# Are two Random Variables Independent if their support has a dependency?

This might be a really dumb question, but in a joint PDF of $$X$$ and $$Y$$, $$f_{XY}(x,y)$$, if the support of a random variable $$Y$$ depends on $$X$$, are the two random variables necessarily dependent? For example, if one has $$f_{XY}(x,y)=1/x$$, where $$0, then can one tell that $$X$$ and $$Y$$ are not independent simply by examining the support and not even looking at the $$1/X$$? My intuition suggests they can't be independent.

• An additional thought: I'm starting to think the they can't be independent since we can incorporate the support as indicator variables in the PDF. Is this correct? Commented May 9, 2017 at 1:53
• $y$ shows up in the support -- the entire point of my question. Commented May 9, 2017 at 2:11
• @SmallChess The support of $Y$ is wherever $Y$ has non-zero density. It's standard mathematical terminology. See en.wikipedia.org/wiki/Support_(mathematics)#Formulation. Commented May 9, 2017 at 2:45
• Support is quite standard, indeed. The terminology is used in nearly every statistics textbook I've ever come across. The terminology is the right term and is used in statistics. Commented May 9, 2017 at 3:16

I've determined that if there is a dependency between $X$ and $Y$ in the support of a bivariate pdf, then $X$ and $Y$ cannot be independent. To be sure, there is a Lemma (4.2.7 in Casella and Berger's Statistical Inference, 2d) that states: Let ($X$,$Y$) be a bivariate random vector with joint pdf or pmf $f(x,y)$. Then $X$ and $Y$ are independent random variables if and only if there exists functions $g(x)$ and $h(y)$ such that for every $x$ $\in\mathbb{R}$ and $y$ $\in\mathbb{R}$:
$f(x,y)=g(x)h(y)$
If we incorporate the support (e.g. $0<y<x<1$) as an indicator function in the joint pdf (e.g. ${f_{XY}(x,y)=xy}I_{(0<y<x<1)}$ then the joint PDF cannot be written as a product of only $g(x)$ and only $h(y)$, so $X$ and $Y$ cannot be independent.)