Aufgaben:Exercise 4.1: Log Likelihood Ratio - Revision history

Hwang at 15:27, 23 January 2023

2023-01-23T15:27:00Z

Guenter at 16:43, 30 November 2022

2022-11-30T16:43:57Z

Guenter at 14:17, 28 November 2022

2022-11-28T14:17:34Z

Guenter at 14:11, 28 November 2022

2022-11-28T14:11:04Z

Guenter at 13:47, 28 November 2022

2022-11-28T13:47:40Z

Guenter at 10:43, 28 November 2022

2022-11-28T10:43:39Z

Guenter at 08:14, 28 November 2022

2022-11-28T08:14:16Z

Guenter at 08:08, 28 November 2022

2022-11-28T08:08:56Z

Noah at 15:45, 27 October 2022

2022-10-27T15:45:56Z

Noah at 15:45, 27 October 2022

2022-10-27T15:45:45Z

@@ Line 74: / Line 74: @@
 + For the uniform distribution only:&nbsp; $\hspace{0.2cm} {\rm Pr}(x = 0) = {\rm Pr}(x = 1) = 1/2$.
-{Let the initial symbol be&nbsp; $y = 1$.&nbsp; What inference LLR is obtained with the corruption probability&nbsp; $\varepsilon = 0.1$&nbsp; for equally probable symbols?
+{Let the initial symbol be&nbsp; $y = 1$.&nbsp; What inference LLR is obtained with the falsification probability&nbsp; $\varepsilon = 0.1$&nbsp; for equally probable symbols?
 |type="{}"}
 $L_{\rm R}(y = 1) = L(x | y = 1) \ = \ ${ -2.26291--2.13109 }

@@ Line 1: / Line 1: @@
 {{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"&nbsp; $\rm (LLR)$&nbsp; we start from the&nbsp; "binary symmetric channel"&nbsp; $\rm (BSC)$&nbsp; as in the&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio|"theory section"]]&nbsp;.
@@ Line 82: / Line 82: @@
 $L_{\rm R}(y = 0) = L(x | y = 0) \ = \ ${ 0.815 3% }
-{Can the result derived in subtask&nbsp; '''(3)''' &nbsp; &#8658; &nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$&nbsp; also be applied to the&nbsp; "2-on $M$-channel"?
+{Can the result derived in subtask&nbsp; '''(3)''' &nbsp; &#8658; &nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$&nbsp; also be applied to the&nbsp; "2-on-''M''"&nbsp; channel?
 |type="()"}
 + Yes.

@@ Line 82: / Line 82: @@
 $L_{\rm R}(y = 0) = L(x | y = 0) \ = \ ${ 0.815 3% }
-{Can the result derived in subtask&nbsp; '''(3)'''' &nbsp; &#8658; &nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$&nbsp; also be applied to the&nbsp; "2-on $M$-channel"?
+{Can the result derived in subtask&nbsp; '''(3)''' &nbsp; &#8658; &nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$&nbsp; also be applied to the&nbsp; "2-on $M$-channel"?
 |type="()"}
 + Yes.
@@ Line 145: / Line 145: @@
-'''(5)'''&nbsp; With the same falsification probabilityf&nbsp; $\varepsilon = 0.1$ $L_{\rm V}(y = 0)$&nbsp; differs from&nbsp; $L_{\rm V}(y = 1)$&nbsp; only by the sign.
+'''(5)'''&nbsp; With the same falsification probability&nbsp; $\varepsilon = 0.1$ &nbsp; &rArr; &nbsp; $L_{\rm V}(y = 0)$&nbsp; differs from&nbsp; $L_{\rm V}(y = 1)$&nbsp; only by the sign.
 *With&nbsp; ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ L_{\rm A}(x) = \, -1.382$&nbsp; we thus obtain:
 :$$L_{\rm R}(y = 0)  = (+)2.197 - 1.382 \hspace{0.15cm}\underline{=+0.815}\hspace{0.05cm}.$$

