Aufgaben:Exercise 1.7: Ternary Markov Chain - Revision history

Guenter at 13:21, 2 December 2021

2021-12-02T13:21:55Z

Guenter at 13:06, 2 December 2021

2021-12-02T13:06:10Z

Noah at 13:42, 29 November 2021

2021-11-29T13:42:20Z

Guenter at 14:17, 29 October 2021

2021-10-29T14:17:59Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.7: Ternäre Markovkette to Aufgaben:Exercise 1.7: Ternary Markov Chain

2021-10-29T14:15:19Z

Guenter moved page Aufgabe 1.7: Ternäre Markovkette to Exercise 1.7: Ternary Markov Chain

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

2021-03-23T12:44:35Z

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

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

2020-07-09T09:52:42Z

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

Guenter at 13:12, 12 November 2019

2019-11-12T13:12:41Z

Guenter at 13:04, 12 November 2019

2019-11-12T13:04:47Z

Guenter at 15:33, 1 August 2018

2018-08-01T15:33:09Z

@@ Line 15: / Line 15: @@
 :$${\rm Pr}(B_0) = 1,$$
 :$${\rm Pr}(C_0) = 0.$$
@@ Line 56: / Line 53: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; In general, or in this special case, it must hold:
+'''(1)'''&nbsp; In general&nbsp; (or in this special case)&nbsp; it must hold:
 :$$p_{\rm AA} = 1 - p_{\rm AB} - p_{\rm AC} \hspace{0.5cm} \Rightarrow  \hspace{0.5cm} p_{\rm AA} = 1 - 0.75 -0 \hspace{0.15cm}\underline {= 0.25},$$
 :$$p_{\rm BB} = 1 - p_{\rm BA} - p_{\rm BC} \hspace{0.5cm} \Rightarrow  \hspace{0.5cm} p_{\rm BB} = 1 - 0.75 -0.25 \hspace{0.15cm}\underline {= 0},$$
 :$$p_{\rm CC} = 1 - p_{\rm CA} - p_{\rm CB} \hspace{0.5cm} \Rightarrow  \hspace{0.5cm} p_{\rm CC} = 1 - 0 - 0.25 \hspace{0.15cm}\underline {= 0.75}.$$
-*Thus, the transition matrix is:
+*Thus,&nbsp; the transition matrix is:
 :$${\mathbf{P}} = \left[ \begin{array}{ccc} 1/4 & 3/4 & 0 \\ 3/4 & 0 & 1/4  \\ 0 & 1/4 & 3/4  \end{array} \right] .$$
-'''(2)'''&nbsp; Because of&nbsp; ${\rm Pr}(B_0) = 1$&nbsp; and&nbsp; $p_\text{BB}  = 0$&nbsp;, at time&nbsp; $\nu = 1$&nbsp; event&nbsp; $B$&nbsp; cannot occur and&nbsp; $A$&nbsp; is much more probable than&nbsp; $C$:
+'''(2)'''&nbsp; Because of&nbsp; ${\rm Pr}(B_0) = 1$&nbsp; and&nbsp; $p_\text{BB}  = 0$,&nbsp; at time&nbsp; $\nu = 1$&nbsp; the event&nbsp; $B$&nbsp; cannot occur and&nbsp; $A$&nbsp; is much more probable than&nbsp; $C$:
 :$$\hspace{0.15cm}\underline {{\rm Pr}(A_1) = 0.75}; \hspace{0.5cm} {\rm Pr}(B_1) = 0; \hspace{0.5cm}{\rm Pr}(C_1) = 0.25.$$
 *The same result is obtained by applying the vector matrix representation.
@@ Line 73: / Line 70: @@
 '''(3)'''&nbsp; For the probability vector at time&nbsp; $\nu = 2$&nbsp; holds:
 :$${\mathbf{p}^{(\nu = 2)}} = {\mathbf{P}}^{\rm T} \cdot {\mathbf{p}^{(\nu =1 )}}= \left[ \begin{array}{ccc} 1/4 & 3/4&  0 \\ 3/4 & 0 & 1/4 \\ 0& 1/4& 3/4  \end{array} \right] \left[ \begin{array}{c} 3/4 \\ 0  \\ 1/4  \end{array} \right] = \left[ \begin{array}{c} 3/16 \\ 10/16  \\ 3/16 \end{array} \right] .$$
-*Thus, the event probability is&nbsp; ${\rm Pr}(A_2) = 3/16\hspace{0.15cm}\underline {=  0.1875}$.
+*Thus,&nbsp; the event probability is&nbsp; ${\rm Pr}(A_2) = 3/16\hspace{0.15cm}\underline {=  0.1875}$.
-'''(4)'''&nbsp; To solve this problem, several possibilities should be given.
+'''(4)'''&nbsp; To solve this problem,&nbsp; several possibilities should be given.
 *One is to solve a system of equations with three unknowns:
 :$${\rm Pr}(A) = 1/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(A)  \hspace{0.1cm} + \hspace{0.1cm} 3/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(B),$$
 :$${\rm Pr}(B) = 3/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(A)  \hspace{2.8cm} + \hspace{0.1cm} 1/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(C),$$
