# Exercise 3.2Z: Relationship between PDF and CDF

From LNTwww

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:

- The exercise belongs to the chapter Cumulative Distribution Function.
- Reference is made to the chapter Probability Density Function.
- The topic of this chapter is illustrated with examples in the (German language) learning video

"Zusammenhang zwischen WDF und VTF" $\Rightarrow$ "Relationship between PDF and CDF".

### Questions

### 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:

- $${\rm Pr}( |\hspace{0.05cm} x\hspace{0.05cm}| >\rm 0.5) \hspace{0.15cm}\underline{= \rm 0.184}.$$

**(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.