Difference between revisions of "Aufgaben:Exercise 3.6: Partitioning Inequality"

Latest revision as of 10:15, 24 September 2021

Two probability functions $P_X$ and $Q_X$

The Kullback–Leibler distance (KLD for short) is also used in the "Partition Unequality":

We assume the set $X = \{ x_1, \hspace{0.15cm} x_2, \text{...} \hspace{0.05cm}, \hspace{0.15cm} x_M \}$ and the probability functions

$$P_X(X) = P_X ( x_1, \hspace{0.15cm} x_2, \text{...} \hspace{0.05cm}, \hspace{0.15cm} x_M )\hspace{0.05cm},$$

$$Q_X(X) =Q_X ( x_1, \hspace{0.15cm} x_2, \text{...} \hspace{0.05cm}, \hspace{0.15cm} x_M ), $$

which are said to be "similar in some way".

We divide the set $X$ into partitions $A_1, \text{...} , A_K$, which are disjoint to each other and result in a complete system :

$$\bigcup_{i=1}^{K} = X, \hspace{0.5cm} A_i \cap A_j = {\it \phi} \hspace{0.25cm}\text{für}\hspace{0.25cm} 1 \le i \ne j \le K .$$

In the following, we denote the probability functions with respect to the partitionings $A_1,\ A_2, \text{...} ,\ A_K$ by

$$P_X^{\hspace{0.15cm}(A)} = \big [ P_X ( A_1 )\hspace{0.05cm}, \hspace{0.05cm}\text{...}\hspace{0.1cm},P_X ( A_K ) \big ],\hspace{0.05cm}\hspace{0.5cm}{\rm where}\hspace{0.15cm} P_X ( A_i ) = \sum_{ x \in A_i} P_X ( x )\hspace{0.05cm},$$

$$Q_X^{\hspace{0.15cm}(A)}= \big [ Q_X ( A_1 )\hspace{0.05cm}, \hspace{0.05cm}\text{...}\hspace{0.1cm},Q_X ( A_K ) \big ],\hspace{0.05cm}\hspace{0.40cm}{\rm where}\hspace{0.15cm} Q_X ( A_i ) = \sum_{ x \in A_i} Q_X ( x )\hspace{0.05cm}. $$

$\text{Please note:}$ The partitioning inequality yields the following size relation with respect to the Kullback-Leibler distances:

$$D(P_X^{\hspace{0.15cm}(A)} \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{\hspace{0.15cm}(A)}) \hspace{0.25cm}\le \hspace{0.25cm}D(P_X \hspace{0.05cm}\vert \vert \hspace{0.05cm} Q_X) \hspace{0.05cm}.$$

In subtask (1) the Kullback-Leibler distance of the two probability functions $P_X(X)$ and $Q_X(X)$ for $X = \{0,\ 1,\ 2\}$ ⇒ $|X| = 3$ is to be determined.

Then the set $X$ is to be partitioned with $K = 2$ according to

$A = \{A_1 ,\ A_2\}$ with $A_1 =\{0\}$ and $A_2 = \{ 1,\ 2 \}$ ,
$B = \{B_1 ,\ B_2\}$ with $B_1 =\{1\}$ and $B_2 = \{ 0,\ 2 \}$,
$C = \{C_1 ,\ C_2\}$ with $C_1 =\{2\}$ and $C_2 = \{ 0,\ 1\}$.

Then the respective Kullback-Leibler distances are to be given:

$D(P_X^{ (A) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (A) } )$,
$D(P_X^{ (B) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (B) } )$,
$D(P_X^{ (C) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (C) } )$.

Finally, subtask (5) asks for the conditions that must be satisfied for the equal sign to be true in the above inequality.

