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

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	[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]		[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]
	[[de:Aufgaben:4.~~4_MAP-_und_ML-Empfänger~~]]		[[de:Aufgaben:Aufgabe 4.4: Maximum–a–posteriori und Maximum–Likelihood]]

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	After transmission, the message is to be estimated by an optimal receiver.  Available are:		After transmission, the message is to be estimated by an optimal receiver.  Available are:
	* the  '''maximum likelihood receiver'''  $\rm (ML$  receiver$)$,  which does not know the occurrence probabilities  ${\rm Pr}(s = s_i)$,  with the decision rule:		* the  '''maximum likelihood receiver'''  $\rm (ML$  receiver$)$,  which does not know the occurrence probabilities  ${\rm Pr}(s = s_i)$,  with the decision rule:
	:$$\hat{m}_{\rm ML} = {\rm arg} \max_i \hspace{0.1cm} \big[ p_{r \|s } \hspace{0.05cm} (\rho		:$$\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},$$
	\|s_i ) \big]\hspace{0.05cm},$$

	* the  '''maximum-a-posteriori receiver'''  $\rm (MAP$  receiver$)$;  this receiver also considers the symbol probabilities of the source in its decision process:		* the  '''maximum-a-posteriori receiver'''  $\rm (MAP$  receiver$)$;  this receiver also considers the symbol probabilities of the source in its decision process:
	:$$\hat{m}_{\rm MAP} = {\rm arg} \max_i \hspace{0.1cm} \big[ {\rm Pr}( s = s_i) \cdot p_{r \|s } \hspace{0.05cm} (\rho		:$$\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}.$$
	\|s_i ) \big ]\hspace{0.05cm}.$$


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	'''(2)'''  For the first inference probability we are looking for holds:		'''(2)'''  For the first inference probability we are looking for holds:
	:$${\rm Pr} (s_0 \hspace{0.05cm}\| \hspace{0.05cm}r = +1) = \frac{{\rm Pr} ( r = +1 \hspace{0.05cm}\|\hspace{0.05cm}s_0 ) \cdot {\rm Pr} ( s_0)}{{\rm Pr} ( r = +1)}		:$${\rm Pr} (s_0 \hspace{0.05cm}\| \hspace{0.05cm}r = +1) = \frac{{\rm Pr} ( r = +1 \hspace{0.05cm}\|\hspace{0.05cm}s_0 ) \cdot {\rm Pr} ( s_0)}{{\rm Pr} ( r = +1)}= \frac{0.8 \cdot 0.75}{0.6} \hspace{0.05cm}\hspace{0.15cm}\underline {= 1}\hspace{0.05cm}.$$
	= \frac{0.8 \cdot 0.75}{0.6} \hspace{0.05cm}\hspace{0.15cm}\underline {= 1}\hspace{0.05cm}.$$

	*Correspondingly,  we obtain for the other probabilities:		*Correspondingly,  we obtain for the other probabilities:

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	[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]		[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]
			[[de:Aufgaben:4.4_MAP-_und_ML-Empfänger]]

Guenter at 14:39, 15 July 2022

2022-07-15T14:39:35Z

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	\|s_i ) \big]\hspace{0.05cm},$$		\|s_i ) \big]\hspace{0.05cm},$$

	* the  '''maximum-a-posteriori receiver'''$\rm (MAP$  receiver$)$;  this receiver also considers the symbol probabilities of the source in its decision process:		* the  '''maximum-a-posteriori receiver'''  $\rm (MAP$  receiver$)$;  this receiver also considers the symbol probabilities of the source in its decision process:
	:$$\hat{m}_{\rm MAP} = {\rm arg} \max_i \hspace{0.1cm} \big[ {\rm Pr}( s = s_i) \cdot p_{r \|s } \hspace{0.05cm} (\rho		:$$\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}.$$		\|s_i ) \big ]\hspace{0.05cm}.$$
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	:$${\rm Pr} ( r = 0) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 - {\rm Pr} ( r = +1) - {\rm Pr} ( r = -1) = 1 - 0.6 - 0.15 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.25}\hspace{0.05cm}.$$		:$${\rm Pr} ( r = 0) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 - {\rm Pr} ( r = +1) - {\rm Pr} ( r = -1) = 1 - 0.6 - 0.15 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.25}\hspace{0.05cm}.$$

	For the last probability also holds:		*For the last probability also holds:
	:$${\rm Pr} ( r = 0) = 0.75 \cdot 0.2 + 0.25 \cdot 0.4 = 0.25\hspace{0.05cm}.$$		:$${\rm Pr} ( r = 0) = 0.75 \cdot 0.2 + 0.25 \cdot 0.4 = 0.25\hspace{0.05cm}.$$

