Aufgaben:Exercise 1.14: Bhattacharyya Bound for BEC - Revision history

Hwang at 16:03, 23 January 2023

2023-01-23T16:03:22Z

Hwang at 17:00, 1 November 2022

2022-11-01T17:00:29Z

Guenter at 15:04, 30 September 2022

2022-09-30T15:04:47Z

Guenter at 14:49, 4 August 2022

2022-08-04T14:49:58Z

Guenter at 14:28, 4 August 2022

2022-08-04T14:28:48Z

Guenter at 14:26, 4 August 2022

2022-08-04T14:26:32Z

Noah at 17:07, 30 July 2022

2022-07-30T17:07:00Z

Noah at 20:36, 29 July 2022

2022-07-29T20:36:00Z

Noah at 20:57, 28 July 2022

2022-07-28T20:57:55Z

Noah at 15:44, 28 July 2022

2022-07-28T15:44:15Z

@@ Line 96: / Line 96: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The code words &nbsp; $\underline{x}_{0}$ &nbsp; and &nbsp; $\underline{x}_{1}$&nbsp; differ in bit $2, \ 4$&nbsp; and $5$.&nbsp; If only one of these three binary values is transmitted correctly,&nbsp; the entire code word is thus uniquely determined.&nbsp; No information about the code word is obtained for the following received vectors&nbsp; (see table on the information page):
+'''(1)'''&nbsp; The code words &nbsp; $\underline{x}_{0}$ &nbsp; and &nbsp; $\underline{x}_{1}$&nbsp; differ in bit $2, \ 4$&nbsp; and $5$.&nbsp; If only one of these three binary values is transmitted correctly,&nbsp; the entire code word is thus uniquely determined.&nbsp; No information about the code word is obtained for the following received vectors&nbsp; (see table in the information section):
 * $\underline{y} = (0, {\rm E}, 0, {\rm E}, {\rm E})$&nbsp; with probability&nbsp; $\lambda^3 \ · \ (1 – \lambda)^2$,
 * $\underline{y} = (0, {\rm E}, {\rm E}, {\rm E}, {\rm E})$&nbsp; with probability&nbsp; $\lambda^4 \ · \ (1 – \lambda)$,

@@ Line 38: / Line 38: @@
 Please note:
-*The event&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$&nbsp; does not necessarily say that at the received vector under consideration&nbsp; $\underline{y}$&nbsp; is actually decided for the codeword&nbsp; $\underline{x}_{1}$,&nbsp;  but only that the decision for&nbsp; $x_{1}$&nbsp; would be more reasonable than the decision for&nbsp; $\underline{x}_{0}$&nbsp; due to statistics.
+*The event&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$&nbsp; does not necessarily say that at the received vector under consideration&nbsp; $\underline{y}$&nbsp; is actually decided for the code word&nbsp; $\underline{x}_{1}$,&nbsp;  but only that the decision for&nbsp; $x_{1}$&nbsp; would be more reasonable than the decision for&nbsp; $\underline{x}_{0}$&nbsp; due to statistics.
 *But it could also be decided for&nbsp; $\underline{x}_{2}$&nbsp; or&nbsp; $\underline{x}_{3}$&nbsp; if the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_.C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-Likelihood.C2.AB|"maximum-likelihood criterion"]]&nbsp; is in favor.

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Limits_for_Block_Error_Probability}}
-[[File:P_ID2411__KC_A_1_13.png|right|frame|Possible received vectors for the&nbsp; $(5, 2)$ code and BEC '''Korrektur''']]
+[[File:EN_KC_A_1_14.png|right|frame|Possible received vectors for the&nbsp; $(5, 2)$ code and BEC ]]
 In this exercise,&nbsp; we consider the systematic&nbsp; $(5, 2)$&nbsp; code

