Aufgaben:Exercise 3.2: Expected Value Calculations - 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.~~02_Erwartungswertberechnungen~~]]		[[de:Aufgaben:Aufgabe 3.2: Erwartungswertberechnungen]]

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	Often, for such discrete random variables, one must have to calculate different expected values of the form		Often, for such discrete random variables, one must have to calculate different expected values of the form
	:$${\rm E} \big [ F(X)\big ] =\hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm}\hspace{-0.03cm} {\rm supp} (P_X)} \hspace{-0.1cm}		:$${\rm E} \big [ F(X)\big ] =\hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm}\hspace{-0.03cm} {\rm supp} (P_X)} \hspace{-0.1cm}P_{X}(x) \cdot F(x). $$
	P_{X}(x) \cdot F(x). $$

	Here, denote:		Here, denote:
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  In general, the following applies to the expected value of the function  $F(X)$  with respect to the random variable  $X$:		'''(1)'''  In general, the following applies to the expected value of the function  $F(X)$  with respect to the random variable  $X$:
	:$${\rm E} \left [ F(X)\right ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} {\rm supp} (P_X)} \hspace{-0.2cm}		:$${\rm E} \left [ F(X)\right ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} {\rm supp} (P_X)} \hspace{-0.2cm}P_{X}(x) \cdot F(x) \hspace{0.05cm}.$$
	P_{X}(x) \cdot F(x) \hspace{0.05cm}.$$
	In the present example,  $X = \{0,\ 1,\ 2,\ 3\}$  and  $P_X(X) = \big [1/2, \ 1/8, \ 0, \ 3/8\big ]$.		In the present example,  $X = \{0,\ 1,\ 2,\ 3\}$  and  $P_X(X) = \big [1/2, \ 1/8, \ 0, \ 3/8\big ]$.
	*Because of  $P_X(X = 2) = 0$ , the quantity to be taken into account  (the "support")  in the above summation thus results in		*Because of  $P_X(X = 2) = 0$ , the quantity to be taken into account  (the "support")  in the above summation thus results in
	:$${\rm supp} (P_X) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 3 \} \hspace{0.05cm}.$$		:$${\rm supp} (P_X) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 3 \} \hspace{0.05cm}.$$
	*With  $F(X) = 1/P_X(X)$  one further obtains:		*With  $F(X) = 1/P_X(X)$  one further obtains:
	:$${\rm E} \big [ 1/P_X(X)\big ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.4cm} P_{X}(x) \cdot {1}/{P_X(x)}		:$${\rm E} \big [ 1/P_X(X)\big ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.4cm} P_{X}(x) \cdot {1}/{P_X(x)}= \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} 1\hspace{0.15cm}\underline{ = 3} \hspace{0.05cm}.$$
	= \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} 1
	\hspace{0.15cm}\underline{ = 3} \hspace{0.05cm}.$$
	*The second expected value gives the same result with  ${\rm supp} (P_Y) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 2 \} $ :		*The second expected value gives the same result with  ${\rm supp} (P_Y) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 2 \} $ :
	:$${\rm E} \left [ 1/P_Y(Y)\right ] \hspace{0.15cm}\underline{ = 3}.$$		:$${\rm E} \left [ 1/P_Y(Y)\right ] \hspace{0.15cm}\underline{ = 3}.$$
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	'''(2)'''  In both cases, the index of the probability mass function is identical with the random variable  $(X$  or   $Y)$  and we obtain		'''(2)'''  In both cases, the index of the probability mass function is identical with the random variable  $(X$  or   $Y)$  and we obtain
	:$${\rm E} \big [ P_X(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_X(x)}		:$${\rm E} \big [ P_X(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_X(x)}= (1/2)^2 + (1/8)^2 + (3/8)^2 = 13/32\hspace{0.15cm}\underline{ \approx 0.406} \hspace{0.05cm},$$
	= (1/2)^2 + (1/8)^2 + (3/8)^2 = 13/32		:$${\rm E} \big [ P_Y(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_Y(y) \cdot P_Y(y) = (1/2)^2 + (1/4)^2 + (1/4)^2\hspace{0.15cm}\underline{ = 0.375} \hspace{0.05cm}.$$
	\hspace{0.15cm}\underline{ \approx 0.406} \hspace{0.05cm},$$
	:$${\rm E} \big [ P_Y(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_Y(y) \cdot P_Y(y) = (1/2)^2 + (1/4)^2 + (1/4)^2
	\hspace{0.15cm}\underline{ = 0.375} \hspace{0.05cm}.$$



