Difference between revisions of "Aufgaben:Exercise 3.2Z: Two-dimensional Probability Mass Function"

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{{quiz-Header|Buchseite=Informationstheorie/Einige Vorbemerkungen zu zweidimensionalen Zufallsgrößen
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{{quiz-Header|Buchseite=Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables
 
}}
 
}}
  
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[[File:P_ID2752__Inf_Z_3_2_neu.png|right|frame|Wahrscheinlichkeitsfunktion der 2D–Zufallsgröße  $XY$]]
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[[File:P_ID2752__Inf_Z_3_2_neu.png|right|frame|$\rm PMF$ of the two-dimensional  random variable  $XY$]]
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Wir betrachten die Zufallsgrößen  $X =  \{ 0,\ 1,\ 2,\ 3 \}$  und  $Y =  \{ 0,\ 1,\ 2 \}$, deren gemeinsame Wahrscheinlichkeitsfunktion  $P_{XY}(X,\ Y)$  gegeben ist.  
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We consider the random variables  $X =  \{ 0,\ 1,\ 2,\ 3 \}$  and  $Y =  \{ 0,\ 1,\ 2 \}$, whose joint probability mass function  $P_{XY}(X,\ Y)$  is given.
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*Aus dieser 2D–Wahrscheinlichkeitsfunktion sollen die eindimensionalen Wahrscheinlichkeitsfunktionen  $P_X(X)$  und  $P_Y(Y)$  ermittelt werden.
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*From this two-dimensional  probability mass function  $\rm (PMF)$,  the one-dimensional probability mass functions  $P_X(X)$  and  $P_Y(Y)$  are to be determined.
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*Man nennt eine solche 1D–Wahrscheinlichkeitsfunktion manchmal auch Randwahrscheinlichkeit  (englisch:  ''Marginal Probability'').
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*Such a one-dimensional probability mass function is sometimes also called  "marginal probability".
  
  
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Gilt   $P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, so sind die beiden Zufallsgrößen  $X$  und  $Y$  statistisch unabhängig.  Andernfalls bestehen zwischen diesen statistische Bindungen.
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If  $P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables  $X$  and  $Y$  are statistically independent.  Otherwise, there are statistical dependencies between them.
  
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Im zweiten Teil der Aufgabe betrachten wir die Zufallsgrößen  $U=  \big \{ 0,\ 1 \big \}$  und  $V= \big  \{ 0,\ 1 \big \}$, die sich aus  $X$  und  $Y$  durch Modulo–2–Operationen ergeben:
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In the second part of the task we consider the random variables  $U=  \big \{ 0,\ 1 \big \}$  and  $V= \big  \{ 0,\ 1 \big \}$,  which result from  $X$  and  $Y$  by modulo-2 operations:
  
 
:$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm}  V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$
 
:$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm}  V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$
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''Hinweise:''
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<u>Hints:</u>
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*Die Aufgabe gehört zum  Kapitel&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Einige Vorbemerkungen zu den 2D-Zufallsgrößen]].
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*The exercise belongs to the chapter&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on two-dimensional random variables]].
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*Ausgegangen wird hier von der gleichen Konstellation wie in&nbsp; [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen|Aufgabe 3.2]].
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*The same constellation is assumed here as in&nbsp; [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen|Exercise 3.2]].
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*Dort wurde die Zufallsgrößen&nbsp;  $Y = \{ 0,\ 1,\ 2,\ 3 \}$&nbsp;  betrachtet, allerdings mit dem Zusatz&nbsp; ${\rm Pr}(Y = 3) = 0$.  
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*There the random variables&nbsp;  $Y = \{ 0,\ 1,\ 2,\ 3 \}$&nbsp;  were considered, but with the addition&nbsp; ${\rm Pr}(Y = 3) = 0$.  
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*Die so erzwungene Eigenschaft&nbsp; $|X| = |Y|$&nbsp;   war in der vorherigen Aufgabe zur formalen Berechnung des Erwartungswertes&nbsp; ${\rm E}[P_X(X)]$&nbsp; von Vorteil.
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*The property&nbsp; $|X| = |Y|$&nbsp; forced in this way was advantageous in the previous task for the formal calculation of the expected value.
 