@@ Line 77: / Line 77: @@
 - It holds&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ \lambda ^{d_{{\rm H},\hspace{0.05cm}i}} \  · \  (1 – \lambda)^{n \hspace{0.05cm}– \hspace{0.05cm}d_{{\rm H},\hspace{0.05cm}i}}$.
 + It holds&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ 1/2 · \lambda ^{d_{{\rm H},\hspace{0.05cm}i}}.$
-- ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$ is the falsification probability from&nbsp; $x_{0}$&nbsp; to&nbsp; $x_{i}$.
+- ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$&nbsp; is the falsification probability from&nbsp; $x_{0}$&nbsp; to&nbsp; $x_{i}$.
 {What are the following probabilities?
@@ Line 96: / Line 96: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The code words $\underline{x}_{0}$ and $\underline{x}_{1}$ differ in bit $2, \ 4$ and $5$. If only one of these three binary values is transmitted correctly, the entire codeword is thus uniquely determined. No information about the codeword is obtained for the following received vectors (see table on the information page):
+'''(1)'''&nbsp; The code words &nbsp; $\underline{x}_{0}$ &nbsp; and &nbsp; $\underline{x}_{1}$&nbsp; differ in bit $2, \ 4$&nbsp; and $5$.&nbsp; If only one of these three binary values is transmitted correctly,&nbsp; the entire code word is thus uniquely determined.&nbsp; No information about the code word is obtained for the following received vectors&nbsp; (see table on the information page):
-* $\underline{y} = (0, {\rm E}, 0, {\rm E}, {\rm E})$ The code words $\underline{x}_{0}$ and $\underline{x}_{1}$ differ in bit $2, \ 4$ and $5$. If only one of these three binary values is transmitted correctly, the entire codeword is thus uniquely determined. No information about the codeword is obtained for the following received vectors (see table on the information page): $\lambda^3 \ · \ (1 – \lambda)^2$,
+* $\underline{y} = (0, {\rm E}, 0, {\rm E}, {\rm E})$&nbsp; with probability&nbsp; $\lambda^3 \ · \ (1 – \lambda)^2$,
-* $\underline{y} = (0, {\rm E}, {\rm E}, {\rm E}, {\rm E})$ with probability $\lambda^4 \ · \ (1 – \lambda)$,
+* $\underline{y} = (0, {\rm E}, {\rm E}, {\rm E}, {\rm E})$&nbsp; with probability&nbsp; $\lambda^4 \ · \ (1 – \lambda)$,
-* $\underline{y} = ({\rm E}, {\rm E}, 0, {\rm E}, {\rm E})$ with probability $\lambda^4 \ · \ (1 – \lambda)$,
+* $\underline{y} = ({\rm E}, {\rm E}, 0, {\rm E}, {\rm E})$&nbsp; with probability $\lambda^4 \ · \ (1 – \lambda)$,
-* $\underline{y} = ({\rm E}, {\rm E}, {\rm E}, {\rm E}, {\rm E})$ with probability $\lambda^5$.
+* $\underline{y} = ({\rm E}, {\rm E}, {\rm E}, {\rm E}, {\rm E})$ with probability&nbsp; $\lambda^5$.
-The probability that, due to the specific received vector $\underline{y}$, the codeword $\underline{x}_{1}$ is just as likely as $\underline{x}_{0}$, is given by
+The probability that due to the specific received vector&nbsp;  $\underline{y}$,&nbsp; the code word&nbsp; $\underline{x}_{1}$&nbsp; is just as likely as&nbsp; $\underline{x}_{0}$,&nbsp; is given by
 :$$\ {\rm Pr}\ [\underline{x}_0 \hspace{0.12cm}{\rm and}\hspace{0.12cm} \underline{x}_1 \hspace{0.15cm}{\rm are \hspace{0.15cm}equally \hspace{0.15cm}probable}] = \lambda^3 \cdot (1- \lambda)^2 + 2 \cdot \lambda^4 \cdot (1- \lambda) + \lambda^5 =\lambda^3 \cdot \left [ (1- \lambda)^2 + 2 \cdot \lambda \cdot (1- \lambda) + \lambda^2 \right ] = \lambda^3 \hspace{0.05cm}.$$
-In this case, one decides at random for $\underline{x}_{0}$ (would be correct) or for $\underline{x}_{1}$ (unfortunately wrong), with equal probability. From this follows:
+In this case,&nbsp; one decides at random for&nbsp; $\underline{x}_{0}$&nbsp; (would be correct)&nbsp; or for&nbsp; $\underline{x}_{1}$&nbsp; (unfortunately wrong),&nbsp; with equal probability.&nbsp; From this follows:
 :$${\rm Pr} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}1}] = 1/2 \cdot \lambda^3 \hspace{0.15cm} \underline{= 0.5 \cdot 10^{-3}} \hspace{0.05cm}.$$
@@ Line 113: / Line 113: @@
-'''(2)'''&nbsp; According to subtask '''(1)''', <u>answer 2</u> is correct and not answer 1. Statement 3 is also incorrect:
+'''(2)'''&nbsp; According to subtask&nbsp; '''(1)''',&nbsp; <u>only answer 2</u>&nbsp; is correct:
-*${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}]$ does not state that with this probability the codeword $\underline{x}_{0}$ actually transitions to the incorrect codeword $\underline{x}_{1}$, but only that it could transition to $\underline{x}_{1}$ with this probability.
+*${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}]$&nbsp; does not state that with this probability the code word&nbsp; $\underline{x}_{0}$&nbsp; actually transitions to the incorrect code word&nbsp; $\underline{x}_{1}$,&nbsp; but only that it could transition to&nbsp; $\underline{x}_{1}$&nbsp; with this probability.
 * ${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}]$ also includes constellations where the decision is actually for $\underline{x}_{2}$ or $\underline{x}_{3}$.
@@ Line 122: / Line 123: @@
 '''(3)'''&nbsp; Because of&nbsp; $d_{\rm H}(\underline{x}_{0}, \underline{x}_{2}) = 3$&nbsp; and&nbsp; $d_{\rm H}(\underline{x}_{0}, \underline{x}_{3}) = 4$&nbsp; it follows that
-:$${\rm Pr} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}2}] = 1/2 \cdot \lambda^3 \hspace{0.15cm} \underline{= 0.5 \cdot 10^{-3}} \hspace{0.05cm},\hspace{0.2cm} {\rm Pr} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}3}] = 1/2 \cdot \lambda^4 \hspace{0.15cm} \underline{= 0.05 \cdot 10^{-3}} \hspace{0.05cm}.$$
+:$${\rm Pr} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}2}] = 1/2 \cdot \lambda^3 \hspace{0.15cm} \underline{= 0.5 \cdot 10^{-3}} \hspace{0.05cm},$$
-'''(4)'''&nbsp; The block error probability is never larger (with a certain probability rather smaller) than the so-called ''Union Bound'':
+'''(4)'''&nbsp; The block error probability is never larger&nbsp; (with a certain probability rather smaller)&nbsp; than the so-called&nbsp; "Union Bound":
 :$${\rm Pr(Union \hspace{0.15cm}Bound)} \hspace{-0.15cm}\ = \ \hspace{-0.15cm} {\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}1}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}2}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}3}] = 2 \cdot \lambda^3/2 + \lambda^4/2 = 0.001 + 0.00005 \hspace{0.15cm} \underline{= 1.05 \cdot 10^{-3}} \hspace{0.05cm}.$$
@@ Line 135: / Line 137: @@
 :$${\rm Pr(block\:error) ≤ {\rm Pr(Union \hspace{0.15cm}Bound)}  \le Pr(Bhattacharyya)} = W(\beta) - 1.$$
-*For the distance spectrum or the weight function, we obtain in the present case:
+*For the distance spectrum or the enumerator weight function,&nbsp; we obtain in the present case:
 :$$W_0 = 1 \hspace{0.05cm}, \hspace{0.2cm} W_3 = 2 \hspace{0.05cm}, \hspace{0.2cm}W_4 = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} W(X) = 1+ 2 \cdot X^{3} +X^{4} \hspace{0.05cm}.$$
-*For the BEC channel, $\beta = \lambda$ also applies. From this follows as a final result for $\lambda = 0.001$:
+*For the BEC channel,&nbsp; $\beta = \lambda$&nbsp; also applies.&nbsp; From this follows as a final result for&nbsp; $\lambda = 0.001$:
 :$${\rm Pr(Bhattacharyya)} = 2 \cdot \lambda^3 + \lambda^4 \hspace{0.15cm} \underline{= 2.1 \cdot 10^{-3}} \hspace{0.05cm}.$$
-Note that in the BEC model, the ''Bhattacharyya bound'' is always twice as large as the ''Union Bound,'' which is itself an upper bound on the block error probability.
+*Note that with the BEC model,&nbsp; the&nbsp; "Bhattacharyya Bound"&nbsp; is always twice as large as the&nbsp; "Union Bound",&nbsp; which is itself an upper bound on the block error probability.
 {{ML-Fuß}}