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	= \frac{0.8 \cdot 0.75}{0.6} \hspace{0.05cm}\hspace{0.15cm}\underline {= 1}\hspace{0.05cm}.$$		= \frac{0.8 \cdot 0.75}{0.6} \hspace{0.05cm}\hspace{0.15cm}\underline {= 1}\hspace{0.05cm}.$$

	Correspondingly, we obtain for the other probabilities:		*Correspondingly,  we obtain for the other probabilities:
	:$${\rm Pr} (s_1 \hspace{0.05cm}\| \hspace{0.05cm}r = +1) \hspace{-0.1cm} \ = \ 1 - {\rm Pr} (s_0 \hspace{0.05cm}\| \hspace{0.05cm}r = +1) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0}\hspace{0.05cm},$$		:$${\rm Pr} (s_1 \hspace{0.05cm}\| \hspace{0.05cm}r = +1) \hspace{-0.1cm} \ = \ 1 - {\rm Pr} (s_0 \hspace{0.05cm}\| \hspace{0.05cm}r = +1) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0}\hspace{0.05cm},$$
	:$${\rm Pr} (s_0 \hspace{0.05cm}\| \hspace{0.05cm}r = -1) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0}\hspace{0.05cm},$$		:$${\rm Pr} (s_0 \hspace{0.05cm}\| \hspace{0.05cm}r = -1) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0}\hspace{0.05cm},$$
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	'''(3)'''  Let $r = +1$. Then decides		'''(3)'''  Let  $r = +1$.  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  $s_0$,  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 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  $s_0$,  since ${\rm Pr} ( r = +1 \hspace{0.05cm}\| \hspace{0.05cm}s_0) = 0.8 > {\rm Pr} ( r = +1 \hspace{0.05cm}\| \hspace{0.05cm}s_1) = 0 \hspace{0.05cm}.$


	So the correct answer is <u>NO</u>.		So the correct answer is  <u>NO</u>.


	'''(4)'''  <u>NO</u> is also true under the condition "$r = \, –1$", since there is no connection between $s_0$ and "$r = \, –1$".		'''(4)'''  <u>NO</u>  is also true under the condition  "$r = \, –1$",  since there is no connection between  $s_0$  and  "$r = \, –1$".


	'''(5)'''  <u>Solutions 1 and 4</u> are correct:		'''(5)'''  <u>Solutions 1 and 4</u>  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  $s_0$,  since ${\rm Pr} (s_0 \hspace{0.05cm}\| \hspace{0.05cm}r = 0) = 0.6 > {\rm Pr} (s_1 \hspace{0.05cm}\| \hspace{0.05cm}r = 0) = 0.4 \hspace{0.05cm}.$
	*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,  the ML receiver will choose  $s_1$,  since  ${\rm Pr} ( r = 0 \hspace{0.05cm}\| \hspace{0.05cm}s_1) = 0.4 > {\rm Pr} ( r = 0 \hspace{0.05cm}\| \hspace{0.05cm}s_0) = 0.2 \hspace{0.05cm}.$


	'''(6)'''  The maximum likelihood receiver		'''(6)'''  The maximum likelihood receiver
	* decides for $s_0$ only if $r = +1$,		* decides for  $s_0$  only if  $r = +1$,
	* thus makes no error if $s_1$ was sent,
	* only makes an error when "$s_0$" and "$r = 0$" are combined:
	~~:$${\rm Pr} ({\rm symbol error} ) = {\rm Pr} ({\cal E } ) = 0.75 \cdot 0.2 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.15} \hspace{0.05cm}.$$~~

			* thus makes no error if  $s_1$  was sent,

	'''(7)'''  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  "$s_0$"  and  "$r = 0$"  are combined:
	:$${\rm Pr} ({\rm symbol error} ) = {\rm Pr} ({\cal E } ) = 0.25 \cdot 0.4 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.1} \hspace{0.05cm}.$$		:$${\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Pr} ({\cal E } ) = 0.75 \cdot 0.2 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.15} \hspace{0.05cm}.$$


			'''(7)'''  The MAP receiver,  on the other hand,  decides for  $s_0$ when  "$r = 0$".  So there is a symbol error only in the combination  "$s_1$"  and  "$r = 0$".  From this follows:
			:$${\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Pr} ({\cal E } ) = 0.25 \cdot 0.4 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.1} \hspace{0.05cm}.$$

	*The error probability here is lower than for the ML receiver,		*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  ${\rm Pr}(s_0)$ and ${\rm Pr}(s_1)$  are considered.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:24, 15 July 2022

