Aufgaben:Exercise 4.2Z: Mixed Random Variables

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PDF of  $X$  (top),  and
CDF of  $Y$  (bottom)

One speaks of a  "mixed random variable",  if the random variable contains discrete components in addition to a continuous component.

• For example, the random variable  $Y$  with  cumulative distribution function  $F_Y(y)$  as shown in the sketch below has both a continuous and a discrete component.
• The  probability density function  $f_Y(y)$  is obtained from  $F_Y(y)$  by differentiation.
• The jump at  $y= 1$  in the CDF thus becomes a "Dirac" in the probability density function.

• In subtask  (4)  the differential entropy  $h(Y)$  of  $Y$  is to be determined (in bit), assuming the following equation:

$$h(Y) = \hspace{0.1cm} - \hspace{-0.45cm} \int\limits_{{\rm supp}\hspace{0.03cm}(\hspace{-0.03cm}f_Y)} \hspace{-0.35cm} f_Y(y) \cdot {\rm log}_2 \hspace{0.1cm} \big[ f_Y(y) \big] \hspace{0.1cm}{\rm d}y \hspace{0.05cm}.$$

• In subtask  (2),  calculate the differential entropy  $h(X)$  of the random variable  $X$  whose PDF  $f_X(x)$  is sketched above.  If one performs a suitable boundary transition, the random variable  $X$  also becomes a mixed random variable.

Hints:

• The exercise belongs to the chapter  Differential Entropy.
• Further information on mixed random variables can be found in the chapter   Cumulative Distribution Function  of the book "Theory of Stochastic Signals".

Questions

1

What is the PDF height  $A$  of  $f_X(x)$  around  $x = 1$?

	$A = 0.5/\varepsilon$,
	$A = 0.5/\varepsilon+0.25$,
	$A = 1/\varepsilon$.

2

Calculate the differential entropy for different  $\varepsilon$–values.

$ε = 10^{-1}\text{:} \ \ h(X) \ = \ $ $\ \rm bit$

$ε = 10^{-2}\text{:} \ \ h(X) \ = \ $ $\ \rm bit$

$ε = 10^{-3}\text{:} \ \ h(X) \ = \ $ $\ \rm bit$

3

What is the result of the limit  $ε \to 0$?

	$f_X(x)$  now has a continuous and a discrete component.
	The differential energy  $h(X)$  is negative.
	The magnitude  $|h(X)|$  is infinite.

4

Which statements are true for the random variable  $Y$?

	The CDF value at the point  $y = 1$  is  $0.5$.
	$Y$  contains a discrete and a continuous component.
	The discrete component at   $Y = 1$  occurs with  $10\%$  probability.
	The continuous component of  $Y$  is uniformly distributed.
	The differential entropies of  $X$  and  $Y$  are equal.

Solution

Solution

(1)  Proposed solution 2 is correct because the integral over the PDF  $1$  must yield:

$$f_X(x) \hspace{0.1cm}{\rm d}x = 0.25 \cdot 2 + (A - 0.25) \cdot \varepsilon \stackrel{!}{=} 1 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}(A - 0.25) \cdot \varepsilon \stackrel{!}{=} 0.5 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} A = 0.5/\varepsilon +0.25\hspace{0.05cm}.$$

(2)  The differential entropy (in "bit") is given as follows:

$$h(X) = \hspace{0.1cm} \hspace{-0.45cm} \int\limits_{{\rm supp}(f_X)} \hspace{-0.35cm} f_X(x) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{f_X(x)} \hspace{0.1cm}{\rm d}x \hspace{0.05cm}.$$

