# Thread: how to tell whether a function is dependent or independent problem

1. ## how to tell whether a function is dependent or independent problem

In the green box in the attachment I have calculated the integral of the pdf of 24xy in the same manner as in the theory that leads up to 3.1 in the attachment. But I cant see any difference in that calculation that would explain that if 24xy is dependent why is not also the theory that leads up to 3.1 dependent. In the green box I would believe that $\frac{1}{2}(1-y)^2$ is $F_X(a-y)$ for a=1. I dont see any distinctions. Why is 24xy dependent while the theory up until 3.1 is independent?

The theory is taken from Ross, A first course in probability.

Tape image twice to read it.

2. ## Re: how to tell whether a function is dependent or independent problem

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

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

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

$f_X(x) \cdot f_Y(y) \neq f_{X,Y}(x,y)$ 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

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

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

$f_X(x) \cdot f_Y(y) = f_{X,Y}(x,y)$ 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