Aufgaben:Exercise 1.6: Transition Probabilities - Revision history

Noah at 20:33, 19 December 2021

2021-12-19T20:33:18Z

Noah at 20:32, 19 December 2021

2021-12-19T20:32:44Z

Noah at 20:32, 19 December 2021

2021-12-19T20:32:32Z

Guenter at 16:50, 1 December 2021

2021-12-01T16:50:25Z

Guenter at 16:39, 1 December 2021

2021-12-01T16:39:17Z

Noah at 11:31, 29 November 2021

2021-11-29T11:31:52Z

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Guenter at 14:17, 29 October 2021

2021-10-29T14:17:08Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.6: Übergangswahrscheinlichkeiten to Aufgaben:Exercise 1.6: Transition Probabilities

2021-10-29T14:12:49Z

Guenter moved page Aufgabe 1.6: Übergangswahrscheinlichkeiten to Exercise 1.6: Transition Probabilities

Javier: Text replacement - "„" to """

2021-05-28T14:27:00Z

Text replacement - "„" to """

Javier: Text replacement - "”" to """

2021-05-28T14:26:59Z

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@@ Line 23: / Line 23: @@
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
-*You can check your results with the interactive SWF applet
+*You can check your results with the (German language) interactive SWF applet
 : [[Applets:Markovketten|Ereigniswahrscheinlichkeiten einer Markov-Kette erster Ordnung]] &nbsp; &rArr; &nbsp; "Event Probabilities of a First Order Markov Chain".

@@ Line 23: / Line 23: @@
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
-*You can check your results with the &nbsp; &nbsp; interactive SWF applet
+*You can check your results with the interactive SWF applet
 : [[Applets:Markovketten|Ereigniswahrscheinlichkeiten einer Markov-Kette erster Ordnung]] &nbsp; &rArr; &nbsp; "Event Probabilities of a First Order Markov Chain".

@@ Line 23: / Line 23: @@
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
-*You can check your results with the &nbsp;  (German language)&nbsp; interactive SWF applet
+*You can check your results with the &nbsp; &nbsp; interactive SWF applet
 : [[Applets:Markovketten|Ereigniswahrscheinlichkeiten einer Markov-Kette erster Ordnung]] &nbsp; &rArr; &nbsp; "Event Probabilities of a First Order Markov Chain".

