# Independence of Random Variables

1. Dec 6, 2008

### kingwinner

1. The problem statement, all variables and given/known data
Suppose X is a discrete random variable with probability mass function
pX(x)=1/5, if x=-2,-1,0,1,2
pX(x)=0, otherwise
Let Y=X2. Are X and Y independent? Prove using definitions and theorems.

2. Relevant equations
3. The attempt at a solution
The random variables X and Y are independent <=> pX,Y(x,y)=pX(x)pY(y) for ALL x,y E R

But the trouble here is that we don't have pX,Y(x,y) and pY(y). What can we do?

Thanks for any help!

2. Dec 6, 2008

### Pere Callahan

You will first have to ccalculate pY and pX,Y.
For example
$$p_Y(1)=P(Y=1)=P(X=1\vee X=-1)=P(X=1)+P(X=-1)=1/5+1/5=2/5$$
IN the same way calculate pY(y) for the remaining values in the range of Y, and similarly for pX,Y. Of course you would suspect X and Y not to be independent, so it would suffice to find one case where the joint distribution does not factorize.

3. Dec 6, 2008

### kingwinner

How can I find the joint mass function pX,Y from their marginal mass functions?

4. Dec 7, 2008

### Pere Callahan

$$p_{X,Y}(1,2)=P(X=1\vee Y=2)=P(X=1\vee X^2=2)...?$$
Which X values contribute to this probabbility. They should be such that X=1 and X^2=2

5. Dec 7, 2008

### kingwinner

How is this possible? Is the probability 0?
What other cases do I have to do?

6. Dec 7, 2008

### Pere Callahan

Yes, the prob. is zero. Either do all remaining cases or just compare
$$p_{X,Y}(1,2)=0$$
to
$$p_{X}(1)p_{Y}(2)=?$$