Aufgaben:Exercise 4.2: Triangular PDF

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Two triangular PDFs

Two probability density functions  $\rm (PDF)$  with triangular shapes are considered.

• The random variable  $X$  is limited to the range from  $0$  to  $1$ ,  and it holds for the PDF (upper sketch):

$$f_X(x) = \left\{ \begin{array}{c} 2x \\ 0 \\ \end{array} \right. \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \hspace{0.1cm} 0 \le x \le 1 \\ {\rm else} \\ \end{array} \hspace{0.05cm}.$$

• According to the lower sketch, the random variable  $Y$  has the following PDF:

$$f_Y(y) = \left\{ \begin{array}{c} 1 - |\hspace{0.03cm}y\hspace{0.03cm}| \\ 0 \\ \end{array} \right. \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \hspace{0.1cm} |\hspace{0.03cm}y\hspace{0.03cm}| \le 1 \\ {\rm else} \\ \end{array} \hspace{0.05cm}.$$

For both random variables, the  differential entropy  is to be determined in each case.

For example, the corresponding equation for the random variable  $X$  is:

$$h(X) = \hspace{0.1cm} - \hspace{-0.45cm} \int\limits_{{\rm supp}\hspace{0.03cm}(\hspace{-0.03cm}f_X)} \hspace{-0.35cm} f_X(x) \cdot {\rm log} \hspace{0.1cm} \big [ f_X(x) \big ] \hspace{0.1cm}{\rm d}x \hspace{0.6cm}{\rm mit}\hspace{0.6cm} {\rm supp}(f_X) = \{ x\text{:} \ f_X(x) > 0 \} \hspace{0.05cm}.$$

• If the  "natural logarithm",  the pseudo-unit  "nat"  must be added.
• If, on the other hand, the result is asked in  "bit"  then the  "dual logarithm"   ⇒   "$\log_2$"  is to be used.

In the fourth subtask, the new random variable  $Z = A \cdot Y$  is considered. Here,  the PDF parameter  $A$  is to be determined in such a way that the differential entropy of the new random variable  $Z$  yields exactly  $1$  bit :

$$h(Z) = h (A \cdot Y) = h (Y) + {\rm log}_2 \hspace{0.1cm} (A) = 1\ {\rm bit} \hspace{0.05cm}.$$

Hints:

• The task belongs to the chapter  Differential Entropy.
• Useful hints for solving this task and further information on value-continuous random variables can be found in the third chapter "Continuous Random Variables" of the book  Theory of Stochastic Signals.

• Given the following indefinite integral:

$$\int \xi \cdot {\rm ln} \hspace{0.1cm} (\xi)\hspace{0.1cm}{\rm d}\xi = \xi^2 \cdot \big [1/2 \cdot {{\rm ln} \hspace{0.1cm} (\xi)} - {1}/{4}\big ] \hspace{0.05cm}.$$

Questions

1

Calculate the differential entropy of the random variable  $X$  in  "nat".

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

2

What result is obtained with the pseudo-unit  "bit"?

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

3

Calculate the differential entropy of the random variable  $Y$.

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

4

Determine the PDF parameter  $A$  such that  $\underline{h(Z) = h (A \cdot Y) = 1 \ \rm bit}$ .

$A\ = $

Solution

Solution

(1)  For the probability density function, in the range  $0 \le X \le 1$ , it is agreed that:

$$f_X(x) = 2x = C \cdot x \hspace{0.05cm}.$$

• Here we have replaced "2" by  $C$    ⇒   generalization in order to be able to use the following calculation again in subtask  $(3)$ .

• Since the differential entropy is sought in "nat", we use the natural logarithm.  With the substitution  $\xi = C \cdot x$  we obtain:

$$h_{\rm nat}(X) = \hspace{0.1cm} - \int_{0}^{1} \hspace{0.1cm} C \cdot x \cdot {\rm ln} \hspace{0.1cm} \big[ C \cdot x \big] \hspace{0.1cm}{\rm d}x = \hspace{0.1cm} - \hspace{0.1cm}\frac{1}{C} \cdot \int_{0}^{C} \hspace{0.1cm} \xi \cdot {\rm ln} \hspace{0.1cm} [ \xi ] \hspace{0.1cm}{\rm d}\xi = - \hspace{0.1cm}\frac{\xi^2}{C} \cdot \left [ \frac{{\rm ln} \hspace{0.1cm} (\xi)}{2} - \frac{1}{4}\right ]_{\xi = 0}^{\xi = C} \hspace{0.05cm}$$

• Here the indefinite integral given in the front was used.  After inserting the limits, considering  $C=2$, we obtain::

$$h_{\rm nat}(X) = - C/2 \cdot \big [ {\rm ln} \hspace{0.1cm} (C) - 1/2 \big ] = - {\rm ln} \hspace{0.1cm} (2) + 1/2 = - {\rm ln} \hspace{0.1cm} (2) + 1/2 \cdot {\rm ln} \hspace{0.1cm} ({\rm e}) = {\rm ln} \hspace{0.1cm} (\sqrt{\rm e}/2)\hspace{0.05cm} = - 0.193 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} h(X) \hspace{0.15cm}\underline {= - 0.193\,{\rm nat}} \hspace{0.05cm}.$$

(2)  In general:

$$h_{\rm bit}(X) = \frac{h_{\rm nat}(X)}{{\rm ln} \hspace{0.1cm} (2)\,{\rm nat/bit}} = - 0.279 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} h(X) \hspace{0.15cm}\underline {= - 0.279\,{\rm bit}} \hspace{0.05cm}.$$

• You can save this conversion if you directly replace  $(1)$  direkt "ln" by "log₂" already in the analytical result of subtask:

$$h(X) = \ {\rm log}_2 \hspace{0.1cm} (\sqrt{\rm e}/2)\hspace{0.05cm}, \hspace{1.3cm} {\rm pseudo-unit\hspace{-0.1cm}:\hspace{0.15cm} bit} \hspace{0.05cm}.$$

To calculate  $h(Y)$

(3)  We again use the natural logarithm and divide the integral into two partial integrals:

$$h(Y) = \hspace{0.1cm} - \hspace{-0.45cm} \int\limits_{{\rm supp} \hspace{0.03cm}( \hspace{-0.03cm}f_Y)} \hspace{-0.35cm} f_Y(y) \cdot {\rm ln} \hspace{0.1cm} \big[ f_Y(y) \big] \hspace{0.1cm}{\rm d}y = I_{\rm neg} + I_{\rm pos} \hspace{0.05cm}.$$

• The first integral for the range  $-1 \le y \le 0$  is identical in form to that of subtask  $(1)$  and only shifted with respect to it, which does not affect the result.
• Now the heighte  $C = 1$  instead of  $C = 2$ has to be considered::

$$I_{\rm neg} =- C/2 \cdot \big [ {\rm ln} \hspace{0.1cm} (C) - 1/2 \big ] = -1/2 \cdot \big [ {\rm ln} \hspace{0.1cm} (1) - 1/2 \cdot {\rm ln} \hspace{0.1cm} ({\rm e}) \big ]= 1/4 \cdot {\rm ln} \hspace{0.1cm} ({\rm e}) \hspace{0.05cm}.$$

• The second integrand is identical to the first except for a shift and reflection.  Moreover, the integration intervals do not overla   ⇒   $I_{\rm pos} = I_{\rm neg}$:

$$h_{\rm nat}(Y) = 2 \cdot I_{\rm neg} = 1/2 \cdot {\rm ln} \hspace{0.1cm} ({\rm e}) = {\rm ln} \hspace{0.1cm} (\sqrt{\rm e}) \hspace{0.3cm} \Rightarrow\hspace{0.3cm}h_{\rm bit}(Y) = {\rm log}_2 \hspace{0.1cm} (\sqrt{\rm e}) \hspace{0.3cm} \Rightarrow\hspace{0.3cm} h(Y) = {\rm log}_2 \hspace{0.1cm} (1.649)\hspace{0.15cm}\underline {= 0.721\,{\rm bit}}\hspace{0.05cm}.$$

(4)  For the differential entropy of the random variable  $Z = A \cdot Y$  holds in general:

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

• Thus, from the requiremen  $h(Z) = 1 \ \rm bit$  and the result of subta  $(3)$  follows:

$${\rm log}_2 \hspace{0.1cm} (A) = 1\,{\rm bit} - 0.721 \,{\rm bit} = 0.279 \,{\rm bit} \hspace{0.3cm} \Rightarrow\hspace{0.3cm} A = 2^{0.279}\hspace{0.15cm}\underline {= 1.213} \hspace{0.05cm}.$$