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Math Help - multivariate dependance question

  1. #1
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    multivariate dependance question

    is the given function dependent or independent:

     f(x,y) = \left\{ \begin{array}{rcl}<br />
3x & \mbox{for} & 0 \leq y \leq x \leq 1\\ <br />
0 & \mbox{for} & \mbox{other}2<br />
\end{array}\right.

    so far I have:

     f(x) = \int^{x}_{0} 3x \ dy \  \ y \leq x \leq 1

     f(x) = \left\{ \begin{array}{rcl}<br />
3x^2 & \mbox{for} &  y \leq x \leq 1\\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

     f(y) = \int^{1}_{y} 3x \ dy \  \ 0 \leq y \leq x


     f(y) = \left\{ \begin{array}{rcl}<br />
3x-\frac{3}{2}y^2 & \mbox{for} &  y \leq x \leq 1\\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    now for f(x,y)

     \int_{0}^{1} \int_{0}^{x} 3x \ dy \ dx= \int^{1}_{0} 3xy \bigg{|}^{x}_{0} \ dx = \int^{1}_{0} 3x^2 \ dx = 3

    now to test independence to I just multiply f(x)f(y) and very if it equals the the integrated form of  f(x,y)
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  2. #2
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    Quote Originally Posted by lllll View Post
    is the given function dependent or independent:

     f(x,y) = \left\{ \begin{array}{rcl}<br />
3x & \mbox{for} & 0 \leq y \leq x \leq 1\\ <br />
0 & \mbox{for} & \mbox{other}2<br />
\end{array}\right.

    so far I have:

     f(x) = \int^{x}_{0} 3x \ dy \ \ y \leq x \leq 1

     f(x) = \left\{ \begin{array}{rcl}<br />
3x^2 & \mbox{for} & y \leq x \leq 1\\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

     f(y) = \int^{1}_{y} 3x \ dy \ \ 0 \leq y \leq x Mr F says: This is not right. {\color{red} f(y) = \int^{x=1}_{x=y} 3x \, {\color{blue}dx}}.Then {\color{red} f(y) = \frac{3 x^2}{2} \bigg{|}_y^1 = \frac{3}{2} - \frac{3 y^2}{2} = \frac{3}{2} (1 - y^2)} for {\color{red}0 \leq y \leq x}.


     f(y) = \left\{ \begin{array}{rcl}<br />
3x-\frac{3}{2}y^2 & \mbox{for} & y \leq x \leq 1\\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    now for f(x,y)

     \int_{0}^{1} \int_{0}^{x} 3x \ dy \ dx= \int^{1}_{0} 3xy \bigg{|}^{x}_{0} \ dx = \int^{1}_{0} 3x^2 \ dx = 3 Mr F says: This is a worry since it should equal 1 if f(x, y) is a valid pdf! Luckily it does: {\color{red}\int^{1}_{0} 3x^2 \ dx = x^3 \bigg{|}_{0}^{1} = 1}. Note: This calculation is NOT relevant for testing independence, but it does confirm that f(x,y) is a valid pdf.

    now to test independence to I just multiply f(x)f(y) and very if it equals the the integrated form of  f(x,y)
    To test independence, you look to see if f(x, y) = f(x) f(y) ....
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