Aufgaben:Exercise 3.6: Partitioning Inequality

• Return to book

Probability functions  $P_X$,  $Q_X$

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

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

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

$\text{Please note:}$  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 setnbsp; $X$  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 task belongs to the chapter  Some preliminary remarks on 2D random variables.
• In particular, reference is made to the page  Informational Divergence – 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

1

Calculate the Kullback-Leibler distance (KLD) in general.

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

2

WWhat is the Kullback-Leibler distance for the partition  $ A_1 = \{0\},\ A_2 = \{1, 2\}$?

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

3

What is the Kullback-Leibler distance for the partition  $ B_1 = \{1\}, \ B_2 = \{0, 2\}$?

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

4

What is the Kullback-Leibler distance for the partition  $ C_1 = \{2\},\ C_2 = \{0, 1\}$?

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

5

Under which conditions does equality result for genera  $K$ ?

	$|X|$  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

Solution

(1)  For the Kullback-Leibler distance 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  $\text{partition C}$  has led to the result  $D(P_X^{ (B) } \hspace{0.05cm} || \hspace{0.05cm}Q_X^{ (B) } ) = D(P_X \hspace{0.05cm} || \hspace{0.05cm}Q_X)$ geführt.

• 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  $x \in X$  :

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

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