Aufgaben:Exercise 1.6Z: Ternary Markov Source: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Information_Theory/Discrete_Sources_with_Memory | ||
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[[File:Inf_Z_1_6_vers2.png|right|frame| | [[File:Inf_Z_1_6_vers2.png|right|frame|Ternary Markov source]] | ||
The diagram on the right shows a Markov source with $M = 3$ states $\rm A$, $\rm B$ and $\rm C$. | |||
Let the two parameters of this Markov process be: | |||
:$$0 \le p \le 0.5 \hspace{0.05cm},\hspace{0.2cm}0 \le q \le 1 \hspace{0.05cm}.$$ | :$$0 \le p \le 0.5 \hspace{0.05cm},\hspace{0.2cm}0 \le q \le 1 \hspace{0.05cm}.$$ | ||
Due to the Markov property of this source, the entropy can be determined in different ways: | |||
* | *One calculates the first two entropy approximations $H_1$ and $H_2$. Then the following applies to the actual entropy: | ||
:$$H | :$$H= H_{k \to \infty} = 2 \cdot H_{\rm 2} - H_{\rm 1} \hspace{0.05cm}.$$ | ||
* | *However, according to the "direct calculation method", the entropy can also be written as follows (nine terms in total): | ||
:$$H = p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm AB} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + \ \text{...} | :$$H = p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm AB} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + \ \text{...}\hspace{0.05cm}, \\text{wobei} \ p_{\rm AA} = p_{\rm A} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.05cm},\hspace{0.2cm}p_{\rm AB} = p_{\rm A} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.05cm}, \ \text{...}$$ | ||
\text{wobei} \ p_{\rm AA} = p_{\rm A} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.05cm},\hspace{0.2cm} | |||
p_{\rm AB} = p_{\rm A} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.05cm}, \ \text{...}$$ | |||
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''Hint:'' | |||
*The exercise belongs to the chapter [[Information_Theory/Discrete_Sources_with_Memory|Discrete Sources with Memory]]. | |||
*Reference is made in particular to the page [[Information_Theory/Discrete_Sources_with_Memory#Non-binary_Markov_sources|Non-binary Markov sources]]. | |||
*For all entropies, add the pseudo-unit "bit/symbol". | |||
=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
{ | {For which parameters $p$ and $q$ does the maximum entropy per symbol result? | ||
|type="{}"} | |type="{}"} | ||
$p \ = \ $ { 0.333 1% } | $p \ = \ $ { 0.333 1% } | ||
$q\ = \ $ { 1 1% } | $q\ = \ $ { 1 1% } | ||
$H_\text{max} \ = \ $ { 1.585 1% } $\ \rm bit/ | $H_\text{max} \ = \ $ { 1.585 1% } $\ \rm bit/symbol$ | ||
{ | {Let $p = 1/4$ and $q = 1$. What is the value of the first entropy approximation in this case? | ||
|type="{}"} | |type="{}"} | ||
$H_1 = \ \ $ { 1.585 3% } $\ \rm bit/ | $H_1 = \ \ $ { 1.585 3% } $\ \rm bit/symbol$ | ||
{ | {Furthermore, let $p = 1/4$ and $q = 1$. What value results in this case for the second entropy approximation? | ||
|type="{}"} | |type="{}"} | ||
$H_2 = \ \ $ { 1.5425 1% } $\ \rm bit/ | $H_2 = \ \ $ { 1.5425 1% } $\ \rm bit/symbol$ | ||
{ | {What is the actual source entropy $H= H_{k \to \infty}$ with $p = 1/4$ and $q = 1$? | ||
|type="{}"} | |type="{}"} | ||
$H = \ \ $ { 1.5 1% } $\ \rm bit/ | $H = \ \ $ { 1.5 1% } $\ \rm bit/symbol$ | ||
{ | {What is the actual source entropy $H= H_{k \to \infty}$ with $p = 1/2$ and $q = 0$? | ||
|type="{}"} | |type="{}"} | ||
$H = \ \ $ { 0.667 1% } $\ \rm bit/ | $H = \ \ $ { 0.667 1% } $\ \rm bit/symbol$ | ||
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</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | '''(1)''' The maximum entropy results when the symbols $\rm A$, $\rm B$ and $\rm C$ are equally probable and the symbols within the sequence are statistically independent of each other. Then the following must apply: | ||
[[File:Inf_Z_1_6_vers2.png|right|frame| | [[File:Inf_Z_1_6_vers2.png|right|frame|Transition diagram for $p = 1/4$, $q = 1$]] | ||
* $p_{\rm A} = p_{\rm A|A} = p_{\rm A|B} = p_{\rm A|C} = 1/3$, | * $p_{\rm A} = p_{\rm A|A} = p_{\rm A|B} = p_{\rm A|C} = 1/3$, | ||
* $p_{\rm B} = p_{\rm B|A} = p_{\rm B|B} = p_{\rm B|C} = 1/3$, | * $p_{\rm B} = p_{\rm B|A} = p_{\rm B|B} = p_{\rm B|C} = 1/3$, | ||
| Line 75: | Line 72: | ||
From this, the values we are looking for can be determined: | |||
* | *For example, from $p_{\rm C|C} = 1/3$ one obtains the value $p \hspace{0.15cm}\underline{= 1/3}$. | ||
* | *If one also takes into account the relationship $p_{\rm A|A} = p \cdot q$, then $q \hspace{0.15cm}\underline{= 1}$. | ||
* | *This gives the maximum entropy $H_\text{max} ={\rm log_2} \ 3\hspace{0.15cm}\underline{= 1.585\ \rm bit/symbol}$. | ||
'''(2)''' With the parameter values $p = 1/4$ and $q = 1$ , we obtain the adjacent transition diagram, which has the following symmetries: | |||
* $p_{\rm A|A} = p_{\rm B|B} = p_{\rm C|C} = 1/4$ (marked in red), | |||
* $p_{\rm A|B} = p_{\rm B|C} = p_{\rm C|A} = 1/2$ (marked in green), | |||
*$p_{\rm A|C} = p_{\rm B|A} = p_{\rm C|CB} = 1/4$ (marked in blue). | |||
It is obvious that the symbol probabilities are all equal: | |||
:$$p_{\rm A} = p_{\rm B} = p_{\rm C} = 1/3 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H_1 = {\rm log_2}\hspace{0.1cm} 3 \hspace{0.15cm} \underline {= 1.585 \,{\rm bit/symbol}}\hspace{0.05cm}.$$ | |||
:$$p_{\rm | |||
'''(3)''' For the second entropy approximation one needs $3^2 = 9$ joint probabilities. | |||
'''( | *Using the result from '''(2)''' , one obtains for this: | ||
:$$p_{\rm AA} = p_{\rm BB}= p_{\rm CC}= p_{\rm AC}=p_{\rm BA}=p_{\rm CB}=1/12 \hspace{0.05cm},\hspace{0.5cm}p_{\rm AB} = p_{\rm BC}=p_{\rm CA}=1/6$$ | |||
:$$\Rightarrow \hspace{0.2cm} H_2 = \frac{1}{2} \cdot \left [ 6 \cdot \frac{1}{12} \cdot {\rm log_2}\hspace{0.1cm} 12 +3 \cdot \frac{1}{6} \cdot {\rm log_2}\hspace{0.1cm} 6 \right ] = \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 4 + \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 3 + \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 2 + \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 3= \frac{3}{4} + \frac{{\rm log_2}\hspace{0.1cm} 3}{2} \hspace{0.15cm} \underline {= 1.5425 \,{\rm bit/symbol}}\hspace{0.05cm}.$$ | |||
\ | |||
* | |||
'''(4)''' Due to the Markov property of the source, the following holds true: | |||
:$$H = H_{k \to \infty}= 2 \cdot H_2 - H_1 = \big [ {3}/{2} + {\rm log_2}\hspace{0.1cm} 3 \big ] - {\rm log_2}\hspace{0.1cm} 3\hspace{0.15cm} \underline {= 1.5 \,{\rm bit/symbol}}\hspace{0.05cm}.$$ | |||
*One would arrive at the same result according to the following calculation: | |||
:$$H= p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm AB} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + ... \hspace{0.1cm}= 6 \cdot \frac{1}{12} \cdot {\rm log_2}\hspace{0.1cm} 4 + 3 \cdot \frac{1}{16} \cdot {\rm log_2}\hspace{0.1cm} 2\hspace{0.15cm} \underline {= 1.5 \,{\rm bit/symbol}}\hspace{0.05cm}.$$ | |||
[[File:Inf_Z_1_6e_vers2.png|right|frame|Transition diagram for $p = 1/4$, $q = 0$]] | |||
'''(5)''' From the adjacent transition diagram with the current parameters, one can see that in the case of stationarity $p_{\rm B} = 0$ will apply, since $\rm B$ can occur at most once at the starting time. | |||
*So there is a binary Markov chain with the symbols $\rm A$ and $\rm C$ . | |||
*The symbol probabilities are therefore given by: | |||
:$$p_{\rm A} = 0.5 \cdot p_{\rm C} \hspace{0.05cm}, \hspace{0.2cm}p_{\rm A} + p_{\rm C} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm A} = 1/3 \hspace{0.05cm}, \hspace{0.2cm} p_{\rm C} = 2/3\hspace{0.05cm}. $$ | |||
*This gives the following probabilities: | |||
:$$p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.1cm} = \hspace{0.1cm}0\hspace{0.7cm} \Rightarrow \hspace{0.3cm} p_{\rm AA} = 0 \hspace{0.05cm},$$ | :$$p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.1cm} = \hspace{0.1cm}0\hspace{0.7cm} \Rightarrow \hspace{0.3cm} p_{\rm AA} = 0 \hspace{0.05cm},$$ | ||
:$$ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C} =1/2\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm CA} = | :$$ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C} =1/2\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm CA} =p_{\rm C} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C} = 2/3 \cdot 1/2 = 1/3 \hspace{0.05cm},\hspace{0.2cm}{\rm log_2}\hspace{0.1cm}(1/p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C} )= 1\hspace{0.05cm},$$ | ||
:$$ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A} =1\hspace{0.7cm} \Rightarrow \hspace{0.3cm} p_{\rm AC} =p_{\rm A} \cdot p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A} = 1/3 \cdot 1 = 1/3 \hspace{0.05cm},\hspace{0.61cm}{\rm log_2}\hspace{0.1cm}(1/p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A} )= 0\hspace{0.05cm},$$ | |||
:$$ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A} =1\hspace{0.7cm} \Rightarrow \hspace{0.3cm} p_{\rm AC} = | :$$ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C} =1/2\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm CC} =p_{\rm C} \cdot p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C} = 2/3 \cdot 1/2 = 1/3\hspace{0.05cm},\hspace{0.2cm}{\rm log_2}\hspace{0.1cm}(1/p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C} )= 1 $$ | ||
:$$\Rightarrow \hspace{0.25cm} H = p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} +p_{\rm CA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C}}+ p_{\rm AC} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A}} +p_{\rm CC} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C}}=0 + 1/3 \cdot 1 + 1/3 \cdot 0 + 1/3 \cdot 1\hspace{0.15cm} \underline {= 0.667 \,{\rm bit/symbol}}\hspace{0.05cm}.$$ | |||
:$$ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C} =1/2\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm CC} = | |||
:$$\Rightarrow \hspace{0.25cm} H = p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} +p_{\rm CA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C}}+ p_{\rm AC} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A}} + | |||
0 + 1/3 \cdot 1 + 1/3 \cdot 0 + 1/3 \cdot 1 | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
[[Category: | [[Category:Information Theory: Exercises|^1.2 Sources with Memory^]] | ||
[[de:Aufgaben:Aufgabe 1.6Z: Ternäre Markovquelle]] | |||
Latest revision as of 17:57, 16 March 2026

The diagram on the right shows a Markov source with $M = 3$ states $\rm A$, $\rm B$ and $\rm C$.
Let the two parameters of this Markov process be:
- $$0 \le p \le 0.5 \hspace{0.05cm},\hspace{0.2cm}0 \le q \le 1 \hspace{0.05cm}.$$
Due to the Markov property of this source, the entropy can be determined in different ways:
- One calculates the first two entropy approximations $H_1$ and $H_2$. Then the following applies to the actual entropy:
- $$H= H_{k \to \infty} = 2 \cdot H_{\rm 2} - H_{\rm 1} \hspace{0.05cm}.$$
- However, according to the "direct calculation method", the entropy can also be written as follows (nine terms in total):
- $$H = p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm AB} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + \ \text{...}\hspace{0.05cm}, \\text{wobei} \ p_{\rm AA} = p_{\rm A} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.05cm},\hspace{0.2cm}p_{\rm AB} = p_{\rm A} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.05cm}, \ \text{...}$$
Hint:
- The exercise belongs to the chapter Discrete Sources with Memory.
- Reference is made in particular to the page Non-binary Markov sources.
- For all entropies, add the pseudo-unit "bit/symbol".
Questions
Solution

- $p_{\rm A} = p_{\rm A|A} = p_{\rm A|B} = p_{\rm A|C} = 1/3$,
- $p_{\rm B} = p_{\rm B|A} = p_{\rm B|B} = p_{\rm B|C} = 1/3$,
- $p_{\rm C} = p_{\rm C|A} = p_{\rm C|B} = p_{\rm C|C} = 1/3$.
From this, the values we are looking for can be determined:
- For example, from $p_{\rm C|C} = 1/3$ one obtains the value $p \hspace{0.15cm}\underline{= 1/3}$.
- If one also takes into account the relationship $p_{\rm A|A} = p \cdot q$, then $q \hspace{0.15cm}\underline{= 1}$.
- This gives the maximum entropy $H_\text{max} ={\rm log_2} \ 3\hspace{0.15cm}\underline{= 1.585\ \rm bit/symbol}$.
(2) With the parameter values $p = 1/4$ and $q = 1$ , we obtain the adjacent transition diagram, which has the following symmetries:
- $p_{\rm A|A} = p_{\rm B|B} = p_{\rm C|C} = 1/4$ (marked in red),
- $p_{\rm A|B} = p_{\rm B|C} = p_{\rm C|A} = 1/2$ (marked in green),
- $p_{\rm A|C} = p_{\rm B|A} = p_{\rm C|CB} = 1/4$ (marked in blue).
It is obvious that the symbol probabilities are all equal:
- $$p_{\rm A} = p_{\rm B} = p_{\rm C} = 1/3 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H_1 = {\rm log_2}\hspace{0.1cm} 3 \hspace{0.15cm} \underline {= 1.585 \,{\rm bit/symbol}}\hspace{0.05cm}.$$
(3) For the second entropy approximation one needs $3^2 = 9$ joint probabilities.
- Using the result from (2) , one obtains for this:
- $$p_{\rm AA} = p_{\rm BB}= p_{\rm CC}= p_{\rm AC}=p_{\rm BA}=p_{\rm CB}=1/12 \hspace{0.05cm},\hspace{0.5cm}p_{\rm AB} = p_{\rm BC}=p_{\rm CA}=1/6$$
- $$\Rightarrow \hspace{0.2cm} H_2 = \frac{1}{2} \cdot \left [ 6 \cdot \frac{1}{12} \cdot {\rm log_2}\hspace{0.1cm} 12 +3 \cdot \frac{1}{6} \cdot {\rm log_2}\hspace{0.1cm} 6 \right ] = \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 4 + \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 3 + \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 2 + \frac{1}{4} \cdot {\rm log_2}\hspace{0.1cm} 3= \frac{3}{4} + \frac{{\rm log_2}\hspace{0.1cm} 3}{2} \hspace{0.15cm} \underline {= 1.5425 \,{\rm bit/symbol}}\hspace{0.05cm}.$$
(4) Due to the Markov property of the source, the following holds true:
- $$H = H_{k \to \infty}= 2 \cdot H_2 - H_1 = \big [ {3}/{2} + {\rm log_2}\hspace{0.1cm} 3 \big ] - {\rm log_2}\hspace{0.1cm} 3\hspace{0.15cm} \underline {= 1.5 \,{\rm bit/symbol}}\hspace{0.05cm}.$$
- One would arrive at the same result according to the following calculation:
- $$H= p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm AB} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + ... \hspace{0.1cm}= 6 \cdot \frac{1}{12} \cdot {\rm log_2}\hspace{0.1cm} 4 + 3 \cdot \frac{1}{16} \cdot {\rm log_2}\hspace{0.1cm} 2\hspace{0.15cm} \underline {= 1.5 \,{\rm bit/symbol}}\hspace{0.05cm}.$$

(5) From the adjacent transition diagram with the current parameters, one can see that in the case of stationarity $p_{\rm B} = 0$ will apply, since $\rm B$ can occur at most once at the starting time.
- So there is a binary Markov chain with the symbols $\rm A$ and $\rm C$ .
- The symbol probabilities are therefore given by:
- $$p_{\rm A} = 0.5 \cdot p_{\rm C} \hspace{0.05cm}, \hspace{0.2cm}p_{\rm A} + p_{\rm C} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm A} = 1/3 \hspace{0.05cm}, \hspace{0.2cm} p_{\rm C} = 2/3\hspace{0.05cm}. $$
- This gives the following probabilities:
- $$p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \hspace{0.1cm} = \hspace{0.1cm}0\hspace{0.7cm} \Rightarrow \hspace{0.3cm} p_{\rm AA} = 0 \hspace{0.05cm},$$
- $$ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C} =1/2\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm CA} =p_{\rm C} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C} = 2/3 \cdot 1/2 = 1/3 \hspace{0.05cm},\hspace{0.2cm}{\rm log_2}\hspace{0.1cm}(1/p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C} )= 1\hspace{0.05cm},$$
- $$ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A} =1\hspace{0.7cm} \Rightarrow \hspace{0.3cm} p_{\rm AC} =p_{\rm A} \cdot p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A} = 1/3 \cdot 1 = 1/3 \hspace{0.05cm},\hspace{0.61cm}{\rm log_2}\hspace{0.1cm}(1/p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A} )= 0\hspace{0.05cm},$$
- $$ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C} =1/2\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm CC} =p_{\rm C} \cdot p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C} = 2/3 \cdot 1/2 = 1/3\hspace{0.05cm},\hspace{0.2cm}{\rm log_2}\hspace{0.1cm}(1/p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C} )= 1 $$
- $$\Rightarrow \hspace{0.25cm} H = p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} +p_{\rm CA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}C}}+ p_{\rm AC} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}A}} +p_{\rm CC} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm C\hspace{0.01cm}|\hspace{0.01cm}C}}=0 + 1/3 \cdot 1 + 1/3 \cdot 0 + 1/3 \cdot 1\hspace{0.15cm} \underline {= 0.667 \,{\rm bit/symbol}}\hspace{0.05cm}.$$