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Homework Help: Simple marginal distribution problem

  1. Jun 4, 2007 #1
    Hey guys,

    Doin revision for my maths exam and i came across this question from a past exam:

    1. The problem statement, all variables and given/known data
    Find fx(x,y) of
    [tex]f(x,y) = \frac{(1+4xy)}{2} for 0 \leq x, y \leq 1[/tex] and zero otherwise

    2. Relevant equations
    Now this should equal[tex] \int \frac{(1+4xy)}{2} dy [/tex]over all y but that leads to infinities ( as y goes from minus infinity to 1)which obviously we can't have. Im sure im missing something simple and stupid i just need someone to point ito out :)


    NOTE: Sorry this latex is stuffing up, tryin to fix it
    Last edited: Jun 4, 2007
  2. jcsd
  3. Jun 4, 2007 #2


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    You might want to split the integral into a couple of pieces. Remeber:
    [tex]f(x,y) \neq \frac{1+4xy}{2}[/tex]
  4. Jun 4, 2007 #3
    I think what Nate is trying to say is that f(x,y) is only defined on a certain domain. Since f(x,y) is a pdf, y cannot be arbitrarily negative as this would make f(x,y) negative. Remember, the integral of f(x,y) over the domain must be 1.

    Try to make sense of your domains. Draw them. X and Y can sometimes be dependant on each other...which can make things complicated.
  5. Jun 4, 2007 #4


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    Doesn't this expression,

    [tex]0 \leq x, y \leq 1[/tex]

    mean x and y are both in [0,1]? If y runs to minus infinity the question doesn't make much sense.
  6. Jun 4, 2007 #5


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    Actually, [itex]f(x,y)[/itex] is only non-zero on a certain domain, it's defined on the entire plane.
    Last edited: Jun 4, 2007
  7. Jun 4, 2007 #6
    A good thing to do, is first draw your "support". Sketch where the function is non-zero. This allows you to easily setup the bounds on the integral.
  8. Jun 4, 2007 #7
    Oh come on! Grant me some liberties. Although you're right, I probably shouldn't have used those words.
  9. Jun 5, 2007 #8
    Ah of course! i did draw it, i just drew it wrong :P thanks guys
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