	'''(3)'''  The following equations apply here:		'''(3)'''  The following equations apply here:
	:$${\rm E} \big [ P_Y(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_Y(x)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{4} + \frac{3}{8} \cdot 0 = 9/32		:$${\rm E} \big [ P_Y(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_Y(x)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{4} + \frac{3}{8} \cdot 0 = 9/32\hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm},$$
	\hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm},$$
	*The expected value formation here refers to  $P_X(·)$, i.e. to the random variable  $X$.		*The expected value formation here refers to  $P_X(·)$, i.e. to the random variable  $X$.
	*$P_Y(·)$ is the formal function without (direct) reference to the random variable  $Y$.		*$P_Y(·)$ is the formal function without (direct) reference to the random variable  $Y$.
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	'''(4)'''  We first calculate the three expected values:		'''(4)'''  We first calculate the three expected values:
	:$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_U(U)\big ]		:$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_U(U)\big ]= \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{1} + \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{1} \hspace{0.15cm}\underline{ = 1\ {\rm bit}} \hspace{0.05cm},$$
	= \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{1} + \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{1} \hspace{0.15cm}\underline{ = 1\ {\rm bit}} \hspace{0.05cm},$$		:$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_V(V)\big ]= \frac{3}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{3} + \frac{1}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{1} \hspace{0.15cm}\underline{ = 0.811\ {\rm bit}} \hspace{0.05cm},$$
	:$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_V(V)\big ]		:$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_V(U)\big ]= \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{3} + \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{1} \hspace{0.15cm}\underline{ = 1.208\ {\rm bit}} \hspace{0.05cm}.$$
	= \frac{3}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{3} + \frac{1}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{1} \hspace{0.15cm}\underline{ = 0.811\ {\rm bit}} \hspace{0.05cm},$$
	:$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_V(U)\big ]
	= \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{3} + \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{1} \hspace{0.15cm}\underline{ = 1.208\ {\rm bit}} \hspace{0.05cm}.$$
	Accordingly, the <u>first two statements</u> are correct:		Accordingly, the <u>first two statements</u> are correct:
	* The entropy  $H(U) = 1$  bit  can be calculated according to the first equation.  It applies to the binary random variable  $U$  with equal probabilities.		* The entropy  $H(U) = 1$  bit  can be calculated according to the first equation.  It applies to the binary random variable  $U$  with equal probabilities.

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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:3.02_Erwartungswertberechnungen]]

Noah at 08:11, 24 September 2021

2021-09-24T08:11:36Z

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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 09:35, 17 August 2021

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Guenter at 09:23, 17 August 2021

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	}}		}}

	[[File:P_ID2751__Inf_A_3_2.png\|right\|frame\|2D probability function]]		[[File:P_ID2751__Inf_A_3_2.png\|right\|frame\|Two-dimensional <br>probability mass function]]
	We consider the following probability functions:		We consider the following probability mass functions  $\rm (PMF)$:

	:$$P_X(X) = \big[1/2,\ 1/8,\ 0,\ 3/8 \big],$$		:$$P_X(X) = \big[1/2,\ 1/8,\ 0,\ 3/8 \big],$$
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	: $X= \{0,\ 1,\ 2,\ 3\}$,     $Y= \{0,\ 1,\ 2,\ 3\}$,    $U = \{0,\ 1\}$,     $V = \{0, 1\}$.		: $X= \{0,\ 1,\ 2,\ 3\}$,     $Y= \{0,\ 1,\ 2,\ 3\}$,    $U = \{0,\ 1\}$,     $V = \{0, 1\}$.

