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.
X, Y and Z combinations are –
$$\left\{\raise{5ex}{(1,1), (1,2), (1,3), (1,4), \\(2,1), (2,2), (2,3), (2,4),\\ (3,1), (3,2), (3,3), (3,4), \\(4,1), (4,2), (4,3), (4,4)}\right\}$$
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.
Best Answer
The answers should probably be
For $X$ and $Y$:
(a)
$$\boxed{\begin{array}{c|ccc} X \backslash Y & 0 & 1 & 2 \\[1ex]\hline 0 & 0 & 0 & 1/4 \\[1ex] 1 & 0 & 1/2 & 0 \\[1ex] 2 & 1/4 & 0 & 0\end{array}}$$
(b) they are not independent: e.g. $\Bbb P(X{=}0,Y{=}0)=0 \ne \tfrac{1}{16} = \Bbb P(X{=}0)\;\Bbb P(Y{=}0)$
(c) $\Bbb E(XY) = 0\times 2 \times \tfrac{1}{4}+1\times 1 \times\tfrac{1}{2}+ 2\times 0 \times\tfrac{1}{4}=\tfrac{1}{2}$.
For $X$ and $Z$:
(a)
$$\boxed{\begin{array}{c|ccc} X\backslash Z & 0 & 1 & 2 \\[1ex] \hline 0 & 1/16 & 1/8 & 1/16\\[1ex] 1 & 1/8 & 1/4 & 1/8\\[1ex] 2 & 1/16 & 1/8 & 1/16\end{array}}$$
(b) they are independent: e.g. $\Bbb P(X{=}0,Z{=}0)= \tfrac{1}{16} = \Bbb P(X{=}0)\,\Bbb P(Z{=}0)$ and you can test the others
(c) $\Bbb E(XZ)= 1$ by doing the multiplications and sum, or using independence to have $\Bbb E(XZ)= \Bbb E(X)\,\Bbb E(Z)$.