Aufgaben:Exercise 4.1: Log Likelihood Ratio - Revision history

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	[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]		[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]
	[[de:Aufgaben:4.~~1_Zum_„Log_Likelihood_Ratio”~~]]		[[de:Aufgaben:Aufgabe 4.1: Zum „Log Likelihood Ratio”]]

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	For the binary random variables at the channel input and output holds:		For the binary random variables at the channel input and output holds:
	:$$x \in \{0\hspace{0.05cm}, 1\} \hspace{0.05cm},\hspace{0.25cm}y \in \{0\hspace{0.05cm}, 1\}		:$$x \in \{0\hspace{0.05cm}, 1\} \hspace{0.05cm},\hspace{0.25cm}y \in \{0\hspace{0.05cm}, 1\}\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$

	This model is shown in the upper graph.  The following applies to the conditional probabilities in the forward direction:		This model is shown in the upper graph.  The following applies to the conditional probabilities in the forward direction:
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	From the BSC model,  it is possible to determine the  L–value of the conditional probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  in forward direction  $($German:  "Vorwärtsrichtung"   ⇒   subscript "V"$)$,  which is denoted by  $L_{\rm V}(y)$  in the present exercise:		From the BSC model,  it is possible to determine the  L–value of the conditional probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  in forward direction  $($German:  "Vorwärtsrichtung"   ⇒   subscript "V"$)$,  which is denoted by  $L_{\rm V}(y)$  in the present exercise:
	:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) =		:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) ={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 0)}{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 1)} =\left\{ \begin{array}{c} {\rm ln} \hspace{0.15cm} [(1 - \varepsilon)/\varepsilon]\\{\rm ln} \hspace{0.15cm} [\varepsilon/(1 - \varepsilon)] \end{array} \right.\hspace{0.15cm}\begin{array}{*{1}c} {\rm f\ddot{u}r} \hspace{0.15cm} y = 0,\\ {\rm f\ddot{u}r} \hspace{0.15cm} y = 1. \\ \end{array}$$
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 0)}{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 1)} =
	\left\{ \begin{array}{c} {\rm ln} \hspace{0.15cm} [(1 - \varepsilon)/\varepsilon]\\
	{\rm ln} \hspace{0.15cm} [\varepsilon/(1 - \varepsilon)] \end{array} \right.\hspace{0.15cm}
	\begin{array}{*{1}c} {\rm f\ddot{u}r} \hspace{0.15cm} y = 0,
	\\ {\rm f\ddot{u}r} \hspace{0.15cm} y = 1. \\ \end{array}$$

	For example,  for  $\varepsilon = 0.1$:		For example,  for  $\varepsilon = 0.1$:
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	The corresponding    L–value in backward direction  $($German:  "Rückwärtsrichtung"   ⇒   subscript "R"$)$  is denoted in this exercise  by  $L_{\rm R}(y)$:		The corresponding    L–value in backward direction  $($German:  "Rückwärtsrichtung"   ⇒   subscript "R"$)$  is denoted in this exercise  by  $L_{\rm R}(y)$:
	:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) =		:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) ={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}\|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}\|\hspace{0.05cm}y)} \hspace{0.05cm} .$$
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}\|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}\|\hspace{0.05cm}y)} \hspace{0.05cm} .$$


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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the conditional probabilities,  according to the  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]]  with intersection  $A ∩ B$:		'''(1)'''  For the conditional probabilities,  according to the  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]]  with intersection  $A ∩ B$:
	:$${\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(A)}\hspace{0.05cm},		:$${\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(A)}\hspace{0.05cm},\hspace{0.3cm} {\rm Pr}(A \hspace{0.05cm}\|\hspace{0.05cm} B) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(B)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Pr}(A \hspace{0.05cm}\|\hspace{0.05cm} B) ={\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) \cdot \frac{{\rm Pr}(A)}{{\rm Pr}(B)}\hspace{0.05cm}.$$
	\hspace{0.3cm} {\rm Pr}(A \hspace{0.05cm}\|\hspace{0.05cm} B) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(B)}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}{\rm Pr}(A \hspace{0.05cm}\|\hspace{0.05cm} B) =
	{\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) \cdot \frac{{\rm Pr}(A)}{{\rm Pr}(B)}\hspace{0.05cm}.$$

