Aufgaben:Exercise 1.4: 2S/3E Channel Model - Revision history

Guenter at 14:17, 30 November 2021

2021-11-30T14:17:04Z

Guenter at 14:12, 30 November 2021

2021-11-30T14:12:42Z

Guenter at 14:06, 30 November 2021

2021-11-30T14:06:50Z

Noah at 19:56, 28 November 2021

2021-11-28T19:56:06Z

Guenter at 16:07, 12 October 2021

2021-10-12T16:07:18Z

Guenter at 16:06, 12 October 2021

2021-10-12T16:06:20Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.4: 2S/3E-Kanalmodell to Aufgaben:Exercise 1.4: 2S/3E Channel Model

2021-10-12T15:50:26Z

Guenter moved page Aufgabe 1.4: 2S/3E-Kanalmodell to Exercise 1.4: 2S/3E Channel Model

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"

Guenter at 15:30, 3 August 2020

2020-08-03T15:30:52Z

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

2020-07-09T09:48:22Z

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

@@ Line 64: / Line 64: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Only if the symbol&nbsp; $\rm L$&nbsp; was sent, the receiver can decide for the symbol&nbsp; $\rm L$&nbsp; at the given channel.
+'''(1)'''&nbsp; Only if the symbol&nbsp; $\rm L$&nbsp; was sent,&nbsp; the receiver can decide for the symbol&nbsp; $\rm L$&nbsp; at the given channel.
-*However, the probability for a received&nbsp; $\rm L$&nbsp; is smaller by a factor of&nbsp; $0.7$&nbsp; than for a sent one. From this follows:
+*However,&nbsp; the probability for a received&nbsp; $\rm L$&nbsp; is smaller by a factor of&nbsp; $0.7$&nbsp; than for a sent one.&nbsp; From this follows:
 :$${\rm Pr} (E_{\rm L}) = {\rm Pr}(S_{\rm L}) \cdot {\rm Pr} (E_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}S_{\rm L}) = 0.3 \cdot 0.7 \hspace{0.15cm}\underline {= \rm 0.21}.$$
-'''(2)'''&nbsp; To the event&nbsp; $E_{\rm H}$&nbsp; one comes from&nbsp;  $S_{\rm H}$&nbsp; as well as from&nbsp; $S_{\rm L}$&nbsp;. Therefore holds:
+'''(2)'''&nbsp; To the event&nbsp; $E_{\rm H}$&nbsp; one comes from&nbsp;  $S_{\rm H}$&nbsp; as well as from&nbsp; $S_{\rm L}$.&nbsp; Therefore holds:
 :$${\rm Pr} (E_{\rm H}) = {\rm Pr} (S_{\rm H}) \cdot {\rm Pr}  (E_{\rm H}\hspace{0.05cm}|\hspace{0.05cm}S_{\rm H}) + {\rm Pr} (S_{\rm L}) \cdot {\rm Pr}  (E_{\rm H}\hspace{0.05cm}|\hspace{0.05cm} S_{\rm L})= \rm 0.7 \cdot 0.9 + 0.3 \cdot 0.1\hspace{0.15cm}\underline { =  \rm 0.66}.$$
-'''(3)'''&nbsp; The events&nbsp; $E_{\rm H}$,&nbsp; $E_{\rm L}$&nbsp; and&nbsp; $E_{\rm K}$&nbsp; together form a complete system. It follows that:
+'''(3)'''&nbsp; The events&nbsp; $E_{\rm H}$,&nbsp; $E_{\rm L}$&nbsp; and&nbsp; $E_{\rm K}$&nbsp; together form a complete system.&nbsp; It follows that:
 :$${\rm Pr} (E_{\rm K}) = 1 - {\rm Pr}  (E_{\rm L}) - {\rm Pr} (E_{\rm H}) \hspace{0.15cm}\underline {= \rm 0.13}.$$
@@ Line 87: / Line 87: @@
