Aufgaben:Exercise 4.4: Maximum–a–posteriori and Maximum–Likelihood - Revision history

Guenter at 14:39, 15 July 2022

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Guenter at 14:24, 15 July 2022

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Guenter at 15:09, 8 July 2022

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Hwang at 13:26, 20 June 2022

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Guenter at 11:59, 8 June 2022

2022-06-08T11:59:57Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.4: Maximum–a–posteriori und Maximum–Likelihood to Aufgaben:Exercise 4.4: Maximum–a–posteriori and Maximum–Likelihood

2022-06-08T11:58:02Z

Guenter moved page Aufgabe 4.4: Maximum–a–posteriori und Maximum–Likelihood to Exercise 4.4: Maximum–a–posteriori and Maximum–Likelihood

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2021-05-28T14:37:29Z

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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:48:11Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

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

2020-07-09T09:49:42Z

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

@@ Line 21: / Line 21: @@
 |s_i ) \big]\hspace{0.05cm},$$
-* the &nbsp;'''maximum-a-posteriori receiver'''$\rm (MAP$&nbsp; receiver$)$;&nbsp; this receiver also considers the symbol probabilities of the source in its decision process:
+* the &nbsp;'''maximum-a-posteriori receiver'''&nbsp; $\rm (MAP$&nbsp; receiver$)$;&nbsp; this receiver also considers the symbol probabilities of the source in its decision process:
 :$$\hat{m}_{\rm MAP} = {\rm arg} \max_i \hspace{0.1cm} \big[ {\rm Pr}( s = s_i) \cdot p_{r |s } \hspace{0.05cm} (\rho
 |s_i ) \big ]\hspace{0.05cm}.$$
@@ Line 89: / Line 89: @@
 :$${\rm Pr} ( r = 0) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 - {\rm Pr} ( r = +1) - {\rm Pr} ( r = -1) = 1 - 0.6 - 0.15  \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.25}\hspace{0.05cm}.$$
-For the last probability also holds:
+*For the last probability also holds:
 :$${\rm Pr} ( r = 0) = 0.75 \cdot 0.2 + 0.25 \cdot 0.4 = 0.25\hspace{0.05cm}.$$
@@ Line 97: / Line 97: @@
 = \frac{0.8 \cdot 0.75}{0.6} \hspace{0.05cm}\hspace{0.15cm}\underline {= 1}\hspace{0.05cm}.$$
-Correspondingly, we obtain for the other probabilities:
+*Correspondingly,&nbsp; we obtain for the other probabilities:
 :$${\rm Pr} (s_1 \hspace{0.05cm}| \hspace{0.05cm}r = +1) \hspace{-0.1cm} \ = \ 1 - {\rm Pr} (s_0 \hspace{0.05cm}| \hspace{0.05cm}r = +1)  \hspace{0.05cm}\hspace{0.15cm}\underline {= 0}\hspace{0.05cm},$$
 :$${\rm Pr} (s_0 \hspace{0.05cm}| \hspace{0.05cm}r = -1) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0}\hspace{0.05cm},$$
@@ Line 105: / Line 105: @@
-'''(3)'''&nbsp; Let $r = +1$. Then decides
+'''(3)'''&nbsp; Let&nbsp; $r = +1$.&nbsp; Then decides
-* the MAP receiver for $s_0$, because ${\rm Pr} (s_0 \hspace{0.05cm}| \hspace{0.05cm}r = +1) = 1 > {\rm Pr} (s_1 \hspace{0.05cm}| \hspace{0.05cm}r = +1)= 0\hspace{0.05cm},$