	*Correct is the  <u>proposition 3</u>.		*Correct is the  <u>proposition 3</u>.
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	'''(2)'''  With  $A$   ⇒   "$x = 0$"  and  $B$   ⇒   "$y = 0$"  we immediately get the equation according to  <u>proposition 1</u>:		'''(2)'''  With  $A$   ⇒   "$x = 0$"  and  $B$   ⇒   "$y = 0$"  we immediately get the equation according to  <u>proposition 1</u>:
	:$${\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm} y = 0) =		:$${\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm} y = 0) ={\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm} x = 0) \cdot \frac{{\rm Pr}(x = 0)}{{\rm Pr}(y = 0)}\hspace{0.05cm}.$$
	{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm} x = 0) \cdot \frac{{\rm Pr}(x = 0)}{{\rm Pr}(y = 0)}\hspace{0.05cm}.$$


	'''(3)'''  We compute the L–values of the inference probabilities.  Assuming  $y = 0$  holds:		'''(3)'''  We compute the L–values of the inference probabilities.  Assuming  $y = 0$  holds:
	:$$L_{\rm R}(y= 0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0)=		:$$L_{\rm R}(y= 0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm}y=0)}{{\rm Pr}(x = 1\hspace{0.05cm}\|\hspace{0.05cm}y=0)} = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x=0) \cdot {\rm Pr}(x = 0) / {\rm Pr}(y = 0)}{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x = 1)\cdot {\rm Pr}(x = 1) / {\rm Pr}(y = 0)} $$
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm}y=0)}{{\rm Pr}(x = 1\hspace{0.05cm}\|\hspace{0.05cm}y=0)} = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x=0) \cdot {\rm Pr}(x = 0) / {\rm Pr}(y = 0)}{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x = 1)\cdot {\rm Pr}(x = 1) / {\rm Pr}(y = 0)} $$		:$$\Rightarrow \hspace{0.3cm} L_{\rm R}(y= 0)= {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x=0) }{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x = 1)} +{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x=0) }{{\rm Pr}(x = 1)}$$
	:$$\Rightarrow \hspace{0.3cm} L_{\rm R}(y= 0)= {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x=0) }{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x = 1)} +
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x=0) }{{\rm Pr}(x = 1)}$$
	:$$\Rightarrow \hspace{0.3cm} L_{\rm R}(y= 0) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0) = L_{\rm V}(y= 0) + L_{\rm A}(x)\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} L_{\rm R}(y= 0) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0) = L_{\rm V}(y= 0) + L_{\rm A}(x)\hspace{0.05cm}.$$

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	* as well as the forward log likelihood ratio,		* as well as the forward log likelihood ratio,
	:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x=0) }{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x = 1)}		:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x=0) }{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x = 1)}\hspace{0.05cm},$$
	\hspace{0.05cm},$$

	as follows:		as follows:

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	[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]		[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]
			[[de:Aufgaben:4.1_Zum_„Log_Likelihood_Ratio”]]

Hwang at 15:27, 23 January 2023

2023-01-23T15:27:00Z

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	+ For the uniform distribution only:  $\hspace{0.2cm} {\rm Pr}(x = 0) = {\rm Pr}(x = 1) = 1/2$.		+ For the uniform distribution only:  $\hspace{0.2cm} {\rm Pr}(x = 0) = {\rm Pr}(x = 1) = 1/2$.

	{Let the initial symbol be  $y = 1$.  What inference LLR is obtained with the ~~corruption~~ probability  $\varepsilon = 0.1$  for equally probable symbols?		{Let the initial symbol be  $y = 1$.  What inference LLR is obtained with the falsification probability  $\varepsilon = 0.1$  for equally probable symbols?
	\|type="{}"}		\|type="{}"}
	$L_{\rm R}(y = 1) = L(x \| y = 1) \ = \ ${ -2.26291--2.13109 }		$L_{\rm R}(y = 1) = L(x \| y = 1) \ = \ ${ -2.26291--2.13109 }