-'''(5)'''&nbsp; If the symbol&nbsp; $\rm L$&nbsp; was received, only&nbsp; $\rm L$&nbsp; could have been sent. It follows that:
+'''(5)'''&nbsp; If the symbol&nbsp; $\rm L$&nbsp; was received,&nbsp; only&nbsp; $\rm L$&nbsp; could have been sent. It follows that:
 :$${\rm Pr} (S_{\rm L} \hspace{0.05cm}|\hspace{0.05cm} E_{\rm L}) \hspace{0.15cm}\underline {= \rm 1}.$$
-'''(6)'''&nbsp; For example, Bayes' theorem is suitable for solving this problem:
+'''(6)'''&nbsp; For example,&nbsp; Bayes' theorem is suitable for solving this problem:
 :$${\rm Pr} (S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm} E_{\rm K}) =\frac{ {\rm Pr} ( E_{\rm K} \hspace{0.05cm}|\hspace{0.05cm} S_{\rm L}) \cdot {\rm Pr} (S_{\rm L})}{{\rm Pr} (E_{\rm K})} =\frac{ \rm 0.2 \cdot 0.3}{\rm 0.13} = \frac{\rm 6}{\rm 13}\hspace{0.15cm}\underline { \approx \rm 0.462}.$$
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:EN_Sto_A_1_4.png|right|frame|$\rm 2S/3E$&nbsp; channel model]]
 A transmitter&nbsp; (German:&nbsp; "Sender" &nbsp; &rArr; &nbsp; subscript&nbsp; "S")&nbsp; emits the binary symbols&nbsp; $\rm L$&nbsp; $($event&nbsp; $S_{\rm L})$&nbsp; and&nbsp; $\rm H$&nbsp; $($event&nbsp; $S_{\rm H})$&nbsp;.
-*If conditions are good, the digital receiver&nbsp; (German:&nbsp; "Empfänger" &nbsp; &rArr; &nbsp; subscript&nbsp; "E")&nbsp; also decides only on the binary symbols&nbsp; $\rm L$&nbsp; $($event&nbsp; $E_{\rm L})$&nbsp; or&nbsp; $\rm H$&nbsp; $($event&nbsp; $E_{\rm H})$.
+*If conditions are good,&nbsp; the digital receiver&nbsp; (German:&nbsp; "Empfänger" &nbsp; &rArr; &nbsp; subscript&nbsp; "E")&nbsp; also decides only on the binary symbols&nbsp; $\rm L$&nbsp; $($event&nbsp; $E_{\rm L})$&nbsp; or&nbsp; $\rm H$&nbsp; $($event&nbsp; $E_{\rm H})$.
 *However,&nbsp;  if the receiver can suspect that an error has occurred during transmission,&nbsp; it makes no decision&nbsp; $($event&nbsp; $E_{\rm K})$;&nbsp; <br>$\rm K$&nbsp; here stands for&nbsp; "No decision".
@@ Line 11: / Line 11: @@
 *It can be seen that a transmitted&nbsp; $\rm L$&nbsp; may well be received as a symbol&nbsp; $\rm H$.&nbsp;
 *In contrast,&nbsp; the transition from&nbsp; $\rm H$&nbsp; to&nbsp; $\rm L$&nbsp; is not possible.
-*Let the symbol appearance probabilities at the transmitter be&nbsp; ${\rm Pr}(S_{\rm L}) = 0.3$&nbsp; and&nbsp; ${\rm Pr}(S_{\rm H}) = 0.7$.
+*Let the symbol probabilities at the transmitter be&nbsp; ${\rm Pr}(S_{\rm L}) = 0.3$&nbsp; and&nbsp; ${\rm Pr}(S_{\rm H}) = 0.7$.
@@ Line 23: / Line 23: @@
 *The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video
-:[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] &nbsp; $\Rightarrow$ &nbsp; "Statistical dependence and independence".
+::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] &nbsp; $\Rightarrow$ &nbsp; "Statistical dependence and independence".

