Aufgaben:Exercise 4.4Z: Supplement to Exercise 4.4 - Revision history

Guenter at 14:29, 3 December 2022

2022-12-03T14:29:45Z

Guenter at 14:09, 3 December 2022

2022-12-03T14:09:07Z

Guenter at 14:07, 3 December 2022

2022-12-03T14:07:39Z

Guenter at 14:06, 3 December 2022

2022-12-03T14:06:32Z

Noah at 22:09, 29 October 2022

2022-10-29T22:09:48Z

Noah at 22:09, 29 October 2022

2022-10-29T22:09:25Z

Noah at 22:06, 29 October 2022

2022-10-29T22:06:50Z

Noah at 22:06, 29 October 2022

2022-10-29T22:06:35Z

Noah at 19:47, 29 October 2022

2022-10-29T19:47:10Z

Noah at 19:44, 29 October 2022

2022-10-29T19:44:14Z

Show changes

@@ Line 85: / Line 85: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID2996__KC_Z_4_4a_v1.png|frame|right|Derivation "$w_{\rm H}$ is even" for code length $n = 2$.]]
+[[File:P_ID2996__KC_Z_4_4a_v1.png|frame|right|Derivation&nbsp; "$w_{\rm H}$ is even"&nbsp; for code length&nbsp; $n = 2$.]]
 '''(1)'''&nbsp;  According to the adjacent table applies:
 :$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.10cm}{\rm is \hspace{0.10cm} even}\right ] =
@@ Line 91: / Line 91: @@
 \hspace{0.05cm}. $$
-With the probabilities
+*With the probabilities
 :$$p_1 = {\rm Pr} (x_1 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.2\hspace{0.05cm},\hspace{0.3cm}q_1 = {\rm Pr} (x_1 = 0) = 0.8\hspace{0.05cm},$$
 :$$p_2 = {\rm Pr} (x_2 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.9\hspace{0.05cm},\hspace{0.3cm}q_2 = {\rm Pr} (x_2 = 0) = 0.1$$
-one obtains:
+:one obtains:
 :$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 0\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
 {\rm Pr} \left [(x_1 = 0)\cap (x_2 = 0) \right] = q_1 \cdot q_2 = 0.8 \cdot 0.1
@@ Line 104: / Line 104: @@
 \hspace{0.05cm}.$$
-The Gallager's equation provides for the same set of parameters:
+*The Gallager's equation provides for the same set of parameters:
 :$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  0.5 + 0.5 \cdot \prod\limits_{i =1}^{2} \hspace{0.25cm}(1-2\cdot p_i) = 0.5 + 0.5 \cdot  (1 - 2 \cdot 0.2)\cdot  (1 - 2 \cdot 0.9) = 0.26
 \hspace{0.05cm}.$$
-The equation given by Gallager 1963 was hereby verified for $n = 2$.
+*The equation given by Gallager 1963 was hereby verified for&nbsp; $n = 2$.
 [[File:P_ID2997__KC_Z_4_4b_v1.png|right|frame|Derivation "$w_{\rm H}$ is even" for code length $n = 3$.]]
-'''(2)'''&nbsp; In the second table, the four combinations with an even number of ones are marked in blue. The probabilities of occurrence of each combination are given in the last column. Thus, the result is:
+'''(2)'''&nbsp; In the second table,&nbsp; the four combinations with an even number of&nbsp; "ones"&nbsp; are marked in blue.&nbsp;
 :$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] =  0.056 + 0.216 + 0.006 + 0.126 \hspace{0.15cm} \underline{= 0.404}
 \hspace{0.05cm}.$$
-The red rows provide the complementary event:
+*The red rows provide the complementary event:
 :$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] =  0.024 + 0.504 + 0.014 + 0.054= 0.596
 \hspace{0.05cm}.$$
-The Gallager's equation again gives the exact same result, although it should be noted that this equation is valid for all $n$ and all arbitrary probabilities:
+*Gallager's equation again gives the exact same result,&nbsp; although it should be noted that this equation is valid for all&nbsp; $n$&nbsp; and all arbitrary probabilities:
 :$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  0.5 + 0.5 \cdot \prod\limits_{i =1}^{3} \hspace{0.25cm}(1-2\cdot p_i) $$
 :$$\Rightarrow\hspace{0.3cm}{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot  (+0.6) \cdot  (-0.8) \cdot  (+0.4) = 0.404
@@ Line 131: / Line 134: @@
   \hspace{0.05cm}.$$
-From this can be calculated:
+*From this can be calculated:
 :$${\rm Pr}({\rm blue}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  0.5 + 0.5 \cdot \pi  = 0.5 + 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.5192}\hspace{0.05cm},$$
 :$${\rm Pr}({\rm red}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  0.5 - 0.5 \cdot \pi  = 0.5 - 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.4808}\hspace{0.05cm}. $$
-If you add up the blue and red probabilities on the information page, you get exactly the values calculated here.
+*If you add up the blue and red probabilities on the information page,V you get exactly the values calculated here.&nbsp; For the quotient we get:
 :$$Q = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right]} { {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right]} = \frac{0.5192}{0.4808}\hspace{0.15cm} \underline{= 1.0799}
 \hspace{0.05cm}. $$
-'''(4)'''&nbsp; For the <i>single parity&ndash;check code</i>, the extrinsic $L$ value with respect to the $i$th bit was specified as follows:
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
   \hspace{0.05cm},$$
-or:
+*Or:
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{1+\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}{1-\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}
   \hspace{0.05cm}.$$
-At&nbsp; $\text{SPC (5, 4, 2}$) &nbsp; &#8658; &nbsp; $n = 5$, this product for $i = 5$ results from the following four factors:
+*At&nbsp; $\text{SPC (5, 4, 2}$) &nbsp; &#8658; &nbsp; $n = 5$,&nbsp; this product for&nbsp; $i = 5$&nbsp; results from the following four factors:
 :$$\pi = \prod\limits_{j = 1,  \hspace{0.05cm}2,  \hspace{0.05cm}3,  \hspace{0.05cm}4} \hspace{0.05cm}(1-2\cdot p_j) =
 (1-2\cdot p_1) \cdot (1-2\cdot p_2) \cdot (1-2\cdot p_3) \cdot (1-2\cdot p_4)
   \hspace{0.05cm}.$$
-The comparison with the subtask '''(3)''' shows that $L_{\rm E}(i = 5) = \ln {Q} = \ln {(1.0799)} \ \underline{\approx 0.077}$.
+The comparison with subtask&nbsp; '''(3)'''&nbsp; shows that&nbsp; $L_{\rm E}(i = 5) = \ln {Q} = \ln {(1.0799)} \ \underline{\approx 0.077}$.
-'''(5)'''&nbsp; Correct is <u>proposed solution 3</u> because the result for $L_{\rm E}(i = 5)$ is independent of $p_5$.
+'''(5)'''&nbsp; Correct is the&nbsp; <u>proposed solution 3</u>&nbsp; because the result for&nbsp; $L_{\rm E}(i = 5)$&nbsp; is independent of&nbsp; $p_5$.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^4.1 Soft–in Soft–out Decoder^]]

