Aufgaben:Exercise 3.2Z: Two-dimensional Probability Mass Function

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$\rm PMF$ of the two-dimensional random variable  $XY$

We consider the random variables  $X = \{ 0,\ 1,\ 2,\ 3 \}$  and  $Y = \{ 0,\ 1,\ 2 \}$, whose joint probability mass function  $P_{XY}(X,\ Y)$  is given.

• From this two-dimensional probability mass function  $\rm (PMF)$,  the one-dimensional probability mass functions  $P_X(X)$  and  $P_Y(Y)$  are to be determined.
• Such a one-dimensional probability mass function is sometimes also called  "marginal probability".

If  $P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables  $X$  and  $Y$  are statistically independent.  Otherwise, there are statistical dependencies between them.

In the second part of the task we consider the random variables  $U= \big \{ 0,\ 1 \big \}$  and  $V= \big \{ 0,\ 1 \big \}$,  which result from  $X$  and  $Y$  by modulo-2 operations:

$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm} V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$

Hints:

• The exercise belongs to the chapter  Some preliminary remarks on two-dimensional random variables.
• The same constellation is assumed here as in  Exercise 3.2.
• There the random variables  $Y = \{ 0,\ 1,\ 2,\ 3 \}$  were considered, but with the addition  ${\rm Pr}(Y = 3) = 0$.
• The property  $|X| = |Y|$  forced in this way was advantageous in the previous task for the formal calculation of the expected value.

Questions

1

What is the probability mass function  $P_X(X)$?

$P_X(0) \ = \ $

$P_X(1) \ = \ $

$P_X(2)\ = \ $

$P_X(3) \ = \ $

2

What is the probability mass function  $P_Y(Y)$?

$P_Y(0) \ = \ $

$P_Y(1) \ = \ $

$P_Y(2) \ = \ $

3

Are the random variables  $X$  and  $Y$  statistically independent?

	Yes,
	No.

4

Determine the probabilities  $P_{UV}( U,\ V)$.

$P_{UV}( U = 0,\ V = 0) \ = \ $

$P_{UV}( U = 0,\ V = 1) \ = \ $

$P_{UV}( U = 1,\ V = 0) \ = \ $

$P_{UV}( U =1,\ V = 1) \ = \ $

5

Are the random variables  $U$  and  $V$  statistically independent?

	Yes,
	No.

Solution

Solution

(1)  You get from  $P_{XY}(X,\ Y)$  to the one-dimensional probability mass function  $P_X(X)$ by summing up all  $Y$ probabilities:

$$P_X(X = x_{\mu}) = \sum_{y \hspace{0.05cm} \in \hspace{0.05cm} Y} \hspace{0.1cm} P_{XY}(x_{\mu}, y).$$

• One thus obtains the following numerical values:

$$P_X(X = 0) = 1/4+1/8+1/8 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$

$$P_X(X = 1)= 0+0+1/8 = 1/8 \hspace{0.15cm}\underline{= 0.125},$$

$$P_X(X = 2) = 0+0+0 \hspace{0.15cm}\underline{= 0}$$

$$P_X(X = 3) = 1/4+1/8+0=3/8 \hspace{0.15cm}\underline{= 0.375}\hspace{0.5cm} \Rightarrow \hspace{0.5cm} P_X(X) = \big [ 1/2, \ 1/8 , \ 0 , \ 3/8 \big ].$$

(2)  Analogous to sub-task   (1) , the following now holds:

$$P_Y(Y = y_{\kappa}) = \sum_{x \hspace{0.05cm} \in \hspace{0.05cm} X} \hspace{0.1cm} P_{XY}(x, y_{\kappa})$$

$$P_Y(Y= 0) = 1/4+0+0+1/4 = 1/2 \hspace{0.15cm}\underline{= 0.500},$$

$$P_Y(Y = 1) = 1/8+0+0+1/8 = 1/4 \hspace{0.15cm}\underline{= 0.250},$$

$$P_Y(Y = 2) = 1/8+1/8+0+0 = 1/4 \hspace{0.15cm}\underline{= 0.250} \hspace{0.5cm} \Rightarrow \hspace{0.5cm} P_Y(Y= 0) = \big [ 1/2, \ 1/4 , \ 1/4 ].$$

(3)  With statistical independence,  $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$  should be.

• This does not apply here:     answer   NO.

(4)  Starting from the left-hand table   ⇒   $P_{XY}(X,Y)$,  we arrive at the middle table   ⇒   $P_{UY}(U,Y)$,
by combining certain probabilities according to  $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.

If one also takes into account  $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$, one obtains the probabilities sought according to the right-hand table:

Different probability functions

$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$

$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$

$$P_{UV}( U = 1, V = 0) = 1/8 \hspace{0.15cm}\underline{= 0.125},$$

$$P_{UV}( U = 1, V = 1) = 1/8 \hspace{0.15cm}\underline{= 0.125}.$$

(5)  The correct answer is   YES:

• The corresponding one-dimensional probability mass functions are:  

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

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

• Thus:  $P_{UV}(U,V) = P_U(U) \cdot P_V(V)$   ⇒   $U$  and  $V$  are statistically independent.