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}}
-[[File:EN_Sto_A_1_4.png|right|frame|$\rm 2S/3E$ channel model]]
+[[File:EN_Sto_A_1_4.png|right|frame|$\rm 2S/3E$&nbsp; channel model]]
-A transmitter emits the binary symbols&nbsp; $\rm L$&nbsp; $($event&nbsp; $S_{\rm L})$&nbsp; and&nbsp; $\rm H$&nbsp; $($event&nbsp; $S_{\rm H})$&nbsp;.
+A transmitter&nbsp; (German:&nbsp; "Sender" &nbsp; &rArr; &nbsp; subscript&nbsp; "S")&nbsp; emits the binary symbols&nbsp; $\rm L$&nbsp; $($event&nbsp; $S_{\rm L})$&nbsp; and&nbsp; $\rm H$&nbsp; $($event&nbsp; $S_{\rm H})$&nbsp;.
-*If conditions are good, the digital receiver also decides only on the binary symbols&nbsp; $\rm L$&nbsp; $($Ereignis&nbsp; $E_{\rm L})$&nbsp; or&nbsp; $\rm H$&nbsp; $($event&nbsp; $E_{\rm H})$.
+*If conditions are good, the digital receiver&nbsp; (German:&nbsp; "Empfänger" &nbsp; &rArr; &nbsp; subscript&nbsp; "E")&nbsp; also decides only on the binary symbols&nbsp; $\rm L$&nbsp; $($event&nbsp; $E_{\rm L})$&nbsp; or&nbsp; $\rm H$&nbsp; $($event&nbsp; $E_{\rm H})$.
-*However, if the receiver can suspect that an error has occurred during transmission, it makes no decision&nbsp; $($event&nbsp; $E_{\rm K})$;&nbsp; $\rm K$&nbsp; here stands for "No decision").
+*However,&nbsp;  if the receiver can suspect that an error has occurred during transmission,&nbsp; it makes no decision&nbsp; $($event&nbsp; $E_{\rm K})$;&nbsp; <br>$\rm K$&nbsp; here stands for&nbsp; "No decision".
-The diagram shows a simple channel model in terms of transition probabilities.&nbsp; It can be seen that a transmitted&nbsp; $\rm L$&nbsp; may well be received as a symbol&nbsp; $\rm H$&nbsp;.&nbsp; In contrast, the transition from&nbsp; $\rm H$&nbsp; to&nbsp; $\rm L$&nbsp; is not possible.
+The diagram shows a simple channel model in terms of transition probabilities.&nbsp;
+*It can be seen that a transmitted&nbsp; $\rm L$&nbsp; may well be received as a symbol&nbsp; $\rm H$.&nbsp;
-Let the symbol appearance probabilities at the transmitter be&nbsp; ${\rm Pr}(S_{\rm L}) = 0.3$&nbsp; and&nbsp; ${\rm Pr}(S_{\rm H}) = 0.7$.
+*In contrast,&nbsp; the transition from&nbsp; $\rm H$&nbsp; to&nbsp; $\rm L$&nbsp; is not possible.
@@ Line 28: / Line 29: @@
 <quiz display=simple>
-{What is the probability that the receiver chooses the symbol&nbsp; $\rm L$&nbsp;?
+{What is the probability that the receiver chooses the symbol&nbsp; $\rm L$?
 |type="{}"}
 ${\rm Pr}(E_{\rm L}) \ = \ $  { 0.21 3% }
-{What is the probability that the receiver chooses the symbo&nbsp; $\rm H$&nbsp;?
+{What is the probability that the receiver chooses the symbol&nbsp; $\rm H$?
 |type="{}"}
 ${\rm Pr}(E_{\rm H}) \ = \ $ { 0.66 3% }
@@ Line 45: / Line 46: @@
 {What is the probability that the receiver makes a wrong decision?
 |type="{}"}
-$\text{Pr(falsche Entscheidung)} \ = \ $  { 0.03 3% }
+$\text{Pr(wrong decision)} \ = \ $  { 0.03 3% }
-{What is the probability that symbol&nbsp; $\rm L$&nbsp; was actually sent if the receiver decided to use symbol&nbsp; $\rm L$&nbsp;?
+{What is the probability that symbol&nbsp; $\rm L$&nbsp; was actually sent if the receiver decided to use symbol&nbsp; $\rm L$?
 |type="{}"}
 ${\rm Pr}(S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}E_{\rm L} ) \ = \ $  { 1 3% }

