Aufgaben:Exercise 4.5: Mutual Information from 2D-PDF

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Given joint PDF

Given here are the three different 2D regions  $f_{XY}(x, y)$, which in the task are identified by their fill colors with

• red joint PDF,
• blue joint PDF,  und
• green joint PDF

respectively.  Within each of the regions shown, let  $f_{XY}(x, y) = C = \rm const.$

For example, the transinformation between the continuous-value random variables  $X$  and  $Y$  can be calculated as follows:

$$I(X;Y) = h(X) + h(Y) - h(XY)\hspace{0.05cm}.$$

For the differential entropies used here, the following equations apply:

$$h(X) = -\hspace{-0.7cm} \int\limits_{x \hspace{0.05cm}\in \hspace{0.05cm}{\rm supp}(f_X)} \hspace{-0.55cm} f_X(x) \cdot {\rm log} \hspace{0.1cm} \big[f_X(x)\big] \hspace{0.1cm}{\rm d}x

\hspace{0.05cm},$$

$$h(Y) = -\hspace{-0.7cm} \int\limits_{y \hspace{0.05cm}\in \hspace{0.05cm}{\rm supp}(f_Y)} \hspace{-0.55cm} f_Y(y) \cdot {\rm log} \hspace{0.1cm} \big[f_Y(y)\big] \hspace{0.1cm}{\rm d}y

\hspace{0.05cm},$$

$$h(XY) = \hspace{0.1cm}-\hspace{0.2cm} \int \hspace{-0.9cm} \int\limits_{\hspace{-0.5cm}(x, y) \hspace{0.1cm}\in \hspace{0.1cm}{\rm supp} (f_{XY}\hspace{-0.08cm})}

\hspace{-0.6cm} f_{XY}(x, y) \cdot {\rm log} \hspace{0.1cm} \big[ f_{XY}(x, y) \big]
\hspace{0.15cm}{\rm d}x\hspace{0.15cm}{\rm d}y\hspace{0.05cm}.$$

• For the two marginal probability density functions, the following holds:

$$f_X(x) = \hspace{-0.5cm} \int\limits_{\hspace{-0.2cm}y \hspace{0.1cm}\in \hspace{0.1cm}{\rm supp} (f_{Y}\hspace{-0.08cm})} \hspace{-0.4cm} f_{XY}(x, y)

\hspace{0.15cm}{\rm d}y\hspace{0.05cm},$$

$$f_Y(y) = \hspace{-0.5cm} \int\limits_{\hspace{-0.2cm}x \hspace{0.1cm}\in \hspace{0.1cm}{\rm supp} (f_{X}\hspace{-0.08cm})} \hspace{-0.4cm} f_{XY}(x, y)

\hspace{0.15cm}{\rm d}x\hspace{0.05cm}.$$

Hints:

• The exercise belongs to the chapter  AWGN channel capacitance with continuous-value input.

• Let the following differential entropies also be given:

• If  $X$  is triangularly distributed between  $x_{\rm min}$  and  $x_{\rm max}$, then:

$$h(X) = {\rm log} \hspace{0.1cm} [\hspace{0.05cm}\sqrt{ e} \cdot (x_{\rm max} - x_{\rm min})/2\hspace{0.05cm}]\hspace{0.05cm}.$$

• If  $Y$  is equally distributed between  $y_{\rm min}$  and  $y_{\rm max}$, then holds:

$$h(Y) = {\rm log} \hspace{0.1cm} \big [\hspace{0.05cm}y_{\rm max} - y_{\rm min}\hspace{0.05cm}\big ]\hspace{0.05cm}.$$

• All results should be expressed in "bit".  This is achieved with   $\log$  ⇒  $\log_2$.

Questions

1 What is the mutual information of  the red joint PDF?

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

2 What is the mutual information of  the blue joint PDF?

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

3 What is the mutual information of  the green joint PDF?

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

4 What conditions must the random variables  $X$  and  $Y$  satisfy simultaneously for  $I(X;Y) = 1/2 \cdot \log (\rm e)$  to hold in gener

	The joint WDF  $f_{XY}(x, y)$  results in a parallelogram.
	One of the random variables  $(X$  or  $Y)$  is uniformly distributed.
	The other random variable  $(Y$  or  $X)$  is triangularly distributed.

Solution

Solution

„Red” PDFs

(1)  For the rectangular joint PDF  $f_{XY}(x, y)$  there are no statistical ties between  $X$  and  $Y$    ⇒   $\underline{I(X;Y) = 0}$.

Formally, this result can be proved with the following equation:

$$I(X;Y) = h(X) \hspace{-0.05cm}+\hspace{-0.05cm} h(Y) \hspace{-0.05cm}- \hspace{-0.05cm}h(XY)\hspace{0.02cm}.$$

• The red area 2D-WDF  $f_{XY}(x, y)$  is  $F = 4$.  Since  $f_{XY}(x, y)$  is constant in this area and the volume under  $f_{XY}(x, y)$  must be equal to  $1$ , the height is   $C = 1/F = 1/4$.
• From this follows for the differential joint entropy in "bit":

$$h(XY) \ = \ \hspace{0.1cm}-\hspace{0.2cm} \int \hspace{-0.9cm} \int\limits_{\hspace{-0.5cm}(x, y) \hspace{0.1cm}\in \hspace{0.1cm}{\rm supp} \hspace{0.03cm}(\hspace{-0.03cm}f_{XY}\hspace{-0.08cm})}

\hspace{-0.6cm} f_{XY}(x, y) \cdot {\rm log}_2 \hspace{0.1cm} [ f_{XY}(x, y) ]
\hspace{0.15cm}{\rm d}x\hspace{0.15cm}{\rm d}y$$

$$\Rightarrow \hspace{0.3cm} h(XY) \ = \ \ {\rm log}_2 \hspace{0.1cm} (4) \cdot \hspace{0.02cm} \int \hspace{-0.9cm} \int\limits_{\hspace{-0.5cm}(x, y) \hspace{0.1cm}\in \hspace{0.1cm}{\rm supp} \hspace{0.03cm}(\hspace{-0.03cm}f_{XY}\hspace{-0.08cm})}

\hspace{-0.6cm} f_{XY}(x, y) 
\hspace{0.15cm}{\rm d}x\hspace{0.15cm}{\rm d}y = 2 \,{\rm bit}\hspace{0.05cm}.$$

• It is considered that the double integral is equal to  $1$ .  The pseudo-unit "bit" corresponds to the binary logarithm  ⇒  "log₂".

Furthermore:

• The marginal probability density functions  $f_{X}(x)$  and  $f_{Y}(y)$  are rectangular   ⇒   uniform distribution between  $0$  and  $2$:

$$h(X) = h(Y) = {\rm log}_2 \hspace{0.1cm} (2) = 1 \,{\rm bit}\hspace{0.05cm}.$$

• Substituting these results into the above equation, we obtain:

$$I(X;Y) = h(X) + h(Y) - h(XY) = 1 \,{\rm bit} + 1 \,{\rm bit} - 2 \,{\rm bit} = 0 \,{\rm (bit)}

\hspace{0.05cm}.$$

„Blue” probability density functions

(2)  Also for this parallelogram we get  $F = 4, \ C = 1/4$  as well as  $h(XY) = 2$ bit.

• Here, as in subtask  (1) , the random variable  $Y$  is uniformly distributed between  $0$  and  $2$   ⇒   $h(Y) = 1$ bit.

• In contrast,  $X$  is triangularly distributed between  $0$  and  $4$  $($with maximum at $2)$. 
• This results in the same differential entropy  $h(Y)$  as for a symmetric triangular distribution in the range between  $±2$  (see specification sheet):

$$h(X) = {\rm log}_2 \hspace{0.1cm} \big[\hspace{0.05cm}2 \cdot \sqrt{ e} \hspace{0.05cm}\big ]

= 1.721 \,{\rm bit}$$

$$\Rightarrow \hspace{0.3cm} I(X;Y) = 1.721 \,{\rm bit} + 1 \,{\rm bit} - 2 \,{\rm bit}\hspace{0.05cm}\underline{ = 0.721 \,{\rm (bit)}}

\hspace{0.05cm}.$$

„Green” probability density functions

(3)  The following properties are obtained for the green conditions:

$$F = A \cdot B \hspace{0.3cm} \Rightarrow \hspace{0.3cm} C = \frac{1}{A \cdot B}

\hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} h(XY) = {\rm log}_2 \hspace{0.1cm} (A \cdot B) \hspace{0.05cm}.$$

• The random variable  $Y$  is now uniformly distributed between  $0$  and  $A$  and the random variable  $X$  is triangularly distributed between  $0$  and  $B$ :

$$h(X) \ = \ {\rm log}_2 \hspace{0.1cm} (B \cdot \sqrt{ e})

\hspace{0.05cm},$$ $$

h(Y)  \  =  \   {\rm log}_2 \hspace{0.1cm} (A)\hspace{0.05cm}.$$

• Thus, for the mutual information between  $X$  and  $Y$:

$$I(X;Y) \ = {\rm log}_2 \hspace{0.1cm} (B \cdot \sqrt{ {\rm e}}) + {\rm log}_2 \hspace{0.1cm} (A) - {\rm log}_2 \hspace{0.1cm} (A \cdot B)$$

$$\Rightarrow \hspace{0.3cm} I(X;Y) = \ {\rm log}_2 \hspace{0.1cm} \frac{B \cdot \sqrt{ {\rm e}} \cdot A}{A \cdot B} = {\rm log}_2 \hspace{0.1cm} (\sqrt{ {\rm e}})\hspace{0.15cm}\underline{= 0.721\,{\rm bit}}

\hspace{0.05cm}.$$

Other examples of 2D PDF  $f_{XY}(x, y)$

• $I(X;Y)$  thus independent of WDF parameters  $A$  and  $B$.

(4)  All the above conditions are required. 

• However, the requirements  (2)  and  (3)  are not to be fulfilled for every parallelogram.  The adjacent graphic shows two such constellations, where the random variable  $X$  is in each case equally distributed between  $0$  and  $1$ .
• In the upper graph, the plotted points lie at a height   ⇒   $f_{Y}(y)$  is triangularly distributed   ⇒   $I(X;Y) = 0.721$ bit.
• The lower composite WDF has a different transinformation because the two points are not at the same height  
  ⇒   die WDF  $f_{Y}(y)$  has a trapezoidal shape here.
• Feeling-wise, I guess  $I(X;Y) < 0.721$  bit, since the 2D area more closely approximates a rectangle.  If you still feel like it, please check.