+* the MAP receiver for&nbsp; $s_0$,&nbsp; because ${\rm Pr} (s_0 \hspace{0.05cm}| \hspace{0.05cm}r = +1) = 1 > {\rm Pr} (s_1 \hspace{0.05cm}| \hspace{0.05cm}r = +1)= 0\hspace{0.05cm},$
-* the ML receiver likewise for $s_0$, since ${\rm Pr} ( r = +1 \hspace{0.05cm}| \hspace{0.05cm}s_0) = 0.8 > {\rm Pr} ( r = +1 \hspace{0.05cm}| \hspace{0.05cm}s_1) = 0 \hspace{0.05cm}.$
+* the ML receiver likewise for&nbsp; $s_0$,&nbsp; since ${\rm Pr} ( r = +1 \hspace{0.05cm}| \hspace{0.05cm}s_0) = 0.8 > {\rm Pr} ( r = +1 \hspace{0.05cm}| \hspace{0.05cm}s_1) = 0 \hspace{0.05cm}.$
-So the correct answer is <u>NO</u>.
+So the correct answer is&nbsp; <u>NO</u>.
-'''(4)'''&nbsp; <u>NO</u> is also true under the condition "$r = \, &ndash;1$", since there is no connection between $s_0$ and "$r = \, &ndash;1$".
+'''(4)'''&nbsp; <u>NO</u>&nbsp; is also true under the condition&nbsp; "$r = \, &ndash;1$",&nbsp; since there is no connection between&nbsp; $s_0$&nbsp; and&nbsp; "$r = \, &ndash;1$".
-'''(5)'''&nbsp; <u>Solutions 1 and 4</u> are correct:
+'''(5)'''&nbsp; <u>Solutions 1 and 4</u>&nbsp;  are correct:
-*The MAP receiver will choose event $s_0$, since ${\rm Pr} (s_0 \hspace{0.05cm}| \hspace{0.05cm}r = 0) = 0.6 > {\rm Pr} (s_1 \hspace{0.05cm}| \hspace{0.05cm}r = 0) = 0.4 \hspace{0.05cm}.$
+*The MAP receiver will choose event&nbsp;  $s_0$,&nbsp;  since ${\rm Pr} (s_0 \hspace{0.05cm}| \hspace{0.05cm}r = 0) = 0.6 > {\rm Pr} (s_1 \hspace{0.05cm}| \hspace{0.05cm}r = 0) = 0.4 \hspace{0.05cm}.$
-*In contrast, the ML receiver will choose $s_1$, since ${\rm Pr} ( r = 0 \hspace{0.05cm}| \hspace{0.05cm}s_1) = 0.4 > {\rm Pr} ( r = 0 \hspace{0.05cm}| \hspace{0.05cm}s_0) = 0.2 \hspace{0.05cm}.$
+*In contrast,&nbsp;  the ML receiver will choose&nbsp;  $s_1$,&nbsp;  since&nbsp;  ${\rm Pr} ( r = 0 \hspace{0.05cm}| \hspace{0.05cm}s_1) = 0.4 > {\rm Pr} ( r = 0 \hspace{0.05cm}| \hspace{0.05cm}s_0) = 0.2 \hspace{0.05cm}.$
 '''(6)'''&nbsp; The maximum likelihood receiver
-* decides for $s_0$ only if $r = +1$,
+* decides for&nbsp;  $s_0$&nbsp;  only if&nbsp;  $r = +1$,
-'''(7)'''&nbsp; The MAP receiver, on the other hand, decides to use $s_0$ when "$r = 0$". So there is a symbol error only in the combination "$s_1$" and "$r = 0$". From this follows:
+* only makes an error when&nbsp;  "$s_0$"&nbsp;  and&nbsp;  "$r = 0$"&nbsp;  are combined:
-:$${\rm Pr} ({\rm symbol error} ) = {\rm Pr} ({\cal E } ) = 0.25 \cdot 0.4  \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.1}  \hspace{0.05cm}.$$
+:$${\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Pr} ({\cal E } ) = 0.75 \cdot 0.2  \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.15}  \hspace{0.05cm}.$$
 *The error probability here is lower than for the ML receiver,
-*because now also the different apriori probabilities ${\rm Pr}(s_0)$ and ${\rm Pr}(s_1)$ are considered.