2022-07-15T14:24:54Z

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	[[File:EN_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  $m_0 = 0$  and  $m_1 = 1$, represented by the signal values  $s_0$  ~~and~~  $s_1$~~,  respectively~~:		To illustrate  "maximum–a–posteriori"  $\rm (MAP)$  and  "maximum likelihood"  $\rm (ML)$  decision,  we now construct a very simple example with only two possible messages  $m_0 = 0$  and  $m_1 = 1$,  represented by the signal values  $s_0$  resp.  $s_1$:
	:$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_0 = +1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_0 = 0\hspace{0.05cm},$$		:$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_0 = +1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_0 = 0\hspace{0.05cm},$$
	:$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_1 = -1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_1 = 1\hspace{0.05cm}.$$		:$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_1 = -1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_1 = 1\hspace{0.05cm}.$$
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	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.  Available are:
	* the  '''maximum likelihood receiver'''  (ML receiver), which does not know the occurrence probabilities  ${\rm Pr}(s = s_i)$,  with the decision rule:		* the  '''maximum likelihood receiver'''  $\rm (ML$  receiver$)$,  which does not know the occurrence probabilities  ${\rm Pr}(s = s_i)$,  with the decision rule:
	:$$\hat{m}_{\rm ML} = {\rm arg} \max_i \hspace{0.1cm} \big[ p_{r \|s } \hspace{0.05cm} (\rho		:$$\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},$$		\|s_i ) \big]\hspace{0.05cm},$$

	* the  '''maximum-a-posteriori receiver'''  ~~(MAP~~ receiver); this receiver also considers the symbol probabilities of the source in its decision process:		* the  '''maximum-a-posteriori receiver'''$\rm (MAP$  receiver$)$;  this receiver also considers the symbol probabilities of the source in its decision process:
	:$$\hat{m}_{\rm MAP} = {\rm arg} \max_i \hspace{0.1cm} \big[ {\rm Pr}( s = s_i) \cdot p_{r \|s } \hspace{0.05cm} (\rho		:$$\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}.$$		\|s_i ) \big ]\hspace{0.05cm}.$$
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			Notes:
			*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimal_Receiver_Strategies\|"Optimal Receiver Strategies"]].

			*Reference is also made to the chapter  [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver\|"Structure of the Optimal Receiver]].

	~~''Notes:''~~		* The necessary statistical principles can be found in the chapter  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence\| "Statistical Dependence and Independence"]]  of the book  "Theory of Stochastic Signals".
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimal_Receiver_Strategies\|"Optimal Receiver Strategies"]].
	*Reference is also made to the chapter  [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver\|"Structure of the Optimal Receiver]].
	* The necessary statistical principles can be found in the chapter  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence\| "Statistical Dependence and Independence"]]  of the book "Theory of Stochastic Signals".


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	{Calculate the symbol error probability of the  '''ML receiver'''.		{Calculate the symbol error probability of the  '''ML receiver'''.
	\|type="{}"}		\|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  '''MAP receiver'''.		{Calculate the symbol error probability of the  '''MAP receiver'''.
	\|type="{}"}		\|type="{}"}
	${\rm Pr(symbol error)}\ = \ $ { 0.1 3% }		${\rm Pr(symbol\hspace{0.15cm}error)}\ = \ $ { 0.1 3% }
	</quiz>		</quiz>

Guenter at 15:09, 8 July 2022

2022-07-08T15:09:01Z

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver}}		{{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  $m_0 = 0$  and  $m_1 = 1$, represented by the signal values  $s_0$  and  $s_1$,  respectively:		To illustrate MAP and ML decision, we now construct a very simple example with only two possible messages  $m_0 = 0$  and  $m_1 = 1$, represented by the signal values  $s_0$  and  $s_1$,  respectively:
	:$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_0 = +1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_0 = 0\hspace{0.05cm},$$		:$$s \hspace{-0.15cm} \ = \ \hspace{-0.15cm}s_0 = +1 \hspace{0.2cm} \Longleftrightarrow \hspace{0.2cm}m = m_0 = 0\hspace{0.05cm},$$

Hwang at 13:26, 20 June 2022

2022-06-20T13:26:30Z

Show changes

Guenter at 11:59, 8 June 2022

2022-06-08T11:59:57Z

← Older revision		Revision as of 13:59, 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^]]

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

← Older revision		Revision as of 13:58, 8 June 2022
(No difference)

Javier: Text replacement - "”" to """