   
 
   
  
  
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===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
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{Wie lautet die Wahrscheinlichkeitsfunktion&nbsp; $P_X(X)$?
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{What is the probability mass function&nbsp; $P_X(X)$?
 
|type="{}"}
 
|type="{}"}
 
$P_X(0) \ = \ $ { 0.5 3% }
 
$P_X(0) \ = \ $ { 0.5 3% }
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$P_X(3) \ = \ ${ 0.375 3% }
 
$P_X(3) \ = \ ${ 0.375 3% }
  
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{Wie lautet die Wahrscheinlichkeitsfunktion&nbsp; $P_Y(Y)$?
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{What is the probability mass function&nbsp; $P_Y(Y)$?
 
|type="{}"}
 
|type="{}"}
 
$P_Y(0) \ = \ $ { 0.5 3% }
 
$P_Y(0) \ = \ $ { 0.5 3% }
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$P_Y(2) \ = \ $ { 0.25 3% }
 
$P_Y(2) \ = \ $ { 0.25 3% }
  
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{Sind die Zufallsgrößen&nbsp; $X$&nbsp; und&nbsp; $Y$&nbsp; statistisch unabhängig?
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{Are the random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; statistically independent?
 +
 
 
|type="()"}
 
|type="()"}
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- Ja,
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- Yes,
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+ Nein.
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+ No.
  
  
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{Ermitteln Sie die Wahrscheinlichkeiten&nbsp; $P_{UV}( U,\ V)$.
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{Determine the probabilities&nbsp; $P_{UV}( U,\ V)$.
 
|type="{}"}
 
|type="{}"}
 
$P_{UV}( U = 0,\ V = 0) \ = \ $ { 0.375 3% }
 
$P_{UV}( U = 0,\ V = 0) \ = \ $ { 0.375 3% }
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$P_{UV}( U =1,\ V = 1) \ = \ $ { 0.125 3% }
 
$P_{UV}( U =1,\ V = 1) \ = \ $ { 0.125 3% }
  
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{Sind die Zufallsgrößen&nbsp; $U$&nbsp; und&nbsp; $V$&nbsp; statistisch unabhängig?
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{Are the random variables&nbsp; $U$&nbsp; and&nbsp; $V$&nbsp; statistically independent?
 
|type="()"}
 
|type="()"}
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+ Ja,
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+ Yes,
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- Nein.
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- No.
  
  
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</quiz>
 
</quiz>
  
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===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
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'''(1)'''&nbsp; Man kommt von&nbsp;  $P_{XY}(X,\ Y)$&nbsp; zur 1D–Wahrscheinlichkeitsfunktion&nbsp; $P_X(X)$, indem man alle&nbsp; $Y$-Wahrscheinlichkeiten aufsummiert:
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'''(1)'''&nbsp; You get from&nbsp;  $P_{XY}(X,\ Y)$&nbsp; to the one-dimensional probability mass function&nbsp; $P_X(X)$ by summing up all&nbsp; $Y$ probabilities:
 
:$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$
 
:$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$
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*Man erhält somit folgende Zahlenwerte:
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*One thus obtains the following numerical values:
 
:$$P_X(X = 0) = 1/4+1/8+1/8 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
 
:$$P_X(X = 0) = 1/4+1/8+1/8 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
 
:$$P_X(X = 1)= 0+0+1/8 =  1/8 \hspace{0.15cm}\underline{= 0.125},$$
 
:$$P_X(X = 1)= 0+0+1/8 =  1/8 \hspace{0.15cm}\underline{= 0.125},$$
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'''(2)'''&nbsp; Analog zur Teilaufgabe&nbsp; '''(1)'''&nbsp; gilt nun:  
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'''(2)'''&nbsp; Analogous to sub-task &nbsp; '''(1)'''&nbsp;, the following now holds:
 
:$$P_Y(Y = y_{\kappa}) = \sum_{x \hspace{0.05cm} \in \hspace{0.05cm} X} \hspace{0.1cm} P_{XY}(x, y_{\kappa})$$
 
