Aufgaben:Exercise 4.1: PDF, CDF and Probability

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$\rm (CDF)$  (top),  $\rm (PDF)$  (bottom)

To repeat some important basics from the book "Theory of Stochastic Signals" we are dealing with

• the  probability density function $\rm (PDF)$,
• the  cumulative distribution function $\rm (CDF)$.

The upper plot shows the cumulative distribution function  $F_X(x)$  of a value-discrete random variable  $X$.  The corresponding probability density function  $f_X(x)$  has to be determined in subtask  (1).

The equation

$$ {\rm Pr}(A < X \le B) = F_X(B) - F_X(A) = \lim_{\varepsilon \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} \int_{A+\varepsilon}^{B+\varepsilon} \hspace{-0.15cm} f_X(x) \hspace{0.1cm}{\rm d}x $$

represents two ways to calculate the probability for the event  "The random variable  $X$  lies in a given interval"  from the CDF and the PDF,  respectively.

The lower graph shows the probability density function

$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}l} | y| \le 2, \\ y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$

of a continuous random variable  $Y$,  which is restricted to the range  $|Y| \le 2$ .  In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable  $Y$  as for a value-discrete random variable.  Nevertheless, you will notice some differences in details.

For example, for the continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:

$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y) \hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$

Hints:

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

• Given also is the following indefinite integral:

$$\int \hspace{0.1cm} \cos^2(A \eta) \hspace{0.1cm}{\rm d}\eta = \frac{\eta}{2} + \frac{1}{4A} \cdot \sin(2A \eta).$$

Questions

1

Determine the PDF  $f_X(x)$  of the value-discrete random variable  $X$.  Which of the following statements are true?

	The PDF is composed of five Dirac functions.
	${\rm Pr}(X= 0) = 0.4$   and  ${\rm Pr}(X= 1) = 0.2$  are true.
	${\rm Pr}(X= 2) = 0.4$  is true.

2

Calculate the following probabilities:

${\rm Pr}(X > 0) \ = \ $

${\rm Pr}(|X| ≤ 1) \ = \ $

3

What are the values of the cumulative distribution function  $F_Y(y) ={\rm Pr}(Y \le y)$  of the continuous random variable  $Y$,  in particular:

$F_Y(y = 0) \ = \ $

$F_Y(y = 1) \ = \ $

$F_Y(y = 2) \ = \ $

4

What is the probability that  $Y = 0$ ?

${\rm Pr}(Y = 0) \ = \ $

5

Which of the following statements are correct?|type="[]"

	The result  $Y = 0$  is impossible.
	The result  $Y = 3$  is impossible.

6

What are the following probabilities?

${\rm Pr}(Y > 0) \ = \ $

${\rm Pr}(|Y| ≤ 1) \ = \ $

Solution

Solution

PDF and CDF of the value-discrete random variable  $X$

(1)  Proposed solutions 1 and 2 are correct:

• The cumulative distribution function  $F_X(x)$  is obtained from the probability density function  $f_X(x)$  by integration over the (renamed) random variable in the range from  $- \infty$  to  $x$.
• The inverse is:   Given the CDF, obtain the PDF by differentiation.
• The given CDF contains five discontinuity points, which after differentiation lead to five Dirac functions:

$$f_X(x) = 0.1 \cdot {\rm \delta}( x+2) + 0.2 \cdot {\rm \delta}( x+1) + 0.4 \cdot {\rm \delta}( x) + 0.2 \cdot {\rm \delta}( x-1) + 0.1 \cdot {\rm \delta}( x-2)\hspace{0.05cm}.$$

• The Dirac weights give the occurrence probabilities of the random variable  $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$ , e.g.:

$${\rm Pr}(X = 0) = F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-}) = 0.7 - 0.3 = 0.4\hspace{0.05cm}.$$

• Accordingly, the other probabilities are:

$${\rm Pr}(X = +1) = {\rm Pr}(X = -1) = 0.2\hspace{0.05cm},\hspace{0.3cm} {\rm Pr}(X = +2) = {\rm Pr}(X = -2) = 0.1\hspace{0.05cm}.$$

(2)  From the PDF just calculated, we obtain:

$${\rm Pr}(X >0) = {\rm Pr}(X = +1) + {\rm Pr}(X = +2) \hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$

$${\rm Pr}(|X| \le 1) ={\rm Pr}(X = -1) + {\rm Pr}(X = 0) + {\rm Pr}(X = +1) = 0.2 + 0.4 +0.2 \hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$

The same result is obtained using the CDF.  Here the general equation, which is equally valid for value-discrete and continuous random variables, is:

$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$

• Thus, with  $A= 0$  and  $B = +2$  we obtain:

$${\rm Pr}(0 < X \le +2) = {\rm Pr}(X >0)= F_X(+2) - F_X(0) = 1 - 0.7 \hspace{0.15cm}\underline {= 0.3} \hspace{0.05cm}.$$

• Setting  $A=-2$  and  $B = +1$,  we get:

$${\rm Pr}(-2 < X \le +1) = {\rm Pr}(|X| \le 1)= F_X(+1) - F_X(-2) = 0.9 - 0.1 \hspace{0.15cm}\underline {= 0.8} \hspace{0.05cm}.$$

PDF and CDF of the continuous random variable  $Y$

(3)  The cumulative distribution function  $F_Y(y)$  is obtained from the (renamed) WDF  $f_Y(\eta)$  by integrating  $- \infty$  to  $x$.  Due to symmetry, this can be written in the range  $0 \le y \le +2$:

$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$

$$\Rightarrow \hspace{0.3cm}F_Y(y) = \frac{1}{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{0.1cm}\frac{1}{2} \cdot \cos^2({\pi}/{4} \cdot \eta) \hspace{0.1cm}{\rm d}\eta = \frac{1}{2}+\frac{y}{4} + \frac{1}{2\pi} \cdot \sin({\pi}/{2} \cdot y).$$

The equation holds in the entire range  $0 \le y \le +2$.  The CDF values we are looking for are thus:

• $F_Y(y=0)\hspace{0.15cm}\underline{= 0.5}$  (integral over half the PDF),
• $F_Y(y=1)= 3/4 + 1/(2 \pi)\hspace{0.15cm}\underline{= 0.909}$  (area in red background in the PDF),
• $F_Y(y=2)\hspace{0.15cm}\underline{= 1}$  (integral over the entire PDF).

(4)  The probability that the continuous random variable  $Y$  lies in the range from  $-\varepsilon$  to  $+\varepsilon$  can be calculated using the given equation as follows:

$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$

• It was taken into account that for the random variable  $Y$  the "<"sign can be replaced by the "≤" sign without distortion.
• With the boundary transition  $\varepsilon \to 0$,  the probability we are looking for is obtained:

$${\rm Pr}(Y = 0) =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm}{\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(+\varepsilon) - \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(-\varepsilon) = F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$

• Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.

In general:   The probability  ${\rm Pr}(Y = y_0)$  that a continuous random variable  $Y$  takes a fixed value  $y_0$,  is always zero.

(5)  Proposed solution 2 is correct:

• Based on the PDF at hand, the result  $Y=3$  can be excluded.
• The result  $Y=0$  on the other hand is quite possible, although  ${\rm Pr}(Y = 0) = 0$ .
• For example, if one performs a random experiment  $N \to \infty$  times and obtains the result  $Y= 0$   for  $N_0$  times, then with finite   $N_0$  according to the classical definition of probability:

$${\rm Pr}(Y = 0) = \lim_{N\hspace{0.05cm}\rightarrow\hspace{0.05cm}\infty}\hspace{0.1cm}{N_0}/{N} = 0\hspace{0.05cm}.$$

(6)  We again assume the equation  $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$    valid for the continuous random quantity  $Y$:

• With  $A = 0$  and  $B \to \infty$  $($or  $B = 2)$  we obtain:

$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty) = {\rm Pr}(0 \le Y \le 2) = F_Y(2) - F_Y(0) \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$

• Thus, for the symmetric continuous random variable  $Y$  holds indeed as expected:  ${\rm Pr}( Y > 0) = 1/2$.
• Although the value-discrete random variable  $X$  is also symmetrical about $x= 0$   ⇒   ${\rm Pr}( X > 0) = 0.3$  was determined in subtask  (3), on the other hand.
• Further, with  $A = -1$  and  $B = +1$,  one obtains because of $F_Y(-1) = 1- F_Y(+1)$:

$${\rm Pr}( |Y| \le 1) = {\rm Pr}(-1 \le Y \le +1) = F_Y(+1) - F_Y(-1) = 2 \cdot F_Y(+1) -1 = 2 \cdot 0.909 -1 \hspace{0.15cm}\underline {= 0.818}. $$