+*because now also the different a-priori probabilities&nbsp; ${\rm Pr}(s_0)$ and ${\rm Pr}(s_1)$&nbsp; are considered.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:EN_Dig_A_4_4.png|right|frame|Channel transition probabilities]]
-To illustrate MAP and ML decision, we now construct a very simple example with only two possible messages&nbsp; $m_0 = 0$&nbsp; and&nbsp; $m_1 = 1$, represented by the signal values&nbsp; $s_0$&nbsp; and&nbsp; $s_1$,&nbsp; respectively:
+To illustrate&nbsp; "maximum–a–posteriori"&nbsp; $\rm (MAP)$&nbsp; and&nbsp; "maximum likelihood"&nbsp; $\rm (ML)$&nbsp;  decision,&nbsp; we now construct a very simple example with only two possible messages&nbsp; $m_0 = 0$&nbsp; and&nbsp; $m_1 = 1$,&nbsp; represented by the signal values&nbsp; $s_0$&nbsp; resp.&nbsp; $s_1$:
 :$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_0 = +1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_0 = 0\hspace{0.05cm},$$
 :$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_1 = -1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_1 = 1\hspace{0.05cm}.$$
@@ Line 16: / Line 16: @@
-After transmission, the transmitted message is to be estimated by an optimal receiver. Available are:
+After transmission, the  message is to be estimated by an optimal receiver.&nbsp; Available are:
-* the &nbsp;'''maximum likelihood receiver'''&nbsp; (ML receiver), which does not know the occurrence probabilities&nbsp; ${\rm Pr}(s = s_i)$,&nbsp; with the decision rule:
+* the &nbsp;'''maximum likelihood receiver'''&nbsp; $\rm (ML$&nbsp; receiver$)$,&nbsp; which does not know the occurrence probabilities&nbsp; ${\rm Pr}(s = s_i)$,&nbsp; with the decision rule:
 :$$\hat{m}_{\rm ML} = {\rm arg} \max_i \hspace{0.1cm} \big[ p_{r |s } \hspace{0.05cm} (\rho
 |s_i ) \big]\hspace{0.05cm},$$
-* the &nbsp;'''maximum-a-posteriori receiver'''&nbsp; (MAP receiver); this receiver also considers the symbol probabilities of the source in its decision process:
+* the &nbsp;'''maximum-a-posteriori receiver'''$\rm (MAP$&nbsp; receiver$)$;&nbsp; this receiver also considers the symbol probabilities of the source in its decision process:
 :$$\hat{m}_{\rm MAP} = {\rm arg} \max_i \hspace{0.1cm} \big[ {\rm Pr}( s = s_i) \cdot p_{r |s } \hspace{0.05cm} (\rho
 |s_i ) \big ]\hspace{0.05cm}.$$
@@ Line 28: / Line 28: @@
-''Notes:''
+* The necessary statistical principles can be found in the chapter&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence| "Statistical Dependence and Independence"]]&nbsp; of the book&nbsp; "Theory of Stochastic Signals".
@@ Line 75: / Line 75: @@
 {Calculate the symbol error probability of the &nbsp;'''ML receiver'''.
 |type="{}"}
-${\rm Pr(symbol error)}\ = \ $  { 0.15 3% }
+${\rm Pr(symbol\hspace{0.15cm} error)}\ = \ $  { 0.15 3% }
 {Calculate the symbol error probability of the &nbsp;'''MAP receiver'''.
 |type="{}"}
-${\rm Pr(symbol error)}\ = \ $ { 0.1 3% }
+${\rm Pr(symbol\hspace{0.15cm}error)}\ = \ $ { 0.1 3% }
 </quiz>

