Some issues concerning joint random variables

Question

Let the joint random variable $P[x;y]$ be

$P[x;y] = c[2x^2 + y^2], x=-1;0;1, y=1;2;3;4$

$=0$ $elsewhere$

So I had to find the value of $c$ that makes $P[x;y]$ a joint discrete random variable.

I think I did that right. I just add up all the probabilities where $x = -1;0;1$ and $y = 1;2;3;4$ and made it equal to 1. Then I solved for $c$ and got $c=\frac{1}{106}$. Please check this for me if you think I've done something wrong.

Now they asked to calculate the $E[y]$ and the next question asked if $x$ and $y$ are independent.

So. I was a bit confused with the whole $P[x;y]$ and how to split it up into $P[x]$ and $P[y]$ (if that's possible) and then how to work out the $E[y]$.

Answer 1

To clarify the points above $$ \begin{align} P_Y(y) &=& \sum_{x = \lbrace-1,0,1\rbrace} c\left(2x^2 + y^2\right)\\ &=&c\left[\sum_{x = \lbrace-1,0,1\rbrace} 2x^2 + \sum_{x = \lbrace-1,0,1\rbrace} y^2\right]\\ &=&c\left[2\cdot (-1)^2+ 2\cdot(0)^2+2\cdot(1)^2 + 3\cdot y^2\right] \\ &=& c\left(4+3y^2\right) \end{align} $$

now you can apply the formula as @Did.

Answer 2

$$E(Y)=\sum_{x,y}y\,P_{X,Y}(x,y)=\sum_yy\,P_Y(y),\qquad P_Y(y)=\sum_xP_{X,Y}(x,y)$$

Answer 3

I have worked out the problem for you and I hope it helps to check with your answer. I have done it in EXCEL what other responders have alluded to. To verify your answer.