 :$${\rm Pr}(C) = \hspace{2.8cm} 1/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(B)  \hspace{0.1cm} + \hspace{0.1cm} 3/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(C).$$
-:From the first equation we get ${\rm Pr}(B) = {\rm Pr}(A)$, from the last ${\rm Pr}(C) = {\rm Pr}(A)$. Since the sum of all probabilities is equal to $1$, it follows  $ {\rm Pr}(A) = {\rm Pr}(B) = {\rm Pr}(C) = 1/3 \hspace{0.15cm}\underline {\approx  0.333}$.
+:From the first equation we get&nbsp; ${\rm Pr}(B) = {\rm Pr}(A)$,&nbsp; from the last&nbsp; ${\rm Pr}(C) = {\rm Pr}(A)$.&nbsp; Since the sum of all probabilities is equal to&nbsp; $1$,&nbsp; it follows
-*The same result is obtained by analyzing the transition matrix.&nbsp; Since the sum of each column equals&nbsp; $1$&nbsp;&nbsp; $(that is, the sum of each row of the transposed matrix also equals $1)$, it is obvious that all event probabilities must be equal.
+:$$ {\rm Pr}(A) = {\rm Pr}(B) = {\rm Pr}(C) = 1/3 \hspace{0.15cm}\underline {\approx  0.333}.$$
-*Even by thinking about it briefly, one could have predicted the result without calculating.&nbsp; Since at each event the numerical values at the outgoing arrows&nbsp; (only to other events)&nbsp; are equal with those at the incoming ones, it is not to be seen why one of the events should be preferred.
+*Even by thinking about it briefly,&nbsp; one could have predicted the result without calculating.&nbsp; Since at each event the numerical values at the outgoing arrows&nbsp; (only to other events)&nbsp; are equal with those at the incoming ones,&nbsp; it is not to be seen why one of the events should be preferred.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID453__Sto_A_1_7.png|right|frame|Ternary Markov chain]]
-We consider a Markov chain with three possible events&nbsp; $A$,&nbsp; $B$&nbsp; and&nbsp; $C$:
+We consider a Markov chain with three possible events&nbsp; $A$,&nbsp; $B$,&nbsp; $C$:
 *The transition probabilities are shown in the graph.
-*Thus, a transition from&nbsp; $A$&nbsp; to&nbsp; $C$&nbsp; and vice versa is not possible
+*Thus,&nbsp; a transition from&nbsp; $A$&nbsp; to&nbsp; $C$&nbsp; and vice versa is not possible.
 :$$p_\text{AC} = p_\text{CA} = 0.$$
-The three event probabilities at the starting time&nbsp; $\nu = 0$&nbsp; are given as follows:
+The three event probabilities at the starting time&nbsp; $(\nu = 0)$&nbsp; are given as follows:
 :$${\rm Pr}(A_0) = 0,$$
 :$${\rm Pr}(B_0) = 1,$$
 :$${\rm Pr}(C_0) = 0.$$
@@ Line 26: / Line 24: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
-*In particular, reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains#Matrix_vector_representation|Matrix vector representation]]&nbsp;.
+*In particular,&nbsp; reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains#Matrix_vector_representation|Matrix vector representation]].
@@ Line 33: / Line 31: @@
 <quiz display=simple>
-{Give the transition matrix&nbsp; ${\mathbf{P}}$&nbsp; and the transition probabilities&nbsp; $p_\text{AA}$,&nbsp; $p_\text{BB}$&nbsp; and&nbsp; $p_\text{CC}$ an.
+{Give the transition matrix&nbsp; ${\mathbf{P}}$&nbsp; and the transition probabilities&nbsp; $p_\text{AA}$,&nbsp; $p_\text{BB}$&nbsp; and&nbsp; $p_\text{CC}$.
 |type="{}"}
 $p_\text{AA} \ = \ $ { 0.25 3% }
@@ Line 39: / Line 37: @@
 $p_\text{CC} \ = \ $ { 0.75 3% }
-{Calculate the event probabilities at time&nbsp; $\nu = 1$, in particular
+{Calculate the event probabilities at time&nbsp; $\nu = 1$,&nbsp; in particular
 |type="{}"}
 ${\rm Pr}(A_1) \ = \ $ { 0.75 3% }
-{Calculate the event probabilities at time&nbsp; $\nu = 2$, in particular
+{Calculate the event probabilities at time&nbsp; $\nu = 2$,&nbsp; in particular
 |type="{}"}
 ${\rm Pr}(A_2) \ =  \ ${ 0.1875 3% }

