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?

Any comments/answers will very be appreciated.

For those who may argue that the discrete rv does not have its pdf, the pdf of a discrete rv can be defined via the Dirac-delta function.


1 Answer 1


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.

  • $\begingroup$ Great answer! Thanks! $\endgroup$ Sep 9, 2018 at 1:56
  • $\begingroup$ What can we say if $V$ is another uniform distribution on the same underlying probability space but is independent of $U$? $\endgroup$
    – FShrike
    Mar 14, 2023 at 22:38
  • $\begingroup$ @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. $\endgroup$ Mar 15, 2023 at 1:37

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