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Math Help - Bivariate problem need help asap

  1. #1
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    Bivariate problem need help asap

    The joint Pdf of a bivariate Rv xy

    Fxy(X,Y)= {k(xy).............0<y<2x<2
    0...................otherwise

    K a constant

    Here you will first neeb to identify and delineate properly the triangle above which pdf is not zero????

    What does this mean???

    Also find K
    find marginal pdfs of x and y
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  2. #2
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    Quote Originally Posted by kermit View Post
    The joint Pdf of a bivariate Rv xy

    Fxy(X,Y)= {k(xy).............0<y<2x<2
    0...................otherwise

    K a constant

    Here you will first neeb to identify and delineate properly the triangle above which pdf is not zero????

    What does this mean???

    Also find K
    find marginal pdfs of x and y
    It means draw the region defined by 0 < y < 2x < 2. The region is the area bounded by the x-axis and the lines y = 2x and x = 1.

    This question is essentially the same as the one you asked here: http://www.mathhelpforum.com/math-he...lp-please.html. Please review that thread and learn from it.
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  3. #3
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    Can you let me know if this is correct

    <br />
\int_{y=0}^{y=2}\int_{x= \frac{y}{2}}^{x=1}k(xy) \ dx \ dy = 1<br />

    Thanks
    Last edited by mr fantastic; December 23rd 2009 at 02:03 AM.
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  4. #4
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    Quote Originally Posted by kermit View Post
    Can you let me know if this is correct

    <br />
\int_{y=0}^{y=2}\int_{x= \frac{y}{2}}^{x=1}k(xy) \ dx \ dy = 1<br />

    Thanks
    The setup is correct.
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  5. #5
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    <br />
k\int_{0}^{2}\left[\left({\dfrac{x^2y}{2}}\right)\right]_{\frac{y}{2}}^{1} \ dy=1<br />

    <br />
k\int_{0}^{2} \left( \frac{y}{2} - \frac{(y/2)^2 y}{2}\right) \ dy=1<br />

    Please sort out the next piece of LaTeX yourself - CB

    <br />
k[({\dfrac{y}{\cfrac{2}{2}})^2 }-({\dfrac{y}{\cfrac{2}{\cfrac{2}{3}}}})^3 {\dfrac{y}{2}}^2]_{0}^{2}=1<br />

    can anyone tell me if this is integrated correctly. or have i gone wrong somewhere please
    Last edited by CaptainBlack; December 28th 2009 at 08:51 AM. Reason: sort out dreadful LaTeX
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  6. #6
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    Quote Originally Posted by kermit View Post


    <br />
k\int_{0}^{2}\left[\left({\dfrac{x^2y}{2}}\right)\right]_{\frac{y}{2}}^{1} \ dy=1<br />

    <br />
k\int_{0}^{2} \left( \frac{y}{2} - \frac{(y/2)^2 y}{2}\right) \ dy=1<br />

    Please sort out the next piece of LaTeX yourself - CB

    <br />
k[({\dfrac{y}{\cfrac{2}{2}})^2 }-({\dfrac{y}{\cfrac{2}{\cfrac{2}{3}}}})^3 {\dfrac{y}{2}}^2]_{0}^{2}=1<br />

    can anyone tell me if this is integrated correctly. or have i gone wrong somewhere please
    Not likely as no one is going to be able to parse your final expressions without working out the real answer, and then they still won't know if that is what you intended.

    It would also have been nice if you had said what f(x,y) was earlier.


    CB
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  7. #7
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    Quote Originally Posted by kermit View Post


    <br />
k\int_{0}^{2}\left[\left({\dfrac{x^2y}{2}}\right)\right]_{\frac{y}{2}}^{1} \ dy=1<br />

    <br />
k\int_{0}^{2} \left( \frac{y}{2} - \frac{(y/2)^2 y}{2}\right) \ dy=1<br />
<---

    Please sort out the next piece of LaTeX yourself - CB

    <br />
k[({\dfrac{y}{\cfrac{2}{2}})^2 }-({\dfrac{y}{\cfrac{2}{\cfrac{2}{3}}}})^3 {\dfrac{y}{2}}^2]_{0}^{2}=1<br />

    can anyone tell me if this is integrated correctly. or have i gone wrong somewhere please
    Life would be easier if you simplified your expressions before continuing to integrate.

    After simplifying the line I labelled <--- you have k \int_0^2 \frac{y}{2} - \frac{y^3}{4} \, dy = 1 and you shouldn't need re-assurance on how to finish this.
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