Aufgaben:Exercise 1.7Z: BARBARA Generator - 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.~~7_BARBARA~~-Generator]]		[[de:Aufgaben:Aufgabe 1.7Z: BARBARA-Generator]]

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	:$${\rm Pr}(\rm BARBARA) = p_{\rm A} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$		:$${\rm Pr}(\rm BARBARA) = p_{\rm A} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$
	*This leads to the result:		*This leads to the result:
	:$${\rm Pr}(\rm BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right)		:$${\rm Pr}(\rm BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right)= \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q}= \hspace{-0.15cm} \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3}\hspace{0.15cm}\underline { \approx 0.244 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$
	= \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q}
	= \hspace{-0.15cm} \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3}
	\hspace{0.15cm}\underline { \approx 0.244 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$

	'''(4)'''  The probability calculated in  '''(3)'''  is  $p^5 \cdot (1-p)/3$,  where  $q= 1-p$  is considered.		'''(4)'''  The probability calculated in  '''(3)'''  is  $p^5 \cdot (1-p)/3$,  where  $q= 1-p$  is considered.

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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:1.7_BARBARA-Generator]]

Guenter at 16:29, 2 December 2021

2021-12-02T16:29:21Z

Guenter at 13:44, 2 December 2021

2021-12-02T13:44:16Z

← Older revision		Revision as of 15:44, 2 December 2021
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	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Markov_Chains}}		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Markov_Chains}}

	[[File:P_ID454__Sto_Z_1_7.png\|right\|frame\|$BARBARA$ Generator]]		[[File:P_ID454__Sto_Z_1_7.png\|right\|frame\|$\rm BARBARA$  Generator]]
	Here we consider a ternary random generator with symbols  $A$,  $B$  and  $R$, which can be described by a homogeneous and stationary first order Markov chain.		Here we consider a ternary random generator with symbols  $A$,  $B$  and  $R$,  which can be described by a homogeneous and stationary first order Markov chain.

	*The transition probabilities can be taken from the sketched Markov diagram.		*The transition probabilities can be taken from the sketched Markov diagram.
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	Hints:		Hints:
	*The ~~task~~ 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]].


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	\|type="[]"}		\|type="[]"}
	- The values of  $p > 0$  and  $q < 1$  are largely arbitrary.		- The values of  $p > 0$  and  $q < 1$  are largely arbitrary.
	+ For the transition probabilities, the following must hold:   $p + q = 1$.		+ For the transition probabilities,  the following must hold:   $p + q = 1$.
	+ All symbols have equal ergodic probabilities.		+ All symbols have equal ergodic probabilities.
	- It holds here:  ${\rm Pr}(A) = 1/2, \; {\rm Pr}(B) = 1/3, \; {\rm Pr}(R) = 1/6$.		- It holds here:  ${\rm Pr}(A) = 1/2, \; {\rm Pr}(B) = 1/3, \; {\rm Pr}(R) = 1/6$.

	{What are the conditional probabilities  $p_{\rm A}$,  $p_{\rm B}$  and  $p_{\rm C}$ that at times between  $ν+1$  and  $ν+7$  the sequence  $BARBARA$  is output, <br>if one is in state ~~im Zustand~~  $A$,  $B$  or  $R$ , respectively~~,at time  $ν$ ~~?  Let  $p = 1/4$.		{What are the conditional probabilities  $p_{\rm A}$,  $p_{\rm B}$  and  $p_{\rm C}$ that at times between  $ν+1$  and  $ν+7$  the sequence  "$\rm BARBARA$"  is output, <br>if one is at time  $ν$  in state  $A$,  $B$  or  $R$,  respectively?  Let  $p = 1/4$.
	\|type="{}"}		\|type="{}"}
	$p_{\rm A} \ = \ $ { 0.549 3% } $\ \cdot 10^{-3}$		$p_{\rm A} \ = \ $ { 0.549 3% } $\ \cdot 10^{-3}$
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	$p_{\rm C} \ = \ $ { 0.183 3% } $\ \cdot 10^{-3}$		$p_{\rm C} \ = \ $ { 0.183 3% } $\ \cdot 10^{-3}$

