Aufgaben:Exercise 4.08Z: Error Probability with Three Symbols - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]
	[[de:4.~~08Z_Fehlerwahrscheinlichkeit~~]]		[[de:Aufgaben:Aufgabe 4.08Z: Fehlerwahrscheinlichkeit bei drei Symbolen]]

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	The diagram shows exactly the same signal space constellation as in  [[Aufgaben:Exercise_4.08:_Decision_Regions_at_Three_Symbols\|"Exercise 4.8"]]:		The diagram shows exactly the same signal space constellation as in  [[Aufgaben:Exercise_4.08:_Decision_Regions_at_Three_Symbols\|"Exercise 4.8"]]:
	* the  $M = 3$  possible transmitted signals,  viz.		* the  $M = 3$  possible transmitted signals,  viz.
	:$$\boldsymbol{ s }_0 = (-1, \hspace{0.1cm}1)\hspace{0.05cm}, \hspace{0.2cm}		:$$\boldsymbol{ s }_0 = (-1, \hspace{0.1cm}1)\hspace{0.05cm}, \hspace{0.2cm}\boldsymbol{ s }_1 = (1, \hspace{0.1cm}2)\hspace{0.05cm}, \hspace{0.2cm}\boldsymbol{ s }_2 = (2, \hspace{0.1cm}-1)\hspace{0.05cm}.$$
	\boldsymbol{ s }_1 = (1, \hspace{0.1cm}2)\hspace{0.05cm}, \hspace{0.2cm}
	\boldsymbol{ s }_2 = (2, \hspace{0.1cm}-1)\hspace{0.05cm}.$$

	* the  $M = 3$  decision boundaries		* the  $M = 3$  decision boundaries
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	In contrast,  in this exercise we consider a two–dimensional uniform distribution for the noise PDF:		In contrast,  in this exercise we consider a two–dimensional uniform distribution for the noise PDF:
	:$$\boldsymbol{ p }_{\boldsymbol{ n }} (x,\hspace{0.15cm} y) =		:$$\boldsymbol{ p }_{\boldsymbol{ n }} (x,\hspace{0.15cm} y) =\left\{ \begin{array}{c} K\\0 \end{array} \right.\quad\begin{array}{*{1}c}{\rm for} \hspace{0.15cm}\|x\| <A, \hspace{0.15cm} \|y\| <A \hspace{0.05cm},\\ {\rm else} \hspace{0.05cm}.\\ \end{array}$$
	\left\{ \begin{array}{c} K\\
	0 \end{array} \right.\quad
	\begin{array}{*{1}c}{\rm for} \hspace{0.15cm}\|x\| <A, \hspace{0.15cm} \|y\| <A \hspace{0.05cm},
	\\ {\rm else} \hspace{0.05cm}.\\ \end{array}$$

	*Such an amplitude-limited noise is admittedly without any practical meaning.		*Such an amplitude-limited noise is admittedly without any practical meaning.
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	* To simplify the notation, the following is used:		* To simplify the notation, the following is used:
	:$$x = {\varphi_1(t)}/{\sqrt{E}}\hspace{0.05cm}, \hspace{0.2cm}		:$$x = {\varphi_1(t)}/{\sqrt{E}}\hspace{0.05cm}, \hspace{0.2cm}y = {\varphi_2(t)}/{\sqrt{E}}\hspace{0.05cm}.$$
	y = {\varphi_2(t)}/{\sqrt{E}}\hspace{0.05cm}.$$


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	* $m_0$  can only be falsified to $m_1$.  The  (conditional)  falsification probability is equal to the ratio of the areas of the small yellow triangle  $($area $1/16)$  and the square  $($area  $4)$:		* $m_0$  can only be falsified to $m_1$.  The  (conditional)  falsification probability is equal to the ratio of the areas of the small yellow triangle  $($area $1/16)$  and the square  $($area  $4)$:

	:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) = \frac{1/2 \cdot 1/2 \cdot 1/4}{4}= {1}/{64}		:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) = \frac{1/2 \cdot 1/2 \cdot 1/4}{4}= {1}/{64}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	* For symmetry reasons,  equally:		* For symmetry reasons,  equally:
	:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 )={1}/{64}		:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 )={1}/{64}\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$

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	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]
			[[de:4.08Z_Fehlerwahrscheinlichkeit]]

Guenter at 16:16, 28 July 2022

2022-07-28T16:16:58Z

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	\|type="[]"}		\|type="[]"}
	- All messages  $m_i$  are falsified in the same way.		- All messages  $m_i$  are falsified in the same way.
	+ Conditional error probability  ${\rm Pr(~~Fehler~~} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_0) = 1/64$.		+ Conditional error probability  ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_0) = 1/64$.
	- Conditional error probability  ${\rm Pr(~~Fehler~~} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_1) = 0$.		- Conditional error probability  ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_1) = 0$.
	+ Conditional error probability  ${\rm Pr(~~Fehler~~} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_2) = 0$.		+ Conditional error probability  ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_2) = 0$.

	{What is the error probability with  $A=1$  and  ${\rm Pr}(m_0) = {\rm Pr}(m_1) = {\rm Pr}(m_2) = 1/3$?		{What is the error probability with  $A=1$  and  ${\rm Pr}(m_0) = {\rm Pr}(m_1) = {\rm Pr}(m_2) = 1/3$?
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID2039__Dig_Z_4_8b.png\|right\|frame\|Noise regions with $A = 0.75$]]		[[File:P_ID2039__Dig_Z_4_8b.png\|right\|frame\|Noise regions with  $A = 0.75$]]
	'''(1)'''  The volume of the 2D PDF must give $p_n(x, y) =1$, that is:		'''(1)'''  The volume of the two-dimensional PDF must give  $p_n(x, y) =1$,  that is:
	:$$2A \cdot 2A \cdot K = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \frac{1}{4A^2}\hspace{0.05cm}.$$		:$$2A \cdot 2A \cdot K = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \frac{1}{4A^2}\hspace{0.05cm}.$$

	*With $A = 0.75$   ⇒   $2A = 3/2$, we get $K = 4/9 \ \underline {=0.444}$.		*With  $A = 0.75$   ⇒   $2A = 3/2$,  we get $K = 4/9 \ \underline {=0.444}$.


	'''(2)'''  In the accompanying graph, the noise component $\boldsymbol{n}$ is plotted by the squares of edge length $1.5$ around the 2D signal space points $\boldsymbol{s}_i$.
			'''(2)'''  In the accompanying graph,  the noise component  $\boldsymbol{n}$ is plotted by the squares of edge length  $1.5$  around the signal space points  $\boldsymbol{s}_i$.
	*It can be seen that no decision boundary is exceeded by noise components.		*It can be seen that no decision boundary is exceeded by noise components.
	*It follows:  The symbol error probability is $p_{\rm S}\ \underline { \equiv 0}$ under the conditions given here.

			*It follows:  The symbol error probability is  $p_{\rm S}\ \underline { \equiv 0}$  under the conditions given here.
			<br clear=all>
			[[File:P_ID2040__Dig_Z_4_8c.png\|right\|frame\|Noise regions with  $A = 1$]]
			'''(3)'''  <u>Statements 2 and 4</u>  are correct,  as can be seen from the second graph:
			* The message  $m_2$  cannot be falsified because the square around  $\boldsymbol{s}_2$  lies entirely in the lower right quadrant and thus in the decision region  $I_2$.

