{0.05cm}.$$


However, these approximations are not very important.  Mostly only the limit value  $H$  is of interest.  For sources without Markov properties the approximations  $H_k$  are calculated only to be able to estimate this limit value, i.e. the actual entropy.


Notes:

  • In the  Exercise 1.5  the above equations are applied to the more general case of an asymmetric binary source.
  • All equations on this page also apply to non-binary Markov sources  $(M > 2)$ as shown on the next page.


Non-binary Markov sources


Ternary and quaternary first order Markov source

The following equations apply to each Markov source regardless of the symbol set size  $M$:

$$H = 2 \cdot H_2 - H_{\rm 1} \hspace{0.05cm},$$
$$H_k = {1}/{k} \cdot \big [ H_{\rm 1} + (k-1) \cdot H_{\rm M}\big ] \hspace{0.05cm},$$
$$ \lim_{k \rightarrow \infty} H_k = H \hspace{0.05cm}.$$

These equations allow the simple calculation of the entropy  $H$  from the approximations  $H_1$  and  $H_2$.

We now look at the transition diagrams sketched on the right:

  • A ternary Markov source  $\rm MQ3$  $(M = 3$,  blue coloring$)$,  and
  • a quaternary Markov source  $\rm MQ4$  $(M = 4$,  red color$)$.


In  Exercise 1.6  the entropy approximations  $H_k$  and the total entropy  $H$  are calculated as the limit of  $H_k$  for  $k \to \infty$ .   The results are shown in the following figure.  All entropies specified there have the unit "bit/symbol".

Entropies for  $\rm MQ3$,  $\rm MQ4$  and the  $\rm AMI$  code

These results can be interpreted as follows:

  • For the ternary markov source  $\rm MQ3$  the entropy approximations of  $H_1 = 1.500$  above  $H_2 = 1.375$  up to the limit  $H = 1.250$  continuously decreasing.  Because of  $M = 3$  ⇒   $H_0 = 1.585$.
  • For the quaternary Markov source  $\rm MQ4$  one receives  $H_0 = H_1 = 2.000$  (since four equally probable states) and  $H_2 = 1.5$.   From the  $H_1$-  and  $H_2$-values all entropy approximations  $H_k$  and the final value  $H = 1.000$  can be calculated.
  • The two models  $\rm MQ3$  and  $\rm MQ4$  were created during the attempt to calculate the  AMI code  to be described information–theoretically by Markov sources.  The symbols  $\rm M$,  $\rm N$  and  $\rm P$  stand for  "minus",  "zero"  and  "plus".
  • The entropy approximations  $H_1$,  $H_2$  and  $H_3$  of the AMI code (green markers) were calculated in the  Exercise 1.4.  On the calculation of  $H_4$,  $H_5$, ... had to be omitted for reasons of effort.  But the final value of  $H_k$  for  $k \to \infty$   ⇒   $H = 1.000$  is known.
  • You can see that the Markov model  $\rm MQ3$  for  $H_0 = 1.585$,  $H_1 = 1.500$  and  $H_2 = 1.375$  yields exactly the same numerical values as the AMI code.   On the other hand  $H_3$  ⇒   $(1.333$  instead of  $1.292)$  and especially the final value  $H$  ⇒   $(1.250$  compared to  $1.000)$.
  • The model  $\rm MQ4$  $(M = 4)$  differs from the AMI code  $(M = 3)$  with respect to the decision content  $H_0$  and also with respect to all entropy approximations  $H_k$.   Nevertheless, $\rm MQ4$  is the more suitable model for the AMI code, since the final value  $H = 1,000$  is the same.
  • The model  $\rm MQ3$  yields entropy values that are too large, since the sequences  $\rm PNP$  and  $\rm MNM$  are possible here, which cannot occur in the AMI code.   Already with the approximation  $H_3$  the difference is slightly noticeable, in the final value  $H$  even clearly  $(1.25$  instead of  $1)$.


In the model  $\rm MQ4$  the state  "Zero"  was split into two states  $\rm N$  and  $\rm O$  (see upper right figure on this page):

  • Here applies to the state  $\rm N$:   The current binary symbol  $\rm L$  is encoded with the amplitude value  $0$  (zero), as per the AMI rule.  The next occurring  $\rm H$ symbol, on the other hand, is displayed as  $-1$  (minus), because the last symol  $\rm H$  was encoded as  $+1$  (plus).
  • Also with the state  $\rm O$  the current binary symbol  $\rm L$  is represented with the ternary value  $0$.  In contrast to the state  $\rm N$  however, the next occurring  $\rm H$ symbol is now encoded as  $+1$  (plus) since the last  $\rm H$ symbol was encoded as  $-1$  (minus).


$\text{Conclusion:}$ 

  • The  $\rm MQ4$ output sequence actually follows the rules of the AMI code and has entropy  $H = 1.000 \hspace{0.15cm} \rm bit/symbol$.
  • But because of the new state  $\rm O$  now  $H_0 = 2.000 \hspace{0.15cm} \rm bit/symbol$  $($against  $1.585 \hspace{0.15cm} \rm bit/symbol)$  is clearly too large.
  • Also all  $H_k$ approximations are larger than with the AMI code.  First for  $k \to \infty$  the model  $\rm MQ4$  and the AMI code match exactly:   $H = 1.000 \hspace{0.15cm} \rm bit/symbol$.


Exercises for the chapter


Exercise 1.3: Entropy Approximations

Exercise 1.4: Entropy Approximations for the AMI Code

Exercise 1.4Z: Entropy of the AMI Code

Exercise 1.5: Binary Markov Source

Exercise 1.5Z: Symmetrical Markov Source

Exercise 1.6: Non-Binary Markov Sources

Exercise 1.6Z:Ternary Markov Source