Aufgaben:Exercise 4.1: PDF, CDF and Probability - Revision history

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	[[Category:Information Theory: Exercises\|^4.1 Differential Entropy^]]		[[Category:Information Theory: Exercises\|^4.1 Differential Entropy^]]
	[[de:Aufgaben:4.~~01_WDF~~,~~_VTF_und_Wahrscheinlichkeit~~]]		[[de:Aufgaben:Aufgabe 4.1: WDF, VTF und Wahrscheinlichkeit]]

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	The lower graph shows the probability density function		The lower graph shows the probability density function
	:$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{{20}l} \| y\| \le 2, \\		:$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{{20}l} \| y\| \le 2, \\y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
	y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
	of a continuous random variable  $Y$,  which is restricted to the range  $\|Y\| \le 2$ .		of a continuous random variable  $Y$,  which is restricted to the range  $\|Y\| \le 2$ .
	In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable  $Y$  as for a discrete random variable.  Nevertheless, you will notice some differences in details.		In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable  $Y$  as for a discrete random variable.  Nevertheless, you will notice some differences in details.

	For example, for the continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:		For example, for the continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:
	:$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y)		:$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y)\hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$
	\hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$


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	*The inverse is:   Given the CDF, obtain the PDF by differentiation.		*The inverse is:   Given the CDF, obtain the PDF by differentiation.
	*The given CDF contains five discontinuity points, which after differentiation lead to five Dirac functions:		*The given CDF contains five discontinuity points, which after differentiation lead to five Dirac functions:
	:$$f_X(x) = 0.1 \cdot {\rm \delta}( x+2)		:$$f_X(x) = 0.1 \cdot {\rm \delta}( x+2)+ 0.2 \cdot {\rm \delta}( x+1)+ 0.4 \cdot {\rm \delta}( x) + 0.2 \cdot {\rm \delta}( x-1)+ 0.1 \cdot {\rm \delta}( x-2)\hspace{0.05cm}.$$
	+ 0.2 \cdot {\rm \delta}( x+1)
	+ 0.4 \cdot {\rm \delta}( x) + 0.2 \cdot {\rm \delta}( x-1)
	+ 0.1 \cdot {\rm \delta}( x-2)\hspace{0.05cm}.$$
	*The Dirac weights give the occurrence probabilities of the random variable  $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$ , e.g.:		*The Dirac weights give the occurrence probabilities of the random variable  $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$ , e.g.:
	:$${\rm Pr}(X = 0) = F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-}) =		:$${\rm Pr}(X = 0) = F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-}) =0.7 - 0.3 = 0.4\hspace{0.05cm}.$$
	0.7 - 0.3 = 0.4\hspace{0.05cm}.$$
	*Accordingly, the other probabilities are:		*Accordingly, the other probabilities are:
	:$${\rm Pr}(X = +1) = {\rm Pr}(X = -1) = 0.2\hspace{0.05cm},\hspace{0.3cm}		:$${\rm Pr}(X = +1) = {\rm Pr}(X = -1) = 0.2\hspace{0.05cm},\hspace{0.3cm}{\rm Pr}(X = +2) = {\rm Pr}(X = -2) = 0.1\hspace{0.05cm}.$$
	{\rm Pr}(X = +2) = {\rm Pr}(X = -2) = 0.1\hspace{0.05cm}.$$



	'''(2)'''  From the PDF just calculated, we obtain:		'''(2)'''  From the PDF just calculated, we obtain:
	:$${\rm Pr}(X >0) = {\rm Pr}(X = +1) + {\rm Pr}(X = +2)		:$${\rm Pr}(X >0) = {\rm Pr}(X = +1) + {\rm Pr}(X = +2)\hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$
	\hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$		:$${\rm Pr}(\|X\| \le 1) ={\rm Pr}(X = -1) + {\rm Pr}(X = 0) + {\rm Pr}(X = +1) = 0.2 + 0.4 +0.2\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$
	:$${\rm Pr}(\|X\| \le 1) ={\rm Pr}(X = -1) + {\rm Pr}(X = 0) + {\rm Pr}(X = +1) = 0.2 + 0.4 +0.2
	\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$

	The same result is obtained using the CDF.  Here the general equation, which is equally valid for discrete and continuous random variables, is:		The same result is obtained using the CDF.  Here the general equation, which is equally valid for discrete and continuous random variables, is:
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	*It was taken into account that for the random variable  $Y$  the "<"sign can be replaced by the "≤" sign without distortion.		*It was taken into account that for the random variable  $Y$  the "<"sign can be replaced by the "≤" sign without distortion.
	*With the boundary transition  $\varepsilon \to 0$,  the probability we are looking for is obtained:		*With the boundary transition  $\varepsilon \to 0$,  the probability we are looking for is obtained:
	:$${\rm Pr}(Y = 0) =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm}{\rm Pr}(-\varepsilon \le Y \le +\varepsilon) =		:$${\rm Pr}(Y = 0) =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm}{\rm Pr}(-\varepsilon \le Y \le +\varepsilon) =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(+\varepsilon) - \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(-\varepsilon) =F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$
	\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(+\varepsilon) - \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(-\varepsilon) =
	F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$

	*Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.		*Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.
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	'''(6)'''  We again assume the equation  $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$    valid for the continuous random quantity  $Y$:		'''(6)'''  We again assume the equation  $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$    valid for the continuous random quantity  $Y$:
	*With  $A = 0$  and  $B \to \infty$  $($or  $B = 2)$  we obtain:		*With  $A = 0$  and  $B \to \infty$  $($or  $B = 2)$  we obtain:
	:$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)		:$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)= {\rm Pr}(0 \le Y \le 2) = F_Y(2) - F_Y(0)\hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$
	= {\rm Pr}(0 \le Y \le 2) = F_Y(2) - F_Y(0)
	\hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$
	*Thus, for the symmetric continuous random variable  $Y$  holds indeed as expected:  ${\rm Pr}( Y > 0) = 1/2$.		*Thus, for the symmetric continuous random variable  $Y$  holds indeed as expected:  ${\rm Pr}( Y > 0) = 1/2$.
	*Although the discrete random variable  $X$  is also symmetrical about $x= 0$   ⇒   ${\rm Pr}( X > 0) = 0.3$  was determined in subtask  '''(3)''', on the other hand.		*Although the discrete random variable  $X$  is also symmetrical about $x= 0$   ⇒   ${\rm Pr}( X > 0) = 0.3$  was determined in subtask  '''(3)''', on the other hand.
	*Further, with  $A = -1$  and  $B = +1$,  one obtains because of $F_Y(-1) = 1- F_Y(+1)$:		*Further, with  $A = -1$  and  $B = +1$,  one obtains because of $F_Y(-1) = 1- F_Y(+1)$:
	:$${\rm Pr}( \|Y\| \le 1) = {\rm Pr}(-1 \le Y \le +1)		:$${\rm Pr}( \|Y\| \le 1) = {\rm Pr}(-1 \le Y \le +1)= F_Y(+1) - F_Y(-1) = 2 \cdot F_Y(+1) -1 = 2 \cdot 0.909 -1 \hspace{0.15cm}\underline {= 0.818}. $$
	= F_Y(+1) - F_Y(-1) = 2 \cdot F_Y(+1) -1 = 2 \cdot 0.909 -1 \hspace{0.15cm}\underline {= 0.818}. $$

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	[[Category:Information Theory: Exercises\|^4.1 Differential Entropy^]]		[[Category:Information Theory: Exercises\|^4.1 Differential Entropy^]]
			[[de:Aufgaben:4.01_WDF,_VTF_und_Wahrscheinlichkeit]]

Javier: Text replacement - "value-discrete" to "discrete"

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Javier: Text replacement - "value-continuous" to "continuous"

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Noah at 10:37, 27 September 2021

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	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Differentielle Entropie~~		{{quiz-Header\|Buchseite=Information_Theory/Differential_Entropy
	}}		}}

Guenter at 15:30, 24 September 2021

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Guenter at 15:09, 24 September 2021

