Aufgaben:Exercise 2.2Z: Discrete Random Variables - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]
	[[de:~~Zusatzaufgaben~~:2.~~2_Diskrete_Zufallsgrößen~~]]		[[de:Aufgaben:Aufgabe 2.2Z: Diskrete Zufallsgrößen]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]
			[[de:Zusatzaufgaben:2.2_Diskrete_Zufallsgrößen]]

Hwang at 12:20, 18 January 2023

2023-01-18T12:20:05Z

← Older revision		Revision as of 14:20, 18 January 2023
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	{What is the ~~quadratic mean~~ value  (second order moment)  $m_{2d}$  of this random variable?		{What is the second moment value  (second order moment)  $m_{2d}$  of this random variable?
	\|type="{}"}		\|type="{}"}
	$m_{2d}\ = \ $ { 2.5 3% }		$m_{2d}\ = \ $ { 2.5 3% }

Guenter at 13:54, 16 February 2022

2022-02-16T13:54:47Z

← Older revision		Revision as of 15:54, 16 February 2022
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	<quiz display=simple>		<quiz display=simple>

	{What is the ~~rms value~~ of the random variable  $a$?		{What is the standard deviation of the random variable  $a$?
	\|type="{}"}		\|type="{}"}
	$\sigma_a \ = \ $ { 1 3% }		$\sigma_a \ = \ $ { 1 3% }


	{What is the ~~rms value~~ of the random variable  $b$?  Set  $p = 0.25$.		{What is the standard deviation of the random variable  $b$?  Set  $p = 0.25$.
	\|type="{}"}		\|type="{}"}
	$\sigma_b \ = \ $ { 0.433 3% }		$\sigma_b \ = \ $ { 0.433 3% }


	{What is the ~~rms value~~ of the random variable  $c$?		{What is the standard deviation of the random variable  $c$?
	\|type="{}"}		\|type="{}"}
	$\sigma_c \ = \ $ { 0.707 3% }		$\sigma_c \ = \ $ { 0.707 3% }
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	{What is the ~~rms value~~  $\sigma_d$?		{What is the standard deviation  $\sigma_d$?
	\|type="{}"}		\|type="{}"}
	$\sigma_d\ = \ $ { 1.5 3% }		$\sigma_d\ = \ $ { 1.5 3% }

Guenter at 13:18, 7 December 2021

2021-12-07T13:18:53Z

← Older revision		Revision as of 15:18, 7 December 2021
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	{What is the ~~root~~ mean ~~square~~ value  $m_{2d}$  of this random variable?		{What is the quadratic mean value  (second order moment)  $m_{2d}$  of this random variable?
	\|type="{}"}		\|type="{}"}
	$m_{2d}\ = \ $ { 2.5 3% }		$m_{2d}\ = \ $ { 2.5 3% }


	{What is the ~~dispersion~~  $\sigma_d$?		{What is the rms value  $\sigma_d$?
	\|type="{}"}		\|type="{}"}
	$\sigma_d\ = \ $ { 1.5 3% }		$\sigma_d\ = \ $ { 1.5 3% }
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	*From this one obtains with Steiner's theorem:		*From this one obtains with Steiner's theorem:
	:$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}or \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$		:$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}or \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$



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	:$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$		:$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$

	*From this follows with  $p = 1/4$:		*From this follows with  $p = 1/4$  and  $k=2$:
	:$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$		:$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$



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	:$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$		:$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$
	:$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$		:$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$



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	'''(5)'''  Analogous to the subtask  '''(4)'''  we obtain for the ~~root mean square~~:		'''(5)'''  Analogous to the subtask  '''(4)'''  we obtain for the second order moment:
	:$$m_{2d}= {\rm E}\big[( a-2b+c)^{\rm 2}\big] = {\rm E}\big[a^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[c^{\rm 2}\big]\hspace{0.1cm} - \hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[a\hspace{0.05cm}\cdot \hspace{0.05cm}b\big]\hspace{0.1cm}+\hspace{0.1cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ a\hspace{0.05cm}\cdot \hspace{0.05cm}c\big]\hspace{0.1cm}-\hspace{0.1cm} 4\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ b\hspace{0.05cm}\cdot \hspace{0.05cm}c\big].$$		:$$m_{2d}= {\rm E}\big[( a-2b+c)^{\rm 2}\big] = {\rm E}\big[a^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[c^{\rm 2}\big]\hspace{0.1cm} - \hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[a\hspace{0.05cm}\cdot \hspace{0.05cm}b\big]\hspace{0.1cm}+\hspace{0.1cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ a\hspace{0.05cm}\cdot \hspace{0.05cm}c\big]\hspace{0.1cm}-\hspace{0.1cm} 4\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ b\hspace{0.05cm}\cdot \hspace{0.05cm}c\big].$$


