# Exercise 3.11Z: Extremely Asymmetrical Channel

The channel shown opposite with the following properties is considered:

• The symbol  $X = 0$  is always transmitted correctly and leads always to the result  $Y = 0$.
• The symbol  $X = 1$  is distorted to the maximum.

From the point of view of information theory, this means:

$${\rm Pr}(Y \hspace{-0.05cm} = 0\hspace{-0.05cm}\mid \hspace{-0.05cm} X \hspace{-0.05cm}= 1) ={\rm Pr}(Y \hspace{-0.05cm} = 1\hspace{-0.05cm}\mid \hspace{-0.05cm} X \hspace{-0.05cm}= 1) = 0.5 \hspace{0.05cm}.$$

To be determined in this task are:

• the mutual information  $I(X; Y)$  for  $P_X(0) = p_0 = 0.4$  and  $P_X(1) = p_1 = 0.6$.
The general rule is:
$$I(X;Y) = H(X) - H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y)\hspace{0.05cm}=H(Y) - H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X)\hspace{0.05cm} =\hspace{-0.15cm} H(X) + H(Y)- H(XY)\hspace{0.05cm},$$
• the channel capacity:
$$C = \max_{P_X(X)} \hspace{0.15cm} I(X;Y) \hspace{0.05cm}.$$

Hints:

### Questions

1

Calculate the source entropy in general and for  $\underline{p_0 = 0.4}$.

 $H(X) \ = \$ $\ \rm bit$

2

Calculate the sink entropy in general and for  $p_0 = 0.4$.

 $H(Y) \ = \$ $\ \rm bit$

3

Calculate the joint entropy in general and for  $p_0 = 0.4$.

 $H(XY) \ = \$ $\ \rm bit$

4

Calculate the mutual information in general and for  $p_0 = 0.4$.

 $I(X; Y) \ = \$ $\ \rm bit$

5

What probability  $p_0^{(*)}$  leads to channel capacity  $C$?

 $p_0^{(*)} \ = \$

6

What is the channel capacity of the present channel?

 $C \ = \$ $\ \rm bit$

7

What are the conditional entropies with  $p_0 = p_0^{(*)}$  according to subtask  (5)?

 $H(X|Y) \ = \$ $\ \rm bit$ $H(Y|X) \ = \$ $\ \rm bit$

### Solution

#### Solution

(1)  The source entropy results according to the binary entropy function:

$$H(X) = H_{\rm bin}(p_0)= H_{\rm bin}(0.4) \hspace{0.15cm} \underline {=0.971\,{\rm bit}} \hspace{0.05cm}.$$

(2)  The probabilities of the sink symbols are:

$$P_Y(1) = p_1/2 = (1 - p_0)/2 = 0.3\hspace{0.05cm},\hspace{0.2cm} P_Y(0) = 1-P_Y(1) = p_1/2 = (1 - p_0)/2 = 0.7$$
$$\Rightarrow \hspace{0.3cm} H(Y) = H_{\rm bin}(\frac{1+p_0}{2})= H_{\rm bin}(0.7) \hspace{0.15cm} \underline {=0.881\,{\rm bit}} \hspace{0.05cm}.$$

(3)  The joint probabilities  $p_{μκ} = {\rm Pr}\big[(X = μ) ∩ (Y = κ)\big]$ are obtained as:

$$p_{00} = p_0 \hspace{0.05cm},\hspace{0.3cm} p_{01} = 0 \hspace{0.05cm},\hspace{0.3cm} p_{10} = (1 - p_0)/2 \hspace{0.05cm},\hspace{0.3cm} p_{11} = (1 - p_0)/2$$
$$\Rightarrow \hspace{0.3cm} H(XY) =p_0 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{ p_0} + 2 \cdot \frac{1-p_0}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{ 1- p_0} = p_0 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{ p_0} + (1-p_0) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{ 1- p_0} + (1-p_0) \cdot {\rm log}_2 \hspace{0.1cm} (2)$$
$$\Rightarrow \hspace{0.3cm}H(XY) =H_{\rm bin}(p_0) + 1 - p_0 \hspace{0.05cm}.$$
• The numerical result for  $p_0 = 0.4$  is thus:
$$H(XY) = H_{\rm bin}(0.4) + 0.6 = 0.971 + 0.6 \hspace{0.15cm} \underline {=1.571\,{\rm bit}} \hspace{0.05cm}.$$

