Difference between revisions of "Aufgaben:Exercise 3.10: Mutual Information at the BSC"

Latest revision as of 06:05, 18 September 2022

BSC model considered

We consider the Binary Symmetric Channel $\rm (BSC)$. The parameter values are valid for the whole exercise:

Crossover probability: $\varepsilon = 0.1$,
Probability for $0$: $p_0 = 0.2$,
Probability for $1$: $p_1 = 0.8$.

Thus the probability mass function of the source is: $P_X(X)= (0.2 , \ 0.8)$ and for the source entropy applies:

$$H(X) = p_0 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_0} + p_1\cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_1} = H_{\rm bin}(0.2)={ 0.7219\,{\rm bit}} \hspace{0.05cm}.$$

The task is to determine:

the probability function of the sink:

$$P_Y(Y) = (\hspace{0.05cm}P_Y(0)\hspace{0.05cm}, \ \hspace{0.05cm} P_Y(1)\hspace{0.05cm}) \hspace{0.05cm},$$

the joint probability function:

$$P_{XY}(X, Y) = \begin{pmatrix} p_{00} & p_{01}\\ p_{10} & p_{11} \end{pmatrix} \hspace{0.05cm},$$

the mutual information:

$$I(X;Y) = {\rm E} \hspace{-0.08cm}\left [ \hspace{0.02cm}{\rm log}_2 \hspace{0.1cm} \frac{P_{XY}(X, Y)} {P_{X}(X) \cdot P_{Y}(Y) }\right ] \hspace{0.05cm},$$

the equivocation:

$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = {\rm E} \hspace{0.02cm} \big [ \hspace{0.02cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}X \mid \hspace{0.03cm} Y} (X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y)} \big ] \hspace{0.05cm},$$

the irrelevance:

$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = {\rm E} \hspace{0.02cm} \big [ \hspace{0.02cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}Y \mid \hspace{0.03cm} X} (Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X)} \big ] \hspace{0.05cm}.$$

Hints:

The exercise belongs to the chapter Application to Digital Signal Transmission.
Reference is made in particular to the page Mutual information calculation for the binary channel.
In Exercise 3.10Z the channel capacity $C_{\rm BSC }$ of the BSC model is calculated.
This results as the maximum mutual information $I(X;\ Y)$ by maximization with respect to the probabilities $p_0$ or $p_1 = 1 - p_0$.

Questions

$P_{ XY }(0, 0) \ = \ $

$P_{ XY }(0, 1) \ = \ $

$P_{ XY }(1, 0) \ = \ $

$P_{ XY }(1, 1) \ = \ $

$P_Y(0)\ = \ $

$P_Y(1) \ = \ $

$I(X; Y)\ = \ $

$\ \rm bit$

$H(X|Y) \ = \ $

$\ \rm bit$

	$H(Y)$ is never greater than $H(X)$.
	$H(Y)$ is never smaller than $H(X)$.

	$H(Y\|X)$ is never larger than the equivocation $H(X\|Y)$.
	$H(Y\|X)$ is never smaller than the equivocation $H(X\|Y)$.

Solution

(1) The following applies in general or with the numerical values $p_0 = 0.2$ and $\varepsilon = 0.1$ for the quantities sought:

$$P_{XY}(0, 0) = p_0 \cdot (1 - \varepsilon) \hspace{0.15cm} \underline {=0.18} \hspace{0.05cm}, \hspace{0.5cm} P_{XY}(0, 1) = p_0 \cdot \varepsilon \hspace{0.15cm} \underline {=0.02} \hspace{0.05cm},$$

$$P_{XY}(1, 0) = p_1 \cdot \varepsilon \hspace{0.15cm} \underline {=0.08} \hspace{0.05cm}, \hspace{1.55cm} P_{XY}(1, 1) = p_1 \cdot (1 - \varepsilon) \hspace{0.15cm} \underline {=0.72} \hspace{0.05cm}.$$

(2) In general:

$$P_Y(Y) = \big [ {\rm Pr}( Y = 0)\hspace{0.05cm}, {\rm Pr}( Y = 1) \big ] = \big ( p_0\hspace{0.05cm}, p_1 \big ) \cdot \begin{pmatrix} 1 - \varepsilon & \varepsilon\\ \varepsilon & 1 - \varepsilon \end{pmatrix}.$$