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Statistische Abhängigkeit und Unabhängigkeit
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}}
-[[File:EN_Sto_A_1_4.png|right|frame|$\rm 2S/3E$-Kanalmodell]]
+[[File:EN_Sto_A_1_4.png|right|frame|$\rm 2S/3E$ channel model]]
-Ein Sender gibt die binären Symbole&nbsp; $\rm L$&nbsp; $($Ereignis&nbsp; $S_{\rm L})$&nbsp; und&nbsp; $\rm H$&nbsp; $($Ereignis&nbsp; $S_{\rm H})$&nbsp; ab.
+A transmitter emits the binary symbols&nbsp; $\rm L$&nbsp; $($event&nbsp; $S_{\rm L})$&nbsp; and&nbsp; $\rm H$&nbsp; $($event&nbsp; $S_{\rm H})$&nbsp;.
-*Bei guten Bedingungen entscheidet sich der Digitalempfänger ebenfalls nur für die Binärsymbole&nbsp; $\rm L$&nbsp; $($Ereignis&nbsp; $E_{\rm L})$&nbsp; oder&nbsp; $\rm H$&nbsp; $($Ereignis&nbsp; $E_{\rm H})$.
+*If conditions are good, the digital receiver also decides only on the binary symbols&nbsp; $\rm L$&nbsp; $($Ereignis&nbsp; $E_{\rm L})$&nbsp; or&nbsp; $\rm H$&nbsp; $($event&nbsp; $E_{\rm H})$.
-*Kann der Empfänger allerdings vermuten, dass bei der Übertragung ein Fehler aufgetreten ist, so trifft er keine Entscheidung&nbsp; $($Ereignis&nbsp; $E_{\rm K})$;&nbsp; $\rm K$&nbsp; steht hierbei für „Keine Entscheidung”).
+*However, if the receiver can suspect that an error has occurred during transmission, it makes no decision&nbsp; $($event&nbsp; $E_{\rm K})$;&nbsp; $\rm K$&nbsp; here stands for "No decision").
-Die Grafik zeigt ein einfaches Kanalmodell in Form von Übergangswahrscheinlichkeiten.&nbsp; Es ist zu erkennen, dass ein gesendetes&nbsp; $\rm L$&nbsp; durchaus als Symbol&nbsp; $\rm H$&nbsp; empfangen werden kann.&nbsp; Dagegen ist der Übergang von&nbsp; $\rm H$&nbsp; nach&nbsp; $\rm L$&nbsp; nicht möglich.
+The diagram shows a simple channel model in terms of transition probabilities.&nbsp; It can be seen that a transmitted&nbsp; $\rm L$&nbsp; may well be received as a symbol&nbsp; $\rm H$&nbsp;.&nbsp; In contrast, the transition from&nbsp; $\rm H$&nbsp; to&nbsp; $\rm L$&nbsp; is not possible.
-Die Symbolauftrittswahrscheinlichkeiten am Sender seien&nbsp; ${\rm Pr}(S_{\rm L}) = 0.3$&nbsp; und&nbsp; ${\rm Pr}(S_{\rm H}) = 0.7$.
+Let the symbol appearance probabilities at the transmitter be&nbsp; ${\rm Pr}(S_{\rm L}) = 0.3$&nbsp; and&nbsp; ${\rm Pr}(S_{\rm H}) = 0.7$.
@@ Line 19: / Line 18: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Statistische_Abhängigkeit_und_Unabhängigkeit|Statistische Abhängigkeit und Unabhängigkeit]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence|Statistical dependence and independence]].
-*Eine Zusammenfassung der theoretischen Grundlagen mit Beispielen bringt das Lernvideo&nbsp; [[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]].