:$$P_Y(Y = y_{\kappa}) = \sum_{x \hspace{0.05cm} \in \hspace{0.05cm} X} \hspace{0.1cm} P_{XY}(x, y_{\kappa})$$
 
:$$P_Y(Y= 0) = 1/4+0+0+1/4 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
 
:$$P_Y(Y= 0) = 1/4+0+0+1/4 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
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'''(3)'''&nbsp; Bei statistischer Unabhängigkeit sollte&nbsp;  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$&nbsp; sein.  
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'''(3)'''&nbsp; With statistical independence,&nbsp;  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$&nbsp; should be.
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*Dies trifft hier nicht zu: &nbsp; &nbsp;  Antwort &nbsp; <u>'''Nein'''</u>.
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*This does not apply here: &nbsp; &nbsp;  answer &nbsp; <u>'''NO'''</u>.
  
  
  
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'''(4)'''&nbsp; Ausgehend von  der linken Tabelle &nbsp; &rArr; &nbsp;  $P_{XY}(X,Y)$&nbsp; kommt man zur mittlere Tabelle &nbsp; &rArr; &nbsp; $P_{UY}(U,Y)$, <br>indem man gewisse Wahrscheinlichkeiten entsprechend&nbsp; $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$&nbsp; zusammenfasst.  
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'''(4)'''&nbsp; Starting from the left-hand table &nbsp; &rArr; &nbsp;  $P_{XY}(X,Y)$,&nbsp; we arrive at the middle table &nbsp; &rArr; &nbsp; $P_{UY}(U,Y)$, <br>by combining certain probabilities according to&nbsp; $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.  
  
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[[File:P_ID2753__Inf_Z_3_2d_neu.png|right|frame|Verschiedene Wahrscheinlichkeitsfunktionen]]
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If one also takes into account&nbsp; $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$, one obtains the probabilities sought according to the right-hand table:
 +
[[File:P_ID2753__Inf_Z_3_2d_neu.png|right|frame|Different probability functions]]
  
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Berücksichtigt man noch&nbsp; $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$, so erhält man die gesuchten Wahrscheinlichkeiten entsprechend der rechten Tabelle:
 
 
:$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
 
:$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
 
:$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
 
:$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
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'''(5)'''&nbsp; Richtig ist die Antwort &nbsp; <u>'''Ja'''</u>:
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'''(5)'''&nbsp; The correct answer is &nbsp; <u>'''YES'''</u>:
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*Die zugehörigen 1D–Wahrscheinlichkeitsfunktionen lauten: &nbsp;  
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*The corresponding one-dimensional probability mass functions are:   &nbsp;  
 
:$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$
 
:$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$
 
:$$P_V(V)=\big [3/4, \ 1/4 \big ].$$   
 
:$$P_V(V)=\big [3/4, \ 1/4 \big ].$$   
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*Damit gilt:&nbsp; $P_{UV}(U,V) = P_U(U) \cdot  P_V(V)$  &nbsp; &rArr;  &nbsp;  $U$&nbsp; und&nbsp; $V$&nbsp; sind statistisch unabhängig.
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*Thus:&nbsp; $P_{UV}(U,V) = P_U(U) \cdot  P_V(V)$  &nbsp; &rArr;  &nbsp;  $U$&nbsp; and&nbsp; $V$&nbsp; are statistically independent.  
  
  

Latest revision as of 10:12, 24 September 2021

$\rm PMF$ of the two-dimensional random variable  $XY$

We consider the random variables  $X = \{ 0,\ 1,\ 2,\ 3 \}$  and  $Y = \{ 0,\ 1,\ 2 \}$, whose joint probability mass function  $P_{XY}(X,\ Y)$  is given.

  • From this two-dimensional probability mass function  $\rm (PMF)$,  the one-dimensional probability mass functions  $P_X(X)$  and  $P_Y(Y)$  are to be determined.
  • Such a one-dimensional probability mass function is sometimes also called  "marginal probability".


If  $P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables  $X$  and  $Y$  are statistically independent.  Otherwise, there are statistical dependencies between them.

