Aufgaben:Exercise 1.7Z: BARBARA Generator - Revision history

Guenter at 16:29, 2 December 2021

2021-12-02T16:29:21Z

Guenter at 13:44, 2 December 2021

2021-12-02T13:44:16Z

Noah at 14:33, 29 November 2021

2021-11-29T14:33:22Z

Noah at 14:32, 29 November 2021

2021-11-29T14:32:35Z

Guenter at 14:18, 29 October 2021

2021-10-29T14:18:25Z

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 Aufgabe 1.7Z: BARBARA-Generator to Exercise 1.7Z: BARBARA Generator

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"

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

2020-07-09T09:52:08Z

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

Guenter at 13:30, 12 November 2019

2019-11-12T13:30:55Z

Guenter at 13:22, 12 November 2019

2019-11-12T13:22:08Z

@@ Line 7: / Line 7: @@
 *The transition probabilities can be taken from the sketched Markov diagram.
 *For the first three subtasks,&nbsp; $p = 1/4$&nbsp; should always hold.
@@ Line 47: / Line 43: @@
 ===Musterlösung===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Correct are <u>the second and third suggested solutions</u>:
+'''(1)'''&nbsp; Correct are&nbsp; <u>the second and third suggested solutions</u>:
-*The sum of all outgoing arrows must always be&nbsp; $1$&nbsp;. Therefore&nbsp; $q = 1 - p$ holds.
+*The sum of all outgoing arrows must always be&nbsp; $1$.&nbsp; Therefore&nbsp; $q = 1 - p$ holds.
-*Because of the symmetry of the Markov diagram, the ergodic probabilities are all equal:
+*Because of the symmetry of the Markov diagram,&nbsp; the ergodic probabilities are all equal:
 :$${\rm Pr}(A) ={\rm Pr}(B) ={\rm Pr}(R) = 1/3.$$
-'''(2)'''&nbsp; If one is in the state&nbsp; $B$&nbsp; at the starting time&nbsp; $\nu=1$&nbsp; because of&nbsp; ${\rm Pr}(B\hspace{0.05cm}|\hspace{0.05cm}B) = 0$ the state&nbsp; $B$&nbsp; is not possible.
+'''(2)'''&nbsp; If one is in the state&nbsp; $B$&nbsp; at the starting time&nbsp; $\nu=0$,&nbsp; because of&nbsp; ${\rm Pr}(B\hspace{0.05cm}|\hspace{0.05cm}B) = 0$&nbsp; the state&nbsp; $B$&nbsp; is not possible at time&nbsp; $\nu=1$.
+*One fails here already with the initial letter&nbsp; $B$:
 :$$p_{\rm B} \; \underline{ =0}.$$
-*For the calculation of&nbsp; $p_{\rm A}$&nbsp; it should be noted: &nbsp; Starting from&nbsp; $A$&nbsp; one goes in the Markov diagram first to&nbsp; $B$&nbsp; $($with probability $q)$, then five times clockwise&nbsp; $($each time with probability $p)$&nbsp; and finally from&nbsp; $R$&nbsp; to&nbsp; $A$&nbsp; $($with probability&nbsp; $q)$. &nbsp; Meaning:
+*For the calculation of&nbsp; $p_{\rm A}$&nbsp; it should be noted: &nbsp; Starting from&nbsp; $A$&nbsp; one goes in the Markov diagram first to&nbsp; $B$&nbsp; $($with probability $q)$,&nbsp; then five times clockwise&nbsp; $($each time with probability $p)$&nbsp; and finally from&nbsp; $R$&nbsp; to&nbsp; $A$&nbsp; $($with probability&nbsp; $q)$. &nbsp; Meaning:
 :$$p_{\rm A} = q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 = 3^2 / 4^7 \hspace{0.15cm}\underline {\approx 0.549 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$
-*In a similar way, one obtains:
+*In a similar way,&nbsp; one obtains:
 :$$p_{\rm R} = q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 = 3 / 4^7 \hspace{0.15cm}\underline {\approx 0.183 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$
 '''(3)'''&nbsp; 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:
-:$${\rm Pr}(BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right)
+:$${\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}
 = \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}}.$$
+'''(4)'''&nbsp; The probability calculated in&nbsp; '''(3)'''&nbsp; is&nbsp; $p^5 \cdot (1-p)/3$,&nbsp; where&nbsp; $q= 1-p$&nbsp; is considered.
 *By setting the differential to zero, we obtain the governing equation:
 :$$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}.$$
 *This results in a value that is larger than the subtask&nbsp; '''(3)'''&nbsp; by a factor&nbsp; $90$&nbsp; approximately:
-:$${\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}}

