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Assume that we have a probability space $(\Omega, \mathcal{A}, P)$, and we have two random variables $X,Y: \Omega \rightarrow \mathbb{R}$. On this space.

We can define two measures $\mu_X(B)=P(X^{-1}(B)), \mu_Y(B)=P(Y^{-1}(B)), B \in \mathcal{B}(\mathbb{R})$.

If $\mu_X$ and $\mu_{Y}$ is absolutely continuous with respect to the Lebesgue measure we get that $\mu_X(B)=\int_Bf_Xd\lambda,\mu_Y(B)=\int_Bf_Xd\lambda$, where $\lambda$ is the Lebesgue measure. f is called the probability density. (This follows from the Radon-Nikodym Theorem).

We also have that $\mu_{XY}(B_2)=P((X,Y)\in B_2), B_2\in \mathcal{B}(\mathbb{R}^2)$.

My question is this:

If we know that $\mu_X, \mu_Y$ is absolutely continuous with the respect measure, can we then deduce that $\mu_{XY}$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}^2$, and then there exists a probability density function such that $\mu_{XY}(B_2)=\int_{B_2}f_{XY}(r,t)d(\lambda\times \lambda)?$

I am not able to prove this. I am able to prove the converse that if $\mu_{XY}$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R^2}$ we also have absolute continuity in separate variables, is this correct? Could you please tell me which directions are correct, and which do not hold?

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    $\begingroup$ No we cannot, try X=Y uniform on (0,1). $\endgroup$ – Did Jan 21 '16 at 17:49
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    $\begingroup$ @Did Thanks, I got that $\mu_X=\int_B 1 d\lambda, B \in \mathcal{B}(0,1)$. But $\mu_{XY}(B_2)=P((X,Y)\in B_2)=P((X,X)\in B_2), B_2 \in \mathcal{B}((0,1)\times (0,1)).$ This measure is not absolutely continuous with respect to the lebesgue measure by considering the set $\{(x,x): x \in (0,1)\}$, since this set has lebesgue measure 0, but our probability measure has value 1 on this set. But do you agree that the converse holds. That if the joint distribution is absoltely continous, then each distribution function is? $\endgroup$ – user119615 Jan 21 '16 at 17:58
  • $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$ – Did Jan 21 '16 at 18:04
  • $\begingroup$ @Did Thank you for the help. $\endgroup$ – user119615 Jan 21 '16 at 18:04

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