+*The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie groß ist die Wahrscheinlichkeit dafür, dass sich der Empfänger für das Symbol&nbsp; $\rm L$&nbsp; entscheidet?
+{What is the probability that the receiver chooses the symbol&nbsp; $\rm L$&nbsp;?
 |type="{}"}
 ${\rm Pr}(E_{\rm L}) \ = \ $  { 0.21 3% }
-{Wie groß ist die Wahrscheinlichkeit dafür, dass sich der Empfänger für das Symbol&nbsp; $\rm H$&nbsp; entscheidet?
+{What is the probability that the receiver chooses the symbo&nbsp; $\rm H$&nbsp;?
 |type="{}"}
 ${\rm Pr}(E_{\rm H}) \ = \ $ { 0.66 3% }
-{Wie groß ist die Wahrscheinlichkeit dafür, dass der Empfänger keine Entscheidung trifft?
+{What is the probability that the receiver does not make a decision?
 |type="{}"}
 ${\rm Pr}(E_{\rm K}) \ = \ $  { 0.13 3% }
-{Mit welcher Wahrscheinlichkeit entscheidet der Empfänger falsch?
+{What is the probability that the receiver makes a wrong decision?
 |type="{}"}
 $\text{Pr(falsche Entscheidung)} \ = \ $  { 0.03 3% }
-{Wie groß ist die Wahrscheinlichkeit, dass tatsächlich  das Symbol&nbsp; $\rm L$&nbsp; gesendet wurde, wenn sich der Empfänger für das Symbol&nbsp; $\rm L$&nbsp; entschieden hat?
+{What is the probability that symbol&nbsp; $\rm L$&nbsp; was actually sent if the receiver decided to use symbol&nbsp; $\rm L$&nbsp;?
 |type="{}"}
 ${\rm Pr}(S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}E_{\rm L} ) \ = \ $  { 1 3% }
-{Wie groß ist die Wahrscheinlichkeit, dass das Symbol&nbsp; $\rm L$&nbsp; gesendet wurde, wenn der Empfänger keine Entscheidung trifft?
+{What is the probability that symbol&nbsp; $\rm L$&nbsp; was sent if the receiver does not make a decision?
 |type="{}"}
 ${\rm Pr}(S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}E_{\rm K} ) \ =\ $ { 0.4614 3% }
@@ Line 61: / Line 61: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Nur wenn das Symbol&nbsp; $\rm L$&nbsp; gesendet wurde, kann sich der Empf&auml;nger beim gegebenen Kanal f&uuml;r das Symbol&nbsp; $\rm L$&nbsp; entscheiden.
+'''(1)'''&nbsp; Only if the symbol&nbsp; $\rm L$&nbsp; was sent, the receiver can decide for the symbol&nbsp; $\rm L$&nbsp; at the given channel.
-*Die Wahrscheinlichkeit f&uuml;r ein empfangenes&nbsp; $\rm L$&nbsp; ist allerdings um den Faktor&nbsp; $0.7$&nbsp; kleiner als f&uuml;r ein gesendetes. Daraus folgt:
+*However, the probability for a received&nbsp; $\rm L$&nbsp; is smaller by a factor of&nbsp; $0.7$&nbsp; than for a sent one. From this follows:
 :$${\rm Pr} (E_{\rm L}) = {\rm Pr}(S_{\rm L}) \cdot {\rm Pr} (E_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}S_{\rm L}) = 0.3 \cdot 0.7 \hspace{0.15cm}\underline {= \rm 0.21}.$$