Guenter at 16:43, 30 November 2022

2022-11-30T16:43:57Z

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	{{quiz-Header\|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}		{{quiz-Header\|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}

	[[File:~~P_ID2979__KC_A_4_1_v2~~.png\|right\|frame\|Considered channel models]]		[[File:EN_KC_A_4_1.png\|right\|frame\|Considered channel models]]
	To interpret the "log likelihood ratio"  $\rm (LLR)$  we start from the  "binary symmetric channel"  $\rm (BSC)$  as in the  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio\|"theory section"]] .		To interpret the "log likelihood ratio"  $\rm (LLR)$  we start from the  "binary symmetric channel"  $\rm (BSC)$  as in the  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio\|"theory section"]] .

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	$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }		$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }

	{Can the result derived in subtask  '''(3)'''   ⇒   $L_{\rm R} = L_{\rm V} + L_{\rm A}$  also be applied to the  "2-on $M$-channel"?		{Can the result derived in subtask  '''(3)'''   ⇒   $L_{\rm R} = L_{\rm V} + L_{\rm A}$  also be applied to the  "2-on-''M''"  channel?
	\|type="()"}		\|type="()"}
	+ Yes.		+ Yes.

Guenter at 14:17, 28 November 2022

2022-11-28T14:17:34Z

← Older revision		Revision as of 16:17, 28 November 2022
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	$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }		$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }

	{Can the result derived in subtask  '''(3)''''   ⇒   $L_{\rm R} = L_{\rm V} + L_{\rm A}$  also be applied to the  "2-on $M$-channel"?		{Can the result derived in subtask  '''(3)'''   ⇒   $L_{\rm R} = L_{\rm V} + L_{\rm A}$  also be applied to the  "2-on $M$-channel"?
	\|type="()"}		\|type="()"}
	+ Yes.		+ Yes.
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	'''(5)'''  With the same falsification ~~probabilityf~~  $\varepsilon = 0.1$ $L_{\rm V}(y = 0)$  differs from  $L_{\rm V}(y = 1)$  only by the sign.		'''(5)'''  With the same falsification probability  $\varepsilon = 0.1$   ⇒   $L_{\rm V}(y = 0)$  differs from  $L_{\rm V}(y = 1)$  only by the sign.
	*With  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ L_{\rm A}(x) = \, -1.382$  we thus obtain:		*With  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ L_{\rm A}(x) = \, -1.382$  we thus obtain:
	:$$L_{\rm R}(y = 0) = (+)2.197 - 1.382 \hspace{0.15cm}\underline{=+0.815}\hspace{0.05cm}.$$		:$$L_{\rm R}(y = 0) = (+)2.197 - 1.382 \hspace{0.15cm}\underline{=+0.815}\hspace{0.05cm}.$$

Guenter at 14:11, 28 November 2022

2022-11-28T14:11:04Z

← Older revision		Revision as of 16:11, 28 November 2022
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	- ${\rm Pr}(x = 0 \| y = 0) = {\rm Pr}(y = 0 \| x = 0) \cdot {\rm Pr}(y = 0) / {\rm Pr}(x = 0)$.		- ${\rm Pr}(x = 0 \| y = 0) = {\rm Pr}(y = 0 \| x = 0) \cdot {\rm Pr}(y = 0) / {\rm Pr}(x = 0)$.

	{Under what conditions does the inference log ~~likelihooh~~ ratio hold for all possible output values  $y ∈ \{0, \, 1\}$:     $L(x\hspace{0.05cm}\|\hspace{0.05cm}y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x)$   resp.   $L_{\rm R}(y) = L_{\rm V}(y)$?		{Under what conditions does the inference log likelihood ratio hold for all possible output values  $y ∈ \{0, \, 1\}$:     $L(x\hspace{0.05cm}\|\hspace{0.05cm}y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x)$   resp.   $L_{\rm R}(y) = L_{\rm V}(y)$?
	\|type="()"}		\|type="()"}
	- For any input distribution  ${\rm Pr}(x = 0), \ {\rm Pr}(x = 1)$.		- For any input distribution  ${\rm Pr}(x = 0), \ {\rm Pr}(x = 1)$.
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	$L_{\rm R}(y = 1) = L(x \| y = 1) \ = \ ${ -2.26291--2.13109 }		$L_{\rm R}(y = 1) = L(x \| y = 1) \ = \ ${ -2.26291--2.13109 }