@@ Line 2: / Line 2: @@
 {{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|$\rm BARBARA$&nbsp; Generator]]
-Here we consider a ternary random generator with symbols&nbsp; $A$,&nbsp; $B$&nbsp; and&nbsp; $R$, which can be described by a homogeneous and stationary first order Markov chain.
+Here we consider a ternary random generator with symbols&nbsp; $A$,&nbsp; $B$&nbsp; and&nbsp; $R$,&nbsp; which can be described by a homogeneous and stationary first order Markov chain.
 *The transition probabilities can be taken from the sketched Markov diagram.
@@ Line 14: / Line 14: @@
 Hints:
-*The task belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
@@ Line 24: / Line 24: @@
 |type="[]"}
 - The values of&nbsp; $p > 0$&nbsp; and&nbsp; $q < 1$&nbsp; are largely arbitrary.
-+ For the transition probabilities, the following must hold: &nbsp; $p + q = 1$.
++ For the transition probabilities,&nbsp; the following must hold: &nbsp; $p + q = 1$.
 + All symbols have equal ergodic probabilities.
 - It holds here:&nbsp; ${\rm Pr}(A) = 1/2, \; {\rm Pr}(B) = 1/3, \; {\rm Pr}(R) = 1/6$.
-{What are the conditional probabilities&nbsp; $p_{\rm A}$,&nbsp; $p_{\rm B}$&nbsp; and&nbsp; $p_{\rm C}$ that at times between&nbsp; $ν+1$&nbsp; and&nbsp; $ν+7$&nbsp; the sequence&nbsp; $BARBARA$&nbsp; is output, <br>if one is in state im Zustand&nbsp; $A$,&nbsp; $B$&nbsp; or&nbsp; $R$&nbsp;, respectively,at time&nbsp; $ν$&nbsp;?&nbsp; Let&nbsp; $p = 1/4$.
+{What are the conditional probabilities&nbsp; $p_{\rm A}$,&nbsp; $p_{\rm B}$&nbsp; and&nbsp; $p_{\rm C}$ that at times between&nbsp; $ν+1$&nbsp; and&nbsp; $ν+7$&nbsp; the sequence&nbsp; "$\rm BARBARA$"&nbsp; is output, <br>if one is at time&nbsp; $ν$&nbsp; in state&nbsp; $A$,&nbsp; $B$&nbsp; or&nbsp; $R$,&nbsp; respectively?&nbsp; Let&nbsp; $p = 1/4$.
 |type="{}"}
 $p_{\rm A} \ = \ $  { 0.549 3% } $\ \cdot 10^{-3}$
@@ Line 34: / Line 34: @@
 $p_{\rm C} \ = \ $ { 0.183 3% } $\ \cdot 10^{-3}$
-{What is the overall probability that the generator outputs the sequence&nbsp; $BARBARA$&nbsp; ausgibt?<br>  Let&nbsp; $p = 1/4$ continue to hold.
+{What is the overall probability that the generator outputs the sequence&nbsp; "$\rm BARBARA$"?&nbsp;   Let&nbsp; $p = 1/4$ continue to hold.
 |type="{}"}
-${\rm Pr}(BARBARA)\ = \ $ { 0.244 3% } $\ \cdot 10^{-3}$
+${\rm Pr}(\rm BARBARA)\ = \ $ { 0.244 3% } $\ \cdot 10^{-3}$
-{How should the parameter&nbsp; $p_{\rm opt}$&nbsp; be chosen to make&nbsp; ${\rm Pr}(BARBARA)$&nbsp; as large as possible?  <br>What is the resulting probability for&nbsp; $BARBARA$?
+{How should the parameter&nbsp; $p_{\rm opt}$&nbsp; be chosen to make&nbsp; ${\rm Pr}(\rm BARBARA)$&nbsp; as large as possible?  <br>What is the resulting probability for&nbsp; "$\rm BARBARA$"?
 |type="{}"}
 $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>

@@ Line 69: / Line 69: @@
 '''(3)'''&nbsp; By averaging over the conditional probabilities we obtain:
 :$${\rm Pr}(BARBARA) = p_{\rm A}  \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B}  \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R}  \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$
-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)
   = \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q}