-'''(2)'''&nbsp; Zum Ereignis&nbsp; $E_{\rm H}$&nbsp; kommt man sowohl von&nbsp;  $S_{\rm H}$&nbsp; als auch von&nbsp; $S_{\rm L}$&nbsp; aus. Deshalb gilt:
+'''(2)'''&nbsp; To the event&nbsp; $E_{\rm H}$&nbsp; one comes from&nbsp;  $S_{\rm H}$&nbsp; as well as from&nbsp; $S_{\rm L}$&nbsp;. Therefore holds:
 :$${\rm Pr} (E_{\rm H}) = {\rm Pr} (S_{\rm H}) \cdot {\rm Pr}  (E_{\rm H}\hspace{0.05cm}|\hspace{0.05cm}S_{\rm H}) + {\rm Pr} (S_{\rm L}) \cdot {\rm Pr}  (E_{\rm H}\hspace{0.05cm}|\hspace{0.05cm} S_{\rm L})= \rm 0.7 \cdot 0.9 + 0.3 \cdot 0.1\hspace{0.15cm}\underline { =  \rm 0.66}.$$
-'''(3)'''&nbsp; Die Ereignisse&nbsp; $E_{\rm H}$,&nbsp; $E_{\rm L}$&nbsp; und&nbsp; $E_{\rm K}$&nbsp; bilden zusammen ein vollst&auml;ndiges System. Daraus folgt:
+'''(3)'''&nbsp; The events&nbsp; $E_{\rm H}$,&nbsp; $E_{\rm L}$&nbsp; and&nbsp; $E_{\rm K}$&nbsp; together form a complete system. It follows that:
 :$${\rm Pr} (E_{\rm K}) = 1 - {\rm Pr}  (E_{\rm L}) - {\rm Pr} (E_{\rm H}) \hspace{0.15cm}\underline {= \rm 0.13}.$$
-'''(4)'''&nbsp; Eine falsche Entscheidung kann man mengentheoretisch wie folgt charakterisieren:
+'''(4)'''&nbsp; A wrong decision can be characterized in set-theoretic terms as follows:
-'''(5)'''&nbsp; Wenn das Symbol&nbsp; $\rm L$&nbsp; empfangen wurde, kann nur&nbsp; $\rm L$&nbsp; gesendet worden sein. Daraus folgt:
 :$${\rm Pr} (S_{\rm L} \hspace{0.05cm}|\hspace{0.05cm} E_{\rm L}) \hspace{0.15cm}\underline {= \rm 1}.$$
-'''(6)'''&nbsp; Zur L&ouml;sung dieser Aufgabe eignet sich zum Beispiel der Satz von Bayes:
+'''(6)'''&nbsp; For example, Bayes' theorem is suitable for solving this problem:
 :$${\rm Pr} (S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm} E_{\rm K}) =\frac{ {\rm Pr} ( E_{\rm K} \hspace{0.05cm}|\hspace{0.05cm} S_{\rm L}) \cdot {\rm Pr} (S_{\rm L})}{{\rm Pr} (E_{\rm K})} =\frac{ \rm 0.2 \cdot 0.3}{\rm 0.13} = \frac{\rm 6}{\rm 13}\hspace{0.15cm}\underline { \approx \rm 0.462}.$$
 {{ML-Fuß}}

← Older revision		Revision as of 16:07, 12 October 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence ~~and~~ Independence^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence/Independence^]]

← Older revision		Revision as of 16:06, 12 October 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 ~~Statistische (Un-)Abhängigkeit~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence and Independence^]]

← Older revision	Revision as of 15:50, 12 October 2021
(No difference)

← Older revision		Revision as of 12:44, 23 March 2021
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−	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^1.3 Statistische (Un-)Abhängigkeit^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistische (Un-)Abhängigkeit^]]