(4)  A (possible) equation to calculate the mutual information is:

$$I(X;Y) = H(X) + H(Y)- H(XY)\hspace{0.05cm}.$$
• From this, using the results of the first three subtasks, one obtains:
$$I(X;Y) = H_{\rm bin}(p_0) + H_{\rm bin}(\frac{1+p_0}{2}) - H_{\rm bin}(p_0) -1 + p_0 = H_{\rm bin}(\frac{1+p_0}{2}) -1 + p_0.$$
$$\Rightarrow \hspace{0.3cm} p_0 = 0.4 {\rm :}\hspace{0.5cm} I(X;Y) = H_{\rm bin}(0.7) - 0.6 = 0.881 - 0.6 \hspace{0.15cm} \underline {=0.281\,{\rm bit}}\hspace{0.05cm}.$$

(5)  The channel capacity  $C$  is the mutual information  $I(X; Y)$  at best possible probabilities  $p_0$  and   $p_1$  of the source symbols.

• After differentiation, the determination equation is obtained:
$$\frac{\rm d}{{\rm d}p_0} \hspace{0.1cm} I(X;Y) = \frac{\rm d}{{\rm d}p_0} \hspace{0.1cm} H_{\rm bin}(\frac{1+p_0}{2}) +1 \stackrel{!}{=} 0 \hspace{0.05cm}.$$
• With the differential quotient of the binary entropy function
$$\frac{\rm d}{{\rm d}p} \hspace{0.1cm} H_{\rm bin}(p) = {\rm log}_2 \hspace{0.1cm} \frac{1-p}{ p} \hspace{0.05cm},$$
and corresponding post-differentialisation one obtains:
$${1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{(1-p_0)/2}{1- (1-p_0)/2} +1 \stackrel{!}{=} 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{(1-p_0)/2}{(1+p_0)/2} +1 \stackrel{!}{=} 0$$
$$\Rightarrow \hspace{0.3cm} {\rm log}_2 \hspace{0.1cm} \frac{1+p_0}{1-p_0} \stackrel{!}{=} 2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{1+p_0}{1-p_0} \stackrel{!}{=} 4 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_0 \hspace{0.15cm} \underline {=0.6}=p_0^{(*)}\hspace{0.05cm}.$$

(6)  Accordingly, for the channel capacity:

$$C = I(X;Y) \big |_{p_0 \hspace{0.05cm}=\hspace{0.05cm} 0.6} = H_{\rm bin}(0.8) - 0.4 = 0.722 -0.4 \hspace{0.15cm} \underline {=0.322\,{\rm bit}}\hspace{0.05cm}.$$
• Exercise 3.14 interprets this result in comparison to the BSC channel model.

(7)  For the equivocation holds:

$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm}Y) = H(X) - I(X;Y) = 0.971 -0.322 \hspace{0.15cm} \underline {=0.649\,{\rm bit}}\hspace{0.05cm}.$$
• Because of   $H_{\rm bin}(0.4) = H_{\rm bin}(0.6)$  the source entropy  $H(X)$  is the same as in subtask  (1).
• The sink entropy must be recalculated.  With  $p_0 = 0.6$  we get  $H(Y) = H_{\rm bin}(0.8) = 0.722\ \rm bit$.
• This gives the irrelevance:
$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = H(Y) - I(X;Y) = 0.722 -0.322 \hspace{0.15cm} \underline {=0.400\,{\rm bit}}\hspace{0.05cm}.$$