2021-05-28T14:37:29Z

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← Older revision		Revision as of 16:37, 28 May 2021
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	*Die Aufgabe gehört zum Kapitel  [[Digitalsignal%C3%BCbertragung/Optimale_Empf%C3%A4ngerstrategien\| Optimale Empfängerstrategien]].		*Die Aufgabe gehört zum Kapitel  [[Digitalsignal%C3%BCbertragung/Optimale_Empf%C3%A4ngerstrategien\| Optimale Empfängerstrategien]].
	*Bezug genommen wird auch auf das Kapitel  [[Digitalsignal%C3%BCbertragung/Struktur_des_optimalen_Empf%C3%A4ngers\| Struktur des optimalen Empfängers]].		*Bezug genommen wird auch auf das Kapitel  [[Digitalsignal%C3%BCbertragung/Struktur_des_optimalen_Empf%C3%A4ngers\| Struktur des optimalen Empfängers]].
	* Die notwendigen statistischen Grundlagen finden Sie im Kapitel  [[Theory_of_Stochastic_Signals/Statistische_Abh%C3%A4ngigkeit_und_Unabh%C3%A4ngigkeit\| Statistische Abhängigkeit und Unabhängigkeit]]  des Buches "Stochastische Signaltheorie~~”~~.		* Die notwendigen statistischen Grundlagen finden Sie im Kapitel  [[Theory_of_Stochastic_Signals/Statistische_Abh%C3%A4ngigkeit_und_Unabh%C3%A4ngigkeit\| Statistische Abhängigkeit und Unabhängigkeit]]  des Buches "Stochastische Signaltheorie".


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	${\rm Pr}(s_1\|r = 0) \hspace{0.45cm} = \ $ { 0.4 3% }		${\rm Pr}(s_1\|r = 0) \hspace{0.45cm} = \ $ { 0.4 3% }

	{Unterscheiden sich MAP– und ML–Empfänger unter der Voraussetzung  "$r = +1$~~”~~?		{Unterscheiden sich MAP– und ML–Empfänger unter der Voraussetzung  "$r = +1$"?
	\|type="()"}		\|type="()"}
	- ja,		- ja,
	+ nein.		+ nein.

	{Unterscheiden sich MAP– und ML–Empfänger unter der Voraussetzung  "$r = -1$~~”~~?		{Unterscheiden sich MAP– und ML–Empfänger unter der Voraussetzung  "$r = -1$"?
	\|type="()"}		\|type="()"}
	- ja,		- ja,
	+ nein.		+ nein.

	{Welche Aussagen gelten unter der Voraussetzung  "$r = 0$~~”~~?		{Welche Aussagen gelten unter der Voraussetzung  "$r = 0$"?
	\|type="[]"}		\|type="[]"}
	+ Der MAP–Empfänger entscheidet sich für  $s_0$.		+ Der MAP–Empfänger entscheidet sich für  $s_0$.
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	'''(4)'''  <u>NEIN</u> gilt auch unter der Voraussetzung "$r = \, –1$~~”~~, da keine Verbindung zwischen $s_0$ und "$r = \, –1$~~”~~ besteht.		'''(4)'''  <u>NEIN</u> gilt auch unter der Voraussetzung "$r = \, –1$", da keine Verbindung zwischen $s_0$ und "$r = \, –1$" besteht.


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	* entscheidet sich nur für $s_0$, wenn $r = +1$ ist,		* entscheidet sich nur für $s_0$, wenn $r = +1$ ist,
	* macht also keinen Fehler, wenn $s_1$ gesendet wurde,		* macht also keinen Fehler, wenn $s_1$ gesendet wurde,
	* macht nur einen Fehler bei der Kombination "$s_0$~~”~~ und "$r = 0$~~”~~:		* macht nur einen Fehler bei der Kombination "$s_0$" und "$r = 0$":
	:$${\rm Pr} ({\rm Symbolfehler} ) = {\rm Pr} ({\cal E } ) = 0.75 \cdot 0.2 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.15} \hspace{0.05cm}.$$		:$${\rm Pr} ({\rm Symbolfehler} ) = {\rm Pr} ({\cal E } ) = 0.75 \cdot 0.2 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.15} \hspace{0.05cm}.$$


	'''(7)'''  Der MAP–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:		'''(7)'''  Der MAP–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:
	:$${\rm Pr} ({\rm Symbolfehler} ) = {\rm Pr} ({\cal E } ) = 0.25 \cdot 0.4 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.1} \hspace{0.05cm}.$$		:$${\rm Pr} ({\rm Symbolfehler} ) = {\rm Pr} ({\cal E } ) = 0.25 \cdot 0.4 \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.1} \hspace{0.05cm}.$$