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Math Help - P(x<2y)

  1. #1
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    P(x<2y)

    Pretty simple question:

    f(x,y) = {2 if 0 <= x <= y <= 1 , 0 elsewhere}

    How do I compute P(X<2Y) ?



    Similarly, if I'm given the following CDF,

    F(x,y) = (1-exp(-x^2))(1-exp(-y^2)) on {x,y>0}

    How would I compute P(X<2Y) here?


    Thanks in advance for your help!
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  2. #2
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    Quote Originally Posted by rockyraccoon View Post
    Pretty simple question:

    f(x,y) = {2 if 0 <= x <= y <= 1 , 0 elsewhere}

    How do I compute P(X<2Y) ?



    Similarly, if I'm given the following CDF,

    F(x,y) = (1-exp(-x^2))(1-exp(-y^2)) on {x,y>0}

    How would I compute P(X<2Y) here?


    Thanks in advance for your help!
    In both cases draw a picture and the integral terminals become obvious.

    1. \int_{x = 0}^{1/2} \int_{y = x/2}^{1} f(x, y) \, dy \, dx.

    2. \int_{x = 0}^{+ \infty} \int_{y = x/2}^{+ \infty} f(x, y) \, dy \, dx.
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  3. #3
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    That's what I was doing as well, although the answers in the back of the book don't match up.

    For the first question, the answer is 1/2.

    For the second, the answer is 1/5.

    Thoughts?
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  4. #4
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    Quote Originally Posted by rockyraccoon View Post
    That's what I was doing as well, although the answers in the back of the book don't match up.

    For the first question, the answer is 1/2.

    For the second, the answer is 1/5.

    Thoughts?
    My reply to question 1 is wrong - I misread the support - but my advice remains: Draw a picture and the integral terminals become obvious (provide the question is read properly):

    1. \int_{x = 0}^{1} \int_{y = x/2}^{x} f(x, y) \, dy \, dx.


    My reply to question 2 is also wrong - I misread the equation as the pdf when it is in fact the cdf. You have a number of options for answering the question, including using the given cdf to get the pdf and then integrating the pdf over the region I gave.
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