Aufgaben:Exercise 1.6Z: Ergodic Probabilities - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^1.4 Markov Chains		[[Category:Theory of Stochastic Signals: Exercises\|^1.4 Markov Chains
	^]]		^]]
	[[de:~~Zusatzaufgaben~~:1.~~6_Ergodische_Wahrscheinlichkeiten~~]]		[[de:Aufgaben:Aufgabe 1.6Z: Ergodische Wahrscheinlichkeiten]]

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	'''(3)'''  No statement is made about the time  $\nu-1$ .		'''(3)'''  No statement is made about the time  $\nu-1$ .
	*At this time  $A$  or  $B$  may have occurred. Therefore holds:		*At this time  $A$  or  $B$  may have occurred. Therefore holds:
	:$${\rm Pr}(B_{\nu} \hspace{0.05cm} \| \hspace{0.05cm}A_{\nu -2}) = {\rm Pr}(A \hspace{0.05cm} \| \hspace{0.05cm}A) \hspace{0.05cm} \cdot \hspace{0.05cm}{\rm Pr}(B \hspace{0.05cm} \| \hspace{0.05cm}A) \hspace{0.15cm} +\hspace{0.15cm} {\rm Pr}(B \hspace{0.05cm} \| \hspace{0.05cm}A) \hspace{0.05cm} \cdot \hspace{0.05cm}{\rm Pr}(B \hspace{0.05cm} \| \hspace{0.05cm}B) = p \hspace{0.1cm} \cdot \hspace{0.1cm} (1-p) + q \hspace{0.1cm} \cdot \hspace{0.1cm} (1-p)		:$${\rm Pr}(B_{\nu} \hspace{0.05cm} \| \hspace{0.05cm}A_{\nu -2}) = {\rm Pr}(A \hspace{0.05cm} \| \hspace{0.05cm}A) \hspace{0.05cm} \cdot \hspace{0.05cm}{\rm Pr}(B \hspace{0.05cm} \| \hspace{0.05cm}A) \hspace{0.15cm} +\hspace{0.15cm} {\rm Pr}(B \hspace{0.05cm} \| \hspace{0.05cm}A) \hspace{0.05cm} \cdot \hspace{0.05cm}{\rm Pr}(B \hspace{0.05cm} \| \hspace{0.05cm}B) = p \hspace{0.1cm} \cdot \hspace{0.1cm} (1-p) + q \hspace{0.1cm} \cdot \hspace{0.1cm} (1-p)= {5}/{12} \hspace{0.15cm}\underline {\approx 0.417}.$$
	= {5}/{12} \hspace{0.15cm}\underline {\approx 0.417}.$$



	'''(4)'''  According to Bayes' theorem:		'''(4)'''  According to Bayes' theorem:
	:$${\rm Pr}(A_{\nu -2} \hspace{0.05cm} \| \hspace{0.05cm}B_{\nu}) = \frac{{\rm Pr}(B_{\nu} \hspace{0.05cm} \| \hspace{0.05cm}A_{\nu -2}) \cdot {\rm Pr}(A_{\nu -2} ) }{{\rm Pr}(B_{\nu}) } = \frac{5/12 \cdot 4/7 }{3/7 }		:$${\rm Pr}(A_{\nu -2} \hspace{0.05cm} \| \hspace{0.05cm}B_{\nu}) = \frac{{\rm Pr}(B_{\nu} \hspace{0.05cm} \| \hspace{0.05cm}A_{\nu -2}) \cdot {\rm Pr}(A_{\nu -2} ) }{{\rm Pr}(B_{\nu}) } = \frac{5/12 \cdot 4/7 }{3/7 }= {5}/{9} \hspace{0.15cm}\underline {\approx 0.556}.$$
	= {5}/{9} \hspace{0.15cm}\underline {\approx 0.556}.$$
	Reasoning:		Reasoning:
	*The probability  ${\rm Pr}(B_{\nu}\hspace{0.05cm}\|\hspace{0.05cm}A_{\nu-2})= 5/12$  has already been calculated in subsection  '''(3)'''.		*The probability  ${\rm Pr}(B_{\nu}\hspace{0.05cm}\|\hspace{0.05cm}A_{\nu-2})= 5/12$  has already been calculated in subsection  '''(3)'''.

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	[[Category:Theory of Stochastic Signals: Exercises\|^1.4 Markov Chains		[[Category:Theory of Stochastic Signals: Exercises\|^1.4 Markov Chains
	^]]		^]]
			[[de:Zusatzaufgaben:1.6_Ergodische_Wahrscheinlichkeiten]]

Guenter at 17:07, 1 December 2021

2021-12-01T17:07:20Z

Guenter at 17:01, 1 December 2021

2021-12-01T17:01:01Z

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	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Markov_Chains}}		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Markov_Chains}}

	[[File:P_ID452__Sto_Z_1_6.png\|right\|frame\|~~Binary~~ Markov chain with  $A$~~  and~~  $B$]]		[[File:P_ID452__Sto_Z_1_6.png\|right\|frame\|Markov chain with  $A$,  $B$]]
	We consider a homogeneous stationary first-order Markov chain with events  $A$  and  $B$  and transition probabilities corresponding to the adjacent Markov diagram:		We consider a homogeneous stationary first-order Markov chain with events  $A$  and  $B$  and transition probabilities corresponding to the adjacent Markov diagram:

	For subtasks  '''(1)'''  to  '''(4)''' , assume:		For subtasks  '''(1)'''  to  '''(4)''',  assume:

	*Event   $A$  is followed by  $A$  and  $B$  with equal probability.		*Event  $A$  is followed by  $A$  and  $B$  with equal probability.

	*After  $B$ , event  $A$  is twice as likely as  $B$.		*After  $B$:  The event  $A$  is twice as likely as  $B$.


	From subtask  '''(5)'''  on,  $p$  and  $q$  are free parameters, while the ~~event~~ probabilities  ${\rm Pr}(A) = 2/3$  and  ${\rm Pr}(B) = 1/3$  are fixed.		From subtask  '''(5)'''  on,  $p$  and  $q$  are free parameters,  while the ergodic probabilities  ${\rm Pr}(A) = 2/3$  and  ${\rm Pr}(B) = 1/3$  are fixed.


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	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Markov_Chains\|Markov Chains]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Markov_Chains\|Markov Chains]].

	*You can check your results with the ~~interactive applet~~  [[Applets:Markovketten\|~~Event probabilities of a 1st order~~ Markov ~~chain~~]] .		*You can check your results with the   (German language)  interactive SWF applet
			: [[Applets:Markovketten\|Ereigniswahrscheinlichkeiten einer Markov-Kette erster Ordnung]]   ⇒   "Event Probabilities of a First Order Markov Chain".


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	${\rm Pr}(B) \ = \ $ { 0.429 3% }		${\rm Pr}(B) \ = \ $ { 0.429 3% }

	{What is the conditional probability that event  $B$  occurs if event  $A$  occurred two ~~bars~~ before?		{What is the conditional probability that event  $B$  occurs if event  $A$  occurred two steps before?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(B_{\nu}\hspace{0.05cm}\|\hspace{0.05cm}A_{\nu-2})\ = \ $ { 0.417 3% }		${\rm Pr}(B_{\nu}\hspace{0.05cm}\|\hspace{0.05cm}A_{\nu-2})\ = \ $ { 0.417 3% }

	{What is the inferential probability that event  $A$  occurred two ~~bars~~ before when event  $B$  currently occurs?		{What is the inferential probability that event  $A$  occurred two steps before,  when event  $B$  currently occurs?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(A_{\nu-2}\hspace{0.05cm}\|\hspace{0.05cm}B_{\nu})\ = \ $ { 0.556 3% }		${\rm Pr}(A_{\nu-2}\hspace{0.05cm}\|\hspace{0.05cm}B_{\nu})\ = \ $ { 0.556 3% }
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	$q\ = \ $ { 0. }		$q\ = \ $ { 0. }

	{How must the parameters be chosen so that the sequence elements of the Markov chain are statistically independent and additionally  ${\rm Pr}(A) = 2/3$  ~~gilt~~?		{How must the parameters be chosen so that the sequence elements of the Markov chain are statistically independent and additionally  ${\rm Pr}(A) = 2/3$ ?
	\|type="{}"}		\|type="{}"}
	$p \ = \ $ { 0.667 3% }		$p \ = \ $ { 0.667 3% }

Noah at 13:08, 29 November 2021

2021-11-29T13:08:37Z

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Guenter at 14:17, 29 October 2021

2021-10-29T14:17:34Z

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	[[Category:Theory of Stochastic Signals: Exercises\|^1.4 ~~Markovketten~~		[[Category:Theory of Stochastic Signals: Exercises\|^1.4 Markov Chains
	^]]		^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 1.6Z: Ergodische Wahrscheinlichkeiten to Aufgaben:Exercise 1.6Z: Ergodic Probabilities

2021-10-29T14:14:12Z

Guenter moved page Aufgaben:Aufgabe 1.6Z: Ergodische Wahrscheinlichkeiten to Aufgaben:Exercise 1.6Z: Ergodic Probabilities

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Javier: Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

2021-03-23T12:44:57Z

Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

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	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^1.4 Markovketten		[[Category:Theory of Stochastic Signals: Exercises\|^1.4 Markovketten
	^]]		^]]

Javier: Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

2020-07-09T09:49:53Z

Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

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	''Hinweise:''		''Hinweise:''
	*Die Aufgabe gehört zum Kapitel  [[~~Stochastische_Signaltheorie~~/Markovketten\|Markovketten]].		*Die Aufgabe gehört zum Kapitel  [[Theory_of_Stochastic_Signals/Markovketten\|Markovketten]].