@@ Line 67: / Line 67: @@
 '''(2)'''&nbsp; <u>Proposed solutions 1 and 2</u> are correct:
-*$A$&nbsp; is followed by&nbsp; $B$&nbsp; much more frequently than&nbsp; $A$, that is, it will certainly be&nbsp; ${\rm Pr}(B \hspace{0.05cm} | \hspace{0.05cm}A) > {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm}A)$&nbsp;.
+*$A$&nbsp; is followed by&nbsp; $B$&nbsp; much more frequently than by&nbsp; $A$,&nbsp; that is,&nbsp; it will certainly be&nbsp; ${\rm Pr}(B \hspace{0.05cm} | \hspace{0.05cm}A) > {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm}A)$.
 *All four transitions between the two events&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; are possible.&nbsp; It follows that all four transition probabilities will be nonzero.
-*Because of&nbsp; ${\rm Pr}(B_\text{v=0})  \ne 0$&nbsp; and&nbsp; ${\rm Pr}(B \hspace{0.05cm} | \hspace{0.05cm}B) \ne 0$&nbsp;, the sequence&nbsp; "$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{0.15cm}...$"&nbsp; can of course also be generated, even though it is not present in the twenty Markov chains output here.
+*Because of&nbsp; ${\rm Pr}(B_\text{v=0})  \ne 0$&nbsp; and&nbsp; ${\rm Pr}(B \hspace{0.05cm} | \hspace{0.05cm}B) \ne 0$,&nbsp; the sequence&nbsp; "$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{0.15cm}...$"&nbsp; can of course also be generated,&nbsp; even though it is not present in the twenty Markov chains output here.
-'''(3)'''&nbsp; For a first-order Markov chain, with the abbreviations&nbsp; ${\rm Pr}(A_0) = {\rm Pr}(A_{\nu=0})$&nbsp;  and&nbsp; ${\rm Pr}(A_1) = {\rm Pr}(A_{\nu=1})$:
+'''(3)'''&nbsp; For a first-order Markov chain,&nbsp; with the abbreviations&nbsp; ${\rm Pr}(A_0) = {\rm Pr}(A_{\nu=0})$&nbsp;  and&nbsp; ${\rm Pr}(A_1) = {\rm Pr}(A_{\nu=1})$:
 :$${\rm Pr}(A_1) = {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} A) \cdot {\rm Pr}(A_0) \hspace{0.1cm} + \hspace{0.1cm} {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} B) \cdot {\rm Pr}(B_0).$$
 *The ergodic probabilities are&nbsp; ${\rm Pr}(A) = {\rm Pr}(A_{\nu \hspace{0.05cm} > \hspace{0.05cm}4}) = 0.4$&nbsp; and&nbsp; ${\rm Pr}(B) = {\rm Pr}(B_{\nu \hspace{0.05cm} > \hspace{0.05cm}4}) = 0.6$.&nbsp; The following relationship exists between these:
 :$${\rm Pr}(A) = {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} A) \cdot {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm} {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} B) \cdot {\rm Pr}(B).$$
-*With the numerical values given, we obtain from these last two equations:
+*With the numerical values given,&nbsp; we obtain from these last two equations:
 :$$0.15 = {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} A) \cdot 0.90 \hspace{0.1cm} + \hspace{0.1cm} {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} B) \cdot 0.10 ,$$
 :$$0.40 = {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} A) \cdot 0.40 \hspace{0.1cm} + \hspace{0.1cm} {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} B) \cdot 0.60 .$$
@@ Line 83: / Line 83: @@
 :$$0.5 = 5 \cdot {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} A) \hspace{0.15cm} \Rightarrow
 \hspace{0.15cm}  {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} A) \hspace{0.15cm}\underline {= 0.1}.$$
-*Substituting this result into one of the upper equations, we get&nbsp; $ {\rm Pr}(A\hspace{0.05cm}|\hspace{0.05cm}B) = 0.6$. The other probabilities are:
+*Substituting this result into one of the upper equations,&nbsp; we get&nbsp; $ {\rm Pr}(A\hspace{0.05cm}|\hspace{0.05cm}B) = 0.6$. The other probabilities are:
 :$${\rm Pr}(B\hspace{0.05cm}|\hspace{0.05cm}A) = 1 - {\rm Pr}(A\hspace{0.05cm}|\hspace{0.05cm}A) = 0.9, \hspace{0.3cm}
 {\rm Pr}(B\hspace{0.05cm}|\hspace{0.05cm}B) = 1 - {\rm Pr}(A\hspace{0.05cm}|\hspace{0.05cm}B)\ \underline{= 0.4}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sto_A_1_6.png|right|frame|$20$&nbsp; Realizations of the Considered Markov Chain]]
+[[File:EN_Sto_A_1_6.png|right|frame|$20$&nbsp; Realizations of the considered Markov chain]]
 On the right you see&nbsp; $20$&nbsp; realizations of a binary homogeneous Markov chain of first order with the events&nbsp; $A$&nbsp; and&nbsp; $B$:
 *One can already see from this representation that at the beginning&nbsp; $(ν = 0)$&nbsp; event&nbsp; $A$&nbsp; predominates.
-*However, at later times – approximately from&nbsp; $ν = 4$&nbsp; – event&nbsp; $B$&nbsp; occurs somewhat more frequently.
+*However,&nbsp; at later times,&nbsp; approximately from&nbsp; $ν = 4$:&nbsp; The event&nbsp; $B$&nbsp; occurs somewhat more frequently.
 By averaging over millions of realizations, some event probabilities were determined numerically:
 :$${\rm Pr}(A_{\nu \hspace{0.05cm} = \hspace{0.05cm}0}) \approx 0.9, \hspace{0.3cm}{\rm Pr}(A_{\nu \hspace{0.05cm} = \hspace{0.05cm}1}) \approx 0.15, \hspace{0.3cm} {\rm Pr}(A_{\nu \hspace{0.05cm} > \hspace{0.05cm}4}) \approx 0.4.$$
-These empirical numerical values will be used to determine (approximately) the parameters (transition probabilities) of the Markov chain.
+These empirical numerical values will be used to determine (approximately) the&nbsp; "transition probabilities"&nbsp; of the Markov chain.
@@ Line 22: / Line 23: @@
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains|Markov Chains]].
-*You can check your results with the interactive applet&nbsp; [[Applets:Markovketten|Event probabilities of a 1st order Markov chain]]&nbsp;.
+*You can check your results with the &nbsp;  (German language)&nbsp; interactive SWF applet
@@ Line 29: / Line 31: @@
 <quiz display=simple>
-{What are the probabilities at times&nbsp; $ν = 0$,&nbsp; $ν = 1$&nbsp; and&nbsp; $ν = 9$, given only the&nbsp; $20$&nbsp; realizations shown?
+{What are the probabilities at times&nbsp; $ν = 0$, &nbsp; $ν = 1$ &nbsp; and&nbsp; $ν = 9$, given&nbsp; only the&nbsp; $20$&nbsp; realizations shown?
 |type="{}"}
 ${\rm Pr}(A_{\nu \hspace{0.05cm} = \hspace{0.05cm}0}) \ = \ $  { 0.85 3% }
@@ Line 35: / Line 37: @@
 ${\rm Pr}(A_{\nu \hspace{0.05cm} = \hspace{0.05cm}9}) \ = \ $ { 0.4 3% }
-{Based on the pattern sequences, which of the statements is true?
+{Based on the pattern sequences,&nbsp; which of the statements are true?
 |type="[]"}
-+ After&nbsp; $A$&nbsp;,&nbsp; $B$&nbsp; is more probable than&nbsp; $A$.
++ After&nbsp; $A$:&nbsp; $B$&nbsp; is more probable than&nbsp; $A$.
-+ After both&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp;,&nbsp; $A$&nbsp; or&nbsp; $B$&nbsp; can follow again.
++ After both&nbsp; $A$&nbsp; and&nbsp; $B$:&nbsp; $A$&nbsp; or&nbsp; $B$&nbsp; can follow again.
-- The sequence "$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}\text{...}$" is not possible.
+- The sequence&nbsp; "$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}\text{...}$"&nbsp; is not possible.
-{Calculate all transition probabilities of the Markov chain. In particular, how large are&nbsp; ${\rm Pr}(A\hspace{0.05cm} | \hspace{0.05cm}A)$&nbsp; and&nbsp; ${\rm Pr}(B\hspace{0.05cm} | \hspace{0.05cm}B)$?
+{Calculate all transition probabilities of the Markov chain.&nbsp; In particular,&nbsp; how large are&nbsp; ${\rm Pr}(A\hspace{0.05cm} | \hspace{0.05cm}A)$&nbsp; and&nbsp; ${\rm Pr}(B\hspace{0.05cm} | \hspace{0.05cm}B)$?
 |type="{}"}
 ${\rm Pr}(A\hspace{0.05cm} | \hspace{0.05cm}A) \ = \ $ { 0.1 3% }

