Difference between revisions of "Aufgaben:Exercise 3.2Z: Relationship between PDF and CDF"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Verteilungsfunktion (VTF)
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Cumulative_Distribution_Function
 
}}
 
}}
  
[[File:P_ID117__Sto_Z_3_2.png|right|frame|Verteilungsfunktion  $ F_x(r)$]]
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[[File:P_ID117__Sto_Z_3_2.png|right|frame|Given CDF  $ F_x(r)$]]
Gegeben ist die Zufallsgröße  $x$  mit der Verteilungsfunktion
+
Given is the random variable  $x$  with the cumulative distribution function  $\rm (CDF)$.
:$$ F_x(r)=\left\{\begin{array}{*{4}{c}} 0.25\cdot {\rm e}^{2\it r}  &\rm f\ddot{u}r\hspace{0.1cm}\it r<\rm 0, \\ 1-0.25\cdot {\rm e}^{-2\it r} & \rm f\ddot{u}r\hspace{0.1cm}\it r\ge\rm 0.  \\\end{array}\right.$$
+
:$$ F_x(r)=\left\{\begin{array}{*{4}{c}} 0.25\cdot {\rm e}^{2\it r}  &\rm for\hspace{0.1cm}\it r<\rm 0, \\ 1-0.25\cdot {\rm e}^{-2\it r} & \rm for\hspace{0.1cm}\it r\ge\rm 0.  \\\end{array}\right.$$
  
*Diese Funktion ist rechts dargestellt.  
+
*This function is shown on the right.  
*Es ist zu erkennen, dass an der Sprungstelle&nbsp; $r = 0$&nbsp; der rechtsseitige Grenzwert g&uuml;ltig ist.
+
*It can be seen that at the unit step point&nbsp; $r = 0$&nbsp; the right-hand side limit is valid.
  
  
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 +
Hints:
 +
*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Cumulative_Distribution_Function|Cumulative Distribution Function]].
 +
*Reference is made 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 <br> &nbsp; &nbsp; [[Zusammenhang_zwischen_WDF_und_VTF_(Lernvideo)|"Zusammenhang zwischen WDF und VTF"]] &nbsp; $\Rightarrow$ &nbsp; "Relationship between PDF and CDF".
  
''Hinweise:''
 
*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Verteilungsfunktion|Verteilungsfunktion]].
 
*Bezug genommen wird auch auf das  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion|Wahrscheinlichkeitsdichtefunktion]].
 
*Eine Zusammenfassung der hier behandelten Thematik bietet das Lernvideo&nbsp; [[Zusammenhang_zwischen_WDF_und_VTF_(Lernvideo)|Zusammenhang zwischen WDF und VTF]].
 
 
   
 
   
  
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche Eigenschaften einer Verteilungsfunktion (VTF) gelten, wenn die Zufallsgröße beidseitig unbegrenzt ist?
+
{What properties of the CDF hold when the random variable has no limits?
 
|type="[]"}
 
|type="[]"}
+ Die VTF steigt von&nbsp; $0$&nbsp; auf&nbsp; $1$&nbsp; zumindest schwach monoton an.
+
+ The CDF increases from&nbsp; $0$&nbsp; to&nbsp; $1$&nbsp; at least weakly monotonically.
- Die&nbsp; $F_x(r)$&ndash;Werte&nbsp; $0$&nbsp; und&nbsp; $1$&nbsp; sind f&uuml;r endliche&nbsp; $r$&ndash;Werte möglich.
+
- The&nbsp; $F_x(r)$&nbsp; values&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; are possible for finite&nbsp; $r$&nbsp; values.
+ Ein horizontaler Abschnitt weist darauf hin, dass in diesem Bereich die Zufallsgr&ouml;&szlig;e keine Anteile besitzt.
+
+A horizontal section indicates that in this range the random size has no proportions.
+Vertikale Abschnitte sind m&ouml;glich.
+
+Vertical sections are possible.
  
  
{Wie gro&szlig; ist die Wahrscheinlichkeit, dass&nbsp; $x$&nbsp; positiv ist?
+
{What is the probability that&nbsp; $x$&nbsp; is positive?
 
|type="{}"}
 
|type="{}"}
 
${\rm Pr}(x > 0) \ = \ $ { 0.25 3% }
 
${\rm Pr}(x > 0) \ = \ $ { 0.25 3% }
  
  
{Wie gro&szlig; ist die Wahrscheinlichkeit, dass&nbsp; $|\hspace{0.05cm}x\hspace{0.05cm}|$&nbsp; gr&ouml;&szlig;er ist als&nbsp; $0.5$?
+
{What is the probability that&nbsp; $|\hspace{0.05cm}x\hspace{0.05cm}|$&nbsp; is larger than&nbsp; $0.5$?
 
|type="{}"}
 
|type="{}"}
${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| > 0.5) \ = \ $ { 0.184 3% }
+
${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| > 0.5) \ = \ $ { 0.184 3% }
  