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	}}		}}

	[[File:P_ID2862__Inf_A_4_1_neu.png\|right\|frame\|CDF (top) ~~and~~ PDF (bottom)]]		[[File:P_ID2862__Inf_A_4_1_neu.png\|right\|frame\|$\rm (CDF)$  (top),  $\rm (PDF)$  (bottom)]]
	To repeat some important basics from the book "Theory of Stochastic Signals" we are dealing with		To repeat some important basics from the book "Theory of Stochastic Signals" we are dealing with
	* the  [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion\|probability density function]] $\rm (PDF)$,		* the  [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion\|probability density function]] $\rm (PDF)$,

Guenter at 15:07, 24 September 2021

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	<quiz display=simple>		<quiz display=simple>
	{Determine the PDF  $f_X(x)$  of the discrete ~~value~~ random variable  $X$.  Which of the following statements are true?		{Determine the PDF  $f_X(x)$  of the value-discrete random variable  $X$.  Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	+ The PDF is composed of five Dirac functions.		+ The PDF is composed of five Dirac functions.
	+  ${\rm Pr}(X= 0) = 0.4$  and  ${\rm Pr}(X= 1) = 0.2$ are true.		+  ${\rm Pr}(X= 0) = 0.4$   and  ${\rm Pr}(X= 1) = 0.2$  are true.
	-  ${\rm Pr}(X= 2) = 0.4$ is true.		-  ${\rm Pr}(X= 2) = 0.4$  is true.


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	${\rm Pr}(\|X\| ≤ 1) \ = \ $ { 0.8 3% }		${\rm Pr}(\|X\| ≤ 1) \ = \ $ { 0.8 3% }

	{What are the values of the cumulative distribution function  $F_Y(y) ={\rm Pr}(Y \le y)$  of the continuous ~~value~~ random variable  $Y$,  in particular:		{What are the values of the cumulative distribution function  $F_Y(y) ={\rm Pr}(Y \le y)$  of the value-continuous random variable  $Y$,  in particular:
	\|type="{}"}		\|type="{}"}
	$F_Y(y = 0) \ = \ $ { 0.5 3% }		$F_Y(y = 0) \ = \ $ { 0.5 3% }

Guenter at 13:05, 24 September 2021

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	The upper plot shows the cumulative distribution function  $F_X(x)$  of a ~~value-~~discrete random variable  $X$.  The corresponding probability density function  $f_X(x)$  has to be determined in subtask  '''(1)'''.		The upper plot shows the cumulative distribution function  $F_X(x)$  of a discrete random variable  $X$.  The corresponding probability density function  $f_X(x)$  has to be determined in subtask  '''(1)'''.

	The equation		The equation
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	y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$		y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
	of a continuous random variable  $Y$,  which is restricted to the range  $\|Y\| \le 2$ .		of a continuous random variable  $Y$,  which is restricted to the range  $\|Y\| \le 2$ .
	In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable  $Y$  as for a ~~value-~~discrete random variable.  Nevertheless, you will notice some differences in details.		In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable  $Y$  as for a discrete random variable.  Nevertheless, you will notice some differences in details.

	For example, for the continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:		For example, for the continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:
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	<quiz display=simple>		<quiz display=simple>
	{Determine the PDF  $f_X(x)$  of the ~~value-~~discrete random variable  $X$.  Which of the following statements are true?		{Determine the PDF  $f_X(x)$  of the discrete random variable  $X$.  Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	+ The PDF is composed of five Dirac functions.		+ The PDF is composed of five Dirac functions.
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID2857__Inf_A_4_1a_neu.png\|right\|frame\|PDF and CDF of the ~~value-~~discrete random variable  $X$]]		[[File:P_ID2857__Inf_A_4_1a_neu.png\|right\|frame\|PDF and CDF of the discrete random variable  $X$]]
	'''(1)'''  <u>Proposed solutions 1 and 2</u> are correct:		'''(1)'''  <u>Proposed solutions 1 and 2</u> are correct:
	*The cumulative distribution function  $F_X(x)$  is obtained from the probability density function  $f_X(x)$  by integration over the (renamed) random variable in the range from  $- \infty$  to  $x$.		*The cumulative distribution function  $F_X(x)$  is obtained from the probability density function  $f_X(x)$  by integration over the (renamed) random variable in the range from  $- \infty$  to  $x$.
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	\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$		\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$