← 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^]]

@@ Line 40: / Line 40: @@
 + Nach&nbsp; $A$&nbsp; ist&nbsp; $B$&nbsp; wahrscheinlicher als&nbsp; $A$.
 + Sowohl nach&nbsp; $A$&nbsp; als auch nach&nbsp; $B$&nbsp; kann wieder&nbsp; $A$&nbsp; oder&nbsp; $B$&nbsp; folgen.
-- Die Folge &bdquo;$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}\text{...}$" ist nicht möglich.
+- Die Folge "$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}\text{...}$" ist nicht möglich.
 {Berechnen Sie alle Übergangswahrscheinlichkeiten der Markovkette. Wie groß sind insbesondere&nbsp; ${\rm Pr}(A\hspace{0.05cm} | \hspace{0.05cm}A)$&nbsp; und&nbsp; ${\rm Pr}(B\hspace{0.05cm} | \hspace{0.05cm}B)$?
@@ Line 52: / Line 52: @@
-{Wie groß ist die Wahrscheinlichkeit, dass sehr lange nach Einschalten der Kette die Zeichenfolge&nbsp; &bdquo;$A\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}A$"&nbsp; erzeugt wird?
+{Wie groß ist die Wahrscheinlichkeit, dass sehr lange nach Einschalten der Kette die Zeichenfolge&nbsp; "$A\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}A$"&nbsp; erzeugt wird?
 |type="{}"}
 ${\rm Pr}(A\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}A)\ = \ $ { 8.64 3% } $\ \%$
@@ Line 68: / Line 68: @@
 *Nach&nbsp; $A$&nbsp; folgt&nbsp; $B$&nbsp; sehr viel h&auml;ufiger als&nbsp; $A$, das heißt, es wird sicher&nbsp; ${\rm Pr}(B \hspace{0.05cm} | \hspace{0.05cm}A) > {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm}A)$&nbsp; sein.
 *Alle vier &Uuml;berg&auml;nge zwischen den zwei Ereignissen&nbsp; $A$&nbsp; und&nbsp; $B$&nbsp; sind m&ouml;glich.&nbsp; Daraus folgt, dass alle vier &Uuml;bergangswahrscheinlichkeiten ungleich Null sein werden.
-*Wegen&nbsp; ${\rm Pr}(B_\text{v=0})  \ne 0$&nbsp; und&nbsp; ${\rm Pr}(B \hspace{0.05cm} | \hspace{0.05cm}B) \ne 0$&nbsp; kann nat&uuml;rlich auch die Folge&nbsp; &bdquo;$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{0.15cm}...$"&nbsp; erzeugt werden, auch wenn diese bei den zwanzig hier ausgegebenen Markovketten nicht dabei ist.
+*Wegen&nbsp; ${\rm Pr}(B_\text{v=0})  \ne 0$&nbsp; und&nbsp; ${\rm Pr}(B \hspace{0.05cm} | \hspace{0.05cm}B) \ne 0$&nbsp; kann nat&uuml;rlich auch die Folge&nbsp; "$B\hspace{-0.05cm}-\hspace{-0.05cm}B \hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{-0.05cm}B\hspace{-0.05cm}-\hspace{0.15cm}...$"&nbsp; erzeugt werden, auch wenn diese bei den zwanzig hier ausgegebenen Markovketten nicht dabei ist.