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Double Integral With probability

  1. Dec 7, 2013 #1
    $$f(x,y)=
    \begin{cases}
    4(x+y^2),&\text{if x > 0, y > 0, x + y < 1} \\
    0, &\text{elsewhere} \\
    \end{cases}$$

    Find two different integral expressions for P(y > x)

    (1) $$\int_{0} ^{1/2}\int_{0} ^y 4(x+y^2),dx\,dy + \int_{1/2} ^{1}\int_{0} ^{1-y} 4(x+y^2),dx\,dy $$

    (2) $$\int_{0} ^{1/2}\int_{x} ^{1-x} 4(x+y^2),dy\,dx $$
    Are they right?

    Second , find P(Y > 1/2 | X < 1/2). the set up I have is = P(Y > 1/2, X < 1/2)/P(X<1/2)= $$\int_{1/2} ^{1}\int_{0} ^{1-y} 4(x+y^2),dx\,dy / \int_{0} ^{1/2}\int_{0} ^{1-x} 4(x+y^2),dy\,dx $$ is this set up right?

    Third, find P(Y > 1/2 | X > 1/2) I found that the prob should be 0 because there no intersection for Y > 1/2 and X < 1/2 ? Right?
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2

    vela

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    Looks like you switched the roles of x and y in both cases.

    If x + y < 1 and x > y, then y can't be greater than 1/2, right? Yet in (1), you have an integral for 1/2 < y < 1.
     
  4. Dec 7, 2013 #3

    vela

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    Looks good.

    I assuming that was a typo. Yes, your reasoning is correct.
     
  5. Dec 7, 2013 #4
    Thank you for pointing it out
     
  6. Dec 7, 2013 #5

    My typo it should P(y> x). With this prob., my (1) and (2) should be right?
     
  7. Dec 7, 2013 #6

    vela

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    Looks good then.
     
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