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Markovketten
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Markov_Chains
 }}
-[[File:P_ID453__Sto_A_1_7.png|right|frame|Ternäre Markovkette]]
+[[File:P_ID453__Sto_A_1_7.png|right|frame|Ternary Markov chain]]
-Wir betrachten eine Markovkette mit den drei möglichen Ereignissen&nbsp; $A$,&nbsp; $B$&nbsp; und&nbsp; $C$:
+We consider a Markov chain with three possible events&nbsp; $A$,&nbsp; $B$&nbsp; and&nbsp; $C$:
-*Die Übergangswahrscheinlichkeiten sind der Grafik zu entnehmen.
+*The transition probabilities are shown in the graph.
-*Ein Übergang von&nbsp; $A$&nbsp; nach&nbsp; $C$&nbsp; und umgekehrt ist somit nicht möglich:
+*Thus, a transition from&nbsp; $A$&nbsp; to&nbsp; $C$&nbsp; and vice versa is not possible
 :$$p_\text{AC} = p_\text{CA} = 0.$$
-Die drei Ereigniswahrscheinlichkeiten zum Startzeitpunkt&nbsp; $\nu = 0$&nbsp; sind wie folgt gegeben:
+The three event probabilities at the starting time&nbsp; $\nu = 0$&nbsp; are given as follows:
 :$${\rm Pr}(A_0) = 0,$$
@@ Line 24: / Line 24: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Markovketten|Markovketten]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
-*Insbesondere wird auf die Seite&nbsp; [[Theory_of_Stochastic_Signals/Markovketten#Matrix-Vektordarstellung|Matrix-Vektordarstellung]]&nbsp; Bezug genommen.
+*In particular, reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains#Matrix_vector_representation|Matrix vector representation]]&nbsp;.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Geben Sie die Übergangsmatrix&nbsp; ${\mathbf{P}}$&nbsp; und die Übergangswahrscheinlichkeiten&nbsp; $p_\text{AA}$,&nbsp; $p_\text{BB}$&nbsp; und&nbsp; $p_\text{CC}$ an.
+{Give the transition matrix&nbsp; ${\mathbf{P}}$&nbsp; and the transition probabilities&nbsp; $p_\text{AA}$,&nbsp; $p_\text{BB}$&nbsp; and&nbsp; $p_\text{CC}$ an.
 |type="{}"}
 $p_\text{AA} \ = \ $ { 0.25 3% }
@@ Line 39: / Line 39: @@
 $p_\text{CC} \ = \ $ { 0.75 3% }
-{Berechnen Sie die Ereigniswahrscheinlichkeiten zum Zeitpunkt&nbsp; $\nu = 1$, insbesondere
+{Calculate the event probabilities at time&nbsp; $\nu = 1$, in particular
 |type="{}"}
 ${\rm Pr}(A_1) \ = \ $ { 0.75 3% }
-{Berechnen Sie die Ereigniswahrscheinlichkeiten zum Zeitpunkt&nbsp; $\nu = 2$, insbesondere
+{Calculate the event probabilities at time&nbsp; $\nu = 2$, in particular
 |type="{}"}
 ${\rm Pr}(A_2) \ =  \ ${ 0.1875 3% }
-{Welche Wahrscheinlichkeiten werden sich sehr lange nach Einschalten der Markovkette einstellen&nbsp; $(ν \rightarrow \infty)$? <br>Wie groß ist insbesondere die ergodische Wahrscheinlichkeit&nbsp; ${\rm Pr}(A)$?