	*But since  $a$  and  $b$  are statistically independent of each other,  also holds:		*But since  $a$  and  $b$  are statistically independent of each other,  also holds:
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	*The same holds for the other mixed terms.  Therefore, using  $p = 0.25$, we obtain:		*The same holds for the other mixed terms.  Therefore, using  $p = 0.25$, we obtain:
	:$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$		:$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$

Guenter at 16:58, 6 December 2021

2021-12-06T16:58:50Z

← Older revision		Revision as of 18:58, 6 December 2021
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	[[File:P_ID84__Sto_Z_2_2.png\|right\|frame\|Different rectangular signals]]		[[File:P_ID84__Sto_Z_2_2.png\|right\|frame\|Different rectangular signals]]
	Let be given three discrete random variables  $a$,  $b$  and  $c$,  which are defined as the ~~current~~ values of the represented signals.  These have the following properties:		Let be given three discrete random variables  $a$,  $b$  and  $c$,  which are defined as the instantaneous values of the represented signals.  These have the following properties:

	*The random variable  $a$  can take the values  $+1$  and  $-1$  with equal probability.		*The random variable  $a$  can take the two values  $+1$  and  $-1$  with equal probability.
	*The random variable  $b$  is also two-point distributed, but with  ${\rm Pr}(b = 1) = p$  and  ${\rm Pr}(b = 0) = 1 - p$.		*The random variable  $b$  is also two-point distributed,  but with  ${\rm Pr}(b = 1) = p$  and  ${\rm Pr}(b = 0) = 1 - p$.
	*The probabilities of  $c$  be  ${\rm Pr}(c = 0) = 1/2$  and  ${\rm Pr}(c = +1) = Pr(c = -1) =1/4$.		*The probabilities of the random variable  $c$  be  ${\rm Pr}(c = 0) = 1/2$  and  ${\rm Pr}(c = +1) = Pr(c = -1) =1/4$.
	*There are no statistical dependencies between the three random variables  $a$,  $b$  and  $c$~~ ~~ .		*There are no statistical dependencies between the three random variables  $a$,  $b$  and  $c$.
	*Another random variable  $d=a$,  $b$  and  $c$~~  is formed from the random variables ~~ $d=a-2 b+c$~~ .~~		*Another random variable  $d$  is formed from the random variables  $a$,  $b$  and  $c$:
			:$$d=a-2 b+c.$$

			The graph shows sections of these random variables.  It can be seen that  $d$  can take all integer values between  $-4$  and  $+2$ .
	The graph shows sections of these ~~four~~ random variables.  It can be seen that  $d$  can take all integer values between  $-4$  and  $+2$ .


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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße\|Moments of a Discrete Random Variable]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße\|Moments of a Discrete Random Variable]].
			*The topic of this chapter is illustrated with examples in the  (German language)  learning video<br>     [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)\|"Momentenberechnung bei diskreten Zufallsgrößen"]]   ⇒   "Calculating Moments for Discrete-Valued Random Variables"
	*The topic of this chapter is illustrated with examples in the  (German language)  learning video<br> [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)\|Momentenberechnung bei diskreten Zufallsgrößen]]  ~~$\Rightarrow$~~   Calculating Moments for Discrete-Valued Random Variables


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	<quiz display=simple>		<quiz display=simple>

	{What is the ~~dispersion (standard deviation)~~ of the random variable $a$?		{What is the rms value of the random variable  $a$?
	\|type="{}"}		\|type="{}"}
	$\sigma_a \ = \ $ { 1 3% }		$\sigma_a \ = \ $ { 1 3% }


