Aufgaben:Exercise 3.2Z: Two-dimensional Probability Mass Function - Revision history

Noah at 08:12, 24 September 2021

2021-09-24T08:12:58Z

Guenter at 10:12, 17 August 2021

2021-08-17T10:12:38Z

Guenter at 10:04, 17 August 2021

2021-08-17T10:04:10Z

Noah at 21:06, 11 August 2021

2021-08-11T21:06:23Z

Guenter at 13:35, 16 July 2021

2021-07-16T13:35:35Z

Guenter: Guenter moved page Aufgaben:Exercize 3.2Z: Two-dimensional Probability Mass Function to Aufgaben:Exercise 3.2Z: Two-dimensional Probability Mass Function

2021-07-16T13:35:04Z

Guenter moved page Exercize 3.2Z: Two-dimensional Probability Mass Function to Exercise 3.2Z: Two-dimensional Probability Mass Function

Guenter: Guenter moved page Aufgaben:Aufgabe 3.2Z: 2D–Wahrscheinlichkeitsfunktion to Aufgaben:Exercize 3.2Z: Two-dimensional Probability Mass Function

2021-07-16T13:12:15Z

Guenter moved page Aufgabe 3.2Z: 2D–Wahrscheinlichkeitsfunktion to Exercize 3.2Z: Two-dimensional Probability Mass Function

Javier: Text replacement - "Category:Aufgaben zu Informationstheorie" to "Category:Information Theory: Exercises"

2021-03-23T12:45:51Z

Text replacement - "Category:Aufgaben zu Informationstheorie" to "Category:Information Theory: Exercises"

Javier: Text replacement - "[[Informationstheorie" to "[[Information_Theory"

2020-07-09T12:45:29Z

Text replacement - "[[Informationstheorie" to "[[Information_Theory"

Guenter at 13:51, 30 January 2020

2020-01-30T13:51:49Z

← Older revision		Revision as of 08:12, 24 September 2021
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−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Einige Vorbemerkungen zu zweidimensionalen Zufallsgrößen~~	+	{{quiz-Header\|Buchseite=Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables
	}}		}}

