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Joint distrubtion question

  1. Jan 11, 2010 #1


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    if the joint p.d.f of x,y is given by

    f(x,y)= a |x-y| ,0<=x,y<=1
    f(x,y)= 0 , o.w

    find i)a
  2. jcsd
  3. Jan 12, 2010 #2


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    a must be such that the total probability is 1. [itex]x\ge y[/itex], and so |x-y|= x- y, for (x,y) on the triangle with vertices at (0,0), (1, 1), and (1, 0). [itex]y\ge x[/itex], and so |x- y|= y- x for (x,y) on the triangle with vertices at (0, 0), (1, 1), and (0, 1). That is, we must have
    [tex]\int_{x=0}^1\int_{y= 0}^x a(y- x) dydx+ \int_{x=0}^1\int_{y=x}^1 a(x- y)dydx= 1[/tex]
    [tex]a\left(\int_{x=0}^1\int_{y= 0}^x (y- x) dydx+ \int_{x=0}^1\int_{y= x}^1 (x- y)dydx\right)= 1[/tex]
    Do the integration on the left and take its reciprocal to find a.

    Again, y> x in the triangle with vertices at (0,0), (1, 1), and (, 1).
    [tex]p(y>x)= a\int_{x=0}^1\int_{y=x}^1 (y- x) dy dx[/tex]
    where "a" is the value from (i). (Though the answer should be obvious from symmetry.)

    Any double integral over a line is 0.
    Last edited by a moderator: Jan 12, 2010
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