4
$\begingroup$

For two measures $\mu, \nu$ on the same space say that $\mu$ is absolutely continuous with respect to $\nu$ ($\mu \ll \nu$) whenever $\nu(A)=0$ implies that $\mu(A)=0$ too.

Let $(\Omega, \mathsf P$) be a probability space and let $S$ be a separable metric space. For an $S$-valued random variable $W$ on $\Omega$, denote by $\mathsf P_W$ the distribution of $W$ on $S$ given by $\mathsf P_W(A)=\mathsf P(W^{-1}(A))$.

Consider four random variables $X, X_1, Y, Y_1\colon \Omega\to S$ such that

  1. $X_1, Y_1$ are independent,
  2. $\mathsf{P}_X \ll \mathsf{P}_{X_1}$,
  3. $\mathsf{P}_Y \ll \mathsf{P}_{Y_1}$.

I am looking for a reference to the statement that $$\mathsf{P}_{(X,Y)}\ll \mathsf{P}_{(X_1,Y_1)}.$$ The case where $S=\mathbb R$ would be fine.

Not sure where to start looking. Thank you for your time.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Did you also want $X,Y$ to be independent? As Iosif points out below, it's trivially false the way it's written. $\endgroup$
    – Nate Eldredge
    Commented Jan 19, 2018 at 14:04
  • $\begingroup$ If $X,Y$ are independent, it should be obvious from Radon-Nikodym, since the density of the pair would be the product of the densities. $\endgroup$
    – Nate Eldredge
    Commented Jan 19, 2018 at 14:07
  • 1
    $\begingroup$ Oops, yes :) will update it. Thank you. $\endgroup$
    – Tomasz Kania
    Commented Jan 19, 2018 at 14:09
  • 1
    $\begingroup$ math.stackexchange.com/questions/1009416/… points out an even easier proof. Suppose $P_{(X_1, Y_1)}(A) = 0$. By Fubini, we have that $P_{X_1}$-a.e. section of $A$ is $P_{Y_1}$-null. Absolute continuity implies that $P_X$-a.e. section is $P_Y$-null, so Fubini again gives $P_{(X,Y)}(A) =0$. I think this is the kind of statement that doesn't really need a reference. $\endgroup$
    – Nate Eldredge
    Commented Jan 19, 2018 at 15:28

1 Answer 1

Reset to default
5
$\begingroup$

This statement is false. Example: $Y=X=X_1=U$ and $Y_1=V$, where (say) $U$