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Joint density function

  1. May 16, 2010 #1
    I have a region R = {(x,y): |x| + |y| <= 1}
    I have sketched the graph on the x-y plane and shaded in the region R (I got a diamond).
    How do I find the area of R and hence find the joint density function f(x,y) and how do I specify the region for which f(x,y) is non-zero.
     
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  3. May 16, 2010 #2

    EnumaElish

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    A diamond is 4 triangles: Area(R) = 4 times the area of one triangle.

    Are x and y random variables?
     
  4. May 16, 2010 #3
    They are just random points chosen uniformly on the region
     
  5. May 17, 2010 #4
    So the area is 0.5 x 4 = 2, how does that relate to finding the joint density function.
    So f(x,y) = 2 for -1 < x < 1 and -1 < y < 1, 0 otherwise
     
    Last edited: May 17, 2010
  6. May 17, 2010 #5
    No. Wouldn't it be f(x,y) = x + y for 0 < x < 1 and 0 < y < 1 and 0 otherwise
    because when I do the double integral, the integral equals 1 which it should.
     
  7. May 17, 2010 #6

    LCKurtz

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    You said your region is R = {(x,y): |x| + |y| <= 1}. That is not the same as you are describing as -1 < x < 1 and -1 < y < 1. And a uniform distribution gives a constant density, not a variable like x + y. You need to have f(x,y) be a constant value on R such that the integral of f(x,y) over R is 1.
     
  8. May 17, 2010 #7
    f(x,y) = 1/2 for 0 < x < 1 - |x|, but I don't know the limits of y.
     
  9. May 17, 2010 #8

    LCKurtz

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    No. Can't x be negative?

    Didn't you say you had drawn a picture of your region R and it was diamond shaped?

    Use the description of R itself to say where f = 1/2.
     
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