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Bivariate probabilities

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem statement, all variables and given/known data[/b]
    Let X and Y have the joint pdf
    fXY(x,y) = x + y, 0<x<1, 0<y<1

    determine Pr X+Y , 1/2


    2. Relevant equations



    3. The attempt at a solution
    [ tex ]
    \int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}x+y dx dy

    [ /tex ]
    = 0.125
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 8, 2009 #2
    Sorry I didn't put the code in right

    [itex]
    \int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}x+y dx dy

    [/itex]
    = 0.125
     
  4. May 8, 2009 #3
    it should be determine Pr (X+Y) < 1/2
     
  5. May 9, 2009 #4
    Do you mean P(X+Y<1/2)? Your integral gives P(X<1/2 and Y<1/2). Draw the region in the unit square where x+y<1/2.
     
  6. May 9, 2009 #5
    Yes I mean P(X+Y<1/2)?
    The unit square would be

    (0,0) (0,1/4) (1/4,0) and (1/4,1/4)

    would the limits than be 0< x <= y < 1/4

    And the integarl

    [itex]
    \int_{0}^{\frac{1}{4}} \int_{y}^{\frac{1}{4}} (x+y) dxdy
    [/itex]
     
  7. May 10, 2009 #6

    Mark44

    Staff: Mentor

    That's not a unit square.
     
  8. May 10, 2009 #7
    Isn't the unit square just the square with side length 1?

    ie

    (0,0) (0,1) (1,0) (1,1)

    So we are looking at the area where x+y < 1/2

    Wouldn't that be in the region (0,0) (0,1/4) (1/4,0) (1/4,1/4) of the unit square ?

    so the lmits are

    0 < x < 1/4 and 0<y<1/4

    [itex]
    \int_{0}^{\frac{1}{4}}\int_{0}^{\frac{1}{4}} x+y\text{ } dx dy
    [/itex]
     
  9. May 10, 2009 #8
    No. Draw {(x,y) : 0<x<1, 0<y<1, and x+y<1/2}. Think calc 1 or calc 3, or even college algebra, instead of probability theory. Find the boundary line x+y = 1/2, then shade one side of it.
     
  10. May 11, 2009 #9
    Is this the area bounded by the triangle with vertices (0,0) (0,1/2) (1/2,0) ?

    with limits

    0 < x < 1/2 0 < y < 1/2-x

    [itex]
    \int_{0}^{1/2}\int_{0}^{\frac{1}{2}-x} x+y\text{ }dydx
    [/itex]
     
  11. May 11, 2009 #10
    Sorry is the limit of y 1/2-x < y < 1/2

    making the integral
    [itex]
    \int_{0}^{\frac{1}{2}}\int_{\frac{1}{2}-x}^{\frac{1}{2}}x+y\text{ }dydx
    [/itex]
     
  12. May 11, 2009 #11
    This one is right. Your later post (#10) is not.
     
  13. May 11, 2009 #12
    Thanks alot with your help guys
     
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