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Markovketten}}
+{{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|$BARBARA$ Generator]]
-Betrachtet wird hier ein ternärer Zufallsgenerator mit den Symbolen&nbsp; $A$,&nbsp; $B$&nbsp; und&nbsp; $R$, der durch eine homogene und stationäre Markovkette erster Ordnung beschrieben werden kann.
+Here we consider a ternary random generator with symbols&nbsp; $A$,&nbsp; $B$&nbsp; and&nbsp; $R$, which can be described by a homogeneous and stationary first order Markov chain.
-*Die Übergangswahrscheinlichkeiten können dem skizzierten Markovdiagramm entnommen werden.
+*The transition probabilities can be taken from the sketched Markov diagram.
-*Für die ersten drei Teilaufgaben soll stets&nbsp; $p = 1/4$&nbsp; gelten.
+*For the first three subtasks,&nbsp; $p = 1/4$&nbsp; should always hold.
@@ Line 13: / Line 13: @@
-''Hinweis:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Markovketten|Markovketten]].
+*The task belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche der nachfolgenden Aussagen sind zutreffend?
+{Which of the following statements are true?
 |type="[]"}
-- Die Werte von&nbsp; $p > 0$&nbsp; und&nbsp; $q < 1$&nbsp; sind weitgehend frei wählbar.
+- The values of&nbsp; $p > 0$&nbsp; and&nbsp; $q < 1$&nbsp; are largely arbitrary.
-+ Für die Übergangswahrscheinlichkeiten muss gelten: &nbsp; $p + q = 1$.
++ For the transition probabilities, the following must hold: &nbsp; $p + q = 1$.
-+ Alle Symbole haben gleiche ergodische Wahrscheinlichkeiten.
++ All symbols have equal ergodic probabilities.
-- Es gilt hier:&nbsp; ${\rm Pr}(A) = 1/2, \; {\rm Pr}(B) = 1/3, \; {\rm Pr}(R) = 1/6$.
+- It holds here:&nbsp; ${\rm Pr}(A) = 1/2, \; {\rm Pr}(B) = 1/3, \; {\rm Pr}(R) = 1/6$.
-{Wie groß sind die bedingten Wahrscheinlichkeiten&nbsp; $p_{\rm A}$,&nbsp; $p_{\rm B}$&nbsp; und&nbsp; $p_{\rm C}$, dass zu den Zeiten zwischen&nbsp; $ν+1$&nbsp; und&nbsp; $ν+7$&nbsp;  die Sequenz&nbsp; $BARBARA$&nbsp; ausgegeben wird, <br>wenn man sich zum Zeitpunkt&nbsp; $ν$&nbsp; im Zustand&nbsp; $A$,&nbsp; $B$&nbsp; bzw.&nbsp; $R$&nbsp; befindet?&nbsp; Es gelte&nbsp; $p = 1/4$.
+{What are the conditional probabilities&nbsp; $p_{\rm A}$,&nbsp; $p_{\rm B}$&nbsp; and&nbsp; $p_{\rm C}$ that at times between&nbsp; $ν+1$&nbsp; and&nbsp; $ν+7$&nbsp; the sequence&nbsp; $BARBARA$&nbsp; is output, <br>if one is in state im Zustand&nbsp; $A$,&nbsp; $B$&nbsp; or&nbsp; $R$&nbsp;, respectively,at time&nbsp; $ν$&nbsp;?&nbsp; Let&nbsp; $p = 1/4$.
 |type="{}"}
 $p_{\rm A} \ = \ $  { 0.549 3% } $\ \cdot 10^{-3}$
@@ Line 34: / Line 34: @@
 $p_{\rm C} \ = \ $ { 0.183 3% } $\ \cdot 10^{-3}$
-{Wie groß ist die Wahrscheinlichkeit insgesamt, dass der Generator zu sieben aufeinanderfolgenden Zeitpunkten die Sequenz&nbsp; $BARBARA$&nbsp; ausgibt?<br>  Es gelte weiter&nbsp; $p = 1/4.$
+{What is the overall probability that the generator outputs the sequence&nbsp; $BARBARA$&nbsp; ausgibt?<br>  Let&nbsp; $p = 1/4$ continue to hold.
 |type="{}"}
 ${\rm Pr}(BARBARA)\ = \ $ { 0.244 3% } $\ \cdot 10^{-3}$