	Often, for such discrete random variables, one must have different expected values of the form		Often, for such discrete random variables, one must have to calculate different expected values of the form
	:$${\rm E} \big [ F(X)\big ] =\hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm}\hspace{-0.03cm} {\rm supp} (P_X)} \hspace{-0.1cm}		:$${\rm E} \big [ F(X)\big ] =\hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm}\hspace{-0.03cm} {\rm supp} (P_X)} \hspace{-0.1cm}
	P_{X}(x) \cdot F(x) $$		P_{X}(x) \cdot F(x). $$

	~~calculate.~~ Here, denote:		Here, denote:

	* $P_X(X)$  denotes the ~~<i>~~probability function~~</i>~~ of the discrete random variable   $X$.		* $P_X(X)$  denotes the probability mass function of the discrete random variable   $X$.
	* The  ~~<i>~~support~~</i>~~  of  $P_X$  includes all those realisations  $x$  of the random variable  $X$  with non-vanishing probability.		* The  "support"  of  $P_X$  includes all those realisations  $x$  of the random variable  $X$  with non-vanishing probability.
	*Formally, this can be written		*Formally, this can be written as
	:$${\rm supp} (P_X) = \{ x: \hspace{0.25cm}x \in X \hspace{0.15cm}\underline{\rm and} \hspace{0.15cm} P_X(x) \ne 0 \} \hspace{0.05cm}.$$		:$${\rm supp} (P_X) = \{ x: \hspace{0.25cm}x \in X \hspace{0.15cm}\underline{\rm and} \hspace{0.15cm} P_X(x) \ne 0 \} \hspace{0.05cm}.$$
	* $F(X)$  is an (arbitrary) real-valued function that can be specified in the entire domain of definition of the random variable  $X$ .		* $F(X)$  is an (arbitrary) real-valued function that can be specified in the entire domain of definition of the random variable  $X$ .
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	In the task, the expected values for various functions  $F(X)$  are to be calculated, among others for		In the task, the expected values for various functions  $F(X)$  are to be calculated, among others for

	: $F(X)= 1/P_X(X)$,		#  $F(X)= 1/P_X(X)$,
	: $F(X)= P_X(X)$,		#  $F(X)= P_X(X)$,
	: $F(X)= - \log_2 \ P_X(X)$.		#  $F(X)= - \log_2 \ P_X(X)$.


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	Hints:		Hints:
	*The exercise 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 two 1D probability functions  $P_X(X)$  and  $P_Y(Y)$  result from the presented 2D–PMF  $P_{XY}(X,\ Y)$, as will be shown in  [[Aufgaben:3.2Z_2D–Wahrscheinlichkeitsfunktion\|Exercise 3.2Z]]~~ ~~.		* The two one-dimensional probability mass functions  $P_X(X)$  and  $P_Y(Y)$  result from the presented 2D–PMF  $P_{XY}(X,\ Y)$,  as will be shown in  [[Aufgaben:3.2Z_2D–Wahrscheinlichkeitsfunktion\|Exercise 3.2Z]].
	* The binary probability functions  $P_U(U)$  and  $P_V(V)$  are obtained according to the modulo operations  $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm}2$  and  $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.		* The binary probability mass functions  $P_U(U)$  and  $P_V(V)$  are obtained according to the modulo operations  $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm}2$  and  $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.

Noah at 20:42, 10 August 2021

2021-08-10T20:42:31Z

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	* The  <i>support</i>  of  $P_X$  includes all those realisations  $x$  of the random variable  $X$  with non-vanishing probability.		* The  <i>support</i>  of  $P_X$  includes all those realisations  $x$  of the random variable  $X$  with non-vanishing probability.
	*Formally, this can be written		*Formally, this can be written
	:$${\rm supp} (P_X) = \{ x: \hspace{0.25cm}x \in X \hspace{0.15cm}\underline{\rm ~~und~~} \hspace{0.15cm} P_X(x) \ne 0 \} \hspace{0.05cm}.$$		:$${\rm supp} (P_X) = \{ x: \hspace{0.25cm}x \in X \hspace{0.15cm}\underline{\rm and} \hspace{0.15cm} P_X(x) \ne 0 \} \hspace{0.05cm}.$$
	* $F(X)$  is an (arbitrary) real-valued function that can be specified in the entire domain of definition of the random variable  $X$ .		* $F(X)$  is an (arbitrary) real-valued function that can be specified in the entire domain of definition of the random variable  $X$ .