@@ Line 69: / Line 69: @@
 - ${\rm Pr}(x = 0 | y = 0) = {\rm Pr}(y = 0 | x = 0) \cdot {\rm Pr}(y = 0) / {\rm Pr}(x = 0)$.
-{Under what conditions does the inference log likelihooh ratio hold for all possible output values&nbsp; $y &#8712; \{0, \, 1\}$: &nbsp; &nbsp;  $L(x\hspace{0.05cm}|\hspace{0.05cm}y) = L(y\hspace{0.05cm}|\hspace{0.05cm}x)$ &nbsp; resp. &nbsp; $L_{\rm R}(y) = L_{\rm V}(y)$?
+{Under what conditions does the inference log likelihood ratio hold for all possible output values&nbsp; $y &#8712; \{0, \, 1\}$: &nbsp; &nbsp;  $L(x\hspace{0.05cm}|\hspace{0.05cm}y) = L(y\hspace{0.05cm}|\hspace{0.05cm}x)$ &nbsp; resp. &nbsp; $L_{\rm R}(y) = L_{\rm V}(y)$?
 |type="()"}
 - For any input distribution&nbsp; ${\rm Pr}(x = 0), \ {\rm Pr}(x = 1)$.
@@ Line 78: / Line 78: @@
 $L_{\rm R}(y = 1) = L(x | y = 1) \ = \ ${ -2.26291--2.13109 }
-{Let the initial symbol now be&nbsp; $y = 0$. What inference log likelihooh ratio is obtained for&nbsp; ${\rm Pr}(x = 0) = 0.2$&nbsp; and&nbsp; $\varepsilon = 0.1$?
+{Let the initial symbol now be&nbsp; $y = 0$. What inference log likelihood ratio is obtained for&nbsp; ${\rm Pr}(x = 0) = 0.2$&nbsp; and&nbsp; $\varepsilon = 0.1$?
 |type="{}"}
 $L_{\rm R}(y = 0) = L(x | y = 0) \ = \ ${ 0.815 3% }
@@ Line 95: / Line 95: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; For the conditional probabilities, according to the [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability| "Bayes' theorem"]] with intersection $A &#8745; B$:
+'''(1)'''&nbsp; For the conditional probabilities,&nbsp; according to the&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability| "Bayes' theorem"]]&nbsp; with intersection&nbsp; $A &#8745; B$:
 :$${\rm Pr}(B \hspace{0.05cm}|\hspace{0.05cm}  A) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(A)}\hspace{0.05cm},
 \hspace{0.3cm} {\rm Pr}(A \hspace{0.05cm}|\hspace{0.05cm} B) = \frac{{\rm Pr}(A \cap B)}{{\rm Pr}(B)}\hspace{0.3cm}
@@ Line 101: / Line 101: @@
 {\rm Pr}(B \hspace{0.05cm}|\hspace{0.05cm} A) \cdot \frac{{\rm Pr}(A)}{{\rm Pr}(B)}\hspace{0.05cm}.$$
-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&nbsp; <u>proposition 3</u>.&nbsp;
 '''(2)'''&nbsp; With&nbsp; $A$ &nbsp;&#8658;&nbsp; "$x = 0$" and&nbsp; $B$ &nbsp;&#8658;&nbsp; "$y = 0$" we immediately get the equation according to <u>proposition 1</u>:
 :$${\rm Pr}(x = 0\hspace{0.05cm}|\hspace{0.05cm} y = 0) =
 {\rm Pr}(y = 0\hspace{0.05cm}|\hspace{0.05cm} x = 0)  \cdot \frac{{\rm Pr}(x = 0)}{{\rm Pr}(y = 0)}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; We compute the $L$ value of the inference probabilities. Assuming $y = 0$ holds:
+'''(3)'''&nbsp; We compute the  L&ndash;values of the inference probabilities.&nbsp; Assuming&nbsp; $y = 0$&nbsp; holds:
 :$$L_{\rm R}(y= 0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} L(x\hspace{0.05cm}|\hspace{0.05cm}y= 0)=
 {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0\hspace{0.05cm}|\hspace{0.05cm}y=0)}{{\rm Pr}(x = 1\hspace{0.05cm}|\hspace{0.05cm}y=0)} = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y = 0\hspace{0.05cm}|\hspace{0.05cm}x=0) \cdot {\rm Pr}(x = 0) / {\rm Pr}(y = 0)}{{\rm Pr}(y = 0\hspace{0.05cm}|\hspace{0.05cm}x = 1)\cdot {\rm Pr}(x = 1) / {\rm Pr}(y = 0)} $$
@@ Line 116: / Line 119: @@
 :$$\Rightarrow \hspace{0.3cm} L_{\rm R}(y= 0) =  L(x\hspace{0.05cm}|\hspace{0.05cm}y= 0) = L_{\rm V}(y= 0) + L_{\rm A}(x)\hspace{0.05cm}.$$
-Similarly, assuming $y = 1$, the result is:
+*Similarly,&nbsp; assuming&nbsp; $y = 1$,&nbsp; the result is:
 :$$L_{\rm R}(y= 1)  = L(x\hspace{0.05cm}|\hspace{0.05cm}y= 1) = L_{\rm V}(y= 1) + L_{\rm A}(x)\hspace{0.05cm}.$$
-The two results can be summarized using $y &#8712; \{0, \, 1\}$ and.
+*The two results can be summarized using&nbsp; $y &#8712; \{0, \, 1\}$&nbsp; 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,
 :$$L_{\rm V}(y) = L(y\hspace{0.05cm}|\hspace{0.05cm}x) =  {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y \hspace{0.05cm}|\hspace{0.05cm}x=0) }{{\rm Pr}(y \hspace{0.05cm}|\hspace{0.05cm}x = 1)}
 \hspace{0.05cm},$$
@@ Line 130: / Line 134: @@
 :$$L_{\rm R}(y)  = L(x\hspace{0.05cm}|\hspace{0.05cm}y) = L_{\rm V}(y) + L_{\rm A}(x)\hspace{0.05cm}.$$
-The identity $L_{\rm R}(y) &equiv; L_{\rm V}(y)$ requires $L_{\rm A}(x) = 0$ &nbsp; &#8658; &nbsp; equally probable symbols &nbsp; &#8658; &nbsp; <u>proposition 2</u>.
+*The identity &nbsp; $L_{\rm R}(y) &equiv; L_{\rm V}(y)$ &nbsp; requires $L_{\rm A}(x) = 0$ &nbsp; &#8658; &nbsp; equally probable symbols &nbsp; &#8658; &nbsp; <u>proposition 2</u>.
-*Wegen ${\rm Pr}(x = 0) = 1/2 \ \Rightarrow \ L_{\rm A}(x) = 0$ gilt somit auch:
+'''(4)'''&nbsp; From the exercise description,&nbsp; you can see that with falsification probability&nbsp; $\varepsilon = 0.1$,&nbsp; the initial value&nbsp; $y = 1$&nbsp; leads to forward log likelihood ratio&nbsp; $L_{\rm V}(y = 1) = \, &ndash;2.197$.
 :$$L_{\rm R}(y = 1)  = L_{\rm V}(y = 1)  \hspace{0.15cm}\underline{= -2.197}\hspace{0.05cm}.$$
-'''(5)'''&nbsp; 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)'''&nbsp; With the same falsification probabilityf&nbsp; $\varepsilon = 0.1$ $L_{\rm V}(y = 0)$&nbsp; differs from&nbsp; $L_{\rm V}(y = 1)$&nbsp; only by the sign.
 :$$L_{\rm R}(y = 0)  = (+)2.197 - 1.382 \hspace{0.15cm}\underline{=+0.815}\hspace{0.05cm}.$$
-'''(6)'''&nbsp; As I'm sure you'll be happy to verify, the relation&nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$&nbsp; also holds for the "2 on $M$ channel", regardless of the size (Für Günter: Umfang?)&nbsp; $M$&nbsp; of the output alphabet &nbsp; &#8658; &nbsp; <u>Answer Yes</u>.
+'''(6)'''&nbsp; The relation &nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$ &nbsp; also holds for the&nbsp; "2-on-$M$ channel",&nbsp; regardless of the set size&nbsp; $M$&nbsp; of the output alphabet &nbsp; &#8658; &nbsp; <u>Answer Yes</u>.
-'''(7)'''&nbsp; The AWGN channel is described by the outlined "2&ndash;on&ndash;$M$&ndash;channel" with&nbsp; $M &#8594; &#8734;$&nbsp; also &nbsp; &#8658; &nbsp; <u>Answer Yes</u>.
+'''(7)'''&nbsp; The AWGN channel is described by the outlined&nbsp; "2&ndash;on&ndash;$M$&ndash;channel"&nbsp; with&nbsp; $M &#8594; &#8734;$&nbsp; also &nbsp; &#8658; &nbsp; <u>Answer Yes</u>.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^4.1 Soft–in Soft–out Decoder^]]

