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Independence of Random Variables

  1. Dec 6, 2008 #1
    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. jcsd
  3. Dec 6, 2008 #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.
     
  4. Dec 6, 2008 #3
    How can I find the joint mass function pX,Y from their marginal mass functions?
     
  5. Dec 7, 2008 #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:
     
  6. Dec 7, 2008 #5
    How is this possible? Is the probability 0?
    What other cases do I have to do?
     
  7. Dec 7, 2008 #6
    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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