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Math Help - Joint pdf Problem

  1. #1
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    Joint pdf Problem

    Hi,

    I have been given a joint pdf for 2 random variables:

    f(x,y) = k - y , 0 < y <  x^2 < 1

    and am trying to find the value of the constant k. I was just wondering if anyone could tell me whether i have set up the integral correctly. This is how i have set it up:

     <br />
\int_0^1 \int_0^{x^2} k-y \ dy dx \ = 1<br />

    any help would be really apprectiated
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  2. #2
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    Quote Originally Posted by Number Cruncher 20 View Post
    Hi,

    I have been given a joint pdf for 2 random variables:

    f(x,y) = k - y , 0 < y <  x^2 < 1

    and am trying to find the value of the constant k. I was just wondering if anyone could tell me whether i have set up the integral correctly. This is how i have set it up:

     <br />
\int_0^1 \int_0^{x^2} k-y \ dy dx \ = 1<br />

    any help would be really apprectiated
    Your set up is correct.
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  3. #3
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    Thanks for the help!
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  4. #4
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    Quote Originally Posted by Number Cruncher 20 View Post
    Hi,

    I have been given a joint pdf for 2 random variables:

    f(x,y) = k - y , 0 < y <  x^2 < 1

    and am trying to find the value of the constant k. I was just wondering if anyone could tell me whether i have set up the integral correctly. This is how i have set it up:

     <br />
{\color{red}2}\,  \int_0^1 \int_0^{x^2} k-y \ dy dx \ = 1<br />

    any help would be really apprectiated
    My mistake .... the red 2 should be there.
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  5. #5
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    Thanks for the help. I think ive got it now.
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  6. #6
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    Post the double integral

    Hello. Could you explain please why the red 2 should be there? Could you post the calculation of the integral? Thanks in advance
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  7. #7
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    Quote Originally Posted by dely84 View Post
    Hello. Could you explain please why the red 2 should be there? Could you post the calculation of the integral? Thanks in advance
    Yeah, I didn't think the 2 was needed, but rather what you did in the first place.
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  8. #8
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    Quote Originally Posted by dely84 View Post
    Hello. Could you explain please why the red 2 should be there? [snip]
    Quote Originally Posted by Richnfg View Post
    Yeah, I didn't think the 2 was needed, but rather what you did in the first place.
    0< x^2 < 1 \Rightarrow -1 < x < 1. Therefore the region of the xy-plane for which the pdf is non-zero is the region bounded by x = -1, x = 1 and y = x^2.

    Quote Originally Posted by dely84 View Post
    [snip]Could you post the calculation of the integral? Thanks in advance
    The integration is routine. What part can't you do?
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  9. #9
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    Ah, I see!
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