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Marginal PDF from Joint PDF

  1. Nov 26, 2012 #1
    Hi all,
    I'm looking at the joint pdf F(x,y) = (8+xy^3)/64) for -1<x<1 and -2<y<2
    (A plot of it is here: https://www.wolframalpha.com/input/?i=(8+xy^3)/64+x+from+-1+to+1,+y+from+-2+to+2 ...sorry about the ugly url) and trying to find the marginal PDFs for X and Y.

    I know I want to integrate the joint function with respect to Y and X in order to to get the marginal pdfs for X and Y, respectively. However, I'm running into trouble when I try to set the bounds for these integrals!
    As far as I can tell, X and Y don't seem to depend on each other in this sense; i.e. for marginal(X) i would have Integral([JointPDF]dy), from -2 to 2, which comes out to 1/2.
    (Similarly, integrating with respect to x from -1 to 1 yields 1/4).
    When I integrate these from their respective bounds (x from -1 to 1, y from -2 to 2) both come out to 1, as a proper pdf should. However the fact that both are independent of x and y values makes me think something might be wrong...does anyone have any suggestions as to what I might be doing wrong?
    Thanks so much!
    Jamie

    Edit: My apologies! Posted this to the wrong place! I can't figure out how to delete it though :x
     
    Last edited: Nov 26, 2012
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  3. Nov 27, 2012 #2

    Stephen Tashi

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    How did you get the integral with respect to y to come out to be a constant? It should be a function of x.
     
  4. Nov 27, 2012 #3
    Maybe I'm mistaking, but as far as I can tell the indefinite integral comes out to:

    (8y+(xy^4)/4)/64 + c. If you evaluate this from -2 to 2, the x terms cancel because the y is an even function, i.e. from -2 to 2 we get [(1/4)+16x]-[-(1/4)+16x], so (1/4)+(1/4)+16x-16x = 1/2.
    This is what made me think that perhaps my bounds are incorrectly calculated...
    Am I doing something wrong?
     
    Last edited: Nov 27, 2012
  5. Nov 27, 2012 #4

    Stephen Tashi

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    Not in the integration. (I'm the one who was confused about that.) But the fact that the conditional distributions are constant (and thus "independent" of the values of both variables) doesn't show that the x and y are independent random variables. If x and y were independent random variables then for each pair of sets [itex] A,B [/itex] that define events [itex] Pr(x \in A) = Pr(x \in A | y \in B) [/itex]

    For example, consider the events [itex] A = \{x: x \in [0, 1]\},\ B = \{y: y \in [0,1]\} [/itex]
     
  6. Nov 28, 2012 #5
    Still, what would the bounds be? A marginal pdf should not be a constant
     
  7. Nov 28, 2012 #6

    Stephen Tashi

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    I don't know any theorem that asserts that. Do you?
     
  8. Nov 28, 2012 #7
    I think a marginal pdf would be constant if X and Y are uniformly distributed, but I'm not sure how to tell if they are in this case. Any input?
     
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