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver}}
-[[File:P_ID2013__Dig_A_4_4.png|right|frame|Channel transition probabilities]]
+[[File:EN_Dig_A_4_4.png|right|frame|Channel transition probabilities]]
 To illustrate MAP and ML decision, we now construct a very simple example with only two possible messages&nbsp; $m_0 = 0$&nbsp; and&nbsp; $m_1 = 1$, represented by the signal values&nbsp; $s_0$&nbsp; and&nbsp; $s_1$,&nbsp; respectively:
 :$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_0 = +1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_0 = 0\hspace{0.05cm},$$

← Older revision		Revision as of 11:59, 8 June 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^4.2 ~~Struktur des optimalen Empfängers~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]

@@ Line 33: / Line 33: @@
 *Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignal%C3%BCbertragung/Optimale_Empf%C3%A4ngerstrategien| Optimale Empfängerstrategien]].
 *Bezug genommen wird auch auf das Kapitel&nbsp; [[Digitalsignal%C3%BCbertragung/Struktur_des_optimalen_Empf%C3%A4ngers| Struktur des optimalen Empfängers]].
-* Die notwendigen statistischen Grundlagen finden Sie im Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Statistische_Abh%C3%A4ngigkeit_und_Unabh%C3%A4ngigkeit| Statistische Abhängigkeit und Unabhängigkeit]]&nbsp; des Buches "Stochastische Signaltheorie&rdquo;.
+* Die notwendigen statistischen Grundlagen finden Sie im Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Statistische_Abh%C3%A4ngigkeit_und_Unabh%C3%A4ngigkeit| Statistische Abhängigkeit und Unabhängigkeit]]&nbsp; des Buches "Stochastische Signaltheorie".
@@ Line 56: / Line 56: @@
 ${\rm Pr}(s_1|r = 0) \hspace{0.45cm} = \ $ { 0.4 3% }
-{Unterscheiden sich MAP&ndash; und ML&ndash;Empfänger unter der Voraussetzung&nbsp; "$r = +1$&rdquo;?
+{Unterscheiden sich MAP&ndash; und ML&ndash;Empfänger unter der Voraussetzung&nbsp; "$r = +1$"?
 |type="()"}
 - ja,
 + nein.
-{Unterscheiden sich MAP&ndash; und ML&ndash;Empfänger unter der Voraussetzung&nbsp; "$r = -1$&rdquo;?
+{Unterscheiden sich MAP&ndash; und ML&ndash;Empfänger unter der Voraussetzung&nbsp; "$r = -1$"?
 |type="()"}
 - ja,
 + nein.
-{Welche Aussagen gelten unter der Voraussetzung&nbsp; "$r = 0$&rdquo;?
+{Welche Aussagen gelten unter der Voraussetzung&nbsp; "$r = 0$"?
 |type="[]"}
 + Der MAP&ndash;Empfänger entscheidet sich für&nbsp; $s_0$.
@@ Line 113: / Line 113: @@
-'''(4)'''&nbsp; <u>NEIN</u> gilt auch unter der Voraussetzung "$r = \, &ndash;1$&rdquo;, da keine Verbindung zwischen $s_0$ und "$r = \, &ndash;1$&rdquo; besteht.
+'''(4)'''&nbsp; <u>NEIN</u> gilt auch unter der Voraussetzung "$r = \, &ndash;1$", da keine Verbindung zwischen $s_0$ und "$r = \, &ndash;1$" besteht.
@@ Line 124: / Line 124: @@
 * entscheidet sich nur für $s_0$, wenn $r = +1$ ist,
 * macht also keinen Fehler, wenn $s_1$ gesendet wurde,
-* macht nur einen Fehler bei der Kombination "$s_0$&rdquo; und "$r = 0$&rdquo;:
+* macht nur einen Fehler bei der Kombination "$s_0$" und "$r = 0$":
 :$${\rm Pr} ({\rm Symbolfehler} ) = {\rm Pr} ({\cal E } ) = 0.75 \cdot 0.2  \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.15}  \hspace{0.05cm}.$$
-'''(7)'''&nbsp; Der MAP&ndash;Empfänger entscheidet sich dagegen bei "$r = 0$&rdquo; für $s_0$. Einen Symbolfehler gibt es also nur in der Kombination "$s_1$&rdquo; und "$r = 0$&rdquo;. Daraus folgt:
+'''(7)'''&nbsp; Der MAP&ndash;Empfänger entscheidet sich dagegen bei "$r = 0$" für $s_0$. Einen Symbolfehler gibt es also nur in der Kombination "$s_1$" und "$r = 0$". Daraus folgt:
 :$${\rm Pr} ({\rm Symbolfehler} ) = {\rm Pr} ({\cal E } ) = 0.25 \cdot 0.4  \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.1}  \hspace{0.05cm}.$$

← Older revision	Revision as of 11:58, 8 June 2022
(No difference)

← Older revision		Revision as of 12:48, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.2 Struktur des optimalen Empfängers^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.2 Struktur des optimalen Empfängers^]]