	The same result is obtained using the CDF.  Here the general equation, which is equally valid for ~~value-~~discrete and continuous random variables, is:		The same result is obtained using the CDF.  Here the general equation, which is equally valid for discrete and continuous random variables, is:
	:$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$		:$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$

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	\hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$		\hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$
	*Thus, for the symmetric continuous random variable  $Y$  holds indeed as expected:  ${\rm Pr}( Y > 0) = 1/2$.		*Thus, for the symmetric continuous random variable  $Y$  holds indeed as expected:  ${\rm Pr}( Y > 0) = 1/2$.
	*Although the ~~value-~~discrete random variable  $X$  is also symmetrical about $x= 0$   ⇒   ${\rm Pr}( X > 0) = 0.3$  was determined in subtask  '''(3)''', on the other hand.		*Although the discrete random variable  $X$  is also symmetrical about $x= 0$   ⇒   ${\rm Pr}( X > 0) = 0.3$  was determined in subtask  '''(3)''', on the other hand.
	*Further, with  $A = -1$  and  $B = +1$,  one obtains because of $F_Y(-1) = 1- F_Y(+1)$:		*Further, with  $A = -1$  and  $B = +1$,  one obtains because of $F_Y(-1) = 1- F_Y(+1)$:
	:$${\rm Pr}( \|Y\| \le 1) = {\rm Pr}(-1 \le Y \le +1)		:$${\rm Pr}( \|Y\| \le 1) = {\rm Pr}(-1 \le Y \le +1)

← Older revision		Revision as of 10:27, 11 October 2021
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	:$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{{20}l} \| y\| \le 2, \\		:$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{{20}l} \| y\| \le 2, \\
	y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$		y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
	of a ~~value-~~continuous random variable  $Y$,  which is restricted to the range  $\|Y\| \le 2$ .		of a continuous random variable  $Y$,  which is restricted to the range  $\|Y\| \le 2$ .
	In principle, the same relationship between PDF, CDF and probabilities exists for the ~~value-~~continuous random variable  $Y$  as for a value-discrete random variable.  Nevertheless, you will notice some differences in details.		In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable  $Y$  as for a value-discrete random variable.  Nevertheless, you will notice some differences in details.

	For example, for the ~~value-~~continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:		For example, for the continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:
	:$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y)		:$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y)
	\hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$		\hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$
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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Information_Theory/Differentielle_Entropie\|Differential Entropy]].		*The exercise belongs to the chapter  [[Information_Theory/Differentielle_Entropie\|Differential Entropy]].
	*Useful hints for solving this problem and further information on ~~value-~~continuous random variables can be found in the third chapter  "Continuous Random Variables"  of the book  [[Theory of Stochastic Signals]].		*Useful hints for solving this problem and further information on continuous random variables can be found in the third chapter  "Continuous Random Variables"  of the book  [[Theory of Stochastic Signals]].

	*Given also is the following indefinite integral:		*Given also is the following indefinite integral:
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	${\rm Pr}(\|X\| ≤ 1) \ = \ $ { 0.8 3% }		${\rm Pr}(\|X\| ≤ 1) \ = \ $ { 0.8 3% }

	{What are the values of the cumulative distribution function  $F_Y(y) ={\rm Pr}(Y \le y)$  of the ~~value-~~continuous random variable  $Y$,  in particular:		{What are the values of the cumulative distribution function  $F_Y(y) ={\rm Pr}(Y \le y)$  of the continuous random variable  $Y$,  in particular:
	\|type="{}"}		\|type="{}"}
	$F_Y(y = 0) \ = \ $ { 0.5 3% }		$F_Y(y = 0) \ = \ $ { 0.5 3% }
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	\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$		\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$

	The same result is obtained using the CDF.  Here the general equation, which is equally valid for value-discrete and ~~value-~~continuous random variables, is:		The same result is obtained using the CDF.  Here the general equation, which is equally valid for value-discrete and continuous random variables, is:
	:$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$		:$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$