	{Let the initial symbol now be  $y = 0$. What inference log ~~likelihooh~~ ratio is obtained for  ${\rm Pr}(x = 0) = 0.2$  and  $\varepsilon = 0.1$?		{Let the initial symbol now be  $y = 0$. What inference log likelihood ratio is obtained for  ${\rm Pr}(x = 0) = 0.2$  and  $\varepsilon = 0.1$?
	\|type="{}"}		\|type="{}"}
	$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }		$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the conditional probabilities, according to the [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]] with intersection $A ∩ B$:		'''(1)'''  For the conditional probabilities,  according to the  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]]  with intersection  $A ∩ B$:
	:$${\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(A)}\hspace{0.05cm},		:$${\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(A)}\hspace{0.05cm},
	\hspace{0.3cm} {\rm Pr}(A \hspace{0.05cm}\|\hspace{0.05cm} B) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(B)}\hspace{0.3cm}		\hspace{0.3cm} {\rm Pr}(A \hspace{0.05cm}\|\hspace{0.05cm} B) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(B)}\hspace{0.3cm}
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	{\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) \cdot \frac{{\rm Pr}(A)}{{\rm Pr}(B)}\hspace{0.05cm}.$$		{\rm Pr}(B \hspace{0.05cm}\|\hspace{0.05cm} A) \cdot \frac{{\rm Pr}(A)}{{\rm Pr}(B)}\hspace{0.05cm}.$$

	Correct is the <u>proposition 3</u>. ~~In the special case ${\rm Pr}(B) = {\rm Pr}(A)$ also the suggestion 1 would be correct.~~		*Correct is the  <u>proposition 3</u>.

			*In the special case  ${\rm Pr}(B) = {\rm Pr}(A)$   ⇒   also the suggestion 1 would be correct.

	'''(2)'''  With  $A$  ⇒  "$x = 0$" and  $B$  ⇒  "$y = 0$" we immediately get the equation according to <u>proposition 1</u>:

			'''(2)'''  With  $A$   ⇒   "$x = 0$"  and  $B$   ⇒   "$y = 0$"  we immediately get the equation according to  <u>proposition 1</u>:
	:$${\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm} y = 0) =		:$${\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm} y = 0) =
	{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm} x = 0) \cdot \frac{{\rm Pr}(x = 0)}{{\rm Pr}(y = 0)}\hspace{0.05cm}.$$		{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm} x = 0) \cdot \frac{{\rm Pr}(x = 0)}{{\rm Pr}(y = 0)}\hspace{0.05cm}.$$


	'''(3)'''  We compute the $L~~$ value~~ of the inference probabilities. Assuming $y = 0$ holds:		'''(3)'''  We compute the L–values of the inference probabilities.  Assuming  $y = 0$  holds:
	:$$L_{\rm R}(y= 0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0)=		:$$L_{\rm R}(y= 0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0)=
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm}y=0)}{{\rm Pr}(x = 1\hspace{0.05cm}\|\hspace{0.05cm}y=0)} = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x=0) \cdot {\rm Pr}(x = 0) / {\rm Pr}(y = 0)}{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x = 1)\cdot {\rm Pr}(x = 1) / {\rm Pr}(y = 0)} $$		{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0\hspace{0.05cm}\|\hspace{0.05cm}y=0)}{{\rm Pr}(x = 1\hspace{0.05cm}\|\hspace{0.05cm}y=0)} = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x=0) \cdot {\rm Pr}(x = 0) / {\rm Pr}(y = 0)}{{\rm Pr}(y = 0\hspace{0.05cm}\|\hspace{0.05cm}x = 1)\cdot {\rm Pr}(x = 1) / {\rm Pr}(y = 0)} $$
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	:$$\Rightarrow \hspace{0.3cm} L_{\rm R}(y= 0) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0) = L_{\rm V}(y= 0) + L_{\rm A}(x)\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} L_{\rm R}(y= 0) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 0) = L_{\rm V}(y= 0) + L_{\rm A}(x)\hspace{0.05cm}.$$

	Similarly, assuming $y = 1$, the result is:		*Similarly,  assuming  $y = 1$,  the result is:
	:$$L_{\rm R}(y= 1) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 1) = L_{\rm V}(y= 1) + L_{\rm A}(x)\hspace{0.05cm}.$$		:$$L_{\rm R}(y= 1) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y= 1) = L_{\rm V}(y= 1) + L_{\rm A}(x)\hspace{0.05cm}.$$

	The two results can be summarized using $y ∈ \{0, \, 1\}$ and.		*The two results can be summarized using  $y ∈ \{0, \, 1\}$  and
	* the input ~~LLR~~,
	:$$L_{\rm A}(x) =		* the input log likelihood ratio,
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x=0) }{\rm Pr}(x = 1)}\hspace{0.05cm},$$		:$$L_{\rm A}(x) = {\rm ln} \hspace{0.15cm} \frac{ {\rm Pr}(x=0) }{ {\rm Pr}(x = 1)}\hspace{0.05cm},$$
	* as well as the forward ~~LLR~~,
			* as well as the forward log likelihood ratio,
	:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x=0) }{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x = 1)}		:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x=0) }{{\rm Pr}(y \hspace{0.05cm}\|\hspace{0.05cm}x = 1)}
	\hspace{0.05cm},$$		\hspace{0.05cm},$$
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	:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) = L_{\rm V}(y) + L_{\rm A}(x)\hspace{0.05cm}.$$		:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) = L_{\rm V}(y) + L_{\rm A}(x)\hspace{0.05cm}.$$

	The identity $L_{\rm R}(y) &equiv; L_{\rm V}(y)$ requires $L_{\rm A}(x) = 0$   ⇒   equally probable symbols   ⇒   <u>proposition 2</u>.		*The identity   $L_{\rm R}(y) &equiv; L_{\rm V}(y)$   requires $L_{\rm A}(x) = 0$   ⇒   equally probable symbols   ⇒   <u>proposition 2</u>.


	'''(4)'''  From the exercise description, you can see that with corruption probability $\varepsilon = 0.1$, the initial value $y = 1$ leads to forward–LLR $L_{\rm V}(y = 1) = \, –2.197$.

	*~~Wegen~~ ${\rm Pr}(x = 0) = 1/2 \ \Rightarrow \ L_{\rm A}(x) = 0$ ~~gilt somit auch~~:		'''(4)'''  From the exercise description,  you can see that with falsification probability  $\varepsilon = 0.1$,  the initial value  $y = 1$  leads to forward log likelihood ratio  $L_{\rm V}(y = 1) = \, –2.197$.

			*Because of  ${\rm Pr}(x = 0) = 1/2 \ \Rightarrow \ L_{\rm A}(x) = 0$:
	:$$L_{\rm R}(y = 1) = L_{\rm V}(y = 1) \hspace{0.15cm}\underline{= -2.197}\hspace{0.05cm}.$$		:$$L_{\rm R}(y = 1) = L_{\rm V}(y = 1) \hspace{0.15cm}\underline{= -2.197}\hspace{0.05cm}.$$


	'''(5)'''  With the same ~~corruption probability~~ $\varepsilon = 0.1$ $L_{\rm V}(y = 0)$ differs from $L_{\rm V}(y = 1)$ only by the sign.
	*With ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ L_{\rm A}(x) = \, -1.382$ we thus obtain:		'''(5)'''  With the same falsification probabilityf  $\varepsilon = 0.1$ $L_{\rm V}(y = 0)$  differs from  $L_{\rm V}(y = 1)$  only by the sign.
			*With  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ L_{\rm A}(x) = \, -1.382$  we thus obtain:
	:$$L_{\rm R}(y = 0) = (+)2.197 - 1.382 \hspace{0.15cm}\underline{=+0.815}\hspace{0.05cm}.$$		:$$L_{\rm R}(y = 0) = (+)2.197 - 1.382 \hspace{0.15cm}\underline{=+0.815}\hspace{0.05cm}.$$



	'''(6)'''  ~~As I'm sure you'll be happy to verify, the~~ relation  $L_{\rm R} = L_{\rm V} + L_{\rm A}$  also holds for the "2 on $M$ channel", regardless of the size ~~(Für Günter: Umfang?)~~  $M$  of the output alphabet   ⇒   <u>Answer Yes</u>.		'''(6)'''  The relation   $L_{\rm R} = L_{\rm V} + L_{\rm A}$   also holds for the  "2-on-$M$ channel",  regardless of the set size  $M$  of the output alphabet   ⇒   <u>Answer Yes</u>.



	'''(7)'''  The AWGN channel is described by the outlined "2–on–$M$–channel" with  $M → ∞$  also   ⇒   <u>Answer Yes</u>.		'''(7)'''  The AWGN channel is described by the outlined  "2–on–$M$–channel"  with  $M → ∞$  also   ⇒   <u>Answer Yes</u>.
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]		[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]

Guenter at 13:47, 28 November 2022

2022-11-28T13:47:40Z

← Older revision		Revision as of 15:47, 28 November 2022
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	:$$L_{\rm A}(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)}{{\rm Pr}(x = 1)}\hspace{0.05cm},$$		:$$L_{\rm A}(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)}{{\rm Pr}(x = 1)}\hspace{0.05cm},$$

	where the subscript  $\rm A$  indicates the  "a-priori log likelihood ratio"  or the  "a-priori L–value".		where the subscript  "$\rm A$"  indicates the  "a-priori log likelihood ratio"  or the  "a-priori L–value".

	For example,  for  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$   ⇒   $L_{\rm A}(x) = \, -1.382$.		For example,  for  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$   ⇒   $L_{\rm A}(x) = \, -1.382$.
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	:$$L_{\rm V}(y = 0) = +2.197\hspace{0.05cm}, \hspace{0.3cm}L_{\rm V}(y = 1) = -2.197\hspace{0.05cm}.$$		:$$L_{\rm V}(y = 0) = +2.197\hspace{0.05cm}, \hspace{0.3cm}L_{\rm V}(y = 1) = -2.197\hspace{0.05cm}.$$

	Of particular importance to coding theory are the inference probabilities  ${\rm Pr}(x\hspace{0.05cm}\|\hspace{0.05cm}y)$, which are related to the ~~forward~~ probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  and the input probabilities  ${\rm Pr}(x = 0)$  and  ${\rm Pr}(x = 1)$  via Bayes' theorem.		Of particular importance to coding theory are the inference probabilities  ${\rm Pr}(x\hspace{0.05cm}\|\hspace{0.05cm}y)$,  which are related to the backward probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  and the input probabilities  ${\rm Pr}(x = 0)$  and  ${\rm Pr}(x = 1)$  via Bayes' theorem.

	The corresponding    L–value in ~~forward~~ direction  $($German:  "~~Vorwärtsrichtung~~"   ⇒   subscript "V"$)$ in this exercise  ~~is denoted~~ by $L_{\rm R}(y)$~~ ~~:		The corresponding    L–value in backward direction  $($German:  "Rückwärtsrichtung"   ⇒   subscript "R"$)$  is denoted in this exercise  by  $L_{\rm R}(y)$:
	:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) =		:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) =
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}\|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}\|\hspace{0.05cm}y)} \hspace{0.05cm} .$$		{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}\|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}\|\hspace{0.05cm}y)} \hspace{0.05cm} .$$
Line 44:		Line 44:


			<u>Hints:</u>
			* The exercise belongs to the chapter  [[Channel_Coding/Soft-in_Soft-Out_Decoder\| "Soft–in Soft–out Decoder"]].

			* Reference is made in particular to the section  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio\| "Reliability Information – Log Likelihood Ratio"]].

			*In the last subtasks you have to clarify whether the found relations between  $L_{\rm A}, \ L_{\rm V}$  and  $L_{\rm R}$  can also be transferred to the  "2-on-$M$  channel".

	~~Hints:~~		*For this purpose,  we choose a bipolar approach for the input symbols:   "$0$"  →  "$+1$"  and  "$1$"   →  "$–1$".
	* The exercise belongs to the chapter  [[Channel_Coding/Soft-in_Soft-Out_Decoder\| "Soft–in Soft–out Decoder"]].
	* Reference is made in particular to the page  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio\| "Reliability Information – Log Likelihood Ratio"]].
	*In the last subtasks we have to clarify whether the found relations between  $L_{\rm A}, \ L_{\rm V}$  and  $L_{\rm R}$  can also be transferred to the "2 on $M$ channel".
	*For this purpose, we choose a bipolar approach for the input symbols:  "$0$"  →   "$+1$"  and  "$1$"   →   "$–1$".


Line 70:		Line 69:
	- ${\rm Pr}(x = 0 \| y = 0) = {\rm Pr}(y = 0 \| x = 0) \cdot {\rm Pr}(y = 0) / {\rm Pr}(x = 0)$.		- ${\rm Pr}(x = 0 \| y = 0) = {\rm Pr}(y = 0 \| x = 0) \cdot {\rm Pr}(y = 0) / {\rm Pr}(x = 0)$.

	{Under what conditions does the inference ~~LLR~~ hold for all possible output values  $y ∈ \{0, \, 1\}$: ~~<br>~~    $L(x\hspace{0.05cm}\|\hspace{0.05cm}y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  ~~bzw~~.  $L_{\rm R}(y) = L_{\rm V}(y)$?		{Under what conditions does the inference log likelihooh ratio hold for all possible output values  $y ∈ \{0, \, 1\}$:     $L(x\hspace{0.05cm}\|\hspace{0.05cm}y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x)$   resp.   $L_{\rm R}(y) = L_{\rm V}(y)$?
	\|type="()"}		\|type="()"}
	- For any input distribution  ${\rm Pr}(x = 0), \ {\rm Pr}(x = 1)$.		- For any input distribution  ${\rm Pr}(x = 0), \ {\rm Pr}(x = 1)$.
	+ For the uniform distribution only:  $\hspace{0.2cm} {\rm Pr}(x = 0) = {\rm Pr}(x = 1) = 1/2$.		+ For the uniform distribution only:  $\hspace{0.2cm} {\rm Pr}(x = 0) = {\rm Pr}(x = 1) = 1/2$.

	{Let the initial symbol be  $y = 1$. What inference LLR is obtained with the corruption probability  $\varepsilon = 0.1$  for equally probable symbols?		{Let the initial symbol be  $y = 1$.  What inference LLR is obtained with the corruption probability  $\varepsilon = 0.1$  for equally probable symbols?
	\|type="{}"}		\|type="{}"}
	$L_{\rm R}(y = 1) = L(x \| y = 1) \ = \ ${ -2.26291--2.13109 }		$L_{\rm R}(y = 1) = L(x \| y = 1) \ = \ ${ -2.26291--2.13109 }

	{Let the initial symbol now be  $y = 0$. What inference ~~LLR~~ is obtained for  ${\rm Pr}(x = 0) = 0.2$  and  $\varepsilon = 0.1$?		{Let the initial symbol now be  $y = 0$. What inference log likelihooh ratio is obtained for  ${\rm Pr}(x = 0) = 0.2$  and  $\varepsilon = 0.1$?
	\|type="{}"}		\|type="{}"}
	$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }		$L_{\rm R}(y = 0) = L(x \| y = 0) \ = \ ${ 0.815 3% }

	{Can the result derived in '''(3)''''   ⇒   $L_{\rm R} = L_{\rm V} + L_{\rm A}$  also be applied to the "2 on $M$ channel"?		{Can the result derived in subtask  '''(3)''''   ⇒   $L_{\rm R} = L_{\rm V} + L_{\rm A}$  also be applied to the  "2-on $M$-channel"?
	\|type="()"}		\|type="()"}
	+ Yes.		+ Yes.
	- No.		- No.

	{Can the context be applied to the AWGN~~–~~channel as well?		{Can the context be applied to the AWGN channel as well?
	\|type="()"}		\|type="()"}
	+ Yes.		+ Yes.

Guenter at 10:43, 28 November 2022

2022-11-28T10:43:39Z

← Older revision		Revision as of 12:43, 28 November 2022
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	For example,  for  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$   ⇒   $L_{\rm A}(x) = \, -1.382$.		For example,  for  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$   ⇒   $L_{\rm A}(x) = \, -1.382$.

	From the BSC&~~ndash~~;~~model,~~ it is ~~also~~ possible to determine the  $L$ value ~~(LLR)~~ of the conditional probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  in ~~the~~ forward direction, which is ~~also~~ denoted by  $L_{\rm V}(y)$  in the present exercise:		From the BSC model,  it is possible to determine the  L–value of the conditional probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  in forward direction  $($German:  "Vorwärtsrichtung"   ⇒   subscript "V"$)$,  which is denoted by  $L_{\rm V}(y)$  in the present exercise:
	:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) =		:$$L_{\rm V}(y) = L(y\hspace{0.05cm}\|\hspace{0.05cm}x) =
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 0)}{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 1)} =		{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 0)}{{\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x = 1)} =
Line 31:		Line 31:
	\\ {\rm f\ddot{u}r} \hspace{0.15cm} y = 1. \\ \end{array}$$		\\ {\rm f\ddot{u}r} \hspace{0.15cm} y = 1. \\ \end{array}$$

	For example, for  $\varepsilon = 0.1$:		For example,  for  $\varepsilon = 0.1$:
	:$$L_{\rm V}(y = 0) = +2.197\hspace{0.05cm}, \hspace{0.3cm}L_{\rm V}(y = 1) = -2.197\hspace{0.05cm}.$$		:$$L_{\rm V}(y = 0) = +2.197\hspace{0.05cm}, \hspace{0.3cm}L_{\rm V}(y = 1) = -2.197\hspace{0.05cm}.$$

	Of particular importance to coding theory are the inference probabilities  ${\rm Pr}(x\hspace{0.05cm}\|\hspace{0.05cm}y)$, which are related to the forward probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  and the input probabilities  ${\rm Pr}(x = 0)$  and  ${\rm Pr}(x = 1)$  via Bayes' theorem.		Of particular importance to coding theory are the inference probabilities  ${\rm Pr}(x\hspace{0.05cm}\|\hspace{0.05cm}y)$, which are related to the forward probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  and the input probabilities  ${\rm Pr}(x = 0)$  and  ${\rm Pr}(x = 1)$  via Bayes' theorem.

	The corresponding  $L$ ~~value~~ (~~LLR~~) in this exercise  is denoted by $L_{\rm R}(y)$ :		The corresponding    L–value in forward direction  $($German:  "Vorwärtsrichtung"   ⇒   subscript "V"$)$ in this exercise  is denoted by $L_{\rm R}(y)$ :
	:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) =		:$$L_{\rm R}(y) = L(x\hspace{0.05cm}\|\hspace{0.05cm}y) =
	{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}\|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}\|\hspace{0.05cm}y)} \hspace{0.05cm} .$$		{\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}\|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}\|\hspace{0.05cm}y)} \hspace{0.05cm} .$$

Guenter at 08:14, 28 November 2022

2022-11-28T08:14:16Z

← Older revision		Revision as of 10:14, 28 November 2022
Line 19:		Line 19:
	:$$L_{\rm A}(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)}{{\rm Pr}(x = 1)}\hspace{0.05cm},$$		:$$L_{\rm A}(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)}{{\rm Pr}(x = 1)}\hspace{0.05cm},$$

	where the subscript  $\rm A$  indicates the  "a-priori log likelihood ratio".		where the subscript  $\rm A$  indicates the  "a-priori log likelihood ratio"  or the  "a-priori L–value".

	For example, for  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$  ~~the apriori LLR~~  $L_{\rm A}(x) = \, -1.382$.		For example,  for  ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$   ⇒   $L_{\rm A}(x) = \, -1.382$.

	From the BSC–model, it is also possible to determine the  $L$ value (LLR) of the conditional probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  in the forward direction, which is also denoted by  $L_{\rm V}(y)$  in the present exercise:		From the BSC–model, it is also possible to determine the  $L$ value (LLR) of the conditional probabilities  ${\rm Pr}(y\hspace{0.05cm}\|\hspace{0.05cm}x)$  in the forward direction, which is also denoted by  $L_{\rm V}(y)$  in the present exercise: