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Finding c in a joint PDF

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    E = { (x,y) | |x| + |y| ≤ 1}

    fx,y (x,y) =
    {
    c (x,y) ∈ E
    0 otherwise
    }

    Find C.
    Find the Marginal PDFs
    Find the conditional X given Y=y, where -1 ≤ y ≤ 1.
    Are X and Y independent.
    2. Relevant equations
    I'm taking a guess here in the solution...
    but F(x,y) = F(x)F(y)
    and f(x,y) = f(x)f(y)
    These will be used later, when I'm wishing to find the Marginals and the independence.

    3. The attempt at a solution
    So, this is a uniform distribution (if it's not stated in my pdf, it's stated in the problem's text.)
    Considering that this an "area" I should just be able to integrate this with respect to the bondaries correct?
    That would be ∫0,1∫00,1cdxdy? Then c = 1, or do I base it off of E? Then it should be bounded from [-1,1]?
    This is what I believe it to be, but I'm not entirely sure. My professor gave a solution that was probably more general where he found something else first, but i didn't quite get it because he was trying to rush it at the end of class.

    After finding C, the marginals are the integrals with respect to y and x to give us the "trace" of the density function.
     
  2. jcsd
  3. Nov 8, 2015 #2

    andrewkirk

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    Because it's a pdf, we must have ##\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y} (x,y)dx dy =\int_E c dA=c\int_E dA=1## where the ##dA## indicates integrating by area.

    So just work out that last integral, which is the area of ##E##, and then figure out ##c## by the requirement for the final equality to hold.

    You'll find it easier to work out E if you first draw a picture.
     
  4. Nov 8, 2015 #3
    Okay, that's what I thought after reviewing a bit of what this meant.
    Drawing E, we have what looks essentially like a diamond where each point is at y = 1, y = -1, x = 1, x = -1. This would bound the area from [-1,1] for dx and dy.
    My question now is when evaluating the area, do we use E as the function of integration or do we use the PDF? I'm trying to grapple the idea of two things that are related by not the same. We have a distribution of probabilities... across a geometric area E?
     
  5. Nov 8, 2015 #4

    andrewkirk

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    You are trying to get the cumulative probability of all (x,y) pairs, which requires integrating the pdf. So it's the latter. E is a set - a region in the number plane - not a function that can be integrated in this context. The relevance of E is that you know that the pdf is only nonzero inside E, so you can restrict your integration to inside E without changing the result.
     
  6. Nov 8, 2015 #5
    Oh okay, and within that set it has a uniform probability (0,1) where each point is smaller in probability by a factor of 1/4 because E takes the shape it does?
     
  7. Nov 8, 2015 #6

    LCKurtz

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    Remember that ##\iint_E c~dydx = c\iint_E 1~dydx = c\cdot \text{Area of }E##. You shouldn't need calculus and integrals to figure out that last expression.
     
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