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Math Help - Joint PDF Probability

  1. #1
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    Joint PDF Probability

    The joint PDF is:
    f(x,y)= c(x^2 + y^2) for o<_ y <_ 1-x^2
    0 otherwise

    I found c to be 5/4 which I know is correct from the back of the book.
    Now I need to find Pr(Y<_ x+1)
    I know the answer is 13/16 but I can't figure out how. I am doing a double integral...x from -1 to 1 and y from 0 to x+1 of 5/4 (x^2 + y). But this is not giving me 13/16
    Anyone know where my error is? Help! Thanks!
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  2. #2
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    Quote Originally Posted by andyaddition View Post
    The joint PDF is:
    f(x,y)= c(x^2 + y^2) for o<_ y <_ 1-x^2
    0 otherwise

    I found c to be 5/4 which I know is correct from the back of the book.
    Now I need to find Pr(Y<_ x+1)
    I know the answer is 13/16 but I can't figure out how. I am doing a double integral...x from -1 to 1 and y from 0 to x+1 of 5/4 (x^2 + y). But this is not giving me 13/16
    Anyone know where my error is? Help! Thanks!
    Your support "o<_ y <_ 1-x^2" says nothing about values of x .... I assume -1 \leq x \leq 1 ....?

    Using a clearly labelled diagram, I have \displaystyle \frac{5}{4} \int_{y = 0}^{y = 1} \int_{x = y - 1}^{x = +\sqrt{1 - y}} x^2 + y^2 \, dx \, dy (but I don't get 13/16).
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