← Older revision		Revision as of 14:28, 4 August 2022
Line 40:		Line 40:
	*The event  $[\underline{x}_{0} → \underline{x}_{1}]$  does not necessarily say that at the received vector under consideration  $\underline{y}$  is actually decided for the codeword  $\underline{x}_{1}$,  but only that the decision for  $x_{1}$  would be more reasonable than the decision for  $\underline{x}_{0}$  due to statistics.		*The event  $[\underline{x}_{0} → \underline{x}_{1}]$  does not necessarily say that at the received vector under consideration  $\underline{y}$  is actually decided for the codeword  $\underline{x}_{1}$,  but only that the decision for  $x_{1}$  would be more reasonable than the decision for  $\underline{x}_{0}$  due to statistics.

−	*But it could also be decided for  $\underline{x}_{2}$  or  $\underline{x}_{3}$  if the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_. C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-~~likelihood~~.C2.AB\|"maximum-likelihood criterion"]]  is in favor.	+	*But it could also be decided for  $\underline{x}_{2}$  or  $\underline{x}_{3}$  if the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_.C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-Likelihood.C2.AB\|"maximum-likelihood criterion"]]  is in favor.

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Limits_for_Block_Error_Probability}}
-[[File:P_ID2411__KC_A_1_13.png|right|frame|Possible received vectors for the&nbsp; $(5, 2)$ code and BEC]]
+[[File:P_ID2411__KC_A_1_13.png|right|frame|Possible received vectors for the&nbsp; $(5, 2)$ code and BEC '''Korrektur''']]
-In this exercise, we consider the systematic&nbsp; $(5, 2)$ code
+In this exercise,&nbsp; we consider the systematic&nbsp; $(5, 2)$&nbsp; code
 *with the&nbsp; $2×5$ generator matrix
@@ Line 16: / Line 16: @@
 :$$\underline{x}_0 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} (0, 0, 0, 0, 0) \hspace{0.05cm},\hspace{0.2cm}\underline{x}_1 = (0, 1, 0, 1, 1)\hspace{0.05cm},\hspace{0.2cm}\underline{x}_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} (1, 0, 1, 1, 0) \hspace{0.05cm},\hspace{0.2cm}\underline{x}_3 = (1, 1, 1, 0, 1)\hspace{0.05cm}.$$
-At the output of the digital channel defined by the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC|"BEC model"]]&nbsp; (''Binary Erasure Channel'') with the erasure probability&nbsp; $\lambda = 0.001$&nbsp;the received vector
+At the output of the digital channel defined by the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC|"BEC model"]]&nbsp; ("Binary Erasure Channel")&nbsp; with the erasure probability&nbsp; $\lambda = 0.001$, &nbsp;the received vector
 :$$\underline{y} = (y_1, \hspace{0.05cm}y_2, \hspace{0.05cm}y_3, \hspace{0.05cm}y_4, \hspace{0.05cm}y_5)$$
-occurs, where for&nbsp; $i = 1, \ \text{...} \ , 5$&nbsp; holds: &nbsp; $y_{i} \in \{0, 1, \rm E\}$.
+occurs,&nbsp; where for&nbsp; $i = 1, \ \text{...} \ , 5$&nbsp; holds: &nbsp;
-The BEC channel is characterized by the fact that.
+The BEC channel is characterized by the fact that
-*corruptions&nbsp; $(0 → 1, 1 → 0)$&nbsp; are excluded,
+*falsifications&nbsp; $(0 → 1,\ 1 → 0)$&nbsp; are excluded,
-The graph explicitly shows all possible received vectors&nbsp; $\underline{y}$&nbsp; with three or more erasures $\rm E$&nbsp; assuming that the all-zero vector&nbsp; $(0, 0, 0, 0, 0)$ was sent.
+*but erasures&nbsp; $(0 → \rm E,\ 1 → E)$&nbsp; may occur.