+{What probabilities will occur very long after the Markov chain is turned on&nbsp; $(ν \rightarrow \infty)$? <br>In particular, what is the ergodic probability&nbsp; ${\rm Pr}(A)$?
 |type="{}"}
 ${\rm Pr}(A) \ = \ $ { 0.333 3% }
@@ Line 56: / Line 56: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Allgemein bzw. in diesem Sonderfall muss gelten:
+'''(1)'''&nbsp; In general, or in this special case, it must hold:
 :$$p_{\rm AA} = 1 - p_{\rm AB} - p_{\rm AC} \hspace{0.5cm} \Rightarrow  \hspace{0.5cm} p_{\rm AA} = 1 - 0.75 -0 \hspace{0.15cm}\underline {= 0.25},$$
 :$$p_{\rm BB} = 1 - p_{\rm BA} - p_{\rm BC} \hspace{0.5cm} \Rightarrow  \hspace{0.5cm} p_{\rm BB} = 1 - 0.75 -0.25 \hspace{0.15cm}\underline {= 0},$$
 :$$p_{\rm CC} = 1 - p_{\rm CA} - p_{\rm CB} \hspace{0.5cm} \Rightarrow  \hspace{0.5cm} p_{\rm CC} = 1 - 0 - 0.25 \hspace{0.15cm}\underline {= 0.75}.$$
-*Damit lautet die &Uuml;bergangsmatrix:
+*Thus, the transition matrix is:
 :$${\mathbf{P}} = \left[ \begin{array}{ccc} 1/4 & 3/4 & 0 \\ 3/4 & 0 & 1/4  \\ 0 & 1/4 & 3/4  \end{array} \right] .$$
-'''(2)'''&nbsp; Wegen&nbsp; ${\rm Pr}(B_0) = 1$&nbsp; und&nbsp; $p_\text{BB}  = 0$&nbsp; kann zum Zeitpunkt&nbsp; $\nu = 1$&nbsp; das Ereignis&nbsp; $B$&nbsp; nicht auftreten und&nbsp; $A$&nbsp; ist sehr viel wahrscheinlicher als&nbsp; $C$:
+'''(2)'''&nbsp; Because of&nbsp; ${\rm Pr}(B_0) = 1$&nbsp; and&nbsp; $p_\text{BB}  = 0$&nbsp;, at time&nbsp; $\nu = 1$&nbsp; event&nbsp; $B$&nbsp; cannot occur and&nbsp; $A$&nbsp; is much more probable than&nbsp; $C$:
 :$$\hspace{0.15cm}\underline {{\rm Pr}(A_1) = 0.75}; \hspace{0.5cm} {\rm Pr}(B_1) = 0; \hspace{0.5cm}{\rm Pr}(C_1) = 0.25.$$
-*Zum gleichen Ergebnis kommt man durch Anwendung der Vektor-Matrixdarstellung.
+*The same result is obtained by applying the vector matrix representation.
-'''(3)'''&nbsp; F&uuml;r den Wahrscheinlichkeitsvektor zum Zeitpunkt&nbsp; $\nu = 2$&nbsp; gilt:
+'''(3)'''&nbsp; For the probability vector at time&nbsp; $\nu = 2$&nbsp; holds:
 :$${\mathbf{p}^{(\nu = 2)}} = {\mathbf{P}}^{\rm T} \cdot {\mathbf{p}^{(\nu =1 )}}= \left[ \begin{array}{ccc} 1/4 & 3/4&  0 \\ 3/4 & 0 & 1/4 \\ 0& 1/4& 3/4  \end{array} \right] \left[ \begin{array}{c} 3/4 \\ 0  \\ 1/4  \end{array} \right] = \left[ \begin{array}{c} 3/16 \\ 10/16  \\ 3/16 \end{array} \right] .$$
-*Damit ist die Ereigniswahrscheinlichkeit&nbsp; ${\rm Pr}(A_2) = 3/16\hspace{0.15cm}\underline {=  0.1875}$.
+*Thus, the event probability is&nbsp; ${\rm Pr}(A_2) = 3/16\hspace{0.15cm}\underline {=  0.1875}$.