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	[[File:P_ID2858__Inf_A_4_1c_neu.png\|right\|frame\|PDF and CDF of the ~~value-~~continuous random variable  $Y$]]		[[File:P_ID2858__Inf_A_4_1c_neu.png\|right\|frame\|PDF and CDF of the continuous random variable  $Y$]]
	'''(3)'''  The cumulative distribution function  $F_Y(y)$  is obtained from the (renamed) WDF  $f_Y(\eta)$  by integrating  $- \infty$  to  $x$.  Due to symmetry, this can be written in the range  $0 \le y \le +2$:		'''(3)'''  The cumulative distribution function  $F_Y(y)$  is obtained from the (renamed) WDF  $f_Y(\eta)$  by integrating  $- \infty$  to  $x$.  Due to symmetry, this can be written in the range  $0 \le y \le +2$:
	:$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$		:$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$
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	'''(4)'''  The probability that the ~~value-~~continuous random variable  $Y$  lies in the range from  $-\varepsilon$  to  $+\varepsilon$  can be calculated using the given equation as follows:		'''(4)'''  The probability that the continuous random variable  $Y$  lies in the range from  $-\varepsilon$  to  $+\varepsilon$  can be calculated using the given equation as follows:
	:$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$		:$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$

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	F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$		F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$

	*Since for a ~~value-~~continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.		*Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.


	'''In general''':   The probability  ${\rm Pr}(Y = y_0)$  that a ~~value-~~continuous random variable  $Y$  takes a fixed value  $y_0$,  is always zero.		'''In general''':   The probability  ${\rm Pr}(Y = y_0)$  that a continuous random variable  $Y$  takes a fixed value  $y_0$,  is always zero.


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	'''(6)'''  We again assume the equation  $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$    valid for the ~~value-~~continuous random quantity  $Y$:		'''(6)'''  We again assume the equation  $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$    valid for the continuous random quantity  $Y$:
	*With  $A = 0$  and  $B \to \infty$  $($or  $B = 2)$  we obtain:		*With  $A = 0$  and  $B \to \infty$  $($or  $B = 2)$  we obtain:
	:$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)		:$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)

← Older revision		Revision as of 17:30, 24 September 2021
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID2857__Inf_A_4_1a_neu.png\|right\|frame\|PDF and CDF of the discrete random variable  $X$]]		[[File:P_ID2857__Inf_A_4_1a_neu.png\|right\|frame\|PDF and CDF of the value-discrete random variable  $X$]]
	'''(1)'''  <u>Proposed solutions 1 and 2</u> are correct:		'''(1)'''  <u>Proposed solutions 1 and 2</u> are correct:
	*The cumulative distribution function ~~(CDF)~~  $F_X(x)$  is obtained from the probability density function  $f_X(x)$  by integration over the (renamed) random variable in the range from  $- \infty$  to  $x$.		*The cumulative distribution function  $F_X(x)$  is obtained from the probability density function  $f_X(x)$  by integration over the (renamed) random variable in the range from  $- \infty$  to  $x$.
	*The inverse is:   ~~given~~ the CDF, obtain the PDF by differentiation.		*The inverse is:   Given the CDF, obtain the PDF by differentiation.
	*The given CDF contains five discontinuity points, which after differentiation lead to five Dirac functions:		*The given CDF contains five discontinuity points, which after differentiation lead to five Dirac functions:
	:$$f_X(x) = 0.1 \cdot {\rm \delta}( x+2)		:$$f_X(x) = 0.1 \cdot {\rm \delta}( x+2)
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	+ 0.4 \cdot {\rm \delta}( x) + 0.2 \cdot {\rm \delta}( x-1)		+ 0.4 \cdot {\rm \delta}( x) + 0.2 \cdot {\rm \delta}( x-1)
	+ 0.1 \cdot {\rm \delta}( x-2)\hspace{0.05cm}.$$		+ 0.1 \cdot {\rm \delta}( x-2)\hspace{0.05cm}.$$
	*The Dirac weights give the occurrence probabilities of the random variable  $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$ an, ~~<br>for example~~:		*The Dirac weights give the occurrence probabilities of the random variable  $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$ , e.g.:
	:$${\rm Pr}(X = 0) = F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-}) =		:$${\rm Pr}(X = 0) = F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-}) =
	0.7 - 0.3 = 0.4\hspace{0.05cm}.$$		0.7 - 0.3 = 0.4\hspace{0.05cm}.$$
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	\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$		\hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$