 *On the other hand, if there are three or more erasures, wrong decisions may occur. In this case, the following applies to the block error probability:
 :$$ {\rm Pr(block\:error)}= {\rm Pr} (\underline{z} \ne \underline{x}) = {\rm Pr}\left \{ \hspace{0.1cm} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}1}] \hspace{0.05cm}\cup\hspace{0.05cm}[\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}2}] \hspace{0.05cm}\cup \hspace{0.05cm}[\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}3}] \hspace{0.1cm}\right \} \hspace{0.05cm}.$$
@@ Line 35: / Line 38: @@
 Please note:
-*The event&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$&nbsp; does not necessarily say that at the received vector under consideration&nbsp; $\underline{y}$&nbsp; is actually decided for the codeword&nbsp; $\underline{x}_{1}$&nbsp; is decided, but only that the decision for&nbsp; $x_{1}$&nbsp; would be more reasonable than the decision for&nbsp; $\underline{x}_{0}$ due to statistics.
+*The event&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$&nbsp; does not necessarily say that at the received vector under consideration&nbsp; $\underline{y}$&nbsp; is actually decided for the codeword&nbsp; $\underline{x}_{1}$,&nbsp;  but only that the decision for&nbsp; $x_{1}$&nbsp; would be more reasonable than the decision for&nbsp; $\underline{x}_{0}$&nbsp; due to statistics.
 *But it could also be decided for&nbsp; $\underline{x}_{2}$&nbsp; or&nbsp; $\underline{x}_{3}$&nbsp; if the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_. C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-likelihood.C2.AB|"maximum-likelihood criterion"]]&nbsp; is in favor.
-Computing the block error probability is difficult because the events&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$ ,&nbsp; $[\underline{x}_{0} → \underline{x}_{2}]$&nbsp;  and&nbsp; $[\underline{x}_{0} → \underline{x}_{3}]$&nbsp; are not necessarily&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Disjoint_sets|"disjoint"]]&nbsp;. An upper bound is provided by&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Union_Bound_of_the_block_error_probability|"Union Bound"]]:
+Computing the block error probability is difficult because the events&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$ ,&nbsp; $[\underline{x}_{0} → \underline{x}_{2}]$&nbsp;  and&nbsp; $[\underline{x}_{0} → \underline{x}_{3}]$&nbsp; are not necessarily&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Disjoint_sets|"disjoint"]]&nbsp;. An upper bound is provided by the&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Union_Bound_of_the_block_error_probability|"Union Bound"]]:
-:$${\rm Pr(Union \hspace{0.15cm}Bound)} = {\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}1}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}2}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}3}] \ge {\rm Pr(Blockfehler)} \hspace{0.05cm}.$$
+:$${\rm Pr(Union \hspace{0.15cm}Bound)} = {\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}1}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}2}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}3}] \ge {\rm Pr(block\hspace{0.15cm}error)} \hspace{0.05cm}.$$
-−
 Another bound was given by Bhattacharyya:
 :$${\rm Pr(Bhattacharyya)} = W(\beta)-1 \ge {\rm Pr(Union \hspace{0.15cm}Bound)} \ge {\rm Pr(block\:error)} \hspace{0.05cm},$$
