Aufgaben:Exercise 3.7: Some Entropy Calculations

• Return to book

Diagram: Entropies and information

We consider the two random variables  $XY$  and  $UV$  with the following two-dimensional probability mass functions:

$$P_{XY}(X, Y) = \begin{pmatrix} 0.18 & 0.16\\ 0.02 & 0.64 \end{pmatrix}\hspace{0.05cm} \hspace{0.05cm}$$

$$P_{UV}(U, V) \hspace{0.05cm}= \begin{pmatrix} 0.068 & 0.132\\ 0.272 & 0.528 \end{pmatrix}\hspace{0.05cm}$$

For the random variable  $XY$  the following are to be calculated in this exercise:

• the joint entropy:

$$H(XY) = -{\rm E}\big [\log_2 P_{ XY }( X,Y) \big ],$$

• the two individual entropies:

$$H(X) = -{\rm E}\big [\log_2 P_X( X)\big ],$$

$$H(Y) = -{\rm E}\big [\log_2 P_Y( Y)\big ].$$

From this, the following descriptive variables can also be determined very easily according to the above scheme – shown for the random variable  $XY$:

• the conditional entropies:

$$H(X \hspace{0.05cm}|\hspace{0.05cm} Y) = -{\rm E}\big [\log_2 P_{ X \hspace{0.05cm}|\hspace{0.05cm}Y }( X \hspace{0.05cm}|\hspace{0.05cm} Y)\big ],$$

$$H(Y \hspace{0.05cm}|\hspace{0.05cm} X) = -{\rm E}\big [\log_2 P_{ Y \hspace{0.05cm}|\hspace{0.05cm} X }( Y \hspace{0.05cm}|\hspace{0.05cm} X)\big ],$$

• the mutual information between $X$ and $Y$:

$$I(X;Y) = {\rm E} \hspace{-0.08cm}\left [ \hspace{0.02cm}{\rm log}_2 \hspace{0.1cm} \frac{P_{XY}(X, Y)}{P_{X}(X) \cdot P_{Y}(Y) }\right ] \hspace{0.05cm}.$$

Finally, verify qualitative statements regarding the second random variable  $UV$ .

Hints:

• The exercise belongs to the chapter  Different entropy measures of two-dimensional random variables.
• In particular, reference is made to the pages 
      Conditional probability and conditional entropy   as well as
      Mutual information between two random variables.

Questions

1 Calculate the joint entropy.

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

2 What are the entropies of the one-dimensional random variables  $X$  and  $Y$ ?

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

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

3 How large is the mutual information between the random variables  $X$  and  $Y$?

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

4 Calculate the two conditional entropies.

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

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

5 Which of the following statements are true for the two-dimensional random variable $UV$?

	The one-dimensional random variables  $U$  and  $V$  are statistically independent.
	The mutual information of  $U$  and  $V$  is  $I(U; V) = 0$.
	For the compound entropy  $H(UV) = H(XY)$ holds.
	The relations  $H(U|V) = H(U)$  and  $H(V|U) = H(V)$.

Solution

Solution

(1)  From the given composite probability we obtain

$$H(XY) = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.18} + 0.16\cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.16}+0.02\cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.02}+0.64\cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.64}\hspace{0.15cm} \underline {= 1.393\,{\rm (bit)}}\hspace{0.05cm}.$$

(2)  The one-dimensional probability functions are  $P_X(X) = \big [0.2, \ 0.8 \big ]$  and  $P_Y(Y) = \big [0.34, \ 0.66 \big ]$.  From this follows:

$$H(X) = 0.2 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.2} + 0.8\cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.8}\hspace{0.15cm} \underline {= 0.722\,{\rm (bit)}} \hspace{0.05cm},$$

$$H(Y) =0.34 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.34} + 0.66\cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.66}\hspace{0.15cm} \underline {= 0.925\,{\rm (bit)}} \hspace{0.05cm}.$$

(3)  From the graph on the information page you can see the relationship:

$$I(X;Y) = H(X) + H(Y) - H(XY) = 0.722\,{\rm (bit)} + 0.925\,{\rm (bit)}- 1.393\,{\rm (bit)}\hspace{0.15cm} \underline {= 0.254\,{\rm (bit)}} \hspace{0.05cm}.$$

(4)  Similarly, according to the graph on the information page:

$$H(X \hspace{-0.1cm}\mid \hspace{-0.08cm} Y) = H(XY) - H(Y) = 1.393- 0.925\hspace{0.15cm} \underline {= 0.468\,{\rm (bit)}} \hspace{0.05cm},$$

$$H(Y \hspace{-0.1cm}\mid \hspace{-0.08cm} X) = H(XY) - H(X) = 1.393- 0.722\hspace{0.15cm} \underline {= 0.671\,{\rm (bit)}} \hspace{0.05cm}$$

Entropy values for the random variables $XY$ and $UV$

• The left diagram summarises the results of subtasks  (1), ... ,  (4)  true to scale.
• The joint entropy is highlighted in grey and the mutual information in yellow.
• A red background refers to the random variable  $X$,  and a green one to  $Y$.  Hatched fields indicate a conditional entropy.

The right graph describes the same situation for the random variable  $UV$   ⇒   subtask  (5).

(5)  According to the diagram on the right,
statements 1, 2 and 4 are correct:

• One recognises the validity of  $P_{ UV } = P_U · P_V$   ⇒   mutual information $I(U; V) = 0$  by the fact that the second row of the  $P_{ UV }$ matrix differs from the first row only by a constant factor  $(4)$ .
• This results in the same one-dimensional probability mass functions as for the random variable  $XY$   ⇒   $P_U(U) = \big [0.2, \ 0.8 \big ]$  and  $P_V(V) = \big [0.34, \ 0.66 \big ]$.
• Therefore  $H(U) = H(X) = 0.722\ \rm bit$  and  $H(V) = H(Y) = 0.925 \ \rm bit$.
• Here, however, the following now applies for the joint entropy:   $H(UV) = H(U) + H(V) ≠ H(XY)$.