1. ## Independent variables

I know and can even prove that two continuous random variables X and Y are independent iff their joint density function f(x,y) can be written as f(x,y) = h(x)g(y).

Maybe I should carefully look at my proof of this but I do not see why the following two RV are not independent by this theorem.

f(x,y) = 4/9 if 0<= x < y < 3-x and 0 otherwise.

I am thinking of 4/9 as (4/9)*1 = f(x)g(y). That is, f(x) = 4/9 and g(y)=1.

So I conclude that X and Y are independent but this is wrong since for example f(1,2) = 4/9 while f_x(1) = 4/9 and f_y(2) = 4/9.

What am I missing in this theorem??????????

2. ## Re: Independent variables

Originally Posted by JaguarXJS
I know and can even prove that two continuous random variables X and Y are independent iff their joint density function f(x,y) can be written as f(x,y) = h(x)g(y).

Maybe I should carefully look at my proof of this but I do not see why the following two RV are not independent by this theorem.

f(x,y) = 4/9 if 0<= x < y < 3-x and 0 otherwise.

I am thinking of 4/9 as (4/9)*1 = f(x)g(y). That is, f(x) = 4/9 and g(y)=1.

So I conclude that X and Y are independent but this is wrong since for example f(1,2) = 4/9 while f_x(1) = 4/9 and f_y(2) = 4/9.

What am I missing in this theorem??????????
The support of $y$ depends on the value of $x$.

Clearly the joint distribution is not the product of two independent marginals.

Independent marginals have to have independent ranges of support, i.e. $h(x), ~a\leq x \leq b,~~g(y),~c\leq y \leq d$

3. ## Re: Independent variables

Originally Posted by romsek
The support of $y$ depends on the value of $x$.

Clearly the joint distribution is not the product of two independent marginals.

Independent marginals have to have independent ranges of support, i.e. $h(x), ~a\leq x \leq b,~~g(y),~c\leq y \leq d$
Professor, thanks for your input. I understand what you said and appreciate your time spent reading and responding to my post.