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Math Help - Joint probability density

  1. #1
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    Joint probability density

    If the joint probability density of X and Y is given by

    f(x,y)=2 for x>0, y>0, x+y<1
    and zero elsewhere

    find P(X>2Y)

    When I draw a picture with that restriction above( y<(1/2)x and y<1-x )
    I get 1/3, but I want to solve this one by using double integral
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  2. #2
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    Quote Originally Posted by Judi View Post
    If the joint probability density of X and Y is given by

    f(x,y)=2 for x>0, y>0, x+y<1
    and zero elsewhere

    find P(X>2Y)

    When I draw a picture with that restriction above( y<(1/2)x and y<1-x )
    I get 1/3, but I want to solve this one by using double integral
    The double integral is just (this is tough without latex)

    double integral over D 2 dydx

    where the set D satisfies your 4 inequalities. Turning this double integral into an iterated integral to evaluate has to go back to the picture. From that picture I get the iterated integral will have two parts depending on whether x/2 < 1 - x or not:

    int (0,2/3) int (0,x/2) 2 dydx + int (2/3,1) int (0,1-x) 2 dydx =

    int (0,2/3) 2(x/2) dx + int (2/3,1) 2(1-x) dx.

    This will equal 1/3 as you say.
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  3. #3
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    Thank you!
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  4. #4
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    Is there another way to do this one?
    I am trying to understand what my professor did:

    P(X>2Y)=2((2/3)(1/3)(1/2)+(1/3)(1/3)(1/2))=1/3
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  5. #5
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    Quote Originally Posted by Judi View Post
    Is there another way to do this one?
    I am trying to understand what my professor did:

    P(X>2Y)=2((2/3)(1/3)(1/2)+(1/3)(1/3)(1/2))=1/3
    It looks like he/she just used the picture: that's 2 times the sum of the areas of the two triangles in the picture.
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  6. #6
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    why is it times 2?
    I can see two triangels inside the big triangle.
    Why not just sum of the two triangles?
    I am curious...
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  7. #7
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    Quote Originally Posted by Judi View Post
    why is it times 2?
    I can see two triangels inside the big triangle.
    Why not just sum of the two triangles?
    I am curious...
    2 is the constant in the density f(x,y) that converts area to probability. The total probability must equal 1; the total area covered when f(x,y) > 0 is 1/2. So multiply by 2 to convert area to probability.
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