	{What is the overall probability that the generator outputs the sequence  $BARBARA$  ~~ausgibt?<br>~~ Let  $p = 1/4$ continue to hold.		{What is the overall probability that the generator outputs the sequence  "$\rm BARBARA$"?  Let  $p = 1/4$ continue to hold.
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(BARBARA)\ = \ $ { 0.244 3% } $\ \cdot 10^{-3}$		${\rm Pr}(\rm BARBARA)\ = \ $ { 0.244 3% } $\ \cdot 10^{-3}$

	{How should the parameter  $p_{\rm opt}$  be chosen to make  ${\rm Pr}(BARBARA)$  as large as possible? <br>What is the resulting probability for  $BARBARA$?		{How should the parameter  $p_{\rm opt}$  be chosen to make  ${\rm Pr}(\rm BARBARA)$  as large as possible? <br>What is the resulting probability for  "$\rm BARBARA$"?
	\|type="{}"}		\|type="{}"}
	$p_{\rm opt} \ = \ $ { 0.8333 3% }		$p_{\rm opt} \ = \ $ { 0.8333 3% }
	$p = p_{\rm opt}\hspace{-0.1cm}: \hspace{0.3cm}{\rm Pr}(BARBARA)\ = \ $ { 22 3% } $\ \cdot 10^{-3}$		$p = p_{\rm opt}\hspace{-0.1cm}: \hspace{0.3cm}{\rm Pr}(\rm BARBARA)\ = \ $ { 22 3% } $\ \cdot 10^{-3}$

	</quiz>		</quiz>

Noah at 14:33, 29 November 2021

2021-11-29T14:33:22Z

← Older revision		Revision as of 16:33, 29 November 2021
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	'''(3)'''  By averaging over the conditional probabilities we obtain:		'''(3)'''  By averaging over the conditional probabilities we obtain:
	:$${\rm Pr}(BARBARA) = p_{\rm A} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$		:$${\rm Pr}(BARBARA) = p_{\rm A} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$
	This leads to the result:		*This leads to the result:
	:$${\rm Pr}(BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right)		:$${\rm Pr}(BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right)
	= \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q}		= \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q}

Noah at 14:32, 29 November 2021

2021-11-29T14:32:35Z

Show changes

Guenter at 14:18, 29 October 2021

2021-10-29T14:18:25Z

← Older revision		Revision as of 16:18, 29 October 2021
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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 1.7Z: BARBARA-Generator to Aufgaben:Exercise 1.7Z: BARBARA Generator

2021-10-29T14:16:21Z

Guenter moved page Aufgaben:Aufgabe 1.7Z: BARBARA-Generator to Aufgaben:Exercise 1.7Z: BARBARA Generator

← Older revision		Revision as of 16:16, 29 October 2021
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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"

← Older revision		Revision as of 14:44, 23 March 2021
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	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^1.4 Markovketten		[[Category:Theory of Stochastic Signals: Exercises\|^1.4 Markovketten
	^]]		^]]

← Older revision		Revision as of 18:29, 2 December 2021
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	*The transition probabilities can be taken from the sketched Markov diagram.		*The transition probabilities can be taken from the sketched Markov diagram.
	*For the first three subtasks,  $p = 1/4$  should always hold.		*For the first three subtasks,  $p = 1/4$  should always hold.






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	===Musterlösung===		===Musterlösung===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct are <u>the second and third suggested solutions</u>:		'''(1)'''  Correct are  <u>the second and third suggested solutions</u>:
	*The sum of all outgoing arrows must always be  $1$ . Therefore  $q = 1 - p$ holds.		*The sum of all outgoing arrows must always be  $1$.  Therefore  $q = 1 - p$ holds.
	*Because of the symmetry of the Markov diagram, the ergodic probabilities are all equal:		*Because of the symmetry of the Markov diagram,  the ergodic probabilities are all equal:
	:$${\rm Pr}(A) ={\rm Pr}(B) ={\rm Pr}(R) = 1/3.$$		:$${\rm Pr}(A) ={\rm Pr}(B) ={\rm Pr}(R) = 1/3.$$


