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 $0<y<x<1$, 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.

  • $\begingroup$ 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? $\endgroup$ Commented May 9, 2017 at 1:53
  • $\begingroup$ $y$ shows up in the support -- the entire point of my question. $\endgroup$ Commented May 9, 2017 at 2:11
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    $\begingroup$ @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. $\endgroup$ Commented May 9, 2017 at 2:45
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    $\begingroup$ 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. $\endgroup$ Commented May 9, 2017 at 3:16

1 Answer 1


I've convinced myself of the answer, so I'm answering my own question.

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 $y$ $\in\mathbb{R}$:


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.)


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