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Math Help - Bounds for integration in joint pdf

  1. #1
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    Bounds for integration in joint pdf

    Hi,

    Let f(x,y) = 3/2 if 0<=x<=1 and x^2<=y<=1; otherwise f(x,y)=0

    Find

    P(X=1/4, 0<y<3)

    So I know I'm suppose to do double integration. My bounds for x are from 0 to 1/4. The less than sign on y is confusing me. Should I take the integration of y from 0+x to 3-x?

    Thanks.
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  2. #2
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    Quote Originally Posted by shogunhd View Post
    Hi,

    Let f(x,y) = 3/2 if 0<=x<=1 and x^2<=y<=1; otherwise f(x,y)=0

    Find

    P(X=1/4, 0<y<3)

    So I know I'm suppose to do double integration. My bounds for x are from 0 to 1/4. The less than sign on y is confusing me. Should I take the integration of y from 0+x to 3-x?

    Thanks.
    \Pr \left( X = \frac{1}{4}, 0 < Y < 3 \right) = \Pr \left( 0 < Y < 3 \bigg{|} X = \frac{1}{4} \right) \Pr \left( X = \frac{1}{4} \right).

    But since you have a continuous distribution, \Pr \left( X = \frac{1}{4} \right) = 0 ......

    So, do you mean find \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) \, ?

    If so:

    First draw the region of integration. Then:


    \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) = \int_{0}^{1/4} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    \Pr \left( X = \frac{1}{4}, 0 < Y < 3 \right) = \Pr \left( 0 < Y < 3 \bigg{|} X = \frac{1}{4} \right) \Pr \left( X = \frac{1}{4} \right).

    But since you have a continuous distribution, \Pr \left( X = \frac{1}{4} \right) = 0 ......

    So, do you mean find \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) \, ?

    If so:

    First draw the region of integration. Then:


    \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) = \int_{0}^{1/4} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....
    Note:

    \Pr \left( X \geq \frac{1}{4}, 0 < y < 3 \right) = \int_{1/4}^{1} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Note:

    \Pr \left( X \geq \frac{1}{4}, 0 < y < 3 \right) = \int_{1/4}^{1} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....

    the question asks for P(x=1/4, 0<y<3). I initially performed the integration using the first double integral you supplied. I was confused about the y bounds because I know that in a continuous distribution "<" sign can mean <=. Why did you change X<=1/4 to x=>1/4?
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  5. #5
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    Quote Originally Posted by shogunhd View Post
    the question asks for P(x=1/4, 0<y<3). Mr F says: Two things then:

    1. Why did you have "My bounds for x are from 0 to 1/4" then?
    2. I have shown why P(x=1/4, 0<y<3) = 0.

    I initially performed the integration using the first double integral you supplied. I was confused about the y bounds because I know that in a continuous distribution "<" sign can mean <=. Why did you change X<=1/4 to x=>1/4?
    Because I thought you had misposted due to the fact that if it wasn't, then the answer is zero and there's no itegrtaion to do at all ......
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