# Existence of Joint probability density distribution

I am trying to understand how the conditional probability is defined. Along the way, I read the following statement from [page 145, Sec4.3] : let $X$ and $Y$ be random variables.

We now consider this question in the case in which $X$ and $Y$ have a joint density $f$.

It made me think that two random variables whose joint density function does not exist. As far as I know, the probability density always exists by the Radon-Nikodym theorem.

So here are two questions:

• Does the probability density always exist by Radon-Nikodym theorem?
• If not, can I have an easy example of two random variables does not have its joint pdf?

Let $U$ be uniform on $[0,1]$; it has a density. Now let $V=U$. The pair $(U,V)$ is supported on the diagonal of the unit square, and does not have a joint density function.
Your instinct to use the Radon-Nikdym theorem is good, but that theorem has hypotheses that need checking. In this case the joint probability measure of the diagonal is $1$ but its 2-dimensional Lebesgue measure is $0$. The joint measure is not absolutely continuous w.r.t. Lebesgue measure, so the R-N plan cannot work.
• What can we say if $V$ is another uniform distribution on the same underlying probability space but is independent of $U$? Mar 14, 2023 at 22:38
• @FShrike If $U$ and $V$ are both $U[0,1]$ and independent, their joint distribution if uniform on the square $[0,1]^2$ and have an uninteresting density function. I don't see what this has to do with the original problem statement, though. Mar 15, 2023 at 1:37