	*Sie können Ihre Ergebnisse mit dem interaktiven Applet  [[Applets:Markovketten\|Ereigniswahrscheinlichkeiten einer Markovkette 1. Ordnung]]  überprüfen.		*Sie können Ihre Ergebnisse mit dem interaktiven Applet  [[Applets:Markovketten\|Ereigniswahrscheinlichkeiten einer Markovkette 1. Ordnung]]  überprüfen.

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  According to the instruction,   $p = 1 - p$   ⇒   $\underline{p =0.500}$  and  $q = (1 - q)/2$,   ⇒   $\underline{q =0.333}$ holds.		'''(1)'''  According to the instruction,   $p = 1 - p$   ⇒   $\underline{p =0.500}$  and  $q = (1 - q)/2$,   ⇒   $\underline{q =0.333}$  holds.


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	:$${\rm Pr}(A_{\nu -2} \hspace{0.05cm} \| \hspace{0.05cm}B_{\nu}) = \frac{{\rm Pr}(B_{\nu} \hspace{0.05cm} \| \hspace{0.05cm}A_{\nu -2}) \cdot {\rm Pr}(A_{\nu -2} ) }{{\rm Pr}(B_{\nu}) } = \frac{5/12 \cdot 4/7 }{3/7 }		:$${\rm Pr}(A_{\nu -2} \hspace{0.05cm} \| \hspace{0.05cm}B_{\nu}) = \frac{{\rm Pr}(B_{\nu} \hspace{0.05cm} \| \hspace{0.05cm}A_{\nu -2}) \cdot {\rm Pr}(A_{\nu -2} ) }{{\rm Pr}(B_{\nu}) } = \frac{5/12 \cdot 4/7 }{3/7 }
	= {5}/{9} \hspace{0.15cm}\underline {\approx 0.556}.$$		= {5}/{9} \hspace{0.15cm}\underline {\approx 0.556}.$$
	''Reasoning:''		Reasoning:
	*The probability  ${\rm Pr}(B_{\nu}\hspace{0.05cm}\|\hspace{0.05cm}A_{\nu-2})= 5/12$  has already been calculated in subsection  '''(3)'''~~ ~~.		*The probability  ${\rm Pr}(B_{\nu}\hspace{0.05cm}\|\hspace{0.05cm}A_{\nu-2})= 5/12$  has already been calculated in subsection  '''(3)'''.
	*Due to stationarity,  ${\rm Pr}(A_{\nu-2})= {\rm Pr}(A) = 4/7$  and  ${\rm Pr}(B_{\nu})= {\rm Pr}(B) = 3/7$ holds.		*Due to stationarity,  ${\rm Pr}(A_{\nu-2})= {\rm Pr}(A) = 4/7$  and  ${\rm Pr}(B_{\nu})= {\rm Pr}(B) = 3/7$  holds.
	*Thus, the value of  $5/9$ is obtained for the sought inference probability according to the above equation..		*Thus,  the value of  $5/9$ is obtained for the sought inference probability according to the above equation.



	'''(5)'''  According to subtask  '''(2)''' , with  ${p =1/2}$  for the probability of  $A$  in general:		'''(5)'''  According to subtask  '''(2)'''  with  ${p =1/2}$  for the probability of  $A$  in general:
	:$${\rm Pr}(A) = \frac{1-q}{1.5 -q}.$$		:$${\rm Pr}(A) = \frac{1-q}{1.5 -q}.$$
	*Thus from  $ {\rm Pr}(A) = 2/3$  follows  $\underline{q =0}$.		*Thus from  $ {\rm Pr}(A) = 2/3$  follows  $\underline{q =0}$.
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	'''(6)'''  In the case of statistical independence, for example, it must hold:		'''(6)'''  In the case of statistical independence,  for example,  it must hold:
	:$${{\rm Pr}(A \hspace{0.05cm} \| \hspace{0.05cm}A)} = {{\rm Pr}(A \hspace{0.05cm} \| \hspace{0.05cm}B)} = {{\rm Pr}(A)}.$$		:$${{\rm Pr}(A \hspace{0.05cm} \| \hspace{0.05cm}A)} = {{\rm Pr}(A \hspace{0.05cm} \| \hspace{0.05cm}B)} = {{\rm Pr}(A)}.$$
	*From this follows  $p = {\rm Pr}(A) \hspace{0.15cm}\underline {= 2/3}$  and accordingly  $q = 1-p \hspace{0.15cm}\underline {= 1/3}$.		*From this follows  $p = {\rm Pr}(A) \hspace{0.15cm}\underline {= 2/3}$  and accordingly  $q = 1-p \hspace{0.15cm}\underline {= 1/3}$.