@@ Line 19: / Line 19: @@
 :$$L_{\rm A}(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)}{{\rm Pr}(x = 1)}\hspace{0.05cm},$$
-where the subscript &nbsp;$\rm A$&nbsp; indicates the&nbsp; "a-priori log likelihood ratio"&nbsp; or the&nbsp; "a-priori L&ndash;value".
+where the subscript &nbsp;"$\rm A$"&nbsp; indicates the&nbsp; "a-priori log likelihood ratio"&nbsp; or the&nbsp; "a-priori L&ndash;value".
 For example,&nbsp; for&nbsp; ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$ &nbsp; &rArr; &nbsp; $L_{\rm A}(x) = \, -1.382$.
@@ Line 34: / Line 34: @@
 :$$L_{\rm V}(y = 0) = +2.197\hspace{0.05cm}, \hspace{0.3cm}L_{\rm V}(y = 1) = -2.197\hspace{0.05cm}.$$
-Of particular importance to coding theory are the inference probabilities&nbsp; ${\rm Pr}(x\hspace{0.05cm}|\hspace{0.05cm}y)$, which are related to the forward probabilities&nbsp; ${\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x)$&nbsp; and the input probabilities&nbsp; ${\rm Pr}(x = 0)$&nbsp; and&nbsp; ${\rm Pr}(x = 1)$&nbsp; via Bayes' theorem.
+Of particular importance to coding theory are the inference probabilities&nbsp; ${\rm Pr}(x\hspace{0.05cm}|\hspace{0.05cm}y)$,&nbsp; which are related to the backward probabilities&nbsp; ${\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x)$&nbsp; and the input probabilities&nbsp; ${\rm Pr}(x = 0)$&nbsp; and&nbsp; ${\rm Pr}(x = 1)$&nbsp; via Bayes' theorem.
-The corresponding&nbsp; &nbsp; L&ndash;value in forward direction&nbsp; $($German:&nbsp; "Vorwärtsrichtung" &nbsp; &rArr; &nbsp; subscript "V"$)$ in this exercise&nbsp; is denoted by $L_{\rm R}(y)$&nbsp;:
+The corresponding&nbsp; &nbsp; L&ndash;value in backward direction&nbsp; $($German:&nbsp; "Rückwärtsrichtung" &nbsp; &rArr; &nbsp; subscript "R"$)$&nbsp; is denoted in this exercise&nbsp; by&nbsp; $L_{\rm R}(y)$:
 :$$L_{\rm R}(y) = L(x\hspace{0.05cm}|\hspace{0.05cm}y) =
 {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}|\hspace{0.05cm}y)} \hspace{0.05cm} .$$
@@ Line 44: / Line 44: @@
+*In the last subtasks you have to clarify whether the found relations between&nbsp; $L_{\rm A}, \ L_{\rm V}$&nbsp; and&nbsp; $L_{\rm R}$&nbsp; can also be transferred to the&nbsp; "2-on-$M$&nbsp; channel".
-−
-Hints:
+*For this purpose,&nbsp; we choose a bipolar approach for the input symbols: &nbsp; "$0$"&nbsp; &#8594; &nbsp;"$+1$"&nbsp; and&nbsp; "$1$" &nbsp; &#8594; &nbsp;"$&ndash;1$".
@@ Line 70: / Line 69: @@
 - ${\rm Pr}(x = 0 | y = 0) = {\rm Pr}(y = 0 | x = 0) \cdot {\rm Pr}(y = 0) / {\rm Pr}(x = 0)$.
-{Under what conditions does the inference LLR hold for all possible output values&nbsp; $y &#8712; \{0, \, 1\}$: <br>&nbsp; &nbsp;  $L(x\hspace{0.05cm}|\hspace{0.05cm}y) = L(y\hspace{0.05cm}|\hspace{0.05cm}x)$&nbsp; bzw.&nbsp; $L_{\rm R}(y) = L_{\rm V}(y)$?
+{Under what conditions does the inference log likelihooh ratio hold for all possible output values&nbsp; $y &#8712; \{0, \, 1\}$: &nbsp; &nbsp;  $L(x\hspace{0.05cm}|\hspace{0.05cm}y) = L(y\hspace{0.05cm}|\hspace{0.05cm}x)$ &nbsp; resp. &nbsp; $L_{\rm R}(y) = L_{\rm V}(y)$?
 |type="()"}
 - For any input distribution&nbsp; ${\rm Pr}(x = 0), \ {\rm Pr}(x = 1)$.
 + For the uniform distribution only:&nbsp; $\hspace{0.2cm} {\rm Pr}(x = 0) = {\rm Pr}(x = 1) = 1/2$.
-{Let the initial symbol be&nbsp; $y = 1$. What inference LLR is obtained with the corruption probability&nbsp; $\varepsilon = 0.1$&nbsp; for equally probable symbols?
+{Let the initial symbol be&nbsp; $y = 1$.&nbsp; What inference LLR is obtained with the corruption probability&nbsp; $\varepsilon = 0.1$&nbsp; for equally probable symbols?
 |type="{}"}
 $L_{\rm R}(y = 1) = L(x | y = 1) \ = \ ${ -2.26291--2.13109 }
-{Let the initial symbol now be&nbsp; $y = 0$. What inference LLR is obtained for&nbsp; ${\rm Pr}(x = 0) = 0.2$&nbsp; and&nbsp; $\varepsilon = 0.1$?
+{Let the initial symbol now be&nbsp; $y = 0$. What inference log likelihooh ratio is obtained for&nbsp; ${\rm Pr}(x = 0) = 0.2$&nbsp; and&nbsp; $\varepsilon = 0.1$?
 |type="{}"}
 $L_{\rm R}(y = 0) = L(x | y = 0) \ = \ ${ 0.815 3% }
-{Can the result derived in '''(3)'''' &nbsp; &#8658; &nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$&nbsp; also be applied to the "2 on $M$ channel"?
+{Can the result derived in subtask&nbsp; '''(3)'''' &nbsp; &#8658; &nbsp; $L_{\rm R} = L_{\rm V} + L_{\rm A}$&nbsp; also be applied to the&nbsp; "2-on $M$-channel"?
 |type="()"}
 + Yes.
 - No.
-{Can the context be applied to the AWGN&ndash;channel as well?
+{Can the context be applied to the AWGN channel as well?
 |type="()"}
 + Yes.

@@ Line 23: / Line 23: @@
 For example,&nbsp; for&nbsp; ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$ &nbsp; &rArr; &nbsp; $L_{\rm A}(x) = \, -1.382$.
-From the BSC&ndash;model, it is also possible to determine the&nbsp; $L$ value (LLR) of the conditional probabilities&nbsp; ${\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x)$&nbsp; in the forward direction, which is also denoted by&nbsp; $L_{\rm V}(y)$&nbsp; in the present exercise:
+From the BSC model,&nbsp; it is possible to determine the&nbsp; L&ndash;value of the conditional probabilities&nbsp; ${\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x)$&nbsp; in forward direction&nbsp; $($German:&nbsp; "Vorwärtsrichtung" &nbsp; &rArr; &nbsp; subscript "V"$)$,&nbsp; which is denoted by&nbsp; $L_{\rm V}(y)$&nbsp; in the present exercise:
 :$$L_{\rm V}(y) = L(y\hspace{0.05cm}|\hspace{0.05cm}x) =
 {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x = 0)}{{\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x = 1)} =
@@ Line 31: / Line 31: @@
 \\  {\rm f\ddot{u}r} \hspace{0.15cm} y = 1. \\ \end{array}$$
-For example, for&nbsp; $\varepsilon = 0.1$:
+For example,&nbsp; for&nbsp; $\varepsilon = 0.1$:
 :$$L_{\rm V}(y = 0) = +2.197\hspace{0.05cm}, \hspace{0.3cm}L_{\rm V}(y = 1) = -2.197\hspace{0.05cm}.$$
 Of particular importance to coding theory are the inference probabilities&nbsp; ${\rm Pr}(x\hspace{0.05cm}|\hspace{0.05cm}y)$, which are related to the forward probabilities&nbsp; ${\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x)$&nbsp; and the input probabilities&nbsp; ${\rm Pr}(x = 0)$&nbsp; and&nbsp; ${\rm Pr}(x = 1)$&nbsp; via Bayes' theorem.
-The corresponding&nbsp; $L$ value (LLR) in this exercise&nbsp; is denoted by $L_{\rm R}(y)$&nbsp;:
+The corresponding&nbsp; &nbsp; L&ndash;value in forward direction&nbsp; $($German:&nbsp; "Vorwärtsrichtung" &nbsp; &rArr; &nbsp; subscript "V"$)$ in this exercise&nbsp; is denoted by $L_{\rm R}(y)$&nbsp;:
 :$$L_{\rm R}(y) = L(x\hspace{0.05cm}|\hspace{0.05cm}y) =
 {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)\hspace{0.05cm}|\hspace{0.05cm}y)}{{\rm Pr}(x = 1)\hspace{0.05cm}|\hspace{0.05cm}y)} \hspace{0.05cm} .$$