This gives the following numerical values:

$$ {\rm Pr}( Y = 0)= p_0 \cdot (1 - \varepsilon) + p_1 \cdot \varepsilon = 0.2 \cdot 0.9 + 0.8 \cdot 0.1 \hspace{0.15cm} \underline {=0.26} \hspace{0.05cm},$$

$${\rm Pr}( Y = 1)= p_0 \cdot \varepsilon + p_1 \cdot (1 - \varepsilon) = 0.2 \cdot 0.1 + 0.8 \cdot 0.9 \hspace{0.15cm} \underline {=0.74} \hspace{0.05cm}.$$

(3) For the mutual information, according to the definition with $p_0 = 0.2$, $p_1 = 0.8$ and $\varepsilon = 0.1$:

$$I(X;Y) = {\rm E} \hspace{-0.08cm}\left [ \hspace{0.02cm}{\rm log}_2 \hspace{0.08cm} \frac{P_{XY}(X, Y)} {P_{X}(X) \hspace{-0.05cm}\cdot \hspace{-0.05cm} P_{Y}(Y) }\right ] \hspace{0.3cm} \Rightarrow$$

$$I(X;Y) = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.18}{0.2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.26} + 0.02 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.02}{0.2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.74} + 0.08 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.08}{0.8 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.26} + 0.72 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.72}{0.8 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.74} \hspace{0.15cm} \underline {=0.3578\,{\rm bit}} \hspace{0.05cm}.$$

(4) With the source entropy $H(X)$ given, we obtain for the equivocation:

$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = H(X) - I(X;Y) = 0.7219 - 0.3578 \hspace{0.15cm} \underline {=0.3642\,{\rm bit}} \hspace{0.05cm}.$$

However, one could also apply the general definition with the inference probabilities $P_{X|Y}(⋅)$ :

$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = {\rm E} \hspace{0.02cm} \left [ \hspace{0.05cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}X \mid \hspace{0.03cm} Y} (X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y)} \hspace{0.05cm}\right ] = {\rm E} \hspace{0.02cm} \left [ \hspace{0.05cm} {\rm log}_2 \hspace{0.1cm} \frac{P_Y(Y)}{P_{XY} (X, Y)} \hspace{0.05cm} \right ] \hspace{0.05cm}$$

In the example, the same result $H(X|Y) = 0.3642 \ \rm bit$ is also obtained according to this calculation rule:

$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.26}{0.18} + 0.02 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.74}{0.02} + 0.08 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.26}{0.08} + 0.72 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.74}{0.72} \hspace{0.05cm}.$$

(5) Correct is the proposed solution 2:

In the case of disturbed transmission $(ε > 0)$ the uncertainty regarding the sink is always greater than the uncertainty regarding the source. One obtains here as a numerical value:

$$H(Y) = H_{\rm bin}(0.26)={ 0.8268\,{\rm bit}} \hspace{0.05cm}.$$

With error-free transmission $(ε = 0)$, on the other hand, $P_Y(⋅) = P_X(⋅)$ and $H(Y) = H(X)$ would apply.

(6) Here, too, the second proposed solution is correct:

Because of $I(X;Y) = H(X) - H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = H(Y) - H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X)$ , $H(Y|X)$ is greater than $H(X|Y)$ by the same magnitude that $H(Y)$ is greater than $H(X)$:

$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = H(Y) -I(X;Y) = 0.8268 - 0.3578 ={ 0.469\,{\rm bit}} \hspace{0.05cm}$$

Direct calculation gives the same result $H(Y|X) = 0.469\ \rm bit$:

$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = {\rm E} \hspace{0.02cm} \left [ \hspace{0.02cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}Y \mid \hspace{0.03cm} X} (Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X)} \right ] = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.9} + 0.02 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.1} + 0.08 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.1} + 0.72 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.9} \hspace{0.05cm}.$$