-'''(4)'''&nbsp; Zur Lösung dieser Aufgabe sollen verschiedene M&ouml;glichkeiten angegeben werden.
+'''(4)'''&nbsp; To solve this problem, several possibilities should be given.
-*Zum einen das L&ouml;sen eines Gleichungssystems mit drei Unbekannten:
+*One is to solve a system of equations with three unknowns:
 :$${\rm Pr}(A) = 1/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(A)  \hspace{0.1cm} + \hspace{0.1cm} 3/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(B),$$
 :$${\rm Pr}(B) = 3/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(A)  \hspace{2.8cm} + \hspace{0.1cm} 1/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(C),$$
 :$${\rm Pr}(C) = \hspace{2.8cm} 1/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(B)  \hspace{0.1cm} + \hspace{0.1cm} 3/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(C).$$
-:Aus der ersten Gleichung erh&auml;lt man ${\rm Pr}(B) = {\rm Pr}(A)$, aus der letzten ${\rm Pr}(C) = {\rm Pr}(A)$. Da die Summe aller Wahrscheinlichkeiten gleich $1$ ist, folgt $ {\rm Pr}(A) = {\rm Pr}(B) = {\rm Pr}(C) = 1/3 \hspace{0.15cm}\underline {\approx  0.333}$.
+:From the first equation we get ${\rm Pr}(B) = {\rm Pr}(A)$, from the last ${\rm Pr}(C) = {\rm Pr}(A)$. Since the sum of all probabilities is equal to $1$, it follows  $ {\rm Pr}(A) = {\rm Pr}(B) = {\rm Pr}(C) = 1/3 \hspace{0.15cm}\underline {\approx  0.333}$.
-*Zum gleichen Ergebnis kommt man durch Analyse der &Uuml;bergangsmatrix.&nbsp; Da die Summe jeder Spalte gleich&nbsp; $1$&nbsp; ist&nbsp; $($das hei&szlig;t: &nbsp; die Summe einer jeden Zeile der transponierten Matrix ergibt ebenfalls $1)$, ist offensichtlich, dass alle Ereigniswahrscheinlichkeiten gleich sein m&uuml;ssen.
+*The same result is obtained by analyzing the transition matrix.&nbsp; Since the sum of each column equals&nbsp; $1$&nbsp;&nbsp; $(that is, the sum of each row of the transposed matrix also equals $1)$, it is obvious that all event probabilities must be equal.
-*Auch durch kurzes Nachdenken h&auml;tte man das Ergebnis ohne Rechnung vorhersagen k&ouml;nnen.&nbsp; Da bei jedem Ereignis die Zahlenwerte bei den abgehenden Pfeilen&nbsp; (nur zu anderen Ereignissen)&nbsp; mit denen bei den ankommenden gleich sind, ist nicht einzusehen, warum eines der Ereignisse bevorzugt sein sollte.
+*Even by thinking about it briefly, one could have predicted the result without calculating.&nbsp; Since at each event the numerical values at the outgoing arrows&nbsp; (only to other events)&nbsp; are equal with those at the incoming ones, it is not to be seen why one of the events should be preferred.
 {{ML-Fuß}}

