# Exercise 1.3: Entropy Approximations

Different binary sequences

The graphic on the right shows four symbol sequences  $\langle q_\nu \rangle$,  each with length  $N = 60$.  The source symbols are  $\rm A$  and  $\rm B$.

• It follows directly that  $H_0 = 1 \; \rm bit/symbol$  applies to the decision content of all sources considered.
• However, the symbols  $\rm A$  and  $\rm B$  do not occur with equal probability, but with the probabilities  $p_{\rm A}$  and  $p_{\rm B}$.

In addition to  $H_0$ , the table below shows the entropy approximations

• $H_1$,  based on  $p_{\rm A}$  und  $p_{\rm B}$  (column 2),
• $H_2$,  based on two-tuples (column 3),
• $H_3$,  based on three-tuples (column 4),
• $H_4$,  based on four-tuples (column 5),
• the actual entropy  $H$, which is obtained from  $H_k$  by the boundary transition for  $k \to \infty$  (last column).

The following size relations exist between these entropies:   $H \le$ ... $\le H_3 \le H_2 \le H_1 \le H_0 \hspace{0.05cm}.$

• What is not known is the correlation between the sources  $\rm Q1$,  $\rm Q2$,  $\rm Q3$,  $\rm Q4$  and the symbol sequences shown in the graph
(black, blue, red, green).
• It is only known that source  $\rm Q4$  contains a repetition code.  Due to the fact that in the corresponding symbol sequence every second symbol does not lier any information,  the final entrpy value is  $H = 0.5 \; \rm bit/symbol$.
• In addition, the entropy approximations  $H_1 = 1 \; \rm bit/symbol$  and  $H_4 \approx 0.789 \; \rm bit/symbol$  are given.

Finally, the entropy approximations  $H_2$  and  $H_3$ are to be determined for the source  $\rm Q4$ .

Source entropy and approximations in "bit/symbol"

Hints:

• This task belongs to the chapter  Discrete Sources with Memory.
• For the  $k$–th entropy approximation, the following holds for binary sources  $(M = 2)$  with the composite probability  $p_i^{(k)}$  of a  $k$–tuple:
$$H_k = \frac{1}{k} \cdot \sum_{i=1}^{2^k} p_i^{(k)} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{p_i^{(k)}} \hspace{0.5cm}({\rm unit\hspace{-0.1cm}: \hspace{0.1cm}bit/Symbol}) \hspace{0.05cm}.$$

### Questions

1

What is the source of the black symbol sequence?

 $\rm Q1$, $\rm Q2$, $\rm Q3$, $\rm Q4$.

2

What is the source of the blue symbol sequence?

 $\rm Q1$, $\rm Q2$, $\rm Q3$, $\rm Q4$.

3

What is the source of the red symbol sequence?

 $\rm Q1$, $\rm Q2$, $\rm Q3$, $\rm Q4$.

4

Calculate the entropy approximation  $H_2$  of the repetition code  $\rm Q4$.

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

5

Calculate the entropy approximation  $H_3$  of the repetition code  $\rm Q4$.

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

### Solution

#### Solution

(1)  The black binary sequence comes from the source  $\underline{\rm Q3}$,

• since the symbols are equally probable   ⇒   $H_1 = H_0$,  and
• there are no statistical bindings between the symbols   ⇒   $H=$ ... $= H_2 = H_1$.

(2)  It can be seen in the blue binary sequence that  $\rm A$  occurs much more frequently than  $\rm B$, so that  $H_1 < H_0$  must hold.

• According to the table, only source  $\underline{\rm Q1}$  fulfils this condition.
• From  $H_1 = 0.5 \; \rm bit/symbol$  one can determine the symbol probabilities  $p_{\rm A} = 0.89$  and  $p_{\rm B} = 0.11$ .

(3)  By exclusion procedure one arrives at the result  $\underline{\rm Q2}$ for the red binary sequence:

• The source  $\rm Q1$ belongs to the blue sequence,  $\rm Q3$  to the black and  $\rm Q4$  to the repetition code and thus obviously to the green symbol sequence.
• The red symbol sequence has the following properties:
• Because of  $H_1 = H_0$ , the symbols are equally probable:   $p_{\rm A} = p_{\rm B} = 0.5$.
• Because of  $H < H_1$,  there are statistical bindings within the sequence.
• This can be recognised by the fact that there are more transitions between  $\rm A$  and  $\rm B$  than with statistical independence.

(4)  In the green symbol sequence  $($source  $\rm Q4)$ , the symbols  $\rm A$  and  $\rm B$  are equally likely:

Symbol sequences of a binary repetition code
$$p_{\rm A} = p_{\rm B} = 0.5 \hspace{0.3cm}\Rightarrow\hspace{0.3cm}H_1 = 1\,{\rm bit/Symbol} \hspace{0.05cm}.$$

To determine  $H_2$, one considers two-tuples.  The composite probabilities  $p_{\rm AA}$,  $p_{\rm AB}$,  $p_{\rm BA}$  and  $p_{\rm BB}$  can be calculated from this.  You can see from the sketch:

• The combinations  $\rm AB$  and  $\rm BA$  are only possible if a tuple starts at even  $\nu$ .  For the composite probabilities  $p_{\rm AB}$  and  $p_{\rm BA}$  then holds:
$$p_{\rm AB} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}(\nu {\rm \hspace{0.15cm}is\hspace{0.15cm}even}) \cdot {\rm Pr}( q_{\nu} = \mathbf{A}) \cdot {\rm Pr}(q_{\nu+1} = \mathbf{B}\hspace{0.05cm} | q_{\nu} = \mathbf{A}) = {1}/{2} \cdot {1}/{2} \cdot {1}/{2} = {1}/{8} = p_{\rm BA} \hspace{0.05cm}.$$
• In contrast, for the two other combinations  $\rm AA$  and  $\rm BB$:
$$p_{\rm AA} ={\rm Pr}(\nu = 1) \cdot {\rm Pr}( q_1 = \mathbf{A}) \cdot {\rm Pr}(q_{2} = \mathbf{A}\hspace{0.05cm} | q_{1} = \mathbf{A}) + {\rm Pr}(\nu=2) \cdot {\rm Pr}( q_{2} = \mathbf{A}) \cdot {\rm Pr}(q_{3} = \mathbf{A}\hspace{0.05cm} | q_{2} = \mathbf{A}) \hspace{0.05cm}.$$
$$\Rightarrow \hspace{0.3cm}p_{\rm AA} = \frac{1}{2} \cdot \frac{1}{2} \cdot 1+ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{8} = p_{\rm BB} \hspace{0.05cm}.$$
Here  $\nu = 1$  stands for all odd indices and  $\nu = 2$  for all even indices.
• This gives for the entropy approximation:
$$H_2 = \frac{1}{2} \cdot \left [ 2 \cdot \frac{3}{8} \cdot {\rm log}_2\hspace{0.1cm}\frac {8}{3} + 2 \cdot \frac{1}{8} \cdot {\rm log}_2\hspace{0.1cm}(8)\right ] = \frac{3}{8} \cdot {\rm log}_2\hspace{0.1cm}(8) - \frac{3}{8} \cdot {\rm log}_2\hspace{0.1cm}(3) + \frac{1}{8} \cdot {\rm log}_2\hspace{0.1cm}(8) \hspace{0.15cm} \underline {= 0.906 \,{\rm bit/symbol}} \hspace{0.05cm}.$$

(5)  Following a similar procedure, we arrive at the composite probabilities for three-tuples

$$p_{\rm AAA} \hspace{0.1cm} = \hspace{0.1cm} p_{\rm BBB} = 1/4 \hspace{0.05cm},\hspace{0.2cm} p_{\rm ABA} = p_{\rm BAB} = 0 \hspace{0.05cm},\hspace{0.2cm} p_{\rm AAB} \hspace{0.1cm} = \hspace{0.1cm} p_{\rm ABB} = p_{\rm BBA} = p_{\rm BAA} = 1/8$$

and from this to the entropy approximation

$$H_3 = \frac{1}{3} \cdot \left [ 2 \cdot \frac{1}{4} \cdot {\rm log}_2\hspace{0.1cm}(4) + 4 \cdot \frac{1}{8} \cdot {\rm log}_2\hspace{0.1cm}(8)\right ] = \frac{2.5}{3} \hspace{0.15cm} \underline {= 0.833 \,{\rm bit/Symbol}} \hspace{0.05cm}.$$

To calculate  $H_4$ , the  $16$  probabilities are as follows:

$$p_{\rm AAAA} \hspace{0.1cm} = \hspace{0.1cm} p_{\rm BBBB} = 3/16 \hspace{0.05cm},\hspace{0.2cm} p_{\rm AABB} = p_{\rm BBAA} = 2/16 \hspace{0.05cm},$$
$$p_{\rm AAAB} \hspace{0.1cm} = \hspace{0.1cm} p_{\rm ABBA} = p_{\rm ABBB} = p_{\rm BBBA} = p_{\rm BAAB} = p_{\rm BAAA}= 1/16 \hspace{0.05cm}$$
$$p_{\rm AABA} \hspace{0.1cm} = \hspace{0.1cm} p_{\rm ABAA} = p_{\rm ABAB} = p_{\rm BBAB} = p_{\rm BABB} = p_{\rm BABA}= 0\hspace{0.05cm}.$$

It follows that:

$$H_4= \frac{1}{4} \hspace{-0.05cm}\cdot \hspace{-0.05cm}\left [ 2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \frac{3}{16} \hspace{-0.05cm}\cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.1cm}\frac{16}{3} + 2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \frac{1}{8} \hspace{-0.05cm}\cdot \hspace{-0.05cm}{\rm log}_2\hspace{0.1cm}(8) + 6 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \frac{1}{16} \hspace{-0.05cm}\cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.1cm}(16)\right ] =\frac{\left [ 6 \hspace{-0.05cm}\cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.01cm}(16) - 6 \hspace{-0.05cm}\cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.01cm}(3) + 4 \hspace{-0.05cm}\cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.01cm}(8) + 6\hspace{-0.05cm}\cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.01cm}(16)\right ]}{32} .$$

One can see:

• Even the approximation  $H_4 = 0.789\,{\rm bit/Symbol}$  still deviates significantly from the final entropy value  $H = 0.5\,{\rm bit/symbol}$ .
• The repetition code obviously cannot be modelled by a Markov source.  If  $\rm Q4$  were a Markov source, then the following would have to apply:
$$H = 2 \cdot H_2 - H_1 \hspace{0.3cm}\Rightarrow\hspace{0.3cm}H_2 = 1/2 \cdot (H+H_1) = 1/2 \cdot (0.5+1) = 0.75 \,{\rm bit/Symbol}\hspace{0.05cm}.$$