@@ Line 22: / Line 22: @@
-Hints:
+<u>Hints:</u>
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on two-dimensional random variables]].
 *The same constellation is assumed here as in&nbsp; [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen|Exercise 3.2]].
@@ Line 72: / Line 72: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; You get from&nbsp;  $P_{XY}(X,\ Y)$&nbsp; to the 1D probability function&nbsp; $P_X(X)$ by summing up all&nbsp; $Y$ probabilities:
+'''(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:
 :$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$
 *One thus obtains the following numerical values:
@@ Line 91: / Line 91: @@
 '''(3)'''&nbsp; With statistical independence,&nbsp;  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$&nbsp; should be.
-*This does not apply here: &nbsp; &nbsp;  answer &nbsp; <u>'''no'''</u>.
+*This does not apply here: &nbsp; &nbsp;  answer &nbsp; <u>'''NO'''</u>.
-'''(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; $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$&nbsp;.
+'''(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; $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.
 [[File:P_ID2753__Inf_Z_3_2d_neu.png|right|frame|Different probability functions]]
 :$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
 :$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
@@ Line 107: / Line 107: @@
-'''(5)'''&nbsp; The correct answer is &nbsp; <u>'''yes'''</u>:
+'''(5)'''&nbsp; The correct answer is &nbsp; <u>'''YES'''</u>:
-*The corresponding 1D probability functions are:   &nbsp;
+*The corresponding one-dimensional probability mass functions are:   &nbsp;
 :$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$
 :$$P_V(V)=\big [3/4, \ 1/4 \big ].$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID2752__Inf_Z_3_2_neu.png|right|frame|Probability function of the 2D random variable&nbsp; $XY$]]
+[[File:P_ID2752__Inf_Z_3_2_neu.png|right|frame|$\rm PMF$ of the two-dimensional  random variable&nbsp; $XY$]]
-We consider the random variables&nbsp; $X =  \{ 0,\ 1,\ 2,\ 3 \}$&nbsp; and&nbsp; $Y =  \{ 0,\ 1,\ 2 \}$, whose joint probability function&nbsp; $P_{XY}(X,\ Y)$&nbsp; is given.
+We consider the random variables&nbsp; $X =  \{ 0,\ 1,\ 2,\ 3 \}$&nbsp; and&nbsp; $Y =  \{ 0,\ 1,\ 2 \}$, whose joint probability mass function&nbsp; $P_{XY}(X,\ Y)$&nbsp; is given.
-*From this 2D probability function, the one-dimensional probability functions&nbsp; $P_X(X)$&nbsp; and&nbsp; $P_Y(Y)$&nbsp; are to be determined.
+*From this two-dimensional  probability mass function&nbsp; $\rm (PMF)$,&nbsp; the one-dimensional probability mass functions&nbsp; $P_X(X)$&nbsp; and&nbsp; $P_Y(Y)$&nbsp; are to be determined.
-*Such a 1D probability function is sometimes also called marginal probability.
+*Such a one-dimensional probability mass function is sometimes also called&nbsp; "marginal probability".
-If &nbsp;$P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are statistically independent.&nbsp; Otherwise, there are statistical ties between them.
+If &nbsp;$P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are statistically independent.&nbsp; Otherwise, there are statistical dependencies between them.
-In the second part of the task we consider the random variables&nbsp; $U=  \big \{ 0,\ 1 \big \}$&nbsp; and&nbsp; $V= \big  \{ 0,\ 1 \big \}$, which result from&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; by modulo-2 operations:
+In the second part of the task we consider the random variables&nbsp; $U=  \big \{ 0,\ 1 \big \}$&nbsp; and&nbsp; $V= \big  \{ 0,\ 1 \big \}$,&nbsp; which result from&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; by modulo-2 operations:
 :$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm}   V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$
@@ Line 23: / Line 23: @@
 Hints:
-*The task belongs to the chapter&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on 2D random variables]].
+*The exercise belongs to the chapter&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on two-dimensional random variables]].
 *The same constellation is assumed here as in&nbsp; [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen|Exercise 3.2]].
 *There the random variables&nbsp;  $Y = \{ 0,\ 1,\ 2,\ 3 \}$&nbsp;  were considered, but with the addition&nbsp; ${\rm Pr}(Y = 3) = 0$.
@@ Line 34: / Line 34: @@
 <quiz display=simple>
-{What is the probability function&nbsp; $P_X(X)$?
+{What is the probability mass function&nbsp; $P_X(X)$?
 |type="{}"}
 $P_X(0) \ = \ $ { 0.5 3% }
@@ Line 41: / Line 41: @@
 $P_X(3) \ = \ ${ 0.375 3% }
-{What is the probability function&nbsp; $P_Y(Y)$?
+{What is the probability mass function&nbsp; $P_Y(Y)$?
 |type="{}"}
 $P_Y(0) \ = \ $ { 0.5 3% }