We now divide the integral into three partial integrals:

$$h(X) = \hspace{-0.25cm} \int\limits_{0}^{1-\varepsilon/2} \hspace{-0.15cm} 0.25 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.25} \hspace{0.1cm}{\rm d}x + \hspace{-0.25cm}\int\limits_{1+\varepsilon/2}^{2} \hspace{-0.15cm} 0.25 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.25} \hspace{0.1cm}{\rm d}x + \hspace{-0.25cm}\int\limits_{1-\varepsilon/2}^{1+\varepsilon/2} \hspace{-0.15cm} \big [0.5/\varepsilon + 0.25 \big ] \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.5/\varepsilon + 0.25} \hspace{0.1cm}{\rm d}x $$

$$ \Rightarrow \hspace{0.3cm} h(X) = 2 \cdot 0.25 \cdot 2 \cdot (2-\varepsilon) - (0.5 + 0.25 \cdot \varepsilon) \cdot {\rm log}_2 \hspace{0.1cm}(0.5/\varepsilon +0.25) \hspace{0.05cm}.$$

In particular, one obtains

• for  $\varepsilon = 0.1$:

$$h(X) =1.9 - 0.525 \cdot {\rm log}_2 \hspace{0.1cm}(5.25) = 1.9 - 1.256 \hspace{0.15cm}\underline{= 0.644\,{\rm bit}} \hspace{0.05cm},$$

• for  $\varepsilon = 0.01$:

$$h(X) =1.99 - 0.5025 \cdot {\rm log}_2 \hspace{0.1cm}(50.25)= 1.99 - 2.84 \hspace{0.15cm}\underline{= -0.850\,{\rm bit}} \hspace{0.05cm}$$

• for  $\varepsilon = 0.001$:

$$h(X) =1.999 - 0.50025 \cdot {\rm log}_2 \hspace{0.1cm}(500.25) = 1.999 - 8.967 \hspace{0.15cm}\underline{= -6.968\,{\rm bit}} \hspace{0.05cm}.$$

(3)  All the proposed solutions are correct:

• After the boundary transition   $\varepsilon → 0$   we obtain for the differential entropy

$$h(X) = \lim\limits_{\varepsilon \hspace{0.05cm}\rightarrow \hspace{0.05cm} 0} \hspace{0.1cm}\big[(2-\varepsilon) - (0.5 + 0.25 \cdot \varepsilon) \cdot {\rm log}_2 \hspace{0.1cm}(0.5/\varepsilon +0.25)\big] = 2\,{\rm bit} - 0.5 \cdot \lim\limits_{\varepsilon \hspace{0.05cm}\rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}{\rm log}_2 \hspace{0.1cm}(0.5/\varepsilon) \hspace{0.3cm}\Rightarrow\hspace{0.3cm} - \infty \hspace{0.05cm}.$$

PDF and CDF of the mixed random variable  $X$

• The probability density function (PDF) in this case is given by.

$$f_X(x) = \left\{ \begin{array}{c} 0.25 + 0.5 \cdot \delta (x-1) \\ 0 \\ \end{array} \right. \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \hspace{0.1cm} 0 \le x \le 2, \\ {\rm sonst} \\ \end{array} \hspace{0.05cm}.$$

Consequently, it is a "mixed" random variable with

• a stochastic, uniformly distributed part in the range  $0 \le x \le 2$, and
• a discrete component at  $x = 1$  with probability  $0.5$.

The graph shows the PDF  $f_X(x)$  on the left and the distribution function  $F_X(x)$ on the right.
(4)  The correct solutions are 2, 3 and 5. The lower graph shows the PDF and the CDF of the random variable  $Y$.  You can see:

PDF and CDF of the mixed random variable $Y$

• Like $X$ ,  $Y$  contains a continuous and a discrete part.
• The discrete part occurs with probability  ${\rm Pr}(Y = 1) = 0.1$.
• Since  $F_Y(y)= {\rm Pr}(Y \le y)$  holds, the right-hand side limit is:

$$F_Y(y = 1) = 0.55.$$

• The continuous component is not uniformly distributed;  rather, there is a triangular distribution.
• The last proposition is also correct:   $h(Y) = h(X) = - \infty$.

Because:   For every random quantity with a discrete part – and it is also so small, the differential entropy is equal minus infinity..