Question

Suppose that X and Y are integer valued random variables with joint probability mass function given by

Answer 1

Answer:

Step-by-step explanation:

1)

Given that:

The joint pmf  $p_{X,Y(a,b)}=\left \{ {{\dfrac{1}{4a} \ for \ 1 \le b \le a \le 4 } \\ \\ \atop {0} \ \ \ \ \ \ otherwise} } \right.$

To emphasize that this is a joint pmf;

We will notice that it obeys two conditions;

• it comprises a non-negative number and which is less than 1
• the sum of all the probabilities adds up and becomes unity.

Except that X and Y are integer value variables with 1 ≤ b ≤ a ≤ 4 and X = 1, 2, 3, 4 and Y = 1, 2, 3, 4 respectively, according to the condition Y ≤ X

The  table below shows the joint probabilities as a result of this:

              y = 1         y = 2        y = 1         y = 4          Total

x = 1          $\dfrac{1}{4}$             0                0             0               $\dfrac{1}{4}$

x = 2         $\dfrac{1}{8}$             $\dfrac{1}{8}$                 0             0               $\dfrac{1}{4}$

x = 3        $\dfrac{1}{12}$           $\dfrac{1}{12}$               $\dfrac{1}{12}$              0              $\dfrac{1}{4}$  

x = 4        $\dfrac{1}{16}$           $\dfrac{1}{16}$               $\dfrac{1}{16}$              $\dfrac{1}{16}$             $\dfrac{1}{4}$

Total       $\dfrac{25}{48}$            $\dfrac{13}{48}$             $\dfrac{7}{48}$                 $\dfrac{3}{48}$           1

In the table, it is obvious that each respective value of the probability is positive and the addition of all the values sums up to unity (1).

Hence, the given probability shows that it is indeed a pmf(probability mass function).

(b)

Marginal Pmf of x = $\dfrac{sum \ of \ all \ prob. \ of (x,y)}{y}$

Marginal Pmf of y =  $\dfrac{sum \ of \ all \ prob. \ of (x,y)}{x}$

Thus, we can locate the respective values of the marginal probability in the last row as well as the last column in the explained table above.

(c)

To find P(X=Y+1):

P(X = Y + 1) = P(X = 2,3,4)

⇒ 1 - P(X=1)

$\implies 1 - \dfrac{1}{4} \\ \\ \implies \dfrac{4-1}{4} \\ \\ \implies \dfrac{3}{4}$