	{What is the ~~dispersion~~ of the random variable  $b$?  Set  $p = 0.25$.		{What is the rms value of the random variable  $b$?  Set  $p = 0.25$.
	\|type="{}"}		\|type="{}"}
	$\sigma_b \ = \ $ { 0.433 3% }		$\sigma_b \ = \ $ { 0.433 3% }


	{What is the ~~spread~~ of the random variable $c$?		{What is the rms value of the random variable  $c$?
	\|type="{}"}		\|type="{}"}
	$\sigma_c \ = \ $ { 0.707 3% }		$\sigma_c \ = \ $ { 0.707 3% }

Noah at 20:33, 3 December 2021

2021-12-03T20:33:24Z

← Older revision		Revision as of 22:33, 3 December 2021
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	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Momente einer diskreten Zufallsgröße~~		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable
	}}		}}

Noah at 20:33, 3 December 2021

2021-12-03T20:33:05Z

← Older revision		Revision as of 22:33, 3 December 2021
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	:$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$		:$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$

	*The maximum variance for  $p = 0.50$ ~~adds up to~~  $\sigma_{\it d}^{\rm 2}=2.50$.		*The maximum variance for  $p = 0.50$ results in  $\sigma_{\it d}^{\rm 2}=2.50$.

	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 20:32, 3 December 2021

2021-12-03T20:32:40Z

← Older revision		Revision as of 22:32, 3 December 2021
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Aufgrund der Symmetrie gilt~~:		'''(1)'''  Due to the symmetry holds:
	:$$\rm \it m_{\it a}=\rm 0; \hspace{0.5cm}\it m_{\rm 2\it a}=\rm 0.5\cdot (-1)^2 + 0.5\cdot (1)^2{ = 1}.$$		:$$\rm \it m_{\it a}=\rm 0; \hspace{0.5cm}\it m_{\rm 2\it a}=\rm 0.5\cdot (-1)^2 + 0.5\cdot (1)^2{ = 1}.$$

	*~~Daraus erhält man mit dem Satz von~~ Steiner:		*From this one obtains with Steiner's theorem:
	:$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}~~bzw.~~ \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$		:$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}or \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$



	'''(2)'''  ~~Allgemein gilt für das Moment~~  $k$–~~ter Ordnung~~:		'''(2)'''  In general, for the  $k$–th order moment:
	:$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$		:$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$

	*~~Daraus folgt mit~~  $p = 1/4$:		*From this follows with  $p = 1/4$:
	:$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$		:$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$



	'''(3)'''  ~~Für die Zufallsgröße~~  $c$  ~~gilt~~:		'''(3)'''  For the random variable  $c$  holds:
	:$$m_{\it c} = 0\hspace{0.3cm} ({\rm ~~symmetrisch~~\hspace{0.1cm}um\hspace{0.1cm}0)},$$		:$$m_{\it c} = 0\hspace{0.3cm} ({\rm symmetric\hspace{0.1cm}um\hspace{0.1cm}0)},$$
	:$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$		:$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$
	:$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$		:$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$
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	'''(4)'''  ~~Nach den allgemeinen Regeln für Erwartungswerte gilt mit~~  $p = 0.25$:		'''(4)'''  According to the general rules for expected values, with  $p = 0.25$:
	:$$m_{\it d} = {\rm E}\big[a-2 b+c\big]= {\rm E}\big[a\big] \hspace{0.1cm} -\hspace{0.1cm}\rm 2 \hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[ c\big] = m_{ a}\hspace{0.1cm}-\hspace{0.1cm}2\hspace{0.05cm}\cdot\hspace{0.05cm} m_{\it b}\hspace{0.1cm}+\hspace{0.1cm} m_{\it c} = 0-2\hspace{0.05cm}\cdot\hspace{0.05cm} p + 0 \hspace{0.15cm} \underline{= -0.5}.$$		:$$m_{\it d} = {\rm E}\big[a-2 b+c\big]= {\rm E}\big[a\big] \hspace{0.1cm} -\hspace{0.1cm}\rm 2 \hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[ c\big] = m_{ a}\hspace{0.1cm}-\hspace{0.1cm}2\hspace{0.05cm}\cdot\hspace{0.05cm} m_{\it b}\hspace{0.1cm}+\hspace{0.1cm} m_{\it c} = 0-2\hspace{0.05cm}\cdot\hspace{0.05cm} p + 0 \hspace{0.15cm} \underline{= -0.5}.$$


