expectation calculation in probability and statistics

expectation calculation in probability and statistics

2 four-sided dice are rolled. X = number of odd dice Y = number of even
dice Z = number of dice showing 1 or 2 So each of X, Y, Z only takes on
the values 0, 1, 2. (a) joint p.m.f. of (X,Y)? joint p.m.f. of (X,Z). You
can give your answers in the form of 3 by 3 tables. (b) Are X and Y
independent? Are X and Z independent? (c) Compute E(XY ) and E(XZ).
WHAT I TRIED IS - The faces on a four-sided die are labeled with the
numbers 1, 2, 3, and 4. Upon throwing the dice, each one will land with
exactly one of its faces upwards. You will therefore see two numbers, one
for each die. Suppose, for the sake of example, that you see a 2 and a 4.
Both of these are even numbers. None of them are odd numbers. Let's count:
no dice show an odd number, two dice show even numbers, and one die shows
a number in the set {1,2}. Therefore, for this throw,, X=0, Y=2, and Z=1.
There are 16 combinations (1,1) (1,2)(1,3) (1,4)(2,1)(2,2)(2,3)(2,4) ...
(4,1 )(4,2 )(4,3 )(4,4 ) There are only 2 case when no dice show an odd
number, two dice show even numbers, and one die shows a number in the set
{1,2}. ---> when (2,2) (2,4 ) occurs. Here X= 0, Y =2, Z =1 Both odd and
one is in {1,2} is --> (1,1)(1,3)(3,1) Here, X= 2, Y =0, Z =1 Just as we
have to in the case with one discrete random variable, in order to find
the "joint probability distribution" of X and Y, we first need to define
the support of X and Y.
The support of X is: S1 = {2, 4} And, the support of Y is: S2 = {1, 3}
X, Y and Z combinations are - {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2),
(2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)} Out
of this, the X, Y, Z values are respectively,
(2,0,1) , (1,1,2) ,
(2,0,1),(1,1,1),(1,1,2),(0,2,1),(1,1,1),(0,2,1),(2,0,1),(1,1,1),(2,0,0),(1,1,0),(1,1,1),(0,2,1),(1,1,0),(0,2,0).
b) The random variables X and Y are independent if and only if: P(X= x, Y
= y) = P(X = x) ~ P(Y = y) for allx¸S1,y¸S2.
if we again take a look back at the representation of our joint p.m.f. in
tabular form, you might notice that the following holds true: P(X=x,Y=y)
for allx¸S1,y¸S2. When this happens, we say that X and Y are
independent.
Similarly for X, Z, they are also independent.
c) I am not sure how to do
help please.