Hints:

The exercise belongs to the chapter Some preliminary remarks on two-dimensional random variables.
In particular, reference is made to the page Relative entropy – Kullback-Leibler distance.
The two probability functions can be read from the above graph as follows:

$$P_X(X) = \big [1/4 , \ 1/2 , \ 1/4 \big ],\hspace{0.5cm} Q_X(X) = \big [1/8, \ 3/4, \ 1/8 \big].$$

Questions

$ D(P_X \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X) \ = \ $

$\ \rm bit$

$D(P_X^{ (A) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (A) } ) \ = \ $

$\ \rm bit$

$D(P_X^{ (B) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (B) } ) \ = \ $

$\ \rm bit$

	The same result as for partition $\rm A$.
	The same result as for partition $\rm B$.
	A completely different result.

	$\|X\|$ equations must be fulfilled.
	For $x \in A_i$ must hold: $P_X(x)/Q_X(x) = P_X(A_i)/ Q_X(A_i)$

Solution

(1) For the Kullback-Leibler distance of the non-partitioned quantities $X$ and $Y$ holds:

$$D(P_X \hspace{0.05cm} || \hspace{0.05cm}P_Y) = {\rm E} \left [ {\rm log}_2 \hspace{0.1cm} \frac{P_X(X)}{P_Y(X)}\right ] \hspace{0.2cm}=\hspace{0.2cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm}X} P_X(x) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_X(x)}{P_Y(x)}$$

$$\Rightarrow D(P_X \hspace{0.05cm} || \hspace{0.05cm}P_Y) = \hspace{-0.15cm} \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/2}{3/4} + 2 \cdot \frac{1}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/4}{1/8} = \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{3} + \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} (2) = 1- \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} (3) \hspace{0.15cm} \underline {=0.2075\,{\rm (bit)}} \hspace{0.05cm}.$$

(2) With $\text{partition }A$ ⇒ $A_1 = \{0\}$ , $A_2 = \{ 1 , 2 \}$ we get $P_X^{ (A) } (X) = \{1/4 , \ 3/4\}$ and $Q_X^{ (A) } (X) = \{1/8 , \ 7/8\}$. From this follows:

$$D(P_X^{\hspace{0.15cm}(A)} \hspace{0.05cm}|| \hspace{0.05cm} Q_X^{\hspace{0.15cm}(A)}) = \frac{1}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/4}{1/8} + \frac{3}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{3/4}{7/8} =\frac{1}{4}\cdot {\rm log}_2 \hspace{0.1cm} (2) + \frac{3}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{6}{7} \hspace{0.15cm} \underline {=0.0832\,{\rm (bit)}} \hspace{0.05cm}.$$

(3) With $\text{partition } B$ ⇒ $B_1 = \{1\}$ , $B_2 = \{ 0 ,\ 2 \}$ the probability functions are $P_X^{ (B) } (X) = \{1/2 , \ 1/2\}$ und $Q_X^{ (B) } (X) = \{3/4 , \ 1/4\}$.

Analogous to subtask (2) one thus obtains:

$$D(P_X^{\hspace{0.15cm}(B)} \hspace{0.05cm}|| \hspace{0.05cm} Q_X^{\hspace{0.15cm}(B)}) = \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/2}{3/4} + \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/2}{1/4} \hspace{0.15cm} \underline {=0.2075\,{\rm (bit)}} \hspace{0.05cm}.$$

The result agrees with that of subtask (1) ⇒ With $\text{partition }B$ the equal sign applies.

(4) With $\text{partition } C$ ⇒ $C_1 = \{2\}$ , $C_2 = \{ 0 , \ 1\}$ one obtains $P_X^{ (C) } (X) = \{1/4, \ 3/4\}$ , $Q_X^{ (C) } (X) = \{1/8, \ 7/8\}$,
i.e. the same functions as for the $\text{partition }A$ ⇒ solution proposal 1.

(5) The $\text{partition }B$ has led to the result $D(P_X^{ (B) } \hspace{0.05cm} || \hspace{0.05cm}Q_X^{ (B) } ) = D(P_X \hspace{0.05cm} || \hspace{0.05cm}Q_X)$.

