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Math Help - Bivariate Rv help please

  1. #1
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    Bivariate Rv help please

    The joint pdf of a bivariate Rv (X,Y) is given by

    Fxy(x,y)={k(x+2y),…………0<x<y<1
    0…………………otherwise



    K a constant

    Find
    1. Obtain value for k
    2. Find marginal pdfs of X and Y
    3. Obtain the conditional density of X given Y=y, What is this density when Y=0.2
    4. Are X and Y independent
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  2. #2
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    Quote Originally Posted by kermit View Post
    The joint pdf of a bivariate Rv (X,Y) is given by

    Fxy(x,y)={k(x+2y),…………0<x<y<1
    0…………………otherwise



    K a constant

    Find
    1. Obtain value for k
    2. Find marginal pdfs of X and Y
    3. Obtain the conditional density of X given Y=y, What is this density when Y=0.2
    4. Are X and Y independent
    The support is the region of the xy-plane enclosed by the lines x = 0, y = 1 and y = x. Did you draw a sketch? So, for example:

    1. Solve \int_{x = 0}^{x = 1} \int_{y=x}^{y = 1} f(x, y) \, dy \, dx = 1 for k.

    The other questions are done using this support and applying the appropriate definitions/formula. If you need more help, please show all your work and say where you are stuck.
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  3. #3
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    thanks for getting back so quick.

    See the attachment, this is how far i got im not sure if its correct can you have a look and let me know if i went wrong.

    thanks

    kermit
    Attached Files Attached Files
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  4. #4
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    Quote Originally Posted by kermit View Post
    thanks for getting back so quick.

    See the attachment, this is how far i got im not sure if its correct can you have a look and let me know if i went wrong.

    thanks

    kermit
    Many people don't open attachments. It's better if you try to learn some basic latex (see my signature) and type your work in the post.
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  5. #5
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    Quote Originally Posted by kermit View Post
    The joint pdf of a bivariate Rv (X,Y) is given by

    Fxy(x,y)={k(x+2y),…………0<x<y<1
    0…………………otherwise



    K a constant

    Find
    1. Obtain value for k
    2. Find marginal pdfs of X and Y
    3. Obtain the conditional density of X given Y=y, What is this density when Y=0.2
    4. Are X and Y independent
    1. i think ur right
    2. For mdf of x, integrate from x to 1 for dy, for mdf of y, integrate from 0 to y for dx
    3. Conditional density of X given Y=y is f(x,y)/f(y) and do same for Y=0.2
    4. X and Y are independent IFF f(xy)=f(x)f(y)--- mdf's
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  6. #6
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    Ok heres how far i have gotten, im just wondering if it is integrated correctly or have i gone wrong anywhere.

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

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

    <br />
k\int_{x=0}^{x=1}(x+1)-(x^2+x^2) \ dx=1<br />

    <br />
k((\dfrac{x^2}{2}+x)-(\dfrac{2x^3}{3}))_{0}^{1} \ dx=1<br />


    <br />
k((\dfrac{1}{2}+1)-(\dfrac{2}{3}))=1<br />
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  7. #7
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    Quote Originally Posted by firebio View Post
    2. For mdf of x, integrate from x to 1 for dy, for mdf of y, integrate from 0 to y for dx
    for the mdf of x i have intergrated from x to 1 for dy. as you mentioned above

    but for the mdf of y i intergrated from 0 to 1?? for dx am i wrong and if so why is it 0 to y could you please explain
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  8. #8
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    Quote Originally Posted by kermit View Post
    Ok heres how far i have gotten, im just wondering if it is integrated correctly or have i gone wrong anywhere.

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

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

    <br />
k\int_{x=0}^{x=1}(x+1)-(x^2+x^2) \ dx=1<br />

    <br />
k((\dfrac{x^2}{2}+x)-(\dfrac{2x^3}{3}))_{0}^{1} \ dx=1<br />


    <br />
k((\dfrac{1}{2}+1)-(\dfrac{2}{3}))=1<br />
    It looks OK.

    Quote Originally Posted by kermit View Post
    for the mdf of x i have intergrated from x to 1 for dy. as you mentioned above

    but for the mdf of y i intergrated from 0 to 1?? for dx am i wrong Mr F says: Yes. Read below.

    and if so why is it 0 to y could you please explain
    In my first reply I asked
    The support is the region of the xy-plane enclosed by the lines x = 0, y = 1 and y = x. Did you draw a sketch?
    Furthermore, you were explicitly told what to do in post #5. Have you been taught how to integrate over a region?
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  9. #9
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    thanks for your help, i have been given a crash coursein it, and cant seem to figure out how to sketch out the region, therfore i have a problem with getting the limits, any help you can give in how to sketvh out this problem would be helpful
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  10. #10
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    Quote Originally Posted by kermit View Post
    thanks for your help, i have been given a crash coursein it, and cant seem to figure out how to sketch out the region, therfore i have a problem with getting the limits, any help you can give in how to sketvh out this problem would be helpful
    Draw the line y = x. Draw the line y = 1. Draw the line x = 1. x < y is satisfied when you're below the line y = x. This gives the region I described in my first reply.
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  11. #11
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    You may want to try to reverse the order of integration.
    The algebra is a bit easier...

    1=k\int_0^1\int_0^y(x+2y)dxdy

    f_X(x)=\int_x^1f(x,y)dy

    f_Y(y)=\int_0^yf(x,y)dx

    The conditional density of X given Y is {f(x,y)\over f_Y(y)} then set y=.2 for the next question.

    X and Y are dependent since the support has x<y, but if you calculate the two marginals

    then you will see that f(x,y)\ne f_X(x)f_Y(y)
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