@@ Line 49: / Line 49: @@
 *This exercise belongs to the chapter&nbsp; [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder|"Soft&ndash;in Soft&ndash;out Decoder"]].
-*Reference is made in particular to the section&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"For calculating the extrinsic log likelihood ratios"]].
+*Reference is made in particular to the section&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios|"For calculating the extrinsic log likelihood ratios"]].
 *The exercise is intended as a supplement to&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| $\text{Exercise 4.4}$]]&nbsp;.

@@ Line 34: / Line 34: @@
 One applies the equation,&nbsp; for example,&nbsp; to calculate the extrinsic&nbsp; L&ndash;values of a&nbsp; "single parity&ndash;check code".
-Indeed,&nbsp; as already pointed out in&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| $\text{Exercise A4.4}$]]&nbsp; the extrinsic log likelihood ratio with Hamming&ndash;weight&nbsp; $w_{\rm H}$&nbsp; is the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
+Indeed,&nbsp; as already pointed out in&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| $\text{Exercise 4.4}$]]&nbsp; the extrinsic log likelihood ratio with Hamming&ndash;weight&nbsp; $w_{\rm H}$&nbsp; is the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
   \hspace{0.05cm}.$$

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}
-[[File:P_ID2994__KC_Z_4_4_v3.png |right|frame|Hamming weights and sequence probabilities ]]
+[[File:P_ID2994__KC_Z_4_4_v3.png |right|frame|Hamming weights&nbsp; $w_{\rm H}(\underline{x})$, <br>sequence probabilities&nbsp; ${\rm Pr}(\underline{x})$ ]]
-The information theorist&nbsp; [https://en.wikipedia.org/wiki/Robert_G._Gallager "Robert G. Gallager"]&nbsp; already in 1963 dealt with the following problem:
+The information theorist&nbsp; [https://en.wikipedia.org/wiki/Robert_G._Gallager $\text{Robert G. Gallager}$]&nbsp; dealt already in 1963 with the following problem:
 * Given a random vector&nbsp; $\underline{x} = (x_1, \, x_2, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , x_n)$&nbsp; with&nbsp; $n$&nbsp; binary elements&nbsp; $x_i &#8712; \{0, \, 1\}$.
 * Known are all probabilities&nbsp; $p_i = {\rm Pr}(x_i = 1)$&nbsp; and&nbsp; $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$&nbsp; with index&nbsp; $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.
 * What we are looking for is the probability that the number of ones in this vector is even.
 * Or expressed using the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]: &nbsp; What is the probability&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$?
@@ Line 12: / Line 15: @@
 * For the row highlighted in green &nbsp; &#8658; &nbsp; $\underline{x} = (1, \, 0, \, 0, \, 1)$&nbsp; holds&nbsp; $w_{\rm H}(\underline{x}) = 2$&nbsp; and
 :$${\rm Pr}(\underline{x}) = p_1 \cdot q_2 \cdot q_3 \cdot p_4 = 0.0084.$$
-* Blue font means "$w_{\rm H}(\underline{x})$ is even". Red font stands for "$w_{\rm H}(\underline{x})$ is odd."
+* Blue font means&nbsp; "$w_{\rm H}(\underline{x})$&nbsp; is even".&nbsp; Red font stands for&nbsp; "$w_{\rm H}(\underline{x})$&nbsp; is odd."
 *The probability&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$&nbsp; is the sum of the blue numbers in the last column.
 *The sum of the red numbers gives&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ odd}] = 1 - {\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$.
@@ Line 27: / Line 32: @@
   \hspace{0.05cm}.$$
