Aufgaben:Exercise 3.9: Conditional Mutual Information

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Result  $W$  as a function
of  $X$,  $Y$,  $Z$

We assume statistically independent random variables  $X$,  $Y$  and  $Z$  with the following properties:

$$X \in \{1,\ 2 \} \hspace{0.05cm},\hspace{0.35cm} Y \in \{1,\ 2 \} \hspace{0.05cm},\hspace{0.35cm} Z \in \{1,\ 2 \} \hspace{0.05cm},\hspace{0.35cm} P_X(X) = P_Y(Y) = \big [ 1/2, \ 1/2 \big ]\hspace{0.05cm},\hspace{0.35cm}P_Z(Z) = \big [ p, \ 1-p \big ].$$

From  $X$,  $Y$  and  $Z$  we form the new random variable  $W = (X+Y) \cdot Z$.

• It is obvious that there are statistical dependencies between  $X$  and  $W$    ⇒   mutual information  $I(X; W) ≠ 0$.
• Furthermore,  $I(Y; W) ≠ 0$  as well as  $I(Z; W) ≠ 0$  will also apply, but this will not be discussed in detail in this task.

Three different definitions of mutual information are used in this task:

• the conventional  mutual information zwischen  $X$  and  $W$:

$$I(X;W) = H(X) - H(X|\hspace{0.05cm}W) \hspace{0.05cm},$$

• the conditional  mutual information between  $X$  and  $W$  with a given fixed value  $Z = z$:

$$I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = z) = H(X\hspace{0.05cm}|\hspace{0.05cm} Z = z) - H(X|\hspace{0.05cm}W ,\hspace{0.05cm} Z = z) \hspace{0.05cm},$$

• the conditional  mutual information between  $X$  and  $W$  for a given random value  $Z$:

$$I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z ) = H(X\hspace{0.05cm}|\hspace{0.05cm} Z ) - H(X|\hspace{0.05cm}W \hspace{0.05cm} Z ) \hspace{0.05cm}.$$

The relationship between the last two definitions is:

$$I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z ) = \sum_{z \hspace{0.1cm}\in \hspace{0.1cm}{\rm supp} (P_{Z})} \hspace{-0.2cm} P_Z(z) \cdot I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = z)\hspace{0.05cm}.$$

Hints:

• How large is the mutual information between  Different entropies of two-dimensional random variables.
• In particular, reference is made to the page  Conditional mutual information.

Questions

1

How large is the transinformation between  $X$  and  $W$,  if  $Z = 1$  always holds?

$I(X; W | Z = 1) \ = \ $ $\ \rm bit$

2

How large is the transinformation between  $X$  and  $W$,  if  $Z = 2$  always holds?

$I(X; W | Z = 2) \ = \ $ $\ \rm bit$

3

Now let  $p = {\rm Pr}(Z = 1)$.  How large is the conditional mutual information between  $X$  and  $W$, if  $z \in Z = \{1,\ 2\}$  is known?

$p = 1/2\text{:} \ \ \ I(X; W | Z) \ = \ $ $\ \rm bit$

$p = 3/4\text{:} \ \ \ I(X; W | Z) \ = \ $ $\ \rm bit$

4

How large is the unconditional mutual information for  $p = 1/2$?

$I(X; W) \ = \ $ $\ \rm bit$

Solution

Solution

2D probability functions for  $Z = 1$

(1)  The upper graph is valid for  $Z = 1$   ⇒   $W = X + Y$. 

• Under the conditions  $P_X(X) = \big [1/2, \ 1/2 \big]$  as well as  $P_Y(Y) = \big [1/2, \ 1/2 \big]$  the joint probabilities  $P_{ XW|Z=1 }(X, W)$  thus result according to the right graph (grey background).

• Thus the following applies to the mutual information under the fixed condition  $Z = 1$:

$$I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = 1) \hspace{-0.05cm} = \hspace{-1.1cm}\sum_{(x,w) \hspace{0.1cm}\in \hspace{0.1cm}{\rm supp} (P_{XW}\hspace{0.01cm}|\hspace{0.01cm} Z\hspace{-0.03cm} =\hspace{-0.03cm} 1)} \hspace{-1.1cm} P_{XW\hspace{0.01cm}|\hspace{0.01cm} Z\hspace{-0.03cm} =\hspace{-0.03cm} 1} (x,w) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_{XW\hspace{0.01cm}|\hspace{0.01cm} Z\hspace{-0.03cm} =\hspace{-0.03cm} 1} (x,w) }{P_X(x) \cdot P_{W\hspace{0.01cm}|\hspace{0.01cm} Z\hspace{-0.03cm} =\hspace{-0.03cm} 1} (w) }$$

$$I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = 1) = 2 \cdot \frac{1}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{1/4}{1/2 \cdot 1/4} + 2 \cdot \frac{1}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{1/4}{1/2 \cdot 1/2}$$

$$\Rightarrow \hspace{0.3cm} I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = 1) \hspace{0.15cm} \underline {=0.5\,{\rm (bit)}} \hspace{0.05cm}.$$

• The first term summarises the two horizontally shaded fields in the graph, the second term the vertically shaded fields.
• The latter do not contribute because of  $\log_2 (1) = 0$ .

2D probability functions for  $Z = 2$

(2)  For  $Z = 2$  $W = \{4,\ 6,\ 8\}$, is valid, but nothing changes with respect to the probability functions compared to subtask  (1) .

• Consequently, the same conditional mutual information is obtained:

$$I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = 2) = I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = 1) \hspace{0.15cm} \underline {=0.5\,{\rm (bit)}} \hspace{0.05cm}.$$

(3)  The equation is for  $Z = \{1,\ 2\}$  with  ${\rm Pr}(Z = 1) =p$  and  ${\rm Pr}(Z = 2) =1-p$:

$$I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z) = p \cdot I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = 1) + (1-p) \cdot I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z = 2)\hspace{0.15cm} \underline {=0.5\,{\rm (bit)}} \hspace{0.05cm}.$$

• It is considered that according to subtasks  (1)  and  (2)  the conditional mutual information for given  $Z = 1$  and given  $Z = 2$  are equal.
• Thus  $I(X; W|Z)$, i.e. under the condition of a stochastic random variable  $Z = \{1,\ 2\}$  with  $P_Z(Z) = \big [p, \ 1 – p\big ]$  is independent of  $p$.
• In particular, the result is also valid for  $\underline{p = 1/2}$  and  $\underline{p = 3/4}$.

To calculate the joint probability for $XW$

(4)  The joint probability  $P_{ XW }$  depends on the  $Z$–probabilites  $p$  and  $1 – p$ .

• For  $Pr(Z = 1) = Pr(Z = 2) = 1/2$  the scheme sketched on the right results.
• Again, only the two horizontally shaded fields contribute to the mutual information:

$$I(X;W) = 2 \cdot \frac{1}{8} \cdot {\rm log}_2 \hspace{0.1cm} \frac{1/8}{1/2 \cdot 1/8} \hspace{0.15cm} \underline {=0.25\,{\rm (bit)}} \hspace{0.35cm} < \hspace{0.35cm} I(X;W \hspace{0.05cm}|\hspace{0.05cm} Z) \hspace{0.05cm}.$$

The result  $I(X; W|Z) > I(X; W)$  is true for this example, but also for many other applications:

• If I know  $Z$, I know more about the 2D random variable  $XW$  than without this knowledge..
• However, one must not generalise this result:

Sometimes  $I(X; W) > I(X; W|Z)$, actually applies, as in  example 3 in the theory section.