-where, for the Binary Erasure Channel, Bhattacharyya parameter&nbsp; $\beta = \lambda$&nbsp; and&nbsp; $W(X)$&nbsp; the&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Distance_spectrum_of_a_linear_code|"weight enumerator function"]]&nbsp; where the pseudo-variable&nbsp; $X$&nbsp; is to be replaced here by the Bhattacharyya parameter&nbsp; $\lambda$&nbsp;.
+where for the binary erasure channel,&nbsp; the Bhattacharyya parameter&nbsp; $\beta = \lambda$&nbsp; and the&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Distance_spectrum_of_a_linear_code|"weight enumerator function"]]&nbsp; $W(X)$&nbsp; where the pseudo-variable&nbsp; $X$&nbsp; is to be replaced here by the Bhattacharyya parameter&nbsp; $\lambda$.
-*The Bhattacharyya bound is more or less far above the&nbsp; ''Union Bound.'' depending on the channel.
+*The&nbsp; "Bhattacharyya Bound"&nbsp; is more or less far above the&nbsp; "Union Bound"&nbsp; depending on the channel.
 *Its importance lies in the fact that the bound can be specified in the same way for different channels.
@@ Line 57: / Line 61: @@
+Hints:&nbsp; The exercise belongs to the chapter&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability|"Bounds on block error probability"]].
@@ Line 67: / Line 68: @@
 ===Questions===
 <quiz display=simple>
-{What is the pairwise error probability between the codewords&nbsp; $\underline{x}_{0} = (0, 0, 0, 0, 0)$&nbsp; and&nbsp; $\underline{x}_{1} = (0, 1, 0, 1, 1)$?
+{What is the pairwise error probability between the code words &nbsp; $\underline{x}_{0} = (0, 0, 0, 0, 0)$ &nbsp; and &nbsp; $\underline{x}_{1} = (0, 1, 0, 1, 1)$?
 |type="{}"}
 ${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}] \ = \ $ { 0.5 3% }$\ \cdot 10^{-3} $
-{Which statements are true regarding&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$&nbsp; with index&nbsp; $i = 1, \ \text{...} \ , 3$? &nbsp; <br>$d_{{\rm H},\hspace{0.05cm}i}$&nbsp; denotes here the Hamming distance between&nbsp; $x_{0}$&nbsp; and&nbsp; $x_{i}$.
+{Which statements are true regarding&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$&nbsp; with index&nbsp; $i = 1, \ \text{...} \ , 3$? &nbsp;  $d_{{\rm H},\hspace{0.05cm}i}$&nbsp; denotes here the Hamming distance between&nbsp; $x_{0}$&nbsp; and&nbsp; $x_{i}$.
 |type="()"}
 - It holds&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ \lambda ^{d_{{\rm H},\hspace{0.05cm}i}} \  · \  (1 – \lambda)^{n \hspace{0.05cm}– \hspace{0.05cm}d_{{\rm H},\hspace{0.05cm}i}}$.
 + It holds&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ 1/2 · \lambda ^{d_{{\rm H},\hspace{0.05cm}i}}.$
-- ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$ is the corruption probability of $x_{0}$ nach $x_{i}$.
+- ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$ is the falsification probability from&nbsp; $x_{0}$&nbsp; to&nbsp; $x_{i}$.
 {What are the following probabilities?
@@ Line 83: / Line 84: @@
 $\ {\rm Pr}[\underline{x}_{0} → \underline{x}_{3}] \ = \ $ { 0.05 3% } $\ \cdot 10^{-3} $
-{Specify the ''Union Bound''&nbsp; for the block error probability.
+{Specify the&nbsp; "Union Bound"&nbsp; for the block error probability.
 |type="{}"}
 $\ {\rm Pr(Union\  Bound)} \ = \ ${ 1.05 3% } $\ \cdot 10^{-3} $
-{What is the ''Bhattacharyya bound'' in the present case?
+{What is the&nbsp; "Bhattacharyya Bound"&nbsp; in the present case?
 |type="{}"}
 $\ {\rm Pr(Bhattacharyya)} \ = \ ${ 2.1 3% } $\ \cdot 10^{-3} $