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID2752__Inf_Z_3_2_neu.png|right|frame|Wahrscheinlichkeitsfunktion der 2D–Zufallsgröße&nbsp; $XY$]]
+[[File:P_ID2752__Inf_Z_3_2_neu.png|right|frame|Probability function of the 2D random variable&nbsp; $XY$]]
-Wir betrachten die Zufallsgrößen&nbsp; $X =  \{ 0,\ 1,\ 2,\ 3 \}$&nbsp; und&nbsp; $Y =  \{ 0,\ 1,\ 2 \}$, deren gemeinsame Wahrscheinlichkeitsfunktion&nbsp; $P_{XY}(X,\ Y)$&nbsp; gegeben ist.
+We consider the random variables&nbsp; $X =  \{ 0,\ 1,\ 2,\ 3 \}$&nbsp; and&nbsp; $Y =  \{ 0,\ 1,\ 2 \}$, whose joint probability function&nbsp; $P_{XY}(X,\ Y)$&nbsp; is given.
-*Aus dieser 2D–Wahrscheinlichkeitsfunktion sollen die eindimensionalen Wahrscheinlichkeitsfunktionen&nbsp; $P_X(X)$&nbsp; und&nbsp; $P_Y(Y)$&nbsp; ermittelt werden.
+*From this 2D probability function, the one-dimensional probability functions&nbsp; $P_X(X)$&nbsp; and&nbsp; $P_Y(Y)$&nbsp; are to be determined.
-*Man nennt eine solche 1D–Wahrscheinlichkeitsfunktion manchmal auch Randwahrscheinlichkeit&nbsp; (englisch:&nbsp; ''Marginal Probability'').
+*Such a 1D probability function is sometimes also called marginal probability.
-Gilt  &nbsp;$P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, so sind die beiden Zufallsgrößen&nbsp; $X$&nbsp; und&nbsp; $Y$&nbsp; statistisch unabhängig.&nbsp; Andernfalls bestehen zwischen diesen statistische Bindungen.
+If &nbsp;$P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are statistically independent.&nbsp; Otherwise, there are statistical ties between them.
-Im zweiten Teil der Aufgabe betrachten wir die Zufallsgrößen&nbsp; $U=  \big \{ 0,\ 1 \big \}$&nbsp; und&nbsp; $V= \big  \{ 0,\ 1 \big \}$, die sich aus&nbsp; $X$&nbsp; und&nbsp; $Y$&nbsp; durch Modulo–2–Operationen ergeben:
+In the second part of the task we consider the random variables&nbsp; $U=  \big \{ 0,\ 1 \big \}$&nbsp; and&nbsp; $V= \big  \{ 0,\ 1 \big \}$, which result from&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; by modulo-2 operations:
 :$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm}   V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$
@@ Line 22: / Line 22: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Einige Vorbemerkungen zu den 2D-Zufallsgrößen]].
+*The task belongs to the chapter&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on 2D random variables]].
-*Ausgegangen wird hier von der gleichen Konstellation wie in&nbsp; [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen|Aufgabe 3.2]].
+*The same constellation is assumed here as in&nbsp; [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen|Exercise 3.2]].
-*Dort wurde die Zufallsgrößen&nbsp;  $Y = \{ 0,\ 1,\ 2,\ 3 \}$&nbsp;  betrachtet, allerdings mit dem Zusatz&nbsp; ${\rm Pr}(Y = 3) = 0$.
+*There the random variables&nbsp;  $Y = \{ 0,\ 1,\ 2,\ 3 \}$&nbsp;  were considered, but with the addition&nbsp; ${\rm Pr}(Y = 3) = 0$.
-*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.
+*The property&nbsp; $|X| = |Y|$&nbsp; forced in this way was advantageous in the previous task for the formal calculation of the expected value.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie lautet die Wahrscheinlichkeitsfunktion&nbsp; $P_X(X)$?
+{What is the probability function&nbsp; $P_X(X)$?
 |type="{}"}
 $P_X(0) \ = \ $ { 0.5 3% }
@@ Line 41: / Line 41: @@
 $P_X(3) \ = \ ${ 0.375 3% }
-{Wie lautet die Wahrscheinlichkeitsfunktion&nbsp; $P_Y(Y)$?
+{What is the probability function&nbsp; $P_Y(Y)$?
 |type="{}"}
 $P_Y(0) \ = \ $ { 0.5 3% }
@@ Line 47: / Line 47: @@
 $P_Y(2) \ = \ $ { 0.25 3% }
-{Sind die Zufallsgrößen&nbsp; $X$&nbsp; und&nbsp; $Y$&nbsp; statistisch unabhängig?