  
{Geben Sie die zugeh&ouml;rige WDF&nbsp; $f_x(x)$&nbsp; allgemein an und den Wert für&nbsp; $x = 1$.
+
{Specify the associated PDF&nbsp; $f_x(x)$&nbsp; in general and the value for&nbsp; $x = 1$.
 
|type="{}"}
 
|type="{}"}
$f_x(x =1)\ = \ $ { 0.0677 3% }
+
$f_x(x =1)\ = \ $ { 0.0677 3% }
  
  
{Wie gro&szlig; ist die Wahrscheinlichkeit, dass&nbsp; $x$&nbsp; genau gleich&nbsp; $1$&nbsp; ist?
+
{What is the probability that&nbsp; $x$&nbsp; is exactly equal to&nbsp; $1$&nbsp;?
 
|type="{}"}
 
|type="{}"}
${\rm Pr}(x = 1)\ = \ $ { 0. }
+
${\rm Pr}(x = 1)\ = \ $ { 0. }
  
  
{Wie gro&szlig; ist die Wahrscheinlichkeit, dass&nbsp; $x$&nbsp; genau gleich&nbsp; $0$&nbsp; ist?
+
{What is the probability that&nbsp; $x$&nbsp; is exactly equal to&nbsp; $0$&nbsp;?
 
|type="{}"}
 
|type="{}"}
${\rm Pr}(x = 0)\ = \ $ { 0.5 3% }
+
${\rm Pr}(x = 0)\ = \ $ { 0.5 3% }
 
 
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die <u>Aussagen 1, 3 und 4</u> sind immer richtig:
+
'''(1)'''&nbsp; The&nbsp; <u>statements 1, 3 and 4</u>&nbsp; are always correct:
*Ein horizontaler Abschnitt in der VTF weist darauf hin, dass die Zufallsgr&ouml;&szlig;e in diesem Bereich keine Werte besitzt.  
+
*A horizontal intercept in the CDF indicates that the random variable has no values in that region.  
*Dagegen weist ein vertikaler Abschnitt in der VTF auf eine Diracfunktion in der WDF&nbsp; $($an gleicher Stelle&nbsp; $x_0)$&nbsp; hin.  
+
*In contrast,&nbsp; a vertical intercept in the CDF indicates a Dirac delta function in the PDF&nbsp; $($at the same location&nbsp; $x_0)$.  
*Dies bedeutet, dass die Zufallsgr&ouml;&szlig;e den Wert&nbsp; $x_0$&nbsp; sehr h&auml;ufig annimmt, n&auml;mlich mit endlicher Wahrscheinlichkeit.  
+
*This means that the random variable takes the value&nbsp; $x_0$&nbsp; very frequently,&nbsp; namely with finite probability.  
*Alle anderen Werte treten exakt mit der Wahrscheinlichkeit&nbsp; $0$&nbsp; auf.
+
*All other values occur exactly with probability&nbsp; $0$.
*Ist jedoch&nbsp; $x$&nbsp; auf den Bereich von&nbsp; $x_{\rm min}$&nbsp; bis&nbsp; $x_{\rm max}$&nbsp; begrenzt, so ist&nbsp; $F_x(r) = 0$ &nbsp;f&uuml;r&nbsp; $r < x_{\rm min}$&nbsp; und&nbsp; $F_x(r) = 1$ &nbsp;f&uuml;r&nbsp; $r > x_{\rm max}$.  
+
*If,&nbsp; however&nbsp; $x$&nbsp; is limited to the range from&nbsp; $x_{\rm min}$&nbsp; to&nbsp; $x_{\rm max}$&nbsp; then&nbsp; $F_x(r) = 0$&nbsp; &nbsp;for&nbsp; $r < x_{\rm min}$&nbsp; and&nbsp; $F_x(r) = 1$ &nbsp;for&nbsp; $r > x_{\rm max}$.  
*In diesem Sonderfall w&auml;re auch die zweite Aussage zutreffend.
+
*In this special case,&nbsp; the second statement would also be true.
  
  
  
'''(2)'''&nbsp; Die gesuchte Wahrscheinlichkeit kann man aus der Differenz der VTF&ndash;Werte an den Grenzen berechnen:
+
'''(2)'''&nbsp; The sought probability can be calculated from the difference of the CDF&nbsp; values at the boundaries:
:$${\rm Pr}( x> 0)= F_x(\infty)- F_x(\rm 0)
+
:$${\rm Pr}( x> 0)= F_x(\infty)- F_x(\rm 0)
 
\hspace{0.15cm}\underline{=\rm 0.25}.$$
 
\hspace{0.15cm}\underline{=\rm 0.25}.$$
  
  
  
'''(3)'''&nbsp; F&uuml;r die Wahrscheinlichkeit, dass&nbsp; $x$&nbsp; gr&ouml;&szlig;er als&nbsp; $0.5$&nbsp; ist, gilt:
+
'''(3)'''&nbsp; For the probability that&nbsp; $x$&nbsp; is greater than&nbsp; $0.5$&nbsp; holds:
:$${\rm Pr}(x> 0.5)=1- F_x(0.5)=\rm 0.25\cdot e^{-1}
+
:$${\rm Pr}(x> 0.5)=1- F_x(0.5)=\rm 0.25\cdot e^{-1}
 