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Kanalcodierung/Channel_Coding/Limits_for_Block_Error_Probability}}
+{{quiz-Header|Buchseite=Channel_Coding/Limits_for_Block_Error_Probability}}
 [[File:P_ID2411__KC_A_1_13.png|right|frame|Possible received vectors for the&nbsp; $(5, 2)$ code and BEC]]
@@ Line 39: / Line 39: @@
-Die Berechnung der Blockfehlerwahrscheinlichkeit ist schwierig, da die Ereignisse&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$ ,&nbsp; $[\underline{x}_{0} → \underline{x}_{2}]$&nbsp;  und&nbsp; $[\underline{x}_{0} → \underline{x}_{3}]$&nbsp;  nicht notwendigerweise&nbsp; [[Theory_of_Stochastic_Signals/Mengentheoretische_Grundlagen#Disjunkte_Mengen|disjunkt]]&nbsp; sind. Eine obere Schranke liefert die&nbsp; [[Channel_Coding/Schranken_für_die_Blockfehlerwahrscheinlichkeit#Union_Bound_der_Blockfehlerwahrscheinlichkeit|Union Bound]]:
+Computing the block error probability is difficult because the events&nbsp; $[\underline{x}_{0} → \underline{x}_{1}]$ ,&nbsp; $[\underline{x}_{0} → \underline{x}_{2}]$&nbsp;  and&nbsp; $[\underline{x}_{0} → \underline{x}_{3}]$&nbsp; are not necessarily&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Disjoint_sets|disjoint]]&nbsp;. An upper bound is provided by&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Union_Bound_of_the_block_error_probability|Union Bound]]:
 :$${\rm Pr(Union \hspace{0.15cm}Bound)} = {\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}1}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}2}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}3}] \ge {\rm Pr(Blockfehler)} \hspace{0.05cm}.$$
-Eine weitere Schranke wurde von Bhattacharyya angegeben:
-:$${\rm Pr(Bhattacharyya)} = W(\beta)-1 \ge {\rm Pr(Union \hspace{0.15cm}Bound)} \ge {\rm Pr(Blockfehler)} \hspace{0.05cm},$$
+:$${\rm Pr(Bhattacharyya)} = W(\beta)-1 \ge {\rm Pr(Union \hspace{0.15cm}Bound)} \ge {\rm Pr(block\:error)} \hspace{0.05cm},$$
-wobei beim Binary Erasure Channel für den Bhattacharyya&ndash;Parameter&nbsp; $\beta = \lambda$&nbsp; gilt und&nbsp;  $W(X)$&nbsp; die&nbsp; [[Channel_Coding/Schranken_für_die_Blockfehlerwahrscheinlichkeit#Distanzspektrum_eines_linearen_Codes|Gewichtsfunktion]]&nbsp; angibt, wobei die Pseudo–Variable&nbsp; $X$&nbsp; hier durch den Bhattacharyya–Parameter&nbsp; $\lambda$&nbsp; zu ersetzen ist.
+where, for the Binary Erasure Channel, Bhattacharyya parameter&nbsp; $\beta = \lambda$&nbsp; and&nbsp; $W(X)$&nbsp; the&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Distance_spectrum_of_a_linear_code|weight enumerator function]]&nbsp; where the pseudo-variable&nbsp; $X$&nbsp; is to be replaced here by the Bhattacharyya parameter&nbsp; $\lambda$&nbsp;.
-*Die Bhattacharyya–Schranke liegt je nach Kanal mehr oder weniger weit oberhalb der&nbsp; ''Union Bound.''
+*The Bhattacharyya bound is more or less far above the&nbsp; ''Union Bound.'' depending on the channel.
-*Ihre Bedeutung liegt darin, dass die Schranke für unterschiedliche Kanäle in gleicher Weise angebbar ist.
+*Its importance lies in the fact that the bound can be specified in the same way for different channels.
@@ Line 58: / Line 59: @@
-''Hinweis:''
+Hints:
-*Die Aufgabe gehört zum Kapitel&nbsp; [[Channel_Coding/Schranken_für_die_Blockfehlerwahrscheinlichkeit|Schranken für die Blockfehlerwahrscheinlichkeit]].