			'''(5)'''  Analogous to the subtask  '''(4)'''  we obtain for the root mean square:
			:$$m_{2d}= {\rm E}\big[( a-2b+c)^{\rm 2}\big] = {\rm E}\big[a^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[c^{\rm 2}\big]\hspace{0.1cm} - \hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[a\hspace{0.05cm}\cdot \hspace{0.05cm}b\big]\hspace{0.1cm}+\hspace{0.1cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ a\hspace{0.05cm}\cdot \hspace{0.05cm}c\big]\hspace{0.1cm}-\hspace{0.1cm} 4\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ b\hspace{0.05cm}\cdot \hspace{0.05cm}c\big].$$

	~~'''(5)'''  Analog zur Teilaufgabe  '''(4)'''  erhält man für den quadratischen Mittelwert:~~
	:$$m_{2d}= {\rm E}\big[( a-2b+c)^{\rm 2}\big] = {\rm E}\big[a^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[c^{\rm 2}\big]\hspace{0.1cm} - \hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[a\hspace{0.05cm}\cdot \hspace{0.05cm}b\big]\hspace{0.1cm}+\hspace{0.1cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ a\hspace{0.05cm}\cdot \hspace{0.05cm}c\big]\hspace{0.1cm}-\hspace{0.1cm} 4\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ b\hspace{0.05cm}\cdot \hspace{0.05cm}c\big].$$

	*~~Da aber~~  $a$  ~~und~~  $b$  ~~statistisch voneinander unabhängig sind~~,  ~~gilt auch~~:		*But since  $a$  and  $b$  are statistically independent of each other,  also holds:
	:$${\rm E}\big[a\cdot b\big] = {\rm E}\big[ a\big] \cdot {\rm E}\big[ b\big]= m_{ a}\cdot m_{ b} = 0, \hspace{0.2cm} {\rm da}\hspace{0.2cm} m_{ a}=\rm 0.$$		:$${\rm E}\big[a\cdot b\big] = {\rm E}\big[ a\big] \cdot {\rm E}\big[ b\big]= m_{ a}\cdot m_{ b} = 0, \hspace{0.2cm} {\rm da}\hspace{0.2cm} m_{ a}=\rm 0.$$

	*~~Gleiches gilt für die anderen gemischten Terme~~.  ~~Daher erhält man mit~~  $p = 0.25$:		*The same holds for the other mixed terms.  Therefore, using  $p = 0.25$, we obtain:
	:$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$		:$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$



	'''(6)'''  ~~Für allgemeines~~  $p$  ~~bzw~~.  ~~für~~  $p = 0.25$  ~~ergibt sich~~:		'''(6)'''  For general  $p$  resp.  for  $p = 0.25$  results:
	:$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$		:$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$

	*~~Die maximale Varianz ergäbe sich für~~  $p = 0.50$  ~~ zu~~  $\sigma_{\it d}^{\rm 2}=2.50$.		*The maximum variance for  $p = 0.50$ adds up to  $\sigma_{\it d}^{\rm 2}=2.50$.

	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 20:25, 3 December 2021

2021-12-03T20:25:16Z

← Older revision		Revision as of 22:25, 3 December 2021
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	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße\|Moments of a Discrete Random Variable]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße\|Moments of a Discrete Random Variable]].

	**The topic of this chapter is illustrated with examples in the  (German language)  learning video<br> [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)\|Momentenberechnung bei diskreten Zufallsgrößen]]  $\Rightarrow$   Calculating Moments for Discrete-Valued Random Variables		*The topic of this chapter is illustrated with examples in the  (German language)  learning video<br> [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)\|Momentenberechnung bei diskreten Zufallsgrößen]]  $\Rightarrow$   Calculating Moments for Discrete-Valued Random Variables