-One applies the equation, for example, to calculate the extrinsic&nbsp; $L$&ndash;values of a&nbsp; <i>single parity&ndash;check code</i>&nbsp;.
+One applies the equation,&nbsp; for example,&nbsp; to calculate the extrinsic&nbsp; L&ndash;values of a&nbsp; "single parity&ndash;check code".
-Indeed, as already pointed out in&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| "Exercise A4.4"]]&nbsp; the extrinsic $L$&ndash;value with Hamming&ndash;weight&nbsp; $w_{\rm H}$&nbsp; is the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
+Indeed,&nbsp; as already pointed out in&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| $\text{Exercise A4.4}$]]&nbsp; the extrinsic log likelihood ratio with Hamming&ndash;weight&nbsp; $w_{\rm H}$&nbsp; is the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
   \hspace{0.05cm}.$$
@@ Line 41: / Line 46: @@
+*The exercise is intended as a supplement to&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| $\text{Exercise 4.4}$]]&nbsp;.
@@ Line 54: / Line 58: @@
 ===Questions===
 <quiz display=simple>
-{We consider the vector&nbsp; $\underline{x} = (x_1, \, x_2) \ \Rightarrow \ n = 2$&nbsp; with&nbsp; $x_i &#8712; \{0, \, 1\}$&nbsp; and&nbsp; $p_1 = 0.2, \ p_2 = 0.9$.  <br>What is the probability that&nbsp; $\underline{x}$&nbsp; contains an even number of ones?
+{We consider the vector&nbsp; $\underline{x} = (x_1, \, x_2) \ \Rightarrow \ n = 2$&nbsp; with&nbsp; $x_i &#8712; \{0, \, 1\}$&nbsp; and&nbsp; $p_1 = 0.2, \ p_2 = 0.9$.  <br>What is the probability that &nbsp; $\underline{x}$ &nbsp; contains an even number of&nbsp; "ones"&nbsp;?
 |type="{}"}
 ${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.26 3% }
-{Berechnen Sie die gleiche Wahrscheinlichkeit für&nbsp; $\underline{x} = (x_1, \, x_2, \, x_3) \ \Rightarrow \ n = 3$&nbsp;   und&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3$.
+{Compute the same probability for&nbsp; $\underline{x} = (x_1, \, x_2, \, x_3) \ \Rightarrow \ n = 3$&nbsp;   and&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3$.
 |type="{}"}
 ${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.404 3% }
-{Now let&nbsp; $n = 4$&nbsp; and&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$. Calculate the following quantities according to Gallager's equation:
+{Now let be&nbsp; $n = 4$&nbsp; and&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$.&nbsp; Calculate the following quantities according to Gallager's equation:
 |type="{}"}
 ${\rm Pr(blue) = Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \hspace{0.33cm} = \ ${ 0.5192 3% }
@@ Line 68: / Line 72: @@
 $\text{Quotient }Q = {\rm Pr(blau)/Pr(rot)} \hspace{0.4cm} = \ ${ 1.0799 3% }
-{What is the extrinsic&nbsp; $L$ value for the symbol&nbsp; $i = 5$&nbsp; at&nbsp; $\text{SPC (5, 4, 2)}$&nbsp; with&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6, \ p_5 = 0.9$?
+{What is the extrinsic log likelihood ratio for the symbol&nbsp; $i = 5$&nbsp; at&nbsp; $\text{SPC (5, 4, 2)}$&nbsp; with&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6, \ p_5 = 0.9$?
 |type="{}"}
 $L_{\rm E}(i = 5) \ = \ ${ 0.077 3% }
-{How does&nbsp; $L_{\rm E}(i = 5)$ change if we assume&nbsp; $p_5 = 0.1$&nbsp; instead?
+{How does&nbsp; $L_{\rm E}(i = 5)$&nbsp; change if we assume&nbsp; $p_5 = 0.1$&nbsp; instead?
 |type="()"}
 - $L_{\rm E}(i = 5)$&nbsp; becomes larger.
 - $L_{\rm E}(i = 5)$&nbsp; becomes smaller.
-+ $L_{\rm E}(i = 5)$&nbsp; is not changed compared to subtask '''(4)'''.
++ $L_{\rm E}(i = 5)$&nbsp; is not changed compared to subtask&nbsp; '''(4)'''.
 </quiz>

@@ Line 30: / Line 30: @@
 Indeed, as already pointed out in&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| "Exercise A4.4"]]&nbsp; the extrinsic $L$&ndash;value with Hamming&ndash;weight&nbsp; $w_{\rm H}$&nbsp; is the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
-:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} gerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} ungerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
+:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
   \hspace{0.05cm}.$$
@@ Line 46: / Line 46: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder|"Soft&ndash;in Soft&ndash;out Decoder"]].
-*Reference is made in particular to the page&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"For calculating the extrinsic&nbsp; $L$&ndash;values"]].
+*Reference is made in particular to the page&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"For calculating the extrinsic&nbsp; $L$ values"]].
 *The exercise is intended as a supplement to&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| "Exercise 4.4"]]&nbsp;.

@@ Line 6: / Line 6: @@
 * Known are all probabilities&nbsp; $p_i = {\rm Pr}(x_i = 1)$&nbsp; and&nbsp; $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$&nbsp; with index&nbsp; $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.
 * What we are looking for is the probability that the number of ones in this vector is even.
-* Or expressed using the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]: &nbsp; What is the probability&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \even}]$?
+* Or expressed using the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]: &nbsp; What is the probability&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$?
@@ Line 13: / Line 13: @@
 :$${\rm Pr}(\underline{x}) = p_1 \cdot q_2 \cdot q_3 \cdot p_4 = 0.0084.$$
 * Blue font means "$w_{\rm H}(\underline{x})$ is even". Red font stands for "$w_{\rm H}(\underline{x})$ is odd."
-*The probability&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \is \even}]$&nbsp; is the sum of the blue numbers in the last column.
+*The probability&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$&nbsp; is the sum of the blue numbers in the last column.
 *The sum of the red numbers gives&nbsp; ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ odd}] = 1 - {\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$.
-Gallager hat das Problem in analytischer Weise gelöst:
+Gallager solved the problem in an analytical way:
-:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm ist \hspace{0.15cm} gerade} \right ]
+:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \right ]
 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 + \pi]\hspace{0.05cm},$$
-:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm ist \hspace{0.15cm} ungerade}
+:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}
 \right ] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 - \pi]\hspace{0.05cm}.$$
-Hierbei ist die folgende Hilfsgröße verwendet:
+Here the following auxiliary variable is used:
 :$$\pi = \prod\limits_{i =1}^{n} \hspace{0.25cm}(1-2p_i)
   \hspace{0.05cm}.$$
-Die Gleichung wendet man zum Beispiel an, um die extrinsischen&nbsp; $L$&ndash;Werte eines&nbsp; <i>Single Parity&ndash;check Codes</i>&nbsp; zu berechnen.
+One applies the equation, for example, to calculate the extrinsic&nbsp; $L$&ndash;values of a&nbsp; <i>single parity&ndash;check code</i>&nbsp;.
-Wie bereits in der&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| "Aufgabe A4.4"]]&nbsp; dargelegt, lautet nämlich der extrinsische $L$&ndash;Wert mit dem Hamming&ndash;Gewicht&nbsp; $w_{\rm H}$&nbsp; der verkürzten Folge&nbsp; $\underline{x}^{(-i)}$:
+Indeed, as already pointed out in&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| "Exercise A4.4"]]&nbsp; the extrinsic $L$&ndash;value with Hamming&ndash;weight&nbsp; $w_{\rm H}$&nbsp; is the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} gerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} ungerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
   \hspace{0.05cm}.$$
-Hierbei ist berücksichtigt, dass man für&nbsp; $L_{\rm E}(i)$&nbsp; nur die anderen Symbole&nbsp; $(j &ne; i)$&nbsp; heranziehen darf:
+Here it is taken into account that for&nbsp; $L_{\rm E}(i)$&nbsp; one may refer only to the other symbols&nbsp; $(j &ne; i)$&nbsp;:
 :$$\underline{x}^{(-i)} = \big ( \hspace{0.03cm}x_1, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm}  , \hspace{0.03cm} x_{i-1}, \hspace{0.43cm} x_{i+1},  \hspace{-0.04cm} \text{ ...} \hspace{0.08cm}  , x_{n} \hspace{0.03cm} \big )\hspace{0.03cm}. $$