+{Are the random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; statistically independent?
 |type="()"}
-- Ja,
+- Yes,
-+ Nein.
++ No.
-{Ermitteln Sie die Wahrscheinlichkeiten&nbsp; $P_{UV}( U,\ V)$.
+{Determine the probabilities&nbsp; $P_{UV}( U,\ V)$.
 |type="{}"}
 $P_{UV}( U = 0,\ V = 0) \ = \ $ { 0.375 3% }
@@ Line 60: / Line 61: @@
 $P_{UV}( U =1,\ V = 1) \ = \ $ { 0.125 3% }
-{Sind die Zufallsgrößen&nbsp; $U$&nbsp; und&nbsp; $V$&nbsp; statistisch unabhängig?
+{Are the random variables&nbsp; $U$&nbsp; and&nbsp; $V$&nbsp; statistically independent?
 |type="()"}
-+ Ja,
++ Yes,
-- Nein.
+- No.
@@ Line 69: / Line 70: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(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:
+'''(1)'''&nbsp; You get from&nbsp;  $P_{XY}(X,\ Y)$&nbsp; to the 1D probability function&nbsp; $P_X(X)$ by summing up all&nbsp; $Y$ probabilities:
 :$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$
-*Man erhält somit folgende Zahlenwerte:
+*One thus obtains the following numerical values:
 :$$P_X(X = 0) = 1/4+1/8+1/8 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
 :$$P_X(X = 1)= 0+0+1/8 =  1/8 \hspace{0.15cm}\underline{= 0.125},$$
@@ Line 81: / Line 82: @@
-'''(2)'''&nbsp; Analog zur Teilaufgabe&nbsp; '''(1)'''&nbsp; gilt nun:
+'''(2)'''&nbsp; Analogous to sub-task &nbsp; '''(1)'''&nbsp;, the following now holds:
 :$$P_Y(Y = y_{\kappa}) = \sum_{x \hspace{0.05cm} \in \hspace{0.05cm} X} \hspace{0.1cm} P_{XY}(x, y_{\kappa})$$
 :$$P_Y(Y= 0) = 1/4+0+0+1/4 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$
@@ Line 89: / Line 90: @@
-'''(3)'''&nbsp; Bei statistischer Unabhängigkeit sollte&nbsp;  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$&nbsp; sein.
+'''(3)'''&nbsp; With statistical independence,&nbsp;  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$&nbsp; should be.
-*Dies trifft hier nicht zu: &nbsp; &nbsp;  Antwort &nbsp; <u>'''Nein'''</u>.
+*This does not apply here: &nbsp; &nbsp;  answer &nbsp; <u>'''no'''</u>.
-'''(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.
+'''(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; $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$&nbsp;.
-[[File:P_ID2753__Inf_Z_3_2d_neu.png|right|frame|Verschiedene Wahrscheinlichkeitsfunktionen]]
+[[File:P_ID2753__Inf_Z_3_2d_neu.png|right|frame|Different probability functions]]
-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:
+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:
 :$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
 :$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{=  0.375},$$
@@ Line 106: / Line 107: @@
-'''(5)'''&nbsp; Richtig ist die Antwort &nbsp; <u>'''Ja'''</u>:
+'''(5)'''&nbsp; The correct answer is &nbsp; <u>'''yes'''</u>:
-*Die zugehörigen 1D–Wahrscheinlichkeitsfunktionen lauten:  &nbsp;
+*The corresponding 1D probability functions are:   &nbsp;
 :$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$
 :$$P_V(V)=\big [3/4, \ 1/4 \big ].$$
-*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.
+*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.

← Older revision		Revision as of 13:35, 16 July 2021
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−	[[Category:Information Theory: Exercises\|^3.1 ~~Allgemeines zu~~ 2D~~-Zufallsgrößen~~^]]	+	[[Category:Information Theory: Exercises\|^3.1 General Information on 2D Random Variables^]]

← Older revision		Revision as of 12:45, 23 March 2021
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−	[[Category:~~Aufgaben zu Informationstheorie~~\|^3.1 Allgemeines zu 2D-Zufallsgrößen^]]	+	[[Category:Information Theory: Exercises\|^3.1 Allgemeines zu 2D-Zufallsgrößen^]]