	The same result is obtained using the ~~distribution function~~.  Here the general equation, which is equally valid for discrete-~~value and~~ continuous~~-value~~ random variables, is:		The same result is obtained using the CDF.  Here the general equation, which is equally valid for value-discrete and value-continuous random variables, is:
	:$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$		:$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$

	* Thus, with  $A= 0$  and  $B = +2$  we obtain:		* Thus, with  $A= 0$  and  $B = +2$  we obtain:
	:$${\rm Pr}(0 < X \le +2) = {\rm Pr}(X >0)= F_X(+2) - F_X(0) = 1 - 0.7 \hspace{0.15cm}\underline {= 0.3} \hspace{0.05cm}.$$		:$${\rm Pr}(0 < X \le +2) = {\rm Pr}(X >0)= F_X(+2) - F_X(0) = 1 - 0.7 \hspace{0.15cm}\underline {= 0.3} \hspace{0.05cm}.$$
	*Setting $A=-2$ and $B = +1$, we get:		*Setting  $A=-2$  and  $B = +1$,  we get:
	:$${\rm Pr}(-2 < X \le +1) = {\rm Pr}(\|X\| \le 1)= F_X(+1) - F_X(-2) = 0.9 - 0.1 \hspace{0.15cm}\underline {= 0.8} \hspace{0.05cm}.$$		:$${\rm Pr}(-2 < X \le +1) = {\rm Pr}(\|X\| \le 1)= F_X(+1) - F_X(-2) = 0.9 - 0.1 \hspace{0.15cm}\underline {= 0.8} \hspace{0.05cm}.$$



	[[File:P_ID2858__Inf_A_4_1c_neu.png\|right\|frame\|PDF and CDF of the continuous random variable  $Y$]]		[[File:P_ID2858__Inf_A_4_1c_neu.png\|right\|frame\|PDF and CDF of the value-continuous random variable  $Y$]]
	'''(3)'''  The cumulative distribution function  $F_Y(y)$  is obtained from the (renamed) WDF  $f_Y(\eta)$  by integrating  $- \infty$  to  $x$.  Due to symmetry, this can be written in the range  $0 \le y \le +2$~~ ~~:		'''(3)'''  The cumulative distribution function  $F_Y(y)$  is obtained from the (renamed) WDF  $f_Y(\eta)$  by integrating  $- \infty$  to  $x$.  Due to symmetry, this can be written in the range  $0 \le y \le +2$:
	:$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$		:$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$
	:$$\Rightarrow \hspace{0.3cm}F_Y(y) = \frac{1}{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{0.1cm}\frac{1}{2} \cdot \cos^2({\pi}/{4} \cdot \eta) \hspace{0.1cm}{\rm d}\eta = \frac{1}{2}+\frac{y}{4} + \frac{1}{2\pi} \cdot \sin({\pi}/{2} \cdot y).$$		:$$\Rightarrow \hspace{0.3cm}F_Y(y) = \frac{1}{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{0.1cm}\frac{1}{2} \cdot \cos^2({\pi}/{4} \cdot \eta) \hspace{0.1cm}{\rm d}\eta = \frac{1}{2}+\frac{y}{4} + \frac{1}{2\pi} \cdot \sin({\pi}/{2} \cdot y).$$
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	'''(4)'''  The probability that the continuous~~-value~~ random variable  $Y$  lies in the range from  $-\varepsilon$  to  $+\varepsilon$  can be calculated using the given equation as follows:		'''(4)'''  The probability that the value-continuous random variable  $Y$  lies in the range from  $-\varepsilon$  to  $+\varepsilon$  can be calculated using the given equation as follows:
	:$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$		:$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$

	*It was taken into account that for the ~~continuous~~ random variable  $Y$  the "<"sign can be replaced by the "≤" sign without distortion.		*It was taken into account that for the random variable  $Y$  the "<"sign can be replaced by the "≤" sign without distortion.
	*With the boundary transition  $\varepsilon \to 0$ , the probability we are looking for is obtained:		*With the boundary transition  $\varepsilon \to 0$,  the probability we are looking for is obtained:
	:$${\rm Pr}(Y = 0) =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm}{\rm Pr}(-\varepsilon \le Y \le +\varepsilon) =		:$${\rm Pr}(Y = 0) =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm}{\rm Pr}(-\varepsilon \le Y \le +\varepsilon) =
	\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(+\varepsilon) - \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(-\varepsilon) =		\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(+\varepsilon) - \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(-\varepsilon) =
	F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$		F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$

	*Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.		*Since for a value-continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.