@@ Line 2: / Line 2: @@
 [[File:P_ID2979__KC_A_4_1_v2.png|right|frame|Considered channel models]]
-To interpret&nbsp; <i>log likelihood ratios</i>&nbsp; (LLRs / L values) we start from the&nbsp; <i>binary symmetric channel</i>&nbsp; as in&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio|"theory section"]]&nbsp;.
+To interpret the "log likelihood ratio"&nbsp; $\rm (LLR)$&nbsp; we start from the&nbsp; "binary symmetric channel"&nbsp; $\rm (BSC)$&nbsp; as in the&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio|"theory section"]]&nbsp;.
-For the binary random variables at the input and output the following is valid
+For the binary random variables at the channel input and output  holds:
 :$$x \in \{0\hspace{0.05cm}, 1\} \hspace{0.05cm},\hspace{0.25cm}y \in \{0\hspace{0.05cm}, 1\}
 \hspace{0.05cm}. $$
-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.&nbsp; The following applies to the conditional probabilities in the forward direction:
-:$${\rm Pr}(y = 1\hspace{0.05cm}|\hspace{0.05cm} x = 0) \hspace{-0.2cm} \ = \ \hspace{-0.2cm} {\rm Pr}(y = 0\hspace{0.05cm}|\hspace{0.05cm} x = 1) = \varepsilon \hspace{0.05cm},$$
+:$${\rm Pr}(y = 1\hspace{0.05cm}|\hspace{0.05cm} x = 0) = {\rm Pr}(y = 0\hspace{0.05cm}|\hspace{0.05cm} x = 1) = \varepsilon \hspace{0.05cm},$$
-:$${\rm Pr}(y = 0\hspace{0.05cm}|\hspace{0.05cm} x = 0) \hspace{-0.2cm} \ = \ \hspace{-0.2cm} {\rm Pr}(y = 1\hspace{0.05cm}|\hspace{0.05cm} x = 1) = 1-\varepsilon \hspace{0.05cm}.$$
+:$${\rm Pr}(y = 0\hspace{0.05cm}|\hspace{0.05cm} x = 0) = {\rm Pr}(y = 1\hspace{0.05cm}|\hspace{0.05cm} x = 1) = 1-\varepsilon \hspace{0.05cm}.$$
-The corruption probability&nbsp; $\varepsilon$&nbsp; is the crucial parameter of the BSC model.
+The falsification probability&nbsp; $\varepsilon$&nbsp; is the crucial parameter of the BSC model.
-Regarding the probability distribution at the input, instead of considering the probabilities&nbsp; ${\rm Pr}(x = 0)$&nbsp; and&nbsp; ${\rm Pr}(x = 1)$&nbsp; it is convenient to consider the&nbsp; <i>log likelihood ratio</i>&nbsp; (LLR).
+Regarding the probability distribution at the input instead of considering the probabilities&nbsp; ${\rm Pr}(x = 0)$&nbsp; and&nbsp; ${\rm Pr}(x = 1)$&nbsp; it is convenient to consider the&nbsp; log likelihood ratio.
-For the unipolar approach used here, the following applies by definition:
+For the unipolar approach used here,&nbsp; the following applies by definition:
 :$$L_{\rm A}(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = 0)}{{\rm Pr}(x = 1)}\hspace{0.05cm},$$
-where the subscript &nbsp;$\rm A$&nbsp; indicates the apriori probability.
+where the subscript &nbsp;$\rm A$&nbsp; indicates the&nbsp; "a-priori log likelihood ratio".
 For example, for&nbsp; ${\rm Pr}(x = 0) = 0.2 \ \Rightarrow \ {\rm Pr}(x = 1) = 0.8$&nbsp; the apriori LLR&nbsp; $L_{\rm A}(x) = \, -1.382$.