# Exercise 1.5Z: Symmetrical Markov Source

Binary Markov diagram

Task 1.5  dealt with a binary Markov source in which the transition probabilities from  $\rm A$  to  $\rm B$  and from  $\rm B$  to  $\rm A$  were different.

In this task, the following shall now apply

$$p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} = p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} = q \hspace{0.8cm} ( 0 \le q \le 1) \hspace{0.05cm}.$$

All equations given in task 1.5 also apply here:

• Entropy:
$$H = p_{\rm AA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm AB} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A}} + p_{\rm BA} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B}} + p_{\rm BB} \cdot {\rm log_2}\hspace{0.1cm}\frac {1}{ p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B}} \hspace{0.05cm}.$$
• First entropy approximation::
$$H_{\rm 1} = p_{\rm A} \cdot {\rm log_2}\hspace{0.1cm} \frac{1}{p_{\rm A}} + p_{\rm B} \cdot {\rm log_2}\hspace{0.1cm} \frac{1}{p_{\rm B}} \hspace{0.05cm}.$$
• k–th entropy approximation $(k = 2, 3, \ \text{...})$:
$$H_k = {1}/{k} \cdot \big [ H_{\rm 1} + (k-1) \cdot H \big] \hspace{0.05cm},\hspace{0.5cm}H = \lim_{k \rightarrow \infty } H_k \hspace{0.05cm}.$$

Hints:

### Questions

1

Calculate the symbol probabilities for the transition probability  $q = 1/4$.

 $p_{\rm A} \ = \$ $p_{\rm B} \ = \$

2

Calculate the source entropy  $H$  for  $q = 1/4$.

 $H \ =$ $\ \rm bit/symbol$

3

What entropy approximations are obtained for  $q = 1/4$?

 $H_1 \ = \$ $\ \rm bit/symbol$ $H_2 \ = \$ $\ \rm bit/symbol$ $H_3 \ = \$ $\ \rm bit/symbol$

4

Determine  $q$  such that  $H$  becomes maximum. Interpretation.

 $q \ = \$

5

Which symbol sequences are possible with  $q = 0$ ?

 $\rm AAAAAA$ ... $\rm BBBBBB$ ... $\rm ABABAB$ ...

6

Which symbol sequences are possible with  $q = 1$ ?

 $\rm AAAAAA$ ... $\rm BBBBBB$ ... $\rm ABABAB$ ...

### Solution

#### Solution

(1)  For a stationary first order binary Markov source holds:

$$p_{\rm A} = p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} \cdot p_{\rm A} + p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} \cdot p_{\rm B} = (1-q) \cdot p_{\rm A} + q \cdot p_{\rm B}$$
$$\Rightarrow \hspace{0.3cm}q \cdot p_{\rm A} = q \cdot p_{\rm B} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm A} = p_{\rm B}\hspace{0.15cm} \underline {= 0.5} \hspace{0.05cm}.$$

(2)  To calculate the entropy  $H$  one needs all four composite probabilities:

$$p_{\rm AA} \hspace{0.1cm} = \hspace{0.1cm} p_{\rm A} \cdot p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} = 1/2 \cdot(1-q) = p_{\rm BB}\hspace{0.05cm},\hspace{1cm} p_{\rm AB} \hspace{0.1cm} = \hspace{0.1cm} p_{\rm A} \cdot p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} = 1/2 \cdot q = p_{\rm BA}\hspace{0.05cm}.$$
• Substituting these values into the given entropy equation, we get
$$H = 2 \cdot \frac{1}{2} \cdot(1-q) \cdot {\rm log}_2\hspace{0.1cm} \frac{1}{1-q} + 2 \cdot \frac{1}{2} \cdot q \cdot {\rm log}_2\hspace{0.1cm} \frac{1}{q} = q \cdot {\rm log}_2\hspace{0.1cm} \frac{1}{q} + (1-q) \cdot {\rm log}_2\hspace{0.1cm} \frac{1}{1-q} = H_{\rm bin}(q) \hspace{0.05cm}.$$
• The numerical value sought is  $H = H_{\rm bin} (0.25) \hspace{0.15cm}\underline{= 0.811 \, \rm bit/symbol}$.

(3)  For equally probable binary symbols,  $H_1 \hspace{0.15cm}\underline{= 1 \, \rm bit/Symbol}$.

• Using the equation valid for Markov sources, it further holds:
$$H_2 \hspace{0.1cm} = \hspace{0.1cm} {1}/{2} \cdot \big[ H_1 + H \big] \hspace{0.15cm} \underline {= 0.906 \,{\rm bit/Symbol}} \hspace{0.05cm},$$
$$H_3 \hspace{0.1cm} = \hspace{0.1cm} {1}/{3} \cdot \big[ H_1 + 2 H \big] \hspace{0.15cm} \underline {= 0.874 \,{\rm bit/Symbol}} \hspace{0.05cm}.$$

(4)  The maximum of the binary entropy function is obtained for  $q\hspace{0.15cm}\underline{= 0.5}$.

• Thus the maximum entropy is  $H = 1 \, \rm bit/symbol$.
• It can be seen from the relationship  $H = H_1$  and from the transition diagram shown at the front that  $q = 0.5$  results in statistically independent symbols:
$$p_{\rm A} = p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}A} = p_{\rm A\hspace{0.01cm}|\hspace{0.01cm}B} = 0.5 \hspace{0.05cm}, \hspace{0.2cm} p_{\rm B} = p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}A} = p_{\rm B\hspace{0.01cm}|\hspace{0.01cm}B}= 0.5 \hspace{0.05cm}.$$

(5)  Proposed solutions 1 and 2 are correct:

• The symbol sequence results either in  $\rm AAAAAA$ ...  or in  $\rm BBBBBB$ ... , depending on which symbol was given as the starting value.
• The entropy of such a source is always  $H = H_{\rm bin}(0) = 0$.

(6)  Only proposed solution 3 is correct:

• Now neither  $\rm A$  can directly follow  $\rm A$  nor  $\rm B$  can directly follow  $\rm B$ .
• The result is always an alternating sequence, depending on the starting value the sequence  $\rm ABABAB$ ...  or  $\rm BABABA$... .
• This source also has the entropy  $H = H_{\rm bin}(1) = 0$ in both cases.