+*The exercise belongs to the chapter&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability|Bounds on block error probability]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie groß ist die paarweise Fehlerwahrscheinlichkeit zwischen den Codeworten&nbsp; $\underline{x}_{0} = (0, 0, 0, 0, 0)$&nbsp; und&nbsp; $\underline{x}_{1} = (0, 1, 0, 1, 1)$?
+{What is the pairwise error probability between the codewords&nbsp; $\underline{x}_{0} = (0, 0, 0, 0, 0)$&nbsp; and&nbsp; $\underline{x}_{1} = (0, 1, 0, 1, 1)$?
 |type="{}"}
 ${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}] \ = \ $ { 0.5 3% }$\ \cdot 10^{-3} $
-{Welche Aussagen stimmen bezüglich&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$&nbsp; mit Laufindex&nbsp; $i = 1, \ \text{...} \ , 3$? &nbsp; <br>$d_{{\rm H},\hspace{0.05cm}i}$&nbsp; bezeichnet hier die Hamming–Distanz zwischen&nbsp; $x_{0}$&nbsp; und&nbsp; $x_{i}$.
+{Which statements are true regarding&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$&nbsp; with index&nbsp; $i = 1, \ \text{...} \ , 3$? &nbsp; <br>$d_{{\rm H},\hspace{0.05cm}i}$&nbsp; bezeichnet hier die Hamming–Distanz zwischen&nbsp; $x_{0}$&nbsp; und&nbsp; $x_{i}$.
 |type="()"}
-- Es gilt&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ \lambda ^{d_{{\rm H},\hspace{0.05cm}i}} \  · \  (1 – \lambda)^{n \hspace{0.05cm}– \hspace{0.05cm}d_{{\rm H},\hspace{0.05cm}i}}$.
+- It holds&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ \lambda ^{d_{{\rm H},\hspace{0.05cm}i}} \  · \  (1 – \lambda)^{n \hspace{0.05cm}– \hspace{0.05cm}d_{{\rm H},\hspace{0.05cm}i}}$.
-+ Es gilt&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ 1/2 · \lambda ^{d_{{\rm H},\hspace{0.05cm}i}}.$
++ It holds&nbsp; ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ 1/2 · \lambda ^{d_{{\rm H},\hspace{0.05cm}i}}.$
-- ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$ ist die Verfälschungswahrscheinlichkeit von $x_{0}$ nach $x_{i}$.
+- ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$ is the corruption probability of $x_{0}$ nach $x_{i}$.
-{Wie groß sind die folgenden Wahrscheinlichkeiten?
+{What are the following probabilities?
 |type="{}"}
 $\ {\rm Pr}[\underline{x}_{0} → \underline{x}_{2}] \ = \ $ { 0.5 3% } $\ \cdot 10^{-3} $
 $\ {\rm Pr}[\underline{x}_{0} → \underline{x}_{3}] \ = \ $ { 0.05 3% } $\ \cdot 10^{-3} $
-{Geben Sie die ''Union Bound''&nbsp; für die Blockfehlerwahrscheinlichkeit an.
+{Specify the ''Union Bound''&nbsp; for the block error probability.
 |type="{}"}
 $\ {\rm Pr(Union\  Bound)} \ = \ ${ 1.05 3% } $\ \cdot 10^{-3} $
-{Wie lautet im vorliegenden Fall die ''Bhattacharyya–Schranke''?
+{What is the ''Bhattacharyya bound'' in the present case?
 |type="{}"}
 $\ {\rm Pr(Bhattacharyya)} \ = \ ${ 2.1 3% } $\ \cdot 10^{-3} $
@@ Line 92: / Line 93: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Die Codeworte $\underline{x}_{0}$ und $\underline{x}_{1}$ unterscheiden sich in Bit $2, \ 4$ und $5$. Wird nur einer dieser drei Binärwerte richtig übertragen, ist damit das gesamte Codewort eindeutig bestimmt. Keine Information über das Codewort erhält man bei folgenden Empfangsvektoren (siehe Tabelle auf der Angabenseite):