Aufgaben:Exercise 3.2: Expected Value Calculations

• Return to book

Two-dimensional
probability mass function

We consider the following probability mass functions  $\rm (PMF)$:

$$P_X(X) = \big[1/2,\ 1/8,\ 0,\ 3/8 \big],$$

$$P_Y(Y) = \big[1/2,\ 1/4,\ 1/4,\ 0 \big],$$

$$P_U(U) = \big[1/2,\ 1/2 \big],$$

$$P_V(V) = \big[3/4,\ 1/4\big].$$

For the associated random variables, let:

$X= \{0,\ 1,\ 2,\ 3\}$,     $Y= \{0,\ 1,\ 2,\ 3\}$,    $U = \{0,\ 1\}$,     $V = \{0, 1\}$.

Often, for such discrete random variables, one must have to calculate different expected values of the form

$${\rm E} \big [ F(X)\big ] =\hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm}\hspace{-0.03cm} {\rm supp} (P_X)} \hspace{-0.1cm} P_{X}(x) \cdot F(x).$$

Here, denote:

• $P_X(X)$  denotes the probability mass function of the discrete random variable   $X$.
• The  "support"  of  $P_X$  includes all those realisations  $x$  of the random variable  $X$  with non-vanishing probability.
• Formally, this can be written as

$${\rm supp} (P_X) = \{ x: \hspace{0.25cm}x \in X \hspace{0.15cm}\underline{\rm and} \hspace{0.15cm} P_X(x) \ne 0 \} \hspace{0.05cm}.$$

• $F(X)$  is an (arbitrary) real-valued function that can be specified in the entire domain of definition of the random variable  $X$ .

In the task, the expected values for various functions  $F(X)$  are to be calculated, among others for

1.   $F(X)= 1/P_X(X)$,
2.   $F(X)= P_X(X)$,
3.   $F(X)= - \log_2 \ P_X(X)$.

Hints:

• The exercise belongs to the chapter  Some preliminary remarks on two-dimensional random variables.
• The two one-dimensional probability mass functions  $P_X(X)$  and  $P_Y(Y)$  result from the presented 2D–PMF  $P_{XY}(X,\ Y)$,  as will be shown in  Exercise 3.2Z.
• The binary probability mass functions  $P_U(U)$  and  $P_V(V)$  are obtained according to the modulo operations  $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm}2$  and  $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.

Questions

1

What are the results of the following expected values?

${\rm E}\big[1/P_X(X)\big] \ = \ $

${\rm E}\big[1/P_{\hspace{0.04cm}Y}(\hspace{0.02cm}Y\hspace{0.02cm})\big] \ = \ $

2

Give the following expected values:

${\rm E}\big[P_X(X)\big] \ = \ $

${\rm E}\big[P_Y(Y)\big] \ = \ $

3

Now calculate the following expected values:

${\rm E}\big[P_Y(X)\big] \ = \ $

${\rm E}\big[P_X(Y)\big] \ = \ $

4

Which of the following statements are true?

	${\rm E}\big[- \log_2 \ P_U(U)\big]$  gives the entropy of the random variable  $U$.
	${\rm E}\big[- \log_2 \ P_V(V)\big]$  gives the entropy of the random variable  $V$.
	${\rm E}\big[- \log_2 \ P_V(U)\big]$  gives the entropy of the random variable  $V$.

Solution

Solution

(1)  In general, the following applies to the expected value of the function  $F(X)$  with respect to the random variable  $X$:

$${\rm E} \left [ F(X)\right ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} {\rm supp} (P_X)} \hspace{-0.2cm} P_{X}(x) \cdot F(x) \hspace{0.05cm}.$$

In the present example,  $X = \{0,\ 1,\ 2,\ 3\}$  and  $P_X(X) = \big [1/2, \ 1/8, \ 0, \ 3/8\big ]$.

• Because of  $P_X(X = 2) = 0$ , the quantity to be taken into account  (the "support")  in the above summation thus results in

$${\rm supp} (P_X) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 3 \} \hspace{0.05cm}.$$

• With  $F(X) = 1/P_X(X)$  one further obtains:

$${\rm E} \big [ 1/P_X(X)\big ] = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.4cm} P_{X}(x) \cdot {1}/{P_X(x)} = \hspace{-0.4cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} 1 \hspace{0.15cm}\underline{ = 3} \hspace{0.05cm}.$$

• The second expected value gives the same result with  ${\rm supp} (P_Y) = \{ 0\hspace{0.05cm}, 1\hspace{0.05cm}, 2 \}$ :

$${\rm E} \left [ 1/P_Y(Y)\right ] \hspace{0.15cm}\underline{ = 3}.$$

(2)  In both cases, the index of the probability mass function is identical with the random variable  $(X$  or   $Y)$  and we obtain

$${\rm E} \big [ P_X(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_X(x)} = (1/2)^2 + (1/8)^2 + (3/8)^2 = 13/32 \hspace{0.15cm}\underline{ \approx 0.406} \hspace{0.05cm},$$

$${\rm E} \big [ P_Y(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_Y(y) \cdot P_Y(y) = (1/2)^2 + (1/4)^2 + (1/4)^2 \hspace{0.15cm}\underline{ = 0.375} \hspace{0.05cm}.$$

(3)  The following equations apply here:

$${\rm E} \big [ P_Y(X)\big ] = \hspace{-0.3cm} \sum_{x \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 3 \}} \hspace{-0.3cm} P_{X}(x) \cdot {P_Y(x)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{4} + \frac{3}{8} \cdot 0 = 9/32 \hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm},$$

• The expected value formation here refers to  $P_X(·)$, i.e. to the random variable  $X$.
• $P_Y(·)$ is the formal function without (direct) reference to the random variable  $Y$.
• The same numerical value is obtained for the second expected value  (this does not have to be the case in general):

$${\rm E} \big [ P_X(Y)\big ] = \hspace{-0.3cm} \sum_{y \hspace{0.05cm}\in \hspace{0.05cm} \{ 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 2 \}} \hspace{-0.3cm} P_{Y}(y) \cdot {P_X(y)} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{8} + \frac{1}{4} \cdot 0 = 9/32 \hspace{0.15cm}\underline{ \approx 0.281} \hspace{0.05cm}.$$

(4)  We first calculate the three expected values:

$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_U(U)\big ] = \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{1} + \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{2}{1} \hspace{0.15cm}\underline{ = 1\ {\rm bit}} \hspace{0.05cm},$$

$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_V(V)\big ] = \frac{3}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{3} + \frac{1}{4} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{1} \hspace{0.15cm}\underline{ = 0.811\ {\rm bit}} \hspace{0.05cm},$$

$${\rm E} \big [-{\rm log}_2 \hspace{0.1cm} P_V(U)\big ] = \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{3} + \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} \frac{4}{1} \hspace{0.15cm}\underline{ = 1.208\ {\rm bit}} \hspace{0.05cm}.$$

Accordingly, the first two statements are correct:

• The entropy  $H(U) = 1$  bit  can be calculated according to the first equation.  It applies to the binary random variable  $U$  with equal probabilities.
• The entropy  $H(V) = 0.811$  bit  is calculated according to the second equation.  Due to the probabilities  $3/4$  and  $1/4$ , the entropy (uncertainty) is smaller here than for the random variable  $U$.
• The third expected value cannot indicate the entropy of a binary random variable, which is always limited to  $1$  (bit) , simply because of the result   $(1.208$  bit$)$ .