So for this case

$$\frac{P_X(1)}{Q_X(1)} = \frac{1/2}{3/4} = \frac{2}{3}, \ \frac{P_X(B_1)}{Q_X(B_1)} = \frac{1/2}{3/4} = {2}/{3},$$

$$\frac{P_X(0)}{Q_X(0)} = \frac{1/4}{1/8} = 2, \ \frac{P_X(B_2)}{Q_X(B_2)} = \frac{1/2}{1/4} = 2,$$

$$\frac{P_X(2)}{Q_X(2)} = \frac{1/4}{1/8} = 2, \ \frac{P_X(B_2)}{Q_X(B_2)} = \frac{1/2}{1/4} = 2.$$

It must therefore hold for all $x \in X$ :

$$\frac{P_X(x)}{Q_X(x)} = \frac{P_X(B_1)}{Q_X(B_1)}, \text{ if } x \in B_1, \hspace{0.5cm}\frac{P_X(x)}{Q_X(x)} = \frac{P_X(B_2)}{Q_X(B_2)}, \text{ if } x \in B_2.$$

By generalization, one can see that both proposed solutions are correct.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Informationstheorie/Einige Vorbemerkungen zu zweidimensionalen Zufallsgrößen
+{{quiz-Header|Buchseite=Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables
 }}
-[[File:P_ID2812__Inf_A_3_5.png|right|Wahrscheinlichkeitsfunktionen <i>P<sub>X</sub></i> und <i>Q<sub>X</sub></i>]]
+[[File:P_ID2812__Inf_A_3_5.png|right|frame|Two probability functions&nbsp; $P_X$&nbsp; and&nbsp; $Q_X$]]
-Die ''Kullback–Leibler–Distanz'' (kurz KLD) wird auch in der „Partitionierungsungleichung” (englisch: Partition Unequality) verwendet:
+The&nbsp; '''Kullback–Leibler distance'''&nbsp; (KLD for short)&nbsp; is also used in the&nbsp;  "Partition Unequality":
-* Wir gehen von der Menge $X = \big \{ \hspace{0.05cm} x_1, \hspace{0.05cm} x_2, ... \hspace{0.05cm}, \hspace{0.05cm} x_M  \hspace{0.05cm} \big \}$ und den Wahrscheinlichkeitsfunktionen
+* We assume the set&nbsp; $X =  \{ x_1, \hspace{0.15cm} x_2, \text{...} \hspace{0.05cm}, \hspace{0.15cm} x_M   \}$&nbsp; and the probability functions
-:$$P_X(X) \hspace{-0.15cm} & = & \hspace{-0.15cm} P_X (  x_1, \hspace{0.05cm} x_2, ... \hspace{0.05cm}, \hspace{0.05cm} x_M   )\hspace{0.05cm},$$
+:$$P_X(X) = P_X (  x_1, \hspace{0.15cm} x_2, \text{...} \hspace{0.05cm}, \hspace{0.15cm} x_M   )\hspace{0.05cm},$$
-:$$Q_X(X) \hspace{-0.15cm} & = & \hspace{-0.15cm} Q_X (  x_1, \hspace{0.05cm} x_2, ... \hspace{0.05cm}, \hspace{0.05cm} x_M   ) $$ ,
+:$$Q_X(X) =Q_X (  x_1, \hspace{0.15cm} x_2, \text{...} \hspace{0.05cm}, \hspace{0.15cm} x_M   ), $$
+:which are said to be&nbsp; "similar in some way".
-aus, die in irgendeiner Form „ähnlich” sein sollen
+* We divide the set&nbsp; $X$&nbsp; into partitions&nbsp; $A_1, \text{...} ,&nbsp; A_K$, which are&nbsp; [[Theory_of_Stochastic_Signals/Mengentheoretische_Grundlagen#Disjunkte_Mengen|disjoint]]&nbsp; to each other and result in a&nbsp; [[Theory_of_Stochastic_Signals/Mengentheoretische_Grundlagen#Vollst.C3.A4ndiges_System|complete system]]&nbsp;:
-* Die Menge $X$ unterteilen wir in die Partitionen $A_1, ..., A_K$ , die zueinander disjunkt sind und ein $vollständiges System$ ergeben:
+:$$\bigcup_{i=1}^{K} = X, \hspace{0.5cm} A_i \cap A_j = {\it \phi} \hspace{0.25cm}\text{für}\hspace{0.25cm} 1 \le i \ne j \le K .$$
-$\bigcup_{i=_1}^K A_i = X$ , $A_i \cap A_j = \phi$    für   $1 \leq i \neq j \leq K$
+* In the following, we denote the probability functions with respect to the partitionings&nbsp; $A_1,\ A_2, \text{...} ,\ A_K$&nbsp; by
-:* Die Wahrscheinlichkeitsfunktionen bezüglich der Partitionierungen $A=\{ A_1,A_2,.....,A_K \}$bezeichnen wir im Folgenden mit
+:$$P_X^{\hspace{0.15cm}(A)} = \big [ P_X (  A_1  )\hspace{0.05cm}, \hspace{0.05cm}\text{...}\hspace{0.1cm},P_X (  A_K  ) \big  ],\hspace{0.05cm}\hspace{0.5cm}{\rm where}\hspace{0.15cm} P_X (  A_i  ) = \sum_{ x \in A_i} P_X ( x  )\hspace{0.05cm},$$
+:$$Q_X^{\hspace{0.15cm}(A)}= \big  [ Q_X (  A_1  )\hspace{0.05cm}, \hspace{0.05cm}\text{...}\hspace{0.1cm},Q_X (  A_K  ) \big  ],\hspace{0.05cm}\hspace{0.40cm}{\rm where}\hspace{0.15cm} Q_X (  A_i  ) = \sum_{ x \in A_i} Q_X ( x  )\hspace{0.05cm}. $$
-$P_X^{ (A) } = [ P_X(A_1),.......,P_X(A_K)]$ ,   wobei    $P_X(A_i) = \sum\limits_{ x \epsilon A_i } P_X(x)$
+{{BlaueBox|TEXT=
+$\text{Please note:}$ &nbsp;The&nbsp; '''partitioning inequality'''&nbsp; yields the following size relation with respect to the Kullback-Leibler distances:
+:$$D(P_X^{\hspace{0.15cm}(A)} \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{\hspace{0.15cm}(A)})
+\hspace{0.25cm}\le \hspace{0.25cm}D(P_X \hspace{0.05cm}\vert \vert \hspace{0.05cm} Q_X) \hspace{0.05cm}.$$}}
-$Q_X^{ (A) } = [ Q_X(A_1),.......,Q_X(A_K)]$ ,   wobei    $Q_X(A_i) = \sum\limits_{ x \epsilon A_i } Q_X(x)$
-Die $Partitionierungsungleichung$ liefert folgende Größenrelation hinsichtlich der Kullback–Leibler–Distanzen:
+In subtask&nbsp; '''(1)'''&nbsp; the Kullback-Leibler distance of the two probability functions&nbsp; $P_X(X)$&nbsp; and&nbsp; $Q_X(X)$&nbsp; for&nbsp; $X = \{0,\ 1,\ 2\}$ &nbsp; &rArr; &nbsp; $|X| = 3$&nbsp; is to be determined.
-$D(P_X^{ (A) } \parallel Q_X^{ (A) } ) \leq  D(P_X \parallel Q_X)$
+*Then the set&nbsp; $X$&nbsp; is to be partitioned with&nbsp; $K = 2$&nbsp; according to
+:* $A = \{A_1 ,\ A_2\}$&nbsp;   with&nbsp;  $A_1 =\{0\}$&nbsp;  and&nbsp; $A_2 = \{ 1,\ 2 \}$ ,
+:* $B = \{B_1 ,\ B_2\}$&nbsp;   with&nbsp;  $B_1 =\{1\}$&nbsp;  and&nbsp; $B_2 = \{ 0,\ 2 \}$,
+:* $C = \{C_1 ,\ C_2\}$&nbsp;   with&nbsp;  $C_1 =\{2\}$&nbsp;  and&nbsp; $C_2 = \{  0,\ 1\}$.
+*Then the respective Kullback-Leibler distances are to be given:
+:* $D(P_X^{ (A) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (A) } )$,
+:* $D(P_X^{ (B) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (B) } )$,