	'''In general''':   The probability  ${\rm Pr}(Y = y_0)$ that a continuous ~~value~~ random variable  $Y$  takes a fixed value  $y_0$  is always zero.		'''In general''':   The probability  ${\rm Pr}(Y = y_0)$  that a value-continuous random variable  $Y$  takes a fixed value  $y_0$,  is always zero.


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	*Based on the PDF at hand, the result  $Y=3$  can be excluded.		*Based on the PDF at hand, the result  $Y=3$  can be excluded.
	*The result  $Y=0$  on the other hand is quite possible, although  ${\rm Pr}(Y = 0) = 0$ .		*The result  $Y=0$  on the other hand is quite possible, although  ${\rm Pr}(Y = 0) = 0$ .
	*For example, if one performs a random experiment  $N \to \infty$  times and obtains the result  $Y= 0$   $N_0$  times, then with finite   $N_0$  according to the classical definition of probability:		*For example, if one performs a random experiment  $N \to \infty$  times and obtains the result  $Y= 0$   for  $N_0$  times, then with finite   $N_0$  according to the classical definition of probability:
	:$${\rm Pr}(Y = 0) = \lim_{N\hspace{0.05cm}\rightarrow\hspace{0.05cm}\infty}\hspace{0.1cm}{N_0}/{N} = 0\hspace{0.05cm}.$$		:$${\rm Pr}(Y = 0) = \lim_{N\hspace{0.05cm}\rightarrow\hspace{0.05cm}\infty}\hspace{0.1cm}{N_0}/{N} = 0\hspace{0.05cm}.$$


			'''(6)'''  We again assume the equation  $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$    valid for the value-continuous random quantity  $Y$:
	'''(6)'''  We again assume the equation~~  $Y$  valid for the continuous random quantity~~   $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$ :		*With  $A = 0$  and  $B \to \infty$  $($or  $B = 2)$  we obtain:
	*With  $A = 0$  and  $B \to \infty$  $($~~bzw.~~  $B = 2)$  we obtain:
	:$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)		:$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)
	= {\rm Pr}(0 \le Y \le 2) = F_Y(2) - F_Y(0)		= {\rm Pr}(0 \le Y \le 2) = F_Y(2) - F_Y(0)
	\hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$		\hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$
	*Thus, for the symmetric continuous random variable  $Y$  is indeed as expected  ${\rm Pr}( Y > 0) = 1/2$.		*Thus, for the symmetric continuous random variable  $Y$  holds indeed as expected:  ${\rm Pr}( Y > 0) = 1/2$.
	*Although the discrete ~~value~~ random variable  $X$  is also ~~symmetric~~ about $x= 0$ , ${\rm Pr}( X > 0) = 0.3$  was determined in subtask  '''(3)'''~~ ~~, on the other hand.		*Although the value-discrete random variable  $X$  is also symmetrical about $x= 0$   ⇒   ${\rm Pr}( X > 0) = 0.3$  was determined in subtask  '''(3)''', on the other hand.
	*Further, with  $A = -1$  and  $B = +1$ , one obtains because of $F_Y(-1) = 1- F_Y(+1)$:		*Further, with  $A = -1$  and  $B = +1$,  one obtains because of $F_Y(-1) = 1- F_Y(+1)$:
	:$${\rm Pr}( \|Y\| \le 1) = {\rm Pr}(-1 \le Y \le +1)		:$${\rm Pr}( \|Y\| \le 1) = {\rm Pr}(-1 \le Y \le +1)
	= F_Y(+1) - F_Y(-1) = 2 \cdot F_Y(+1) -1 = 2 \cdot 0.909 -1 \hspace{0.15cm}\underline {= 0.818}. $$		= F_Y(+1) - F_Y(-1) = 2 \cdot F_Y(+1) -1 = 2 \cdot 0.909 -1 \hspace{0.15cm}\underline {= 0.818}. $$

