Independence of Random Variables

[kingwinner]

Homework Statement

Suppose X is a discrete random variable with probability mass function
p_X(x)=1/5, if x=-2,-1,0,1,2
p_X(x)=0, otherwise
Let Y=X². Are X and Y independent? Prove using definitions and theorems.

Homework Equations

The Attempt at a Solution

The random variables X and Y are independent <=> p_X,Y(x,y)=p_X(x)p_Y(y) for ALL x,y E R

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

Thanks for any help!

 

[Pere Callahan]

You will first have to ccalculate p_Y and p_X,Y.
For example
<br /> p_Y(1)=P(Y=1)=P(X=1\vee X=-1)=P(X=1)+P(X=-1)=1/5+1/5=2/5<br />
IN the same way calculate p_Y(y) for the remaining values in the range of Y, and similarly for p_X,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.

 

[kingwinner]

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

 

[Pere Callahan]

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

 

[kingwinner]

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

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

 

[Pere Callahan]

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

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