+:* $D(P_X^{ (C) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (C) } )$.
-In der Aufgabe (a) soll die Kullback–Leibler–Distanz der beiden Wahrscheinlichkietsfunktionen $P_X(X)$ und $Q_X(X)$ für $X = \{0, 1, 2\} \Rightarrow  |X| = 3$ ermittelt werden. Anschließend soll die Menge $X$ entsprechend
+*Finally, subtask&nbsp; '''(5)'''&nbsp; asks for the conditions that must be satisfied for the equal sign to be true in the above inequality.
-:* $A = \{A_1 , A_2\}$   mit  $A_1 =\{0\}$  und $A_2 = \{ 1,2 \}$ ,
-:* $B = \{B_1 , B_2\}$   mit  $B_1 =\{1\}$  und $B_2 = \{ 0,2 \}$ ,
-:* $C = \{C_1 , C_2\}$   mit  $C_1 =\{2\}$  und $C_2 = \{  0,1\}$ ,
-mit $K = 2$ partitioniert werden und es sollen die jeweiligen Kullback–Leibler–Distanzen
-:* $D(P_X^{ (A) } \parallel Q_X^{ (A) } )$
-:* $D(P_X^{ (B) } \parallel Q_X^{ (B) } )$ ,
-:* $D(P_X^{ (C) } \parallel Q_X^{ (C) } )$
-angegeben werden. In Aufgabe (e) wird schließlich nach den Bedingungen gefragt, damit in der obigen Ungleichung das Gleichheitszeichen zutrifft.
-''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel [[Informationstheorie/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Einige Vorbemerkungen zu den 2D-Zufallsgrößen]].
-*Insbesondere wird Bezug genommen auf die Seite [[Informationstheorie/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen#Relative_Entropie_.E2.80.93_Kullback.E2.80.93Leibler.E2.80.93Distanz|Relative Entropie &ndash; Kullback-Leibler-Distanz]].
-*Die Angaben der Entropie  $H(Y)$ und der Kullback–Leibler–Distanz  $D( P_X \hspace{0.05cm}|| \hspace{0.05cm}P_Y)$  in obiger Grafik sind in „bit” zu verstehen.
-* Die in der Grafik  mit „???"  versehenen Felder sollen von Ihnen in dieser Aufgabe ergänzt werden.
-*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
-'''Hinweis:''' Die Aufgabe gehört zu
-[http://en.lntwww.de/Informationstheorie/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgr%C3%B6%C3%9Fen Kapitel 3.1].Die Wahrscheinlichkeitsfunktionen können aus obiger Grafik abgelesen werden:
-$P_X(X) = [1/4 , 1/2 , 1/4]$
-$Q_X(X) = [1/8 , 3/4, 1/8]$
+Hints:
+*The exercise belongs to the chapter&nbsp; [[Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables|Some preliminary remarks on two-dimensional random variables]].
+*In particular, reference is made to the page&nbsp; [[Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables#Informational_divergence_-_Kullback-Leibler_distance|Relative entropy &ndash; Kullback-Leibler distance]].
+*The two probability functions can be read from the above graph as follows:
+:$$P_X(X) = \big [1/4 , \ 1/2 , \ 1/4 \big ],\hspace{0.5cm} Q_X(X) = \big [1/8, \ 3/4, \ 1/8 \big].$$
+===Questions===
-===Fragebogen===
 <quiz display=simple>