← Older revision		Revision as of 15:05, 24 September 2021
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	[[File:P_ID2862__Inf_A_4_1_neu.png\|right\|frame\|CDF (top) and PDF (bottom)]]		[[File:P_ID2862__Inf_A_4_1_neu.png\|right\|frame\|CDF (top) and PDF (bottom)]]
	To repeat some important basics from the book "Theory of Stochastic Signals" we are dealing with		To repeat some important basics from the book "Theory of Stochastic Signals" we are dealing with
	* the  [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion\|probability density function]] (PDF),		* the  [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion\|probability density function]] $\rm (PDF)$,
	* the  [[Theory_of_Stochastic_Signals/~~Verteilungsfunktion~~\|cumulative distribution function]] (CDF).		* the  [[Theory_of_Stochastic_Signals/Cumulative_Distribution_Function_(CDF)\|cumulative distribution function]] $\rm (CDF)$.


	The upper plot shows the distribution function  $F_X(x)$  of a discrete ~~value~~ random variable  $X$.  The corresponding ~~PDF~~  $f_X(x)$  has to be determined in subtask  '''(1)'''~~ ~~.		The upper plot shows the cumulative distribution function  $F_X(x)$  of a value-discrete random variable  $X$.  The corresponding probability density function  $f_X(x)$  has to be determined in subtask  '''(1)'''.

	The equation		The equation
	:$$ {\rm Pr}(A < X \le B) = F_X(B) - F_X(A) = \lim_{\varepsilon \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} \int_{A+\varepsilon}^{B+\varepsilon} \hspace{-0.15cm} f_X(x) \hspace{0.1cm}{\rm d}x $$		:$$ {\rm Pr}(A < X \le B) = F_X(B) - F_X(A) = \lim_{\varepsilon \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} \int_{A+\varepsilon}^{B+\varepsilon} \hspace{-0.15cm} f_X(x) \hspace{0.1cm}{\rm d}x $$

	represents two ways to calculate the probability for the event  ~~„The~~ random variable  $X$  lies in a given ~~interval”~~  from the CDF and the PDF, respectively.		represents two ways to calculate the probability for the event  "The random variable  $X$  lies in a given interval"  from the CDF and the PDF,  respectively.

	The lower graph shows the probability density function		The lower graph shows the probability density function
	:$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{{20}l} \| y\| \le 2, \\		:$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{{20}l} \| y\| \le 2, \\
	y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$		y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
	of a continuous~~-valued~~ random variable  $Y$, which is restricted to the range  $\|Y\| \le 2$ .		of a value-continuous random variable  $Y$,  which is restricted to the range  $\|Y\| \le 2$ .
	In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable  $Y$  as for a discrete random variable.  Nevertheless, you will notice some differences in details.		In principle, the same relationship between PDF, CDF and probabilities exists for the value-continuous random variable  $Y$  as for a value-discrete random variable.  Nevertheless, you will notice some differences in details.

	For example, for the continuous random variable  $Y$ , the boundary transition can be omitted in the above equation, and we obtain simplified:		For example, for the value-continuous random variable  $Y$,  the boundary transition can be omitted in the above equation, and we obtain simplified:
	:$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y)		:$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y)
	\hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$		\hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$
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	Hints:		Hints:
	*The ~~task~~ belongs to the chapter  [[Information_Theory/Differentielle_Entropie\|Differential Entropy]].		*The exercise belongs to the chapter  [[Information_Theory/Differentielle_Entropie\|Differential Entropy]].
	*Useful hints for solving this problem and further information on continuous~~-valued~~ random variables can be found in the third chapter "Continuous Random Variables" of the book  [[Theory of Stochastic Signals]].		*Useful hints for solving this problem and further information on value-continuous random variables can be found in the third chapter  "Continuous Random Variables"  of the book  [[Theory of Stochastic Signals]].

	*Given also is the following indefinite integral:		*Given also is the following indefinite integral: