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.


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.

| cite | improve this answer | |
  • $\begingroup$ Great answer! Thanks! $\endgroup$ – induction601 Sep 9 '18 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.