Exercise 4.5Z: Again Mutual Information

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Given joint PDF and graph of differential entropies.

The graph above shows the joint PDF  $f_{XY}(x, y)$, to be considered in this task, which is identical to the "green" constellation in  Exercise 4.5.

• $f_{XY}(x, y)$  is enlarged by a factor of  $3$  in the   $y$–direction.
• In the definition area highlighted in green, the joint PDF is constant equal to  $C = 1/F$, where  $F$  indicates the area of the parallelogram.

In Exercise 4.5 the following differential entropies were calculated:

$$h(X) \ = \ {\rm log} \hspace{0.1cm} (\hspace{0.05cm}A\hspace{0.05cm})\hspace{0.05cm},$$

$$h(Y) = {\rm log} \hspace{0.1cm} (\hspace{0.05cm}B \cdot \sqrt{ {\rm e } } \hspace{0.05cm})\hspace{0.05cm},$$

$$h(XY) = {\rm log} \hspace{0.1cm} (\hspace{0.05cm}F \hspace{0.05cm}) = {\rm log} \hspace{0.1cm} (\hspace{0.05cm}A \cdot B \hspace{0.05cm})\hspace{0.05cm}.$$

In this exercise, the parameter values  $A = {\rm e}^{-2}$  and  $B = {\rm e}^{0.5}$  are now to be used.

According to the above diagram, the conditional differential entropies  $h(Y|X)$  and  $h(X|Y)$  should now also be determined and their relation to the mutual information  $I(X; Y)$  given.

Hints:

• The exercise belongs to the chapter  AWGN channel capacity with continuous value input.
• If the results are to be given in "nat", this is achieved with "log"  ⇒  "ln".
• If the results are to be given in "bit", this is achieved with "log"  ⇒  "log₂".

Questions

1

State the following information theoretic quantities in "nat":

$h(X) \ = \ $ $\ \rm nat$

$h(Y) \ \hspace{0.03cm} = \ $ $\ \rm nat$

$h(XY)\ \hspace{0.17cm} = \ $ $\ \rm nat$

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

2

What are the same quantities with the pseudo–unit "bit"?

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

$h(Y) \ \hspace{0.03cm} = \ $ $\ \rm bit$

$h(XY)\ \hspace{0.17cm} = \ $ $\ \rm bit$

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

3

Calculate the conditional differential entropy  $h(Y|X)$.

$h(Y|X) \ = \ $ $\ \rm nat$

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

4

Calculate the conditional differential entropy  $h(X|Y)$.

$h(X|Y) \ = \ $ $\ \rm nat$

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

5

Which of the following quantities is never negative?

	Both  $H(X)$  and  $H(Y)$  in the discrete value case.
	The mutual information  $I(X; Y)$  in the discrete value case.
	The mutual information  $I(X; Y)$  in the continuous value case.
	Both  $h(X)$  and  $h(Y)$  in the continuous value case.
	Both  $h(X|Y)$  and  $h(Y|X)$  in the continuous value case.
	The joint entropy  $h(XY)$  in the continuous value case.

Solution

Solution

(1)  Since the results are required in "nat", it is convenient to use the natural logarithm:

• The random variable  $X$  is uniformly distributed between  $0$  and  $1/{\rm e}^2={\rm e}^{-2}$:

$$h(X) = {\rm ln} \hspace{0.1cm} (\hspace{0.05cm}{\rm e}^{-2}\hspace{0.05cm}) \hspace{0.15cm}\underline{= -2\,{\rm nat}}\hspace{0.05cm}. $$

• The random variable  $Y$  is triangularly distributed between  $±{\rm e}^{-0.5}$:

$$h(Y) = {\rm ln} \hspace{0.1cm} (\hspace{0.05cm}\sqrt{ {\rm e} } \cdot \sqrt{ {\rm e} } ) = {\rm ln} \hspace{0.1cm} (\hspace{0.05cm}{ { \rm e } } \hspace{0.05cm}) \hspace{0.15cm}\underline{= +1\,{\rm nat}}\hspace{0.05cm}.$$

• The area of the parallelogram is given by

$$F = A \cdot B = {\rm e}^{-2} \cdot {\rm e}^{0.5} = {\rm e}^{-1.5}\hspace{0.05cm}.$$

• Thus, the 2D WDF in the area highlighted in green has constant height  $C = 1/F ={\rm e}^{1.5}$  and we obtain for the joint entropy:

$$h(XY) = {\rm ln} \hspace{0.1cm} (F) = {\rm ln} \hspace{0.1cm} (\hspace{0.05cm}{\rm e}^{-1.5}\hspace{0.05cm}) \hspace{0.15cm}\underline{= -1.5\,{\rm nat}}\hspace{0.05cm}.$$

• From this we obtain for the mutual information:

$$I(X;Y) = h(X) + h(Y) - h(XY) = -2 \,{\rm nat} + 1 \,{\rm nat} - (-1.5 \,{\rm nat} ) \hspace{0.15cm}\underline{= 0.5\,{\rm nat}}\hspace{0.05cm}.$$

(2)  In general, the relation  $\log_2(x) = \ln(x)/\ln(2)$ holds.  Thus, using the results of subtask  (1), we obtain:

$$h(X) \ = \ \frac{-2\,{\rm nat}}{0.693\,{\rm nat/bit}}\hspace{0.35cm}\underline{= -2.886\,{\rm bit}}\hspace{0.05cm},$$

$$h(Y) \ = \ \frac{+1\,{\rm nat}}{0.693\,{\rm nat/bit}}\hspace{0.35cm}\underline{= +1.443\,{\rm bit}}\hspace{0.05cm},$$

$$h(XY) \ = \ \frac{-1.5\,{\rm nat}}{0.693\,{\rm nat/bit}}\hspace{0.35cm}\underline{= -2.164\,{\rm bit}}\hspace{0.05cm},$$

$$I(X;Y) \ = \ \frac{0.5\,{\rm nat}}{0.693\,{\rm nat/bit}}\hspace{0.35cm}\underline{= 0.721\,{\rm bit}}\hspace{0.05cm}.$$

• Or also:

$$I(X;Y) = -2.886 \,{\rm bit} + 1.443 \,{\rm bit}+ 2.164 \,{\rm bit}{= 0.721\,{\rm bit}}\hspace{0.05cm}.$$

(3)  The mutual information can also be written in the form  $I(X;Y) = h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) - h(Y)$ :

$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(Y) - I(X;Y) = 1 \,{\rm nat} - 0.5 \,{\rm nat} \hspace{0.15cm}\underline{= 0.5\,{\rm nat}= 0.721\,{\rm bit}}\hspace{0.05cm}.$$

(4)  For the differential inference entropy, it holds correspondingly:

$$h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) = h(X) - I(X;Y) = -2 \,{\rm nat} - 0.5 \,{\rm nat} \hspace{0.15cm}\underline{= -2.5\,{\rm nat}= -3.607\,{\rm bit}}\hspace{0.05cm}.$$

Summary of all results of this exercise

• All quantities calculated here are summarized in the graph. 
• Arrows pointing up indicate a positive contribution, arrows pointing down indicate a negative contribution.

(5)  Correct are the proposed solutions 1 to 3.

Again for clarification:

• For the mutual information  $I(X;Y) \ge 0$ always holds.
• In the discrete value case there is no negative entropy, but in the continuous value case there is.