Let $X: (\Omega, \mathcal A) \to (\mathbb R, \mathcal B)$ be a random variable on the probability space $(\Omega, \mathcal A, P)$ and define its distribution as the probability measure $P_X(B) = P(X \in B)$ on $\mathcal B$. ($\mathcal B$ is the Borel sigma algebra).

A random variable is absolutely continuous with respect to a measure $\mu$ if its distribution is, i.e. $P_X(B)=0$ for all $B \in \mathcal B$ with $\mu(B) = 0.$ $X$ may not be absolutely continuous with respect to the Lebesgue measure $\lambda$ but it is always absolutely continuous with respect to $P_X$.

Now let $(X,Y): (\Omega, \mathcal A) \to (\mathbb R^2, \mathcal B^2)$ be a random vector with distribution $P_{X, Y}((X, Y) \in B)$, $B \in \mathcal B^2$. By the same reasoning as before, $(X,Y)$ is abolutely continous with respect to $P_{X, Y}$ even if it is not absolutely continuous with respect to $\lambda^2$.

Question: Will $(X, Y)$ always be absolutely continuous with respect to the product measure $P_X \otimes P_Y$?

What I did:

We need to verify that $P_{X, Y}(B) = 0$ whenever $(P_X \otimes P_Y)(B)=0$. If $B = B_1 \times B_2$ is a rectangular set then this is clearly true because $(P_X \otimes P_Y)(B)=0$ implies $P_X(B_1) = 0$ or $P_Y(B_2)=0$ ($P_X(B_1) = 0$, say) and then $$P_{X, Y}(B) = P((X, Y) \in B_1 \times B_2) = P(X \in B_1, Y \in B_2) \le P(X \in B_1) = 0.$$

But for non-rectangular sets I'm not sure how to proceed.

  • $\begingroup$ What is the difference between $P_X$ and $\mu$? Which one is the pushforward measure? How is the other one defined? $\endgroup$ Jul 10, 2022 at 1:22

1 Answer 1


This is untrue in general. The problem is that $P_X \otimes P_Y$ is the law of $(\tilde{X}, \tilde{Y})$ where $\tilde{X}$ has the same distribution as $X$ and similarly for $\tilde{Y}$ but $\tilde{X}$ is independent from $\tilde{Y}$. $X$ and $Y$ need not be independent which will allow us to create a counterexample.

For example, let $X$ be a standard one dimensional Gaussian and consider the case $X = Y$. Then $P_{(X,X)}$ is a measure on $\mathbb{R}^2$ such that if $\Delta = \{(x,x) \in \mathbb{R}^2: x \in \mathbb{R}\}$ then $P_{(X,X)}(\Delta) = 1$. However, $P_X \otimes P_X$ is the law of the standard two dimensional Gaussian and in particular $P_X \otimes P_X(\Delta) = 0$.


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