In the second part of the task we consider the random variables  $U= \big \{ 0,\ 1 \big \}$  and  $V= \big \{ 0,\ 1 \big \}$,  which result from  $X$  and  $Y$  by modulo-2 operations:

$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm} V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$





Hints:

  • The exercise belongs to the chapter  Some preliminary remarks on two-dimensional random variables.
  • The same constellation is assumed here as in  Exercise 3.2.
  • There the random variables  $Y = \{ 0,\ 1,\ 2,\ 3 \}$  were considered, but with the addition  ${\rm Pr}(Y = 3) = 0$.
  • The property  $|X| = |Y|$  forced in this way was advantageous in the previous task for the formal calculation of the expected value.


Questions

1

What is the probability mass function  $P_X(X)$?

$P_X(0) \ = \ $

$P_X(1) \ = \ $

$P_X(2)\ = \ $

$P_X(3) \ = \ $

2

What is the probability mass function  $P_Y(Y)$?

$P_Y(0) \ = \ $

$P_Y(1) \ = \ $

$P_Y(2) \ = \ $

3

Are the random variables  $X$  and  $Y$  statistically independent?

Yes,
No.

4

Determine the probabilities  $P_{UV}( U,\ V)$.

$P_{UV}( U = 0,\ V = 0) \ = \ $

$P_{UV}( U = 0,\ V = 1) \ = \ $

$P_{UV}( U = 1,\ V = 0) \ = \ $

$P_{UV}( U =1,\ V = 1) \ = \ $

5

Are the random variables  $U$  and  $V$  statistically independent?

Yes,
No.


Solution

Solution

(1)  You get from  $P_{XY}(X,\ Y)$  to the one-dimensional probability mass function  $P_X(X)$ by summing up all  $Y$ probabilities:

$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$
  • One thus obtains the following numerical values:
$$P_X(X = 0) = 1/4+1/8+1/8 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
$$P_X(X = 1)= 0+0+1/8 = 1/8 \hspace{0.15cm}\underline{= 0.125},$$
$$P_X(X = 2) = 0+0+0 \hspace{0.15cm}\underline{= 0}$$
$$P_X(X = 3) = 1/4+1/8+0=3/8 \hspace{0.15cm}\underline{= 0.375}\hspace{0.5cm} \Rightarrow \hspace{0.5cm} P_X(X) = \big [ 1/2, \ 1/8 , \ 0 , \ 3/8 \big ].$$


(2)  Analogous to sub-task   (1) , the following now holds:

$$P_Y(Y = y_{\kappa}) = \sum_{x \hspace{0.05cm} \in \hspace{0.05cm} X} \hspace{0.1cm} P_{XY}(x, y_{\kappa})$$
$$P_Y(Y= 0) = 1/4+0+0+1/4 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
$$P_Y(Y = 1) = 1/8+0+0+1/8 = 1/4 \hspace{0.15cm}\underline{= 0.250},$$
$$P_Y(Y = 2) = 1/8+1/8+0+0 = 1/4 \hspace{0.15cm}\underline{= 0.250} \hspace{0.5cm} \Rightarrow \hspace{0.5cm} P_Y(Y= 0) = \big [ 1/2, \ 1/4 , \ 1/4 ].$$


(3)  With statistical independence,  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$  should be.

  • This does not apply here:     answer   NO.


(4)  Starting from the left-hand table   ⇒   $P_{XY}(X,Y)$,  we arrive at the middle table   ⇒   $P_{UY}(U,Y)$,
by combining certain probabilities according to  $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.

If one also takes into account  $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$, one obtains the probabilities sought according to the right-hand table:

Different probability functions
$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
$$P_{UV}( U = 1, V = 0) = 1/8 \hspace{0.15cm}\underline{= 0.125},$$
$$P_{UV}( U = 1, V = 1) = 1/8 \hspace{0.15cm}\underline{= 0.125}.$$


(5)  The correct answer is   YES:

  • The corresponding one-dimensional probability mass functions are:  
$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$
$$P_V(V)=\big [3/4, \ 1/4 \big ].$$
  • Thus:  $P_{UV}(U,V) = P_U(U) \cdot P_V(V)$   ⇒   $U$  and  $V$  are statistically independent.





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