	~~'''(3)'''~~  ~~<u>Statements 2 and 4</u> are correct, as can be seen from the graph below:~~		* Likewise,  $m_2$  was sent with certainty if the received value lies in decision region  $I_2$. <br>The reason:  None of the squares around  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$  extends into the region  $I_2$.
	* The message $m_2$ cannot be falsified because the square around $\boldsymbol{s}_2$ lies entirely in the lower right quadrant and thus in the decision area $I_2$.
	* Likewise, $m_2$ was sent with certainty if the received value lies in decision ~~area~~ $I_2$. <br>The reason: ~~none~~ of the squares around $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$ extends into the ~~area~~ $I_2$.		* $m_0$  can only be falsified to $m_1$.  The  (conditional)  falsification probability is equal to the ratio of the areas of the small yellow triangle  $($area $1/16)$  and the square  $($area  $4)$:
	* $m_0$ can only be falsified to $m_1$. The (conditional) falsification probability is equal to the ratio of the areas of the yellow triangle (area $1/16$) and the square (area 4):

	~~[[File:P_ID2040__Dig_Z_4_8c.png\|right\|frame\|Noise regions with <i>A</i> = 1]]~~
	:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) = \frac{1/2 \cdot 1/2 \cdot 1/4}{4}= {1}/{64}		:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) = \frac{1/2 \cdot 1/2 \cdot 1/4}{4}= {1}/{64}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	* For symmetry reasons, equally:		* For symmetry reasons,  equally:
	:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 )={1}/{64}		:$${\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 )={1}/{64}
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$


	'''(4)'''  For equal probability symbols, we obtain for the (average) error probability:		'''(4)'''  For equal probability symbols,  we obtain for the  (average)  error probability:
	:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \big [{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) + {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 )+{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_2 )\big ]$$		:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \big [{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) + {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 )+{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_2 )\big ]$$
	:$$ \Rightarrow \hspace{0.3cm} p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \left [{1}/{64} + {1}/{64} + 0 )\right ]= \frac{2}{3 \cdot 64} = {1}/{96}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 1.04 \%} \hspace{0.05cm}.$$		:$$ \Rightarrow \hspace{0.3cm} p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \left [{1}/{64} + {1}/{64} + 0 )\right ]= \frac{2}{3 \cdot 64} = {1}/{96}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 1.04 \%} \hspace{0.05cm}.$$


	'''(5)'''  Now we obtain a smaller average error probability, viz.		'''(5)'''  Now we obtain a smaller  average error probability, viz.
	:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{4} \cdot {1}/{64} + {1}/{4} \cdot {1}/{64}+ {1}/{2} \cdot0 = {1}/{128}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.78 \% } \hspace{0.05cm}. $$		:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{4} \cdot {1}/{64} + {1}/{4} \cdot {1}/{64}+ {1}/{2} \cdot0 = {1}/{128}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.78 \% } \hspace{0.05cm}. $$

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	'''(6)'''  <u>Correct is YES</u>:		'''(6)'''  <u>Correct is YES</u>:
	*For example,     $I_1$: first quadrant,     $I_0$: second quadrant,     $I_2 \text{:} \ y < 0$     would give zero error probability.		*For example,     $I_1$: first quadrant,     $I_0$: second quadrant,     $I_2 \text{:} \ y < 0$     would give zero error probability.
	*This means that the given bounds are optimal only in the case of circularly symmetric PDF of the noise, for example, the AWGN ~~channel~~.
			*This means that the given bounds are optimal only in the case of circularly symmetric PDF of the noise,  for example, the AWGN model.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:53, 28 July 2022

2022-07-28T15:53:53Z

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}

	[[File:P_ID2037__Dig_Z_4_8.png\|right\|frame\|Decision regions with  $M = 3$ ~~symbols~~]]		[[File:P_ID2037__Dig_Z_4_8.png\|right\|frame\|Decision regions with  $M = 3$]]
	The diagram shows exactly the same signal space constellation as in  [[Aufgaben:Exercise_4.08:_Decision_Regions_at_Three_Symbols\|"Exercise 4.8"]]:		The diagram shows exactly the same signal space constellation as in  [[Aufgaben:Exercise_4.08:_Decision_Regions_at_Three_Symbols\|"Exercise 4.8"]]:
	* the  $M = 3$  possible transmitted signals, viz.		* the  $M = 3$  possible transmitted signals,  viz.
	:$$\boldsymbol{ s }_0 = (-1, \hspace{0.1cm}1)\hspace{0.05cm}, \hspace{0.2cm}		:$$\boldsymbol{ s }_0 = (-1, \hspace{0.1cm}1)\hspace{0.05cm}, \hspace{0.2cm}
	\boldsymbol{ s }_1 = (1, \hspace{0.1cm}2)\hspace{0.05cm}, \hspace{0.2cm}		\boldsymbol{ s }_1 = (1, \hspace{0.1cm}2)\hspace{0.05cm}, \hspace{0.2cm}
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	The two axes of the 2D signal space are simplistically denoted here as  $x$  and  $y$;  actually,   $\varphi_1(t)/\sqrt {E}$  and  $\varphi_2(t)/\sqrt {E}$  should be written for these, respectively.		The two axes of the two-dimensional signal space are simplistically denoted here as  $x$  and  $y$;  actually,   $\varphi_1(t)/\sqrt {E}$  and  $\varphi_2(t)/\sqrt {E}$  should be written for these, respectively.

	These decision boundaries are optimal under the two conditions:		These decision boundaries are optimal under the two conditions:
	* equal probability symbol probabilities,		* equal probability symbol probabilities,
	* circularly–symmetric PDF of the noise (e.g. AWGN).

			* circularly–symmetric PDF of the noise  (e.g. AWGN).

	In contrast, in this exercise we consider a two–dimensional uniform distribution for the noise PDF:
			In contrast,  in this exercise we consider a two–dimensional uniform distribution for the noise PDF:
	:$$\boldsymbol{ p }_{\boldsymbol{ n }} (x,\hspace{0.15cm} y) =		:$$\boldsymbol{ p }_{\boldsymbol{ n }} (x,\hspace{0.15cm} y) =
	\left\{ \begin{array}{c} K\\		\left\{ \begin{array}{c} K\\
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	*Such an amplitude-limited noise is admittedly without any practical meaning.		*Such an amplitude-limited noise is admittedly without any practical meaning.
	*However, it allows an error probability calculation without extensive integrals, from which the principle of the procedure can be seen.


			*However,  it allows an error probability calculation without extensive integrals,  from which the principle of the procedure can be seen.




	''Notes:''		Notes:
	* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability\|"Approximation of the Error Probability"]].		* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability\|"Approximation of the Error Probability"]].

Hwang at 15:40, 11 July 2022

2022-07-11T15:40:06Z

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Guenter at 10:10, 23 June 2022

2022-06-23T10:10:27Z

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	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.08Z: Fehlerwahrscheinlichkeit bei drei Symbolen to Aufgaben:Exercise 4.08Z: Error Probability with Three Symbols

2022-06-23T10:01:50Z

Guenter moved page Aufgaben:Aufgabe 4.08Z: Fehlerwahrscheinlichkeit bei drei Symbolen to Aufgaben:Exercise 4.08Z: Error Probability with Three Symbols

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

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	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.3 BER-Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]

Guenter at 10:21, 14 March 2019

2019-03-14T10:21:07Z

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	'''(2)'''  In nebenstehender Grafik ist die Rauschkomponente $\boldsymbol{n}$ durch die Quadrate der Kantenlänge $1.5$ um die 2D–Signalraumpunkte $\boldsymbol{s}_i$ eingezeichnet. Man erkennt, dass keine Entscheidungsgrenze durch Rauschkomponenten überschritten wird. Daraus folgt: Die Symbolfehlerwahrscheinlichkeit ist unter den hier gegebenen Voraussetzungen $p_{\rm S}\ \underline { \equiv 0}$.		'''(2)'''  In nebenstehender Grafik ist die Rauschkomponente $\boldsymbol{n}$ durch die Quadrate der Kantenlänge $1.5$ um die 2D–Signalraumpunkte $\boldsymbol{s}_i$ eingezeichnet.
			*Man erkennt, dass keine Entscheidungsgrenze durch Rauschkomponenten überschritten wird.
			*Daraus folgt:  Die Symbolfehlerwahrscheinlichkeit ist unter den hier gegebenen Voraussetzungen $p_{\rm S}\ \underline { \equiv 0}$.


	'''(3)'''  Richtig sind die <u>Aussagen 2 und 4</u>, wie aus der unteren Grafik abgelesen werden kann:		'''(3)'''  Richtig sind die <u>Aussagen 2 und 4</u>, wie aus der unteren Grafik abgelesen werden kann:
	* Die Nachricht $m_2$ kann nicht verfälscht werden, da das Quadrat um $\boldsymbol{s}_2$ vollständig im rechten unteren Quadranten und damit im Entscheidungsgebiet $I_2$ liegt.		* Die Nachricht $m_2$ kann nicht verfälscht werden, da das Quadrat um $\boldsymbol{s}_2$ vollständig im rechten unteren Quadranten und damit im Entscheidungsgebiet $I_2$ liegt.
	* Ebenso wurde mit Sicherheit $m_2$ gesendet, wenn der Empfangswert im Entscheidungsgebiet $I_2$ liegt. Der Grund: Keines der Quadrate um $\boldsymbol{s}_0$ und $\boldsymbol{s}_1$ reicht bis in das Gebiet $I_2$ hinein.		* Ebenso wurde mit Sicherheit $m_2$ gesendet, wenn der Empfangswert im Entscheidungsgebiet $I_2$ liegt. <br>Der Grund: Keines der Quadrate um $\boldsymbol{s}_0$ und $\boldsymbol{s}_1$ reicht bis in das Gebiet $I_2$ hinein.
	* $m_0$ kann nur zu $m_1$ verfälscht werden. Die (bedingte) Verfälschungswahrscheinlichkeit ist gleich dem Verhältnis der Flächen des gelben Dreiecks (Fläche $1/16$) und des Quadrats (Fläche 4):		* $m_0$ kann nur zu $m_1$ verfälscht werden. Die (bedingte) Verfälschungswahrscheinlichkeit ist gleich dem Verhältnis der Flächen des gelben Dreiecks (Fläche $1/16$) und des Quadrats (Fläche 4):

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	'''(4)'''  Bei gleichwahrscheinlichen Symbolen erhält man für die (mittlere) Fehlerwahrscheinlichkeit:		'''(4)'''  Bei gleichwahrscheinlichen Symbolen erhält man für die (mittlere) Fehlerwahrscheinlichkeit:
	:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \~~left~~ [{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) + {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 )+{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_2 )\~~right~~ ]$$		:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \big [{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_0 ) + {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_1 )+{\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_2 )\big ]$$
	:$$ \Rightarrow \hspace{0.3cm} p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \left [{1}/{64} + {1}/{64} + 0 )\right ]= \frac{2}{3 \cdot 64} = {1}/{96}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 1.04 \%} \hspace{0.05cm}.$$		:$$ \Rightarrow \hspace{0.3cm} p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \left [{1}/{64} + {1}/{64} + 0 )\right ]= \frac{2}{3 \cdot 64} = {1}/{96}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 1.04 \%} \hspace{0.05cm}.$$

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	'''(6)'''  <u>Richtig ist JA</u>:		'''(6)'''  <u>Richtig ist JA</u>:
	*Beispielsweise ergäbe sich durch $I_1$: erster Quadrant, $I_0$: zweiter Quadrant, $I_2 \text{:} \ y < 0$ die Fehlerwahrscheinlichkeit Null.		*Beispielsweise ergäbe sich durch     $I_1$: erster Quadrant,     $I_0$: zweiter Quadrant,     $I_2 \text{:} \ y < 0$     die Fehlerwahrscheinlichkeit Null.
	*Das bedeutet, dass die vorgegebenen Grenzen nur bei zirkulär symmetrischer WDF des Rauschens optimal sind, zum Beispiel beim AWGN–Kanal.		*Das bedeutet, dass die vorgegebenen Grenzen nur bei zirkulär symmetrischer WDF des Rauschens optimal sind, zum Beispiel beim AWGN–Kanal.
	{{ML-Fuß}}		{{ML-Fuß}}