@@ Line 1: / Line 1: @@
-[http://www.example.com Link-Text]
+{{quiz-Header|Buchseite=Information_Theory/Application_to_Digital_Signal_Transmission
-{{quiz-Header|Buchseite=Informationstheorie/Anwendung auf DSÜ-Kanäle
 }}
-[[File:P_ID2787__Inf_A_3_9.png|right|]]
+[[File:P_ID2787__Inf_A_3_9.png|right|frame|BSC model considered]]
-Wir betrachten den $\text{Binary Symmetric Channel}$ (BSC). Für die gesamte Aufgabe gelten die  Parameterwerte:
+We consider the&nbsp; [[Channel_Coding/Kanalmodelle_und_Entscheiderstrukturen#Binary_Symmetric_Channel_.E2.80.93_BSC|Binary Symmetric Channel]]&nbsp; $\rm (BSC)$. The parameter values are valid for the whole exercise:
-:* Verfälschungswahrscheinlichkeit: $\epsilon = 0.1$
+* Crossover probability: &nbsp; $\varepsilon = 0.1$,
-:* Wahrscheinlichkeit für $0$:   $p_0 = 0.2$,
+* Probability for&nbsp; $0$: &nbsp;  $p_0 = 0.2$,
-:* Wahrscheinlichkeit für $1$:   $p_1 = 0.8$.
+* Probability for&nbsp; $1$: &nbsp;  $p_1 = 0.8$.
-Damit lautet die Wahrscheinlichkeitsfunktion der Quelle:
-$P_X(X)= (0.2 , 0.8)$
+Thus the probability mass function of the source is: &nbsp; $P_X(X)= (0.2 , \ 0.8)$&nbsp; and for the source entropy applies:
+:$$H(X) = p_0 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_0} + p_1\cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_1} = H_{\rm bin}(0.2)={ 0.7219\,{\rm bit}} \hspace{0.05cm}.$$
-und für die Quellenentropie gilt:
+The task is to determine:
+* the probability function of the sink:
+:$$P_Y(Y) = (\hspace{0.05cm}P_Y(0)\hspace{0.05cm}, \ \hspace{0.05cm} P_Y(1)\hspace{0.05cm})
+\hspace{0.05cm},$$
+* the joint probability function:
+:$$P_{XY}(X, Y) = \begin{pmatrix}
+		p_{00}  & p_{01}\\
+		p_{10}  & p_{11}
+		\end{pmatrix}  \hspace{0.05cm},$$
+* the mutual information:
+:$$I(X;Y) = {\rm E} \hspace{-0.08cm}\left [ \hspace{0.02cm}{\rm log}_2 \hspace{0.1cm} \frac{P_{XY}(X, Y)}
+{P_{X}(X) \cdot P_{Y}(Y) }\right ]  \hspace{0.05cm},$$
+*the equivocation:
+:$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) =  {\rm E} \hspace{0.02cm} \big [ \hspace{0.02cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}X \mid \hspace{0.03cm} Y} (X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y)} \big ] \hspace{0.05cm},$$
+*the irrelevance:
+:$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) =  {\rm E} \hspace{0.02cm} \big [ \hspace{0.02cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}Y \mid \hspace{0.03cm} X} (Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X)} \big ] \hspace{0.05cm}.$$
-$H(X) = p_0 \cdot log_2 \frac{1}{p_0} + p_1 \cdot log_2 \frac{1}{p_1} = H_{bin}(0.2) = 0.7219 (\text{bit})$
-In der Aufgabe sollen ermittelt werden:
-:* die Wahrscheinlichkeitsfunktion der Sinke:
-$P_Y(Y) = (P_Y(0) , P_Y(1))$,
-:* die Verbundwahrscheinlichkeitsfunktion :
-$P_{XY}(X, Y) = \begin{pmatrix} p_{00} & p_{01}\\ p_{10} & p_{11} \end{pmatrix} \hspace{0.05cm}$
-:* die Transinformation
-$I(X;Y) = E[ log_2 \frac{P_{ XY }(X,Y)}{P_X(X) . P_Y(Y)}]$,
-:*die Äquivokation:
-$H(X \mid Y) = E[log_2 \frac{1}{P_{ X \mid Y }(X \mid Y)}]$,
-:*die Irrelevanz:
-$H(Y \mid X) = E[log_2 \frac{1}{P_{ Y \mid X }(Y \mid X)}]$