\hspace{0.15cm}{\approx0.092}. $$
 
\hspace{0.15cm}{\approx0.092}. $$
  
*Aus Symmetriegr&uuml;nden ist&nbsp; ${\rm Pr}(x<- 0.5)$&nbsp; genauso gro&szlig;. Daraus folgt:
+
*For reasons of symmetry:&nbsp; ${\rm Pr}(x<- 0.5)$&nbsp; is just as large.&nbsp; From this follows:
 
:$${\rm Pr}( |\hspace{0.05cm} x\hspace{0.05cm}| >\rm 0.5) \hspace{0.15cm}\underline{= \rm 0.184}.$$
 
:$${\rm Pr}( |\hspace{0.05cm} x\hspace{0.05cm}| >\rm 0.5) \hspace{0.15cm}\underline{= \rm 0.184}.$$
  
  
[[File: P_ID116__Sto_Z_3_2_c.png|right|frame|WDF der Laplace-Verteilung]]
+
[[File: P_ID116__Sto_Z_3_2_c.png|right|frame|PDF of Laplace distribution]]
'''(4)'''&nbsp; Die WDF erh&auml;lt man aus der zugeh&ouml;rigen VTF durch Differenzieren der zwei Bereiche.  
+
'''(4)'''&nbsp; The PDF is obtained from the corresponding CDF by differentiating the two areas.  
*Es ergibt sich eine zweiseitige Exponentialfunktion sowie eine Diracfunktion bei&nbsp; $x = 0$&nbsp;:
+
*The result is a two-sided exponential function as well as a Dirac delta function at&nbsp; $x = 0$:
 
:$$f_x(x)=\rm 0.5\cdot \rm e^{-2\cdot |\hspace{0.05cm}\it x\hspace{0.05cm}|} + \rm 0.5\cdot\delta(\it x).$$
 
:$$f_x(x)=\rm 0.5\cdot \rm e^{-2\cdot |\hspace{0.05cm}\it x\hspace{0.05cm}|} + \rm 0.5\cdot\delta(\it x).$$
*Der gesuchte Zahlenwert ist&nbsp; $f_x(x = 1)\hspace{0.15cm}\underline{= \rm 0.0677}$.
+
*The numerical value we are looking for is&nbsp; $f_x(x = 1)\hspace{0.15cm}\underline{= \rm 0.0677}$.
 
+
*Note:&nbsp; The two-sided exponential distribution is also called "Laplace distribution".
 
 
<i>Hinweis:</i> &nbsp; Die zweiseitige Exponentialverteilung nennt man auch "Laplaceverteilung&rdquo;.
 
 
 
  
  
'''(5)'''&nbsp; Im Bereich um&nbsp; $1$&nbsp; beschreibt&nbsp; $x$&nbsp; eine kontinuierliche Zufallsgr&ouml;&szlig;e.
 
*Die Wahrscheinlichkeit, dass&nbsp; $x$&nbsp;  exakt den Wert&nbsp; $1$&nbsp; aufweist, ist deshalb&nbsp; ${\rm Pr}(x = 1)\hspace{0.15cm}\underline{= \rm 0}.$
 
  
 +
'''(5)'''&nbsp; In the range around&nbsp; $1$&nbsp; describes&nbsp; $x$&nbsp; a continuous valued random variable.
 +
*The probability that&nbsp; $x$&nbsp; has exactly the value&nbsp; $1$&nbsp; is therefore&nbsp; ${\rm Pr}(x = 1)\hspace{0.15cm}\underline{= \rm 0}.$
  
  
'''(6)'''&nbsp; In&nbsp; $50\%$&nbsp; der Zeit wird&nbsp; $x = 0$&nbsp; gelten: &nbsp; ${\rm Pr}(x = 0)\hspace{0.15cm}\underline{= \rm 0.5}.$
 
  
 +
'''(6)'''&nbsp; In&nbsp; $50\%$&nbsp; of time&nbsp; $x = 0$&nbsp; will hold: &nbsp; ${\rm Pr}(x = 0)\hspace{0.15cm}\underline{= \rm 0.5}.$
  
<i>Hinweise:</i>
+
*The PDF of a speech signal is often described by a two-sided exponential function.  
*Die WDF eines Sprachsignals wird h&auml;ufig durch eine zweiseitige Exponentialfunktion beschrieben.  
+
*The Dirac delta function at&nbsp; $x = 0$&nbsp; mainly takes into account speech pauses &ndash; here in&nbsp; $50\%$&nbsp; of all times.
*Die Diracfunktion bei&nbsp; $x = 0$&nbsp; ber&uuml;cksichtigt vor allem Sprachpausen &ndash; hier in&nbsp; $50\%$&nbsp; aller Zeiten.
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Theory of Stochastic Signals: Exercises|^3.2 Verteilungsfunktion^]]
+
[[Category:Theory of Stochastic Signals: Exercises|^3.2 Cumulative Distribution Function^]]

Latest revision as of 15:39, 4 January 2022

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

(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.