← Older revision		Revision as of 14:17, 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 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
	^]]		^]]

@@ Line 72: / Line 72: @@
 :$${\mathbf{p}^{(\nu = 2)}} = {\mathbf{P}}^{\rm T} \cdot {\mathbf{p}^{(\nu =1 )}}= \left[ \begin{array}{ccc} 1/4 & 3/4&  0 \\ 3/4 & 0 & 1/4 \\ 0& 1/4& 3/4  \end{array} \right] \left[ \begin{array}{c} 3/4 \\ 0  \\ 1/4  \end{array} \right] = \left[ \begin{array}{c} 3/16 \\ 10/16  \\ 3/16 \end{array} \right] .$$
 Damit ist die Ereigniswahrscheinlichkeit ${\rm Pr}(A_2) = 3/16\hspace{0.15cm}\underline {=  0.1875}$.
 '''(4)'''&nbsp; Zur Lösung dieser Aufgabe sollen verschiedene M&ouml;glichkeiten angegeben werden.
@@ Line 79: / Line 80: @@
 :$${\rm Pr}(C) = \hspace{2.8cm} 1/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(B)  \hspace{0.1cm} + \hspace{0.1cm} 3/4 \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm Pr}(C).$$
 :Aus der ersten Gleichung erh&auml;lt man ${\rm Pr}(B) = {\rm Pr}(A)$, aus der letzten ${\rm Pr}(C) = {\rm Pr}(A)$. Da die Summe aller Wahrscheinlichkeiten gleich $1$ ist, folgt $ {\rm Pr}(A) = {\rm Pr}(B) = {\rm Pr}(C) = 1/3 \hspace{0.15cm}\underline {\approx  0.333}$.
-*Zum gleichen Ergebnis kommt man durch Analyse der &Uuml;bergangsmatrix. Da die Summe jeder Spalte gleich $1$ ist (das hei&szlig;t: die Summe einer jeden Zeile der transponierten Matrix ergibt ebenfalls $1$), ist offensichtlich, dass alle Ereigniswahrscheinlichkeiten gleich sein m&uuml;ssen.
+*Zum gleichen Ergebnis kommt man durch Analyse der &Uuml;bergangsmatrix. Da die Summe jeder Spalte gleich $1$ ist (das hei&szlig;t: &nbsp; die Summe einer jeden Zeile der transponierten Matrix ergibt ebenfalls $1$), ist offensichtlich, dass alle Ereigniswahrscheinlichkeiten gleich sein m&uuml;ssen.
 *Auch durch kurzes Nachdenken h&auml;tte man das Ergebnis ohne Rechnung vorhersagen k&ouml;nnen. Da bei jedem Ereignis die Zahlenwerte bei den abgehenden Pfeilen (nur zu anderen Ereignissen) mit denen bei den ankommenden gleich sind, ist nicht einzusehen, warum eines der Ereignisse bevorzugt sein sollte.