-{Wie ist der Parameter&nbsp; $p_{\rm opt}$&nbsp; zu wählen, damit&nbsp; ${\rm Pr}(BARBARA)$&nbsp; möglichst groß wird?  <br>Welche Wahrscheinlichkeit ergibt sich damit für&nbsp; $BARBARA$?
+{How should the parameter&nbsp; $p_{\rm opt}$&nbsp; be chosen to make&nbsp; ${\rm Pr}(BARBARA)$&nbsp; as large as possible?  <br>What is the resulting probability for&nbsp; $BARBARA$?
 |type="{}"}
 $p_{\rm opt} \ = \ $  { 0.8333 3% }
@@ Line 47: / Line 47: @@
 ===Musterlösung===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig sind <u>der zweite und der dritte Lösungsvorschlag</u>:
+'''(1)'''&nbsp; Correct are <u>the second and third suggested solutions</u>:
-*Die Summe aller abgehenden Pfeile muss immer&nbsp; $1$&nbsp; sein. Deshalb gilt&nbsp; $q = 1 - p$.
+*The sum of all outgoing arrows must always be&nbsp; $1$&nbsp;. Therefore&nbsp; $q = 1 - p$ holds.
-*Aufgrund der Symmetrie des Markovdiagramms sind die ergodischen Wahrscheinlichkeiten alle gleich:
+*Because of the symmetry of the Markov diagram, the ergodic probabilities are all equal:
 :$${\rm Pr}(A) ={\rm Pr}(B) ={\rm Pr}(R) = 1/3.$$
-'''(2)'''&nbsp; Wenn man zum Startzeitpunkt&nbsp; $\nu = 0$&nbsp; im Zustand&nbsp; $B$&nbsp; ist, ist für den Zeitpunkt&nbsp; $\nu=1$&nbsp; wegen&nbsp; ${\rm Pr}(B\hspace{0.05cm}|\hspace{0.05cm}B) = 0$ der Zustand&nbsp; $B$&nbsp; nicht möglich.
-*Man scheitert hier bereits beim Anfangsbuchstaben $B$:
+*One fails here already with the initial letter $B$:
 :$$p_{\rm B} \; \underline{ =0}.$$
-*F&uuml;r die Berechnung von&nbsp; $p_{\rm A}$&nbsp; ist zu beachten: &nbsp; Ausgehend von&nbsp; $A$&nbsp; geht man im Markovdiagramm zun&auml;chst zu&nbsp; $B$&nbsp; $($mit der Wahrscheinlichkeit $q)$, dann f&uuml;nfmal im Uhrzeigersinn&nbsp; $($jeweils mit der Wahrscheinlichkeit $p)$&nbsp; und schlie&szlig;lich noch von&nbsp; $R$&nbsp; nach&nbsp; $A$&nbsp; $($mit der Wahrscheinlichkeit&nbsp;  $q)$.&nbsp; Das bedeutet:
+*For the calculation of&nbsp; $p_{\rm A}$&nbsp; it should be noted: &nbsp; Starting from&nbsp; $A$&nbsp; one goes in the Markov diagram first to&nbsp; $B$&nbsp; $($with probability $q)$, then five times clockwise&nbsp; $($each time with probability $p)$&nbsp; and finally from&nbsp; $R$&nbsp; to&nbsp; $A$&nbsp; $($with probability&nbsp; $q)$. &nbsp; 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 &auml;hnlicher Weise erh&auml;lt man:
+*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)'''&nbsp; Durch Mittelung &uuml;ber die bedingten Wahrscheinlichkeiten erh&auml;lt man:
+'''(4)'''&nbsp; The probability calculated in&nbsp; '''(3)'''&nbsp; is&nbsp; $p^5 \cdot (1-p)/3$, where&nbsp; $q= 1-p$&nbsp; is considered.
 {{ML-Fuß}}

← Older revision		Revision as of 14:18, 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
	^]]		^]]

← Older revision	Revision as of 14:16, 29 October 2021
(No difference)

← Older revision		Revision as of 12:44, 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 09:52, 9 July 2020
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	''Hinweis:''		''Hinweis:''
−	*Die Aufgabe gehört zum Kapitel  [[~~Stochastische_Signaltheorie~~/Markovketten\|Markovketten]].	+	*Die Aufgabe gehört zum Kapitel  [[Theory_of_Stochastic_Signals/Markovketten\|Markovketten]].