-{Berechnen Sie die Kullback–Leibler–Distanz (KLD) allgemein.
+{Calculate the Kullback-Leibler distance in general.
 |type="{}"}
-$ D(P_X \parallel Q_X)$ = { 0.2075 3% } $bit$
+$ D(P_X \hspace{0.05cm} \vert \vert \hspace{0.05cm}  Q_X) \ = \ $ { 0.2075 3% } $\ \rm bit$
-{Welche KLD ergibt sich für die Partitionierung  $ A_1 = \{0\}, A_2 = \{1, 2\}$?
+{What is the Kullback-Leibler distance for the partition&nbsp;  $ A_1 = \{0\},\ A_2 = \{1, 2\}$?
 |type="{}"}
-$D(P_X^{ (A) } \parallel Q_X^{ (A) } )$ = { 0.0832 3% } $bit$
+$D(P_X^{ (A) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (A) } ) \ = \ $ { 0.0832 3% } $\ \rm bit$
-{Welche KLD ergibt sich für die Partitionierung  $ B_1 = \{1\}, A_2 = \{0, 2\}$?
+{What is the Kullback-Leibler distance for the partition&nbsp;  $ B_1 = \{1\}, \ B_2 = \{0, 2\}$?
 |type="{}"}
-$D(P_X^{ (B) } \parallel Q_X^{ (B) } )$ = { 0.2075 3% } $bit$
+$D(P_X^{ (B) } \hspace{0.05cm} \vert \vert \hspace{0.05cm} Q_X^{ (B) } ) \ = \ $ { 0.2075 3% } $\ \rm bit$
+{What is the Kullback-Leibler distance for the partition&nbsp;  $ C_1 = \{2\},\ C_2 = \{0, 1\}$?
+|type="()"}
++ The same result as for partition&nbsp; $\rm A$.
+- The same result as for partition&nbsp; $\rm B$.
+- A completely different result.
-{Welche KLD ergibt sich für die Partitionierung  $ C_1 = \{2\}, A_2 = \{0, 1\}$?
+{Under which conditions does equality result for general&nbsp; $K$&nbsp;?
 |type="[]"}
-+ Das gleiche Ergebnis wie für die Partitionierung $A$.
++ &nbsp; $|X|$&nbsp; equations must be fulfilled.
-- Das gleiche Ergebnis wie für die Partitionierung $B$.
++ For&nbsp; $x \in A_i$&nbsp; must hold: &nbsp; $P_X(x)/Q_X(x) = P_X(A_i)/ Q_X(A_i)$
-- Ein ganz anderes Ergebnis.
-{Unter welchen Bedingungen ergibt sich für allgemeines $K$ die Gleichheit?
-|type="[]"}
-+ Es müssen $|X|$ Gleichungen erfüllt sein.
-+ Für $x \epsilon A_i$ muss gelten : $P_X(x)/Q_X(x) = P_X(A_i)/ Q_X(A_i)$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''1.'''  Für die Kullback–Leibler–Distanz (KLD) gilt:
+'''(1)'''&nbsp;  For the Kullback-Leibler distance of the non-partitioned quantities&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; holds:
-$$D(P_X \parallel P_Y) = E [ log_2 \frac{P_X(X)}{P_Y(X)}] = \sum\limits_{ x \epsilon X} P_X(x) . log_2 \frac{P_X(x)}{P_Y(x)} =$$
-$$\frac{1}{2} . log_2 \frac{1/2}{3/4}  + 2 . \frac{1}{4} . log_2 \frac{1/4}{1/8}   +log_2 \frac{2}{3} + \frac{1}{2} . log_2(2) =$$
-$$1 -  \frac{1}{2} . log_2(3) = 0.2075 (bit)$$
-'''2.'''  $Partitionierung  A  \Rightarrow  A_1 = \{0\}$ ,  $A_2 = \{ 1 , 2 \}$ :  Man erhält die Wahrscheinlichkeitsfunktionen  $P_X^{ (A) } (X) = \{1/4 , 3/4\}$ und $Q_X^{ (A) } (X) = \{1/8 , 7/8\}$.  Daraus folgt:
-$$D(P_X^{ (A) } \parallel Q_X^{ (A) } )$ =   \frac{1}{4} . log_2 \frac{1/4}{1/8} +  \frac{3}{4} . log_2 \frac{3/4}{7/8} =$$
+:$$D(P_X \hspace{0.05cm} ||  \hspace{0.05cm}P_Y) = {\rm E} \left [ {\rm log}_2 \hspace{0.1cm} \frac{P_X(X)}{P_Y(X)}\right ] \hspace{0.2cm}=\hspace{0.2cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm}X}
+ P_X(x) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_X(x)}{P_Y(x)}$$
+:$$\Rightarrow D(P_X \hspace{0.05cm} ||  \hspace{0.05cm}P_Y) = \hspace{-0.15cm} \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/2}{3/4} +
+\cdot \frac{1}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/4}{1/8} =
+\frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{3} + \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} (2) = 1- \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} (3)
+\hspace{0.15cm}
+\underline {=0.2075\,{\rm (bit)}}
+\hspace{0.05cm}.$$
-$$ =  \frac{1}{4} . log_2 (2) +  \frac{3}{4} . log_2 \frac{6}{7} = 0.0832 (bit)$$
-Es ergibt sich also eine kleinere KLD als in Teilaufgabe (a).
-'''3.''' $Partitionierung  B \Rightarrow  B_1 = \{1\}$ ,  $B_2 = \{ 0 , 2 \}$ :  Man erhält die Wahrscheinlichkeitsfunktionen  $P_X^{ (B) } (X) = \{1/2 , 1/2\}$ und $Q_X^{ (B) } (X) = \{3/4 , 1/4\}$.
+'''(2)'''&nbsp;  With&nbsp; $\text{partition }A$ &nbsp; &rArr; &nbsp;  $A_1 = \{0\}$ ,&nbsp;  $A_2 = \{ 1 , 2 \}$&nbsp; we get&nbsp;  $P_X^{ (A) } (X) = \{1/4 , \ 3/4\}$&nbsp; and&nbsp; $Q_X^{ (A) } (X) = \{1/8 , \ 7/8\}$.&nbsp;  From this follows:
-Analog zur Teilaufgabe (b) erhält man nun:
-$$D(P_X^{ (B) } \parallel Q_X^{ (B) } ) =   \frac{1}{2} . log_2 \frac{1/2}{3/4} +  \frac{1}{2} . log_2 \frac{1/2}{1/4} = 0.2075$$
+:$$D(P_X^{\hspace{0.15cm}(A)} \hspace{0.05cm}|| \hspace{0.05cm} Q_X^{\hspace{0.15cm}(A)}) = \frac{1}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/4}{1/8} +
-$\Rightarrow$ gleiches Ergebnis wie in Aufgabe (a) $\Rightarrow$ Bei Partitionierung (B) gilt das Gleichheitszeichen.
+\frac{3}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{3/4}{7/8} =\frac{1}{4}\cdot {\rm log}_2 \hspace{0.1cm} (2) + \frac{3}{4}\cdot {\rm log}_2 \hspace{0.1cm} \frac{6}{7} \hspace{0.15cm} \underline {=0.0832\,{\rm (bit)}}
+\hspace{0.05cm}.$$
-'''4.'''$Partitionierung  C \Rightarrow  C_1 = \{2\}$ ,  $C_2 = \{ 0 , 1\}$ :  Man erhält  $P_X^{ (C) } (X) = \{1/4, 3/4\}$  , $Q_X^{ (C) } (X) = \{1/8, 7/8\}$ . also die gleichen Funktionen wie bei der Partitionierung
-$A \Rightarrow  Lösungsvorschlag 1$.
+'''(3)'''&nbsp; With&nbsp; $\text{partition } B$ &nbsp; &rArr; &nbsp;  $B_1 = \{1\}$ ,&nbsp;  $B_2 = \{ 0 ,\ 2 \}$&nbsp; the probability functions are&nbsp;  $P_X^{ (B) } (X) = \{1/2 , \ 1/2\}$&nbsp; und&nbsp; $Q_X^{ (B) } (X) = \{3/4 , \ 1/4\}$.&nbsp;
+*Analogous to subtask&nbsp; '''(2)'''&nbsp; one thus obtains:
+:$$D(P_X^{\hspace{0.15cm}(B)} \hspace{0.05cm}|| \hspace{0.05cm} Q_X^{\hspace{0.15cm}(B)})  = \frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/2}{3/4} +
+\frac{1}{2}\cdot {\rm log}_2 \hspace{0.1cm} \frac{1/2}{1/4} \hspace{0.15cm} \underline {=0.2075\,{\rm (bit)}}
+\hspace{0.05cm}.$$
+*The result agrees with that of subtask&nbsp; '''(1)'''&nbsp; &nbsp; &rArr; &nbsp; With&nbsp; $\text{partition }B$&nbsp; the equal sign applies.
-'''5.'''  Partitionierung $B$ hat zum Ergebnis $D(P_X^{ (B) } \parallel Q_X^{ (B) } ) = D(P_X \parallel Q_X)$ geführt. Für diesen Fall ist
-$\frac{P_X(1)}{Q_X(1)} =  \frac{1/2}{3/4} = \frac{2}{3}$ , $\frac{P_X(B_1)}{Q_X(B_1)} = \frac{1/2}{3/4} = \frac{2}{3}$,
-$\frac{P_X(0)}{Q_X(0)} =  \frac{1/4}{1/8} = 2$  , $\frac{P_X(B_2)}{Q_X(B_2)} = \frac{1/2}{1/4} = 2$,
+'''(4)'''&nbsp; With&nbsp; $\text{partition } C$ &nbsp; &rArr; &nbsp;  $C_1 = \{2\}$ ,  $C_2 = \{ 0 , \ 1\}$&nbsp; one obtains&nbsp;  $P_X^{ (C) } (X) = \{1/4, \  3/4\}$  , $Q_X^{ (C) } (X) = \{1/8, \ 7/8\}$, <br>i.e. the same functions as for the&nbsp; $\text{partition }A$ &nbsp; &rArr; &nbsp; <u>solution proposal 1</u>.
-$\frac{P_X(2)}{Q_X(2)} =  \frac{1/4}{1/8} = 2$ , $\frac{P_X(B_2)}{Q_X(B_2)} = \frac{1/2}{1/4} = 2$
-Es muss also für alle $x \epsilon X$ gelten :
-$\frac{P_X(x)}{Q_X(x)}  = \frac{P_X(B_1)}{Q_X(B_1)} $ , falls $ x \epsilon B_1$
+'''(5)'''&nbsp; The&nbsp; $\text{partition }B$&nbsp; has led to the result&nbsp; $D(P_X^{ (B) } \hspace{0.05cm} ||  \hspace{0.05cm}Q_X^{ (B) } ) = D(P_X \hspace{0.05cm} ||  \hspace{0.05cm}Q_X)$.
+*So for this case
+:$$\frac{P_X(1)}{Q_X(1)} =  \frac{1/2}{3/4} = \frac{2}{3},  \ \frac{P_X(B_1)}{Q_X(B_1)} = \frac{1/2}{3/4} = {2}/{3},$$
+:$$\frac{P_X(0)}{Q_X(0)} =  \frac{1/4}{1/8} = 2,  \ \frac{P_X(B_2)}{Q_X(B_2)} = \frac{1/2}{1/4} = 2,$$
+:$$\frac{P_X(2)}{Q_X(2)} =  \frac{1/4}{1/8} = 2,  \ \frac{P_X(B_2)}{Q_X(B_2)} = \frac{1/2}{1/4} = 2.$$
-$\frac{P_X(x)}{Q_X(x)}  = \frac{P_X(B_2)}{Q_X(B_2)} $ , falls $ x \epsilon B_2$
+*It must therefore hold for all&nbsp; $x \in X$&nbsp; :
+:$$\frac{P_X(x)}{Q_X(x)}  = \frac{P_X(B_1)}{Q_X(B_1)}, \text{ if } x \in B_1, \hspace{0.5cm}\frac{P_X(x)}{Q_X(x)}  = \frac{P_X(B_2)}{Q_X(B_2)}, \text{ if } x \in B_2.$$
-Durch Verallgemeinerung erhält man, dass $beide Lösungsvorschläge$ richtig sind.
+*By generalization, one can see that&nbsp; <u>both proposed solutions</u>&nbsp; are correct.
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@@ Line 142: / Line 134: @@
-[[Category:Aufgaben zu Informationstheorie |^3.1 Vorbemerkungen zu 2D-Zufallsgrößen^]]
+[[Category:Information Theory: Exercises |^3.1 General Information on 2D Random Variables^]]