multivariate dependance question

lllll · #1 $Old$ April 11th, 2008, 02:43 PM

is the given function dependent or independent:

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

so far I have:

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

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

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

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

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 $f(x,y)$

mr fantastic · #2 $Old$ April 11th, 2008, 04:57 PM

Quote:

Originally Posted by lllll $View Post$

is the given function dependent or independent:

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

so far I have:

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

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

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

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

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) ....