-'''Hinwies:''' Die Aufgabe gehört zu [http://en.lntwww.de/Informationstheorie/Anwendung_auf_die_Digitalsignal%C3%BCbertragung Kapitel 3.3.] In der [http://en.lntwww.de/Aufgaben:3.09Z_BSC%E2%80%93Kanalkapazit%C3%A4t Aufgabe Z3.9] wird die [http://en.lntwww.de/Informationstheorie/Anwendung_auf_die_Digitalsignal%C3%BCbertragung#Kanalkapazit.C3.A4t_eines_Bin.C3.A4rkanals Kanalkapazität] $C_{ BSC }$ des $BSC$–Modells berechnet. Diese ergibt sich als die maximale Transinformation $I(X; Y)$ durch Maximierung bezüglich der Symbolwahrscheinlichkeiten $p_0$ bzw. $p_1 = 1 – p_0$.
+Hints:
+*The exercise belongs to the chapter&nbsp; [[Information_Theory/Anwendung_auf_die_Digitalsignalübertragung|Application to Digital Signal Transmission]].
+*Reference is made in particular to the page&nbsp;    [[Information_Theory/Anwendung_auf_die_Digitalsignalübertragung#Calculation_of_the_mutual_information_for_the_binary_channel|Mutual information calculation for the binary channel]].
+*In&nbsp; [[Aufgaben:Aufgabe_3.10Z:_BSC–Kanalkapazität|Exercise 3.10Z]]&nbsp; the channel capacity&nbsp; $C_{\rm BSC }$&nbsp; of the BSC model is calculated.
+*This results as the maximum mutual information&nbsp; $I(X;\ Y)$&nbsp;  by maximization with respect to the probabilities&nbsp; $p_0$&nbsp; or&nbsp; $p_1 = 1 - p_0$.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Berechnen Sie die Verbundwahrscheinlichkeiten $P_{ XY }(X, Y)$
+{Calculate the joint probabilities&nbsp; $P_{ XY }(X, Y)$
 |type="{}"}
-$P_{ XY }(0, 0)$ = { 0.18 3% }
+$P_{ XY }(0, 0) \ = \ $ { 0.18 3% }
-$P_{ XY }(0, 1)$ = { 0.02 3% }
+$P_{ XY }(0, 1) \ = \ $ { 0.02 3% }
-$P_{ XY }(1, 0)$ = { 0.08 3% }
+$P_{ XY }(1, 0) \ = \ $ { 0.08 3% }
-$P_{ XY }(1, 1)$ = { 0.72 3% }
+$P_{ XY }(1, 1) \ = \ $ { 0.72 3% }
-{Wie lautet die Wahrscheinlichkeitsfunktion $P_Y(Y)$?
+{What is the probability mass function&nbsp; $P_Y(Y)$&nbsp; of the sink?
 |type="{}"}
-$P_Y(0)$ = { 0.26 3% }
+$P_Y(0)\ = \ $ { 0.26 3% }
-$P_Y(1)$ = { 0.74 3% }
+$P_Y(1) \ = \ $ { 0.74 3% }
-{Welcher Wert ergibt sich für die Transinformation?
+{What is the value of the mutual information&nbsp; $I(X;\
+ Y)$?
 |type="{}"}
-$I(X; Y)$ = { 0.3578 3% }
+$I(X; Y)\ = \ $ { 0.3578 3% } $\ \rm bit$
-{Welcher Wert ergibt sich für die Äquivokation?
+{Which value results for the equivocation&nbsp; $H(X|Y)$?
 |type="{}"}
-$H(X|Y)$ = {  0.3642 3% }
+$H(X|Y) \ = \ $ {  0.3642 3% } $\ \rm bit$
-{Welche Aussage trifft für die Sinkenentropie $H(Y)$ zu?
+{Which statement is true for the sink entropy&nbsp; $H(Y)$&nbsp;?
 |type="[]"}
-- $H(Y)$ ist nie größer als $H(X)$.
+- $H(Y)$&nbsp; is never greater than&nbsp; $H(X)$.
-+ $H(Y)$ ist nie kleiner als $H(X)$.
++ $H(Y)$&nbsp; is never smaller than&nbsp; $H(X)$.
-{Welche Aussage trifft für die Irrelevanz $H(Y|X)$ zu?
+{Which statement is true for the irrelevance&nbsp; $H(Y|X)$&nbsp;?
 |type="[]"}
-- $H(Y|X)$ ist nie größer als die Äquivokation $H(X|Y)$.
+- $H(Y|X)$&nbsp; is never larger than the equivocation&nbsp; $H(X|Y)$.
-+ $H(Y|X)$ ist nie kleiner als die Äquivokation $H(X|Y)$.
++ $H(Y|X)$&nbsp; is never smaller than the equivocation&nbsp; $H(X|Y)$.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''1.''' Für die gesuchten Größen gilt allgemein bzw. mit den Zahlenwerten $p_0 = 0.2$ und $ε = 0.1$:
+'''(1)'''&nbsp; The following applies in general or with the numerical values&nbsp; $p_0 = 0.2$&nbsp; and&nbsp; $\varepsilon = 0.1$ for the quantities sought:
+:$$P_{XY}(0, 0) = p_0 \cdot (1 - \varepsilon)
+ \hspace{0.15cm} \underline {=0.18} \hspace{0.05cm}, \hspace{0.5cm}
+P_{XY}(0, 1) = p_0 \cdot \varepsilon
+ \hspace{0.15cm} \underline {=0.02} \hspace{0.05cm},$$
+:$$P_{XY}(1, 0) = p_1 \cdot \varepsilon
+ \hspace{0.15cm} \underline {=0.08} \hspace{0.05cm}, \hspace{1.55cm}
+P_{XY}(1, 1) = p_1 \cdot (1 - \varepsilon)
+ \hspace{0.15cm} \underline {=0.72} \hspace{0.05cm}.$$
-$P_{ XY }(0 , 0) = p_0 . (1 - \varepsilon ) = 0.18 $ , $P_{XY}(0,1) = p_0  . \varepsilon = 0.02$,
-$P_{XY}(1,0) = p_1  . \varepsilon = 0.08$ ,  $P_{ XY }(1 , 1) = p_1 . (1 - \varepsilon ) = 0.72$.
+'''(2)'''&nbsp; In general:
+:$$P_Y(Y) = \big [ {\rm Pr}( Y = 0)\hspace{0.05cm}, {\rm Pr}( Y = 1) \big ] = \big ( p_0\hspace{0.05cm}, p_1 \big ) \cdot \begin{pmatrix} 1 - \varepsilon & \varepsilon\\ \varepsilon & 1 - \varepsilon \end{pmatrix}.$$
+This gives the following numerical values:
+:$$ {\rm Pr}( Y = 0)= p_0 \cdot (1 - \varepsilon) + p_1 \cdot \varepsilon = 0.2 \cdot 0.9 + 0.8 \cdot 0.1 \hspace{0.15cm} \underline {=0.26} \hspace{0.05cm},$$
+:$${\rm Pr}( Y = 1)= p_0 \cdot \varepsilon + p_1 \cdot (1 - \varepsilon) = 0.2 \cdot 0.1 + 0.8 \cdot 0.9 \hspace{0.15cm} \underline {=0.74} \hspace{0.05cm}.$$
-'''2.''' Es gilt:
-$$P_Y(Y) = \big ( {\rm Pr}( Y = 0)\hspace{0.05cm}, {\rm Pr}( Y = 1) \big ) = \big ( p_0\hspace{0.05cm}, p_1 \big ) \cdot \begin{pmatrix} 1 - \varepsilon & \varepsilon\\ \varepsilon & 1 - \varepsilon \end{pmatrix}$$
-$$\Rightarrow \hspace{0.3cm} {\rm Pr}( Y = 0)\hspace{-0.15cm}  =  \hspace{-0.15cm} p_0 \cdot (1 - \varepsilon) + p_1 \cdot \varepsilon = 0.2 \cdot 0.9 + 0.8 \cdot 0.1 \hspace{0.15cm} \underline {=0.26} \hspace{0.05cm},\\ {\rm Pr}( Y = 1)\hspace{-0.15cm}  = \hspace{-0.15cm} p_0 \cdot \varepsilon + p_1 \cdot (1 - \varepsilon) = 0.2 \cdot 0.1 + 0.8 \cdot 0.9 \hspace{0.15cm} \underline {=0.74} \hspace{0.05cm}$$
+'''(3)'''&nbsp; For the mutual information, according to the definition with&nbsp; $p_0 = 0.2$,&nbsp; $p_1 = 0.8$&nbsp; and&nbsp; $\varepsilon = 0.1$:
+:$$I(X;Y) = {\rm E} \hspace{-0.08cm}\left [ \hspace{0.02cm}{\rm log}_2 \hspace{0.08cm} \frac{P_{XY}(X, Y)} {P_{X}(X) \hspace{-0.05cm}\cdot \hspace{-0.05cm} P_{Y}(Y) }\right ] \hspace{0.3cm} \Rightarrow$$
+:$$I(X;Y)  = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.18}{0.2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.26} + 0.02 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.02}{0.2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.74}
++ 0.08 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.08}{0.8 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.26} + 0.72 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.72}{0.8 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.74} \hspace{0.15cm} \underline {=0.3578\,{\rm bit}} \hspace{0.05cm}.$$
-'''3.'''  Für die Transinformation gilt gemäß der Definition mit $p_0 = 0.2$ , $p_1 = 0.8$ und $ε = 0.1$
-$$I(X;Y) \hspace{-0.2cm}  =  \hspace{-0.2cm} {\rm E} \hspace{-0.08cm}\left [ \hspace{0.02cm}{\rm log}_2 \hspace{0.08cm} \frac{P_{XY}(X, Y)} {P_{X}(X) \hspace{-0.05cm}\cdot \hspace{-0.05cm} P_{Y}(Y) }\right ] = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.18}{0.2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.26} + 0.02 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.02}{0.2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.74} +$$
-$$\hspace{-0.2cm} 0.08 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.08}{0.8 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.26} + 0.72 \cdot {\rm log}_2 \hspace{0.08cm} \frac{0.72}{0.8 \hspace{-0.05cm}\cdot \hspace{-0.05cm} 0.74} \hspace{0.15cm} \underline {=0.3578\,{\rm bit}} \hspace{0.05cm}$$
+'''(4)'''&nbsp; With the source entropy&nbsp; $H(X)$&nbsp; given, we obtain for the equivocation:
+:$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = H(X) - I(X;Y) = 0.7219 - 0.3578 \hspace{0.15cm} \underline {=0.3642\,{\rm bit}} \hspace{0.05cm}.$$
+*However, one could also apply the general definition with the inference probabilities&nbsp; $P_{X|Y}(⋅)$&nbsp;:
+:$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = {\rm E} \hspace{0.02cm} \left [ \hspace{0.05cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}X \mid \hspace{0.03cm} Y} (X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y)} \hspace{0.05cm}\right ] = {\rm E} \hspace{0.02cm} \left [ \hspace{0.05cm} {\rm log}_2 \hspace{0.1cm} \frac{P_Y(Y)}{P_{XY} (X, Y)} \hspace{0.05cm} \right ] \hspace{0.05cm}$$
+*In the example, the same result&nbsp; $H(X|Y) = 0.3642 \ \rm bit$&nbsp; is also obtained according to this calculation rule:
+:$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.26}{0.18} + 0.02 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.74}{0.02} + 0.08 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.26}{0.08} + 0.72 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.74}{0.72} \hspace{0.05cm}.$$
-'''4.'''  Mit  der angegebenen Quellenentropie $H(X)$ erhält man für die Äquivokation:
-$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = H(X) - I(X;Y) = 0.7219 - 0.3578 \hspace{0.15cm} \underline {=0.3642\,{\rm bit}} \hspace{0.05cm}$$.
-Man könnte auch die allgemeine Definition mit den Rückschlusswahrscheinlichkeiten $P_{X|Y}(⋅)$
-anwenden:
-$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = {\rm E} \hspace{0.02cm} \big [ \hspace{0.05cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}X \mid \hspace{0.03cm} Y} (X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y)} \hspace{0.05cm}\big ] = {\rm E} \hspace{0.02cm} \big [ \hspace{0.05cm} {\rm log}_2 \hspace{0.1cm} \frac{P_Y(Y)}{P_{XY} (X, Y)} \hspace{0.05cm} \big ] \hspace{0.05cm}$$
+'''(5)'''&nbsp; Correct is the <u>proposed solution 2:</u>
+*In the case of disturbed transmission&nbsp; $(ε > 0)$&nbsp; the uncertainty regarding the sink is always greater than the uncertainty regarding the source.&nbsp; One obtains here as a numerical value:
+:$$H(Y) = H_{\rm bin}(0.26)={ 0.8268\,{\rm bit}} \hspace{0.05cm}.$$
+*With error-free transmission&nbsp; $(ε = 0)$,&nbsp; on the other hand,&nbsp; $P_Y(⋅) = P_X(⋅)$&nbsp; and&nbsp; $H(Y) = H(X)$&nbsp; would apply.
-Im Beispiel erhält man auch nach dieser Berechnungsvorschrift das gleiche Ergebnis $H(X|Y) = 0.3642 bit$ :
-$$H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) \hspace{-0.15cm}  =  \hspace{-0.15cm} 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.26}{0.18} + 0.02 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.74}{0.02} + 0.08 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.26}{0.08} + 0.72 \cdot {\rm log}_2 \hspace{0.1cm} \frac{0.74}{0.72} \hspace{0.05cm}$$
-'''5.'''  Richtig ist der ''Lösungsvorschlag 2.'' Bei gestörter Übertragung $(ε > 0)$ ist die Unsicherheit hinsichtlich der Sinke stets größer als die Unsicherheit bezüglich der Quelle. Man erhält hier als Zahlenwert:
+'''(6)'''&nbsp; Here, too, the <u>second proposed solution</u> is correct:
-$$H(Y) = H_{\rm bin}(0.26)={ 0.8268\,{\rm bit}} \hspace{0.05cm}$$
+*Because of&nbsp; $I(X;Y) = H(X) - H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = H(Y) - H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X)$&nbsp;,&nbsp; $H(Y|X)$&nbsp; is greater than&nbsp; $H(X|Y)$ by the same magnitude that&nbsp; $H(Y)$&nbsp; is greater than&nbsp; $H(X)$:
-Bei fehlerfreier Übertragung $(ε = 0)$ würde dagegen $P_Y(⋅) = P_X(⋅)$ und $H(Y) = H(X)$ gelten
+:$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = H(Y) -I(X;Y) = 0.8268 - 0.3578 ={ 0.469\,{\rm bit}} \hspace{0.05cm}$$
+*Direct calculation gives the same result&nbsp; $H(Y|X) = 0.469\ \rm  bit$:
+:$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = {\rm E} \hspace{0.02cm} \left [ \hspace{0.02cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}Y \mid \hspace{0.03cm} X} (Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X)} \right ] = 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.9} + 0.02 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.1} + 0.08 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.1} + 0.72 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.9} \hspace{0.05cm}.$$
-'''6.''' Auch hier ist der ''zweite Lösungsvorschlag'' richtig. Wegen
-$$I(X;Y) = H(X) - H(X \hspace{-0.1cm}\mid \hspace{-0.1cm} Y) = H(Y) - H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X)$$
-ist $H(Y|X)$ um den gleichen Betrag größer als $H(X|Y)$, um den auch $H(Y)$ größer ist als $H(X)$:
-$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = H(Y) -I(X;Y) = 0.8268 - 0.3578 ={ 0.4690\,{\rm bit}} \hspace{0.05cm}$$
-Bei direkter Berechnung erhält man das gleiche Ergebnis $H(Y|X) = 0.4690 bit$:
-$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) \hspace{-0.15cm}  =  \hspace{-0.15cm} {\rm E} \hspace{0.02cm} \big [ \hspace{0.02cm} {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.03cm}Y \mid \hspace{0.03cm} X} (Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X)} \big ] =$$
-$$=\hspace{-0.15cm} 0.18 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.9} + 0.02 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.1} + 0.08 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.1} + 0.72 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.9} \hspace{0.05cm}$$
 {{ML-Fuß}}
@@ Line 127: / Line 138: @@
-[[Category:Aufgaben zu Informationstheorie|^3.3 Anwendung auf DSÜ-Kanäle^]]
+[[Category:Information Theory: Exercises|^3.3 Application to Digital Signal Transmission^]]