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

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	[[Category:Information Theory: Exercises\|^3.1 General Information on 2D Random Variables^]]		[[Category:Information Theory: Exercises\|^3.1 General Information on 2D Random Variables^]]
	[[de:Aufgaben:3.~~02Z_2D–Wahrscheinlichkeitsfunktion~~]]		[[de:Aufgaben:Aufgabe 3.2Z: 2D–Wahrscheinlichkeitsfunktion]]

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	[[Category:Information Theory: Exercises\|^3.1 General Information on 2D Random Variables^]]		[[Category:Information Theory: Exercises\|^3.1 General Information on 2D Random Variables^]]
			[[de:Aufgaben:3.02Z_2D–Wahrscheinlichkeitsfunktion]]

Noah at 08:12, 24 September 2021

2021-09-24T08:12:58Z

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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
	}}		}}

Guenter at 10:12, 17 August 2021

2021-08-17T10:12:38Z

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	Hints:		<u>Hints:</u>
	*The exercise belongs to the chapter  [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen\|Some preliminary remarks on two-dimensional random variables]].		*The exercise belongs to the chapter  [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen\|Some preliminary remarks on two-dimensional random variables]].
	*The same constellation is assumed here as in  [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen\|Exercise 3.2]].		*The same constellation is assumed here as in  [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen\|Exercise 3.2]].
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  You get from  $P_{XY}(X,\ Y)$  to the 1D probability function  $P_X(X)$ by summing up all  $Y$ probabilities:		'''(1)'''  You get from  $P_{XY}(X,\ Y)$  to the one-dimensional probability mass function  $P_X(X)$ by summing up all  $Y$ probabilities:
	:$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$		:$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$
	*One thus obtains the following numerical values:		*One thus obtains the following numerical values:
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	'''(3)'''  With statistical independence,  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$  should be.		'''(3)'''  With statistical independence,  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$  should be.
	*This does not apply here:     answer   <u>'''no'''</u>.		*This does not apply here:     answer   <u>'''NO'''</u>.



	'''(4)'''  Starting from the left-hand table   ⇒   $P_{XY}(X,Y)$  , we arrive at the middle table   ⇒   $P_{UY}(U,Y)$, <br>by combining certain probabilities according to  $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$~~ ~~.		'''(4)'''  Starting from the left-hand table   ⇒   $P_{XY}(X,Y)$,  we arrive at the middle table   ⇒   $P_{UY}(U,Y)$, <br>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:
	[[File:P_ID2753__Inf_Z_3_2d_neu.png\|right\|frame\|Different probability functions]]		[[File:P_ID2753__Inf_Z_3_2d_neu.png\|right\|frame\|Different probability functions]]

	~~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:~~
	:$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$		:$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
	:$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$		:$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
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	'''(5)'''  The correct answer is   <u>'''~~yes~~'''</u>:		'''(5)'''  The correct answer is   <u>'''YES'''</u>:
	*The corresponding 1D probability functions are:		*The corresponding one-dimensional probability mass functions are:
	:$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$		:$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$
	:$$P_V(V)=\big [3/4, \ 1/4 \big ].$$		:$$P_V(V)=\big [3/4, \ 1/4 \big ].$$

Guenter at 10:04, 17 August 2021

2021-08-17T10:04:10Z

← Older revision		Revision as of 12:04, 17 August 2021
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	}}		}}

	[[File:P_ID2752__Inf_Z_3_2_neu.png\|right\|frame\|~~Probability function~~ of the 2D random variable  $XY$]]		[[File:P_ID2752__Inf_Z_3_2_neu.png\|right\|frame\|$\rm PMF$ 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 function  $P_{XY}(X,\ Y)$  is given.		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 2D probability function, the one-dimensional probability functions  $P_X(X)$  and  $P_Y(Y)$  are to be determined.		*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 1D probability function is sometimes also called marginal probability.		*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 ~~ties~~ between them.		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:		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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	Hints:		Hints:
	*The ~~task~~ belongs to the chapter  [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen\|Some preliminary remarks on 2D random variables]].		*The exercise belongs to the chapter  [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen\|Some preliminary remarks on two-dimensional random variables]].
	*The same constellation is assumed here as in  [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen\|Exercise 3.2]].		*The same constellation is assumed here as in  [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen\|Exercise 3.2]].
	*There the random variables  $Y = \{ 0,\ 1,\ 2,\ 3 \}$  were considered, but with the addition  ${\rm Pr}(Y = 3) = 0$.		*There the random variables  $Y = \{ 0,\ 1,\ 2,\ 3 \}$  were considered, but with the addition  ${\rm Pr}(Y = 3) = 0$.
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	<quiz display=simple>		<quiz display=simple>

	{What is the probability function  $P_X(X)$?		{What is the probability mass function  $P_X(X)$?
	\|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% }

	{What is the probability function  $P_Y(Y)$?		{What is the probability mass function  $P_Y(Y)$?
	\|type="{}"}		\|type="{}"}
	$P_Y(0) \ = \ $ { 0.5 3% }		$P_Y(0) \ = \ $ { 0.5 3% }

Noah at 21:06, 11 August 2021

2021-08-11T21:06:23Z

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Guenter at 13:35, 16 July 2021

2021-07-16T13:35:35Z

← Older revision		Revision as of 15:35, 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^]]

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

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

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2021-03-23T12:45:51Z

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

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	[[Category:~~Aufgaben zu Informationstheorie~~\|^3.1 Allgemeines zu 2D-Zufallsgrößen^]]		[[Category:Information Theory: Exercises\|^3.1 Allgemeines zu 2D-Zufallsgrößen^]]