Noah at 20:42, 10 August 2021

2021-08-10T20:42:09Z

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	* The entropy  $H(U) = 1$  bit  can be calculated according to the first equation.  It applies to the binary random variable  $U$  with equal probabilities.		* The entropy  $H(U) = 1$  bit  can be calculated according to the first equation.  It applies to the binary random variable  $U$  with equal probabilities.
	* The entropy  $H(V) = 0.811$  bit  is calculated according to the second equation.  Due to the probabilities  $3/4$  and  $1/4$ , the entropy (uncertainty) is smaller here than for the random variable  $U$.		* The entropy  $H(V) = 0.811$  bit  is calculated according to the second equation.  Due to the probabilities  $3/4$  and  $1/4$ , the entropy (uncertainty) is smaller here than for the random variable  $U$.
	* The third expected value cannot indicate the entropy of a binary random variable, which is always limited to  $1$  (bit)&nbsp, simply because of the result   $(1.208$  bit$)$ .		* The third expected value cannot indicate the entropy of a binary random variable, which is always limited to  $1$  (bit) , simply because of the result   $(1.208$  bit$)$ .
	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 20:41, 10 August 2021

2021-08-10T20:41:37Z

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Guenter at 13:31, 16 July 2021

2021-07-16T13:31:59Z

← Older revision		Revision as of 15:31, 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^]]

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	:$${\rm supp} (P_X) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 3 \} \hspace{0.05cm}.$$		:$${\rm supp} (P_X) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 3 \} \hspace{0.05cm}.$$
	*With  $F(X) = 1/P_X(X)$  one further obtains:		*With  $F(X) = 1/P_X(X)$  one further obtains:
	:$${\rm E} \big [ 1/P_X(X)\big ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm}, 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.4cm} P_{X}(x) \cdot {1}/{P_X(x)}		:$${\rm E} \big [ 1/P_X(X)\big ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.4cm} P_{X}(x) \cdot {1}/{P_X(x)}
	= \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm}, 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} 1		= \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} 1
	\hspace{0.15cm}\underline{ = 3} \hspace{0.05cm}.$$		\hspace{0.15cm}\underline{ = 3} \hspace{0.05cm}.$$
	*The second expected value gives the same result with  ${\rm supp} (P_Y) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 2 \} $ :		*The second expected value gives the same result with  ${\rm supp} (P_Y) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 2 \} $ :
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	'''(2)'''  In both cases, the index of the probability function is identical with the random variable  $(X$  or   $Y)$  and we obtain		'''(2)'''  In both cases, the index of the probability mass function is identical with the random variable  $(X$  or   $Y)$  and we obtain
	:$${\rm E} \big [ P_X(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm}, 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_X(x)}		:$${\rm E} \big [ P_X(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_X(x)}
	= (1/2)^2 + (1/8)^2 + (3/8)^2 = 13/32		= (1/2)^2 + (1/8)^2 + (3/8)^2 = 13/32
	\hspace{0.15cm}\underline{ \approx 0.406} \hspace{0.05cm},$$		\hspace{0.15cm}\underline{ \approx 0.406} \hspace{0.05cm},$$
	:$${\rm E} \big [ P_Y(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm}, 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_Y(y) \cdot P_Y(y) = (1/2)^2 + (1/4)^2 + (1/4)^2		:$${\rm E} \big [ P_Y(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_Y(y) \cdot P_Y(y) = (1/2)^2 + (1/4)^2 + (1/4)^2
	\hspace{0.15cm}\underline{ = 0.375} \hspace{0.05cm}.$$		\hspace{0.15cm}\underline{ = 0.375} \hspace{0.05cm}.$$

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	'''(3)'''  The following equations apply here:		'''(3)'''  The following equations apply here:
	:$${\rm E} \big [ P_Y(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm}, 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_Y(x)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{4} + \frac{3}{8} \cdot 0 = 9/32		:$${\rm E} \big [ P_Y(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_Y(x)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{4} + \frac{3}{8} \cdot 0 = 9/32
	\hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm},$$		\hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm},$$
	*The expected value formation here refers to  $P_X(·)$, i.e. to the random variable  $X$.		*The expected value formation here refers to  $P_X(·)$, i.e. to the random variable  $X$.
	*$P_Y(·)$ is the formal function without (direct) reference to the random variable  $Y$.		*$P_Y(·)$ is the formal function without (direct) reference to the random variable  $Y$.
			*The same numerical value is obtained for the second expected value  (this does not have to be the case in general):
	*The same numerical value is obtained for the second expected value  (this does not have to be the case):		:$${\rm E} \big [ P_X(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_{Y}(y) \cdot {P_X(y)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{8} + \frac{1}{4} \cdot 0 = 9/32 \hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm}.$$
	:$${\rm E} \big [ P_X(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm}, 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_{Y}(y) \cdot {P_X(y)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{8} + \frac{1}{4} \cdot 0 = 9/32 \hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm}.$$