View source for Aufgaben:Exercise 3.1Z: Triangular PDF


{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Probability_Density_Function
}}

[[File:P_ID109__Sto_Z_3_1.png|right|frame|Triangular PDF and <br>characteristic curve&nbsp; $y(x)$]]
We consider a continuous random variable&nbsp; $x$&nbsp; with the PDF outlined on the right.&nbsp; 
*The minimum value of the signal is&nbsp; $x_{\rm min} = -2\hspace{0.05cm} {\rm V}$.&nbsp; 
*On the other hand,&nbsp; the maximum value&nbsp; $x_{\rm max}$&nbsp; is a free parameter,&nbsp; allowing values between&nbsp; $+2\hspace{0.05cm}\rm V$&nbsp; and&nbsp; $+4\hspace{0.05cm} \rm V$.


The random variable&nbsp; $x$&nbsp; is to be understood here as the instantaneous value of a random signal.&nbsp; If this signal&nbsp; $x(t)$&nbsp; is applied to an amplitude limiter with the characteristic curve&nbsp; (see sketch below)
:$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) &  {\rm if}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le  x(t)\le +2\hspace{0.05cm} {\rm V},  \\ +2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V},  \\\end{array}\right.$$

so the signal&nbsp; $y(t)$&nbsp; or the new random variable&nbsp; $y$,&nbsp; which is considered in the last two subquestions&nbsp; '''(5)'''&nbsp; and&nbsp; '''(6)'''&nbsp; is obtained. <br />

*For the subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)'''&nbsp; apply&nbsp; $x_{\rm max} = 2\hspace{0.05cm} {\rm V} $.
* For all other subtasks,&nbsp; set&nbsp; $x_{\rm max} = 4\hspace{0.05cm} {\rm V} $&nbsp;.





Hints:
*The task belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Probability_Density_Function|Probability Density Function]].
 
*The topic of this chapter is illustrated with examples in the&nbsp; (German language)&nbsp; learning video&nbsp; <br>&nbsp; &nbsp;[[Wahrscheinlichkeit_und_WDF_(Lernvideo)|"Wahrscheinlichkeit und WDF"]] &nbsp; $\Rightarrow$ &nbsp; "Probability and PDF".


===Questions===

<quiz display=simple>
{Let be&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.&nbsp; Calculate the parameter&nbsp; $A = f_x(0)$.
|type="{}"}
$A \ = \ $ { 0.5 3% } $\ \rm 1/V$.


{Let&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability that&nbsp; $|x(t)|$&nbsp; less than&nbsp; $x_{\rm max}/2$?
|type="{}"}
${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| < 1\hspace{0.05cm} {\rm V}) \ = \ $ { 0.75 3% }


{Now let be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability that&nbsp; $x$&nbsp; lies between&nbsp; $+1\hspace{0.05cm} {\rm V}$&nbsp; and&nbsp; $+3\hspace{0.05cm} {\rm V}$?
|type="{}"}
${\rm Pr}(1\hspace{0.05cm} {\rm V} < x <3\hspace{0.05cm} {\rm V}) \ = \ $ { 0.333 3% }


{ Let further be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability that&nbsp; $x$&nbsp; is exactly equal to $+2\hspace{0.05cm} {\rm V}$?
|type="{}"}
${\rm Pr}(x =2\hspace{0.05cm} {\rm V})\ = \ $ { 0. }


{ Let further be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; Which of the following statements is true?
|type="[]"}
- $y$&nbsp; is a continuous random variable.
- $y$&nbsp; is a discrete random variable.
+ $y$&nbsp; is a mixed continuous-discrete random variable.


{Let further be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability  that&nbsp; $y$&nbsp; is exactly equal&nbsp; $+2\hspace{0.05cm} {\rm V}$?
|type="{}"}
${\rm Pr}(y =2\hspace{0.05cm} {\rm V})\ = \ $ { 0.167 3% }




</quiz>

===Solution===
{{ML-Kopf}}
[[File:EN_Sto_Z_3_1_bc.png|right|frame|Height and area of triangular PDF]]
'''(1)'''&nbsp; The area under the PDF must always yield the value&nbsp; $1$.&nbsp; It follows that:
:$${A}/{ 2}\cdot {4\hspace{0.05cm}\rm V}=1\hspace{0.5cm}\Rightarrow\hspace{0.5cm} A\hspace{0.15cm}\underline{=\rm 0.5\;{1}/{V}}.$$



'''(2)'''&nbsp; With&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$&nbsp; the PDF is obtained according to the left graph. 
*The shading marks the probability we are looking for.
*One obtains by simple geometric considerations:
:$${\rm Pr}(|x|<\rm 1\hspace{0.05cm} V)\hspace{0.15cm}\underline{=\rm 0.75}.$$



'''(3)'''&nbsp; With&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$&nbsp; one obtains the PDF shown on the right.
*The maximum value is now &nbsp; $A = 1/(3\hspace{0.05cm} {\rm V})$. 
*The shaded area again indicates the probability we are looking for,&nbsp; which can be determined using the rectangle of equal area: 
:$${\rm Pr}(1\hspace{0.05cm} {\rm V}< x<3\hspace{0.05cm} {\rm V})=\rm \frac{1}{6\hspace{0.05cm} {\rm V}}\cdot 2\hspace{0.05cm} {\rm V}=\hspace{0.15cm}\underline{0.333}.$$



'''(4)'''&nbsp; Since&nbsp; $x$&nbsp; represents a continuous random variable,&nbsp; this probability is by definition zero &nbsp; &rArr; &nbsp; ${\rm Pr}(x =2\hspace{0.05cm} {\rm V}) \;\underline {= 0}$.



[[File:P_ID113__Sto_Z_3_1_f.png|right|frame|Mixed continuous/discrete PDF]]
'''(5)'''&nbsp; <u>Only the last statement</u> of the given answers is true:

*The PDF&nbsp; $f_y(y)$&nbsp; includes a continuous component (drawn in blue), 
*but also the (red) Dirac delta function at&nbsp; $y = +2\hspace{0.05cm} {\rm V}$&nbsp; with weight&nbsp; ${\rm Pr}(x >2\hspace{0.05cm} {\rm V})$.



'''(6)'''&nbsp; The graphic on the right shows the probability density function of the random variable&nbsp; $y$. 
*From the right figure for subtask&nbsp; '''(3)'''&nbsp; one can see the relation:
:$${\rm Pr}( y=2\hspace{0.05cm} {\rm V}) = {\rm Pr}( x> 2\hspace{0.05cm} {\rm V}) =  \frac{1}{2}\cdot\frac{1}{6\hspace{0.05cm} {\rm V}}\cdot2{\hspace{0.05cm} {\rm V}} = {1}/{6}\hspace{0.15cm}\underline{=0.167}.$$
{{ML-Fuß}}



[[Category:Theory of Stochastic Signals: Exercises|^3.1 Probability Density Function^]]
[[de:Aufgaben:Aufgabe 3.1Z: Dreieckförmige WDF]]