	'''(2)'''  If one is in the state  $B$  at the starting time  $\nu=1$  because of  ${\rm Pr}(B\hspace{0.05cm}\|\hspace{0.05cm}B) = 0$ the state  $B$  is not possible.		'''(2)'''  If one is in the state  $B$  at the starting time  $\nu=0$,  because of  ${\rm Pr}(B\hspace{0.05cm}\|\hspace{0.05cm}B) = 0$  the state  $B$  is not possible at time  $\nu=1$.
			*One fails here already with the initial letter  $B$:

	*One fails here already with the initial letter $B$:
	:$$p_{\rm B} \; \underline{ =0}.$$		:$$p_{\rm B} \; \underline{ =0}.$$

	*For the calculation of  $p_{\rm A}$  it should be noted:   Starting from  $A$  one goes in the Markov diagram first to  $B$  $($with probability $q)$, then five times clockwise  $($each time with probability $p)$  and finally from  $R$  to  $A$  $($with probability  $q)$.   Meaning:		*For the calculation of  $p_{\rm A}$  it should be noted:   Starting from  $A$  one goes in the Markov diagram first to  $B$  $($with probability $q)$,  then five times clockwise  $($each time with probability $p)$  and finally from  $R$  to  $A$  $($with probability  $q)$.   Meaning:
	:$$p_{\rm A} = q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 = 3^2 / 4^7 \hspace{0.15cm}\underline {\approx 0.549 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$		:$$p_{\rm A} = q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 = 3^2 / 4^7 \hspace{0.15cm}\underline {\approx 0.549 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$
	*In a similar way, one obtains:		*In a similar way,  one obtains:
	:$$p_{\rm R} = q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 = 3 / 4^7 \hspace{0.15cm}\underline {\approx 0.183 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$		:$$p_{\rm R} = q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 = 3 / 4^7 \hspace{0.15cm}\underline {\approx 0.183 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$




	'''(3)'''  By averaging over the conditional probabilities we obtain:		'''(3)'''  By averaging over the conditional probabilities we obtain:
	:$${\rm Pr}(BARBARA) = p_{\rm A} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$		:$${\rm Pr}(\rm BARBARA) = p_{\rm A} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$
	*This leads to the result:		*This leads to the result:
	:$${\rm Pr}(BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right)		:$${\rm Pr}(\rm BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right)
	= \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q}		= \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q}
	= \hspace{-0.15cm} \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3}		= \hspace{-0.15cm} \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3}
	\hspace{0.15cm}\underline { \approx 0.244 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$		\hspace{0.15cm}\underline { \approx 0.244 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$

			'''(4)'''  The probability calculated in  '''(3)'''  is  $p^5 \cdot (1-p)/3$,  where  $q= 1-p$  is considered.

	'''(4)'''  The probability calculated in  '''(3)'''  is  $p^5 \cdot (1-p)/3$, where  $q= 1-p$  is considered.

	*By setting the differential to zero, we obtain the governing equation:		*By setting the differential to zero, we obtain the governing equation:
	:$$5 \cdot p^4 - 6 \cdot p^5 = 0 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} p_{\rm opt} = 5/6 \hspace{0.15cm}\underline { \approx \rm 0.833}.$$		:$$5 \cdot p^4 - 6 \cdot p^5 = 0 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} p_{\rm opt} = 5/6 \hspace{0.15cm}\underline { \approx \rm 0.833}.$$
	*This results in a value that is larger than the subtask  '''(3)'''  by a factor  $90$  approximately:		*This results in a value that is larger than the subtask  '''(3)'''  by a factor  $90$  approximately:
	:$${\rm Pr}(BARBARA) \hspace{0.15cm}\underline { \approx 22 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$		:$${\rm Pr}\rm (BARBARA) \hspace{0.15cm}\underline { \approx 22 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$



	{{ML-Fuß}}		{{ML-Fuß}}