# Exercise 3.2Z: Relationship between PDF and CDF

Given CDF  $F_x(r)$

Given is the random variable  $x$  with the cumulative distribution function  $\rm (CDF)$.

$$F_x(r)=\left\{\begin{array}{*{4}{c}} 0.25\cdot {\rm e}^{2\it r} &\rm for\hspace{0.1cm}\it r<\rm 0, \\ 1-0.25\cdot {\rm e}^{-2\it r} & \rm for\hspace{0.1cm}\it r\ge\rm 0. \\\end{array}\right.$$
• This function is shown on the right.
• It can be seen that at the unit step point  $r = 0$  the right-hand side limit is valid.

Hints:

### Questions

1

What properties of the CDF hold when the random variable has no limits?

 The CDF increases from  $0$  to  $1$  at least weakly monotonically. The  $F_x(r)$  values  $0$  and  $1$  are possible for finite  $r$  values. A horizontal section indicates that in this range the random size has no proportions. Vertical sections are possible.

2

What is the probability that  $x$  is positive?

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

3

What is the probability that  $|\hspace{0.05cm}x\hspace{0.05cm}|$  is larger than  $0.5$?

 ${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| > 0.5) \ = \$

4

Specify the associated PDF  $f_x(x)$  in general and the value for  $x = 1$.

 $f_x(x =1)\ = \$

5

What is the probability that  $x$  is exactly equal to  $1$ ?

 ${\rm Pr}(x = 1)\ = \$

6

What is the probability that  $x$  is exactly equal to  $0$ ?

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

### Solution

#### Solution

(1)  The  statements 1, 3 and 4  are always correct:

• A horizontal intercept in the CDF indicates that the random variable has no values in that region.
• In contrast,  a vertical intercept in the CDF indicates a Dirac delta function in the PDF  $($at the same location  $x_0)$.
• This means that the random variable takes the value  $x_0$  very frequently,  namely with finite probability.
• All other values occur exactly with probability  $0$.
• If,  however  $x$  is limited to the range from  $x_{\rm min}$  to  $x_{\rm max}$  then  $F_x(r) = 0$   for  $r < x_{\rm min}$  and  $F_x(r) = 1$  for  $r > x_{\rm max}$.
• In this special case,  the second statement would also be true.

(2)  The sought probability can be calculated from the difference of the CDF  values at the boundaries:

$${\rm Pr}( x> 0)= F_x(\infty)- F_x(\rm 0) \hspace{0.15cm}\underline{=\rm 0.25}.$$

(3)  For the probability that  $x$  is greater than  $0.5$  holds:

$${\rm Pr}(x> 0.5)=1- F_x(0.5)=\rm 0.25\cdot e^{-1} \hspace{0.15cm}{\approx0.092}.$$
• For reasons of symmetry:  ${\rm Pr}(x<- 0.5)$  is just as large.  From this follows:
$${\rm Pr}( |\hspace{0.05cm} x\hspace{0.05cm}| >\rm 0.5) \hspace{0.15cm}\underline{= \rm 0.184}.$$

PDF of Laplace distribution

(4)  The PDF is obtained from the corresponding CDF by differentiating the two areas.

• The result is a two-sided exponential function as well as a Dirac delta function at  $x = 0$:
$$f_x(x)=\rm 0.5\cdot \rm e^{-2\cdot |\hspace{0.05cm}\it x\hspace{0.05cm}|} + \rm 0.5\cdot\delta(\it x).$$
• The numerical value we are looking for is  $f_x(x = 1)\hspace{0.15cm}\underline{= \rm 0.0677}$.
• Note:  The two-sided exponential distribution is also called "Laplace distribution".

(5)  In the range around  $1$  describes  $x$  a continuous valued random variable.

• The probability that  $x$  has exactly the value  $1$  is therefore  ${\rm Pr}(x = 1)\hspace{0.15cm}\underline{= \rm 0}.$

(6)  In  $50\%$  of time  $x = 0$  will hold:   ${\rm Pr}(x = 0)\hspace{0.15cm}\underline{= \rm 0.5}.$

• The PDF of a speech signal is often described by a two-sided exponential function.
• The Dirac delta function at  $x = 0$  mainly takes into account speech pauses – here in  $50\%$  of all times.