Originally Posted by

**abender** Have you tried computing the marginal density functions from each joint density function?

Idea: $X$ and $Y$ are independent IFF we can decompose the joint density function into the product of a function of $X$ only and a function of $Y$ only, for the entire range of $x$ and $y$, as given in the joint pdf.

A definition of independence is that $X $ and $Y $ are independent continuous random variables if the product of the marginal pdf of $X $ and the marginal pdf of $Y$ equals the joint PDF of $X$ and $Y$ for all $x,y$. In mathematics, a definition is equivalent to an IFF (if and only if) statement! So we can use this idea to show whether or not $X$ and $Y$ are independent in our two situations.

__Situation 1:__

$f_{X,Y}(x,y)=24xy, 0<x<1, 0<y<1, x+y<1$

$f_Y(x)=\int^{y=1-x}_{y=0} 24xy dy = 24x(\tfrac{y^2}{2})|^{y=1-x}_{y=0} = 12x(1-x)^2$

$f_Y(y)=\int^{x=1-y}_{x=0} 24xy dx = 24y(\tfrac{x^2}{2})|^{x=1-y}_{x=0} = 12y(1-y)^2$

$f_X(x) \cdot f_Y(y) \neq f_{X,Y}(x,y)$ because $12x(1-x)^2\cdot 12y(1-y)^2 \neq 24xy$ for all $x,y$. Therefore, $X$ and $Y$ here are not independent.

__Situation 2:__

$f_{X,Y}(x,y)=6e^{-2x}e^{-3y}, 0<x<\infty, 0<x<\infty$

$f_{X}(x)=6e^{-2x}\int^{y=\infty}_{y=0}e^{-3y} dy= 6e^{-2x}\left(\tfrac{e^{-3y}}{-3}\right)|^{y=\infty}_{y=0} = 6e^{-2x}\left(\tfrac{e^{-3y}}{3}\right)|^{y=0}_{y=\infty} = 2e^{-2x}$

$f_{Y}(y)=6e^{-3y}\int^{x=\infty}_{x=0}e^{-2x} dx = 6e^{-3y}\left(\tfrac{e^{-2x}}{-2}\right)|^{x=\infty}_{x=0} = 6e^{-3y}\left(\tfrac{e^{-2x}}{2}\right)|^{x=0}_{x=\infty} = 3e^{-3y}$

$f_X(x) \cdot f_Y(y) = f_{X,Y}(x,y)$ because $2e^{-2x}\cdot 3e^{-3y} =6e^{-2x}e^{-3y}$. Therefore, $X$ and $Y$ here are independent.

I hope this helps. Best of luck.

-Andy