Independence of Random Variables

  • Thread starter kingwinner
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  • #1
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Homework Statement


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.

Homework Equations


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!
 

Answers and Replies

  • #2
You will first have to ccalculate pY and pX,Y.
For example
[tex]
p_Y(1)=P(Y=1)=P(X=1\vee X=-1)=P(X=1)+P(X=-1)=1/5+1/5=2/5
[/tex]
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
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How can I find the joint mass function pX,Y from their marginal mass functions?
 
  • #4
[tex]
p_{X,Y}(1,2)=P(X=1\vee Y=2)=P(X=1\vee X^2=2)...?
[/tex]
Which X values contribute to this probabbility. They should be such that X=1 and X^2=2:smile:
 
  • #5
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[tex]
p_{X,Y}(1,2)=P(X=1\vee Y=2)=P(X=1\vee X^2=2)...?
[/tex]
Which X values contribute to this probabbility. They should be such that X=1 and X^2=2:smile:
How is this possible? Is the probability 0?
What other cases do I have to do?
 
  • #6
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
[tex]
p_{X,Y}(1,2)=0
[/tex]
to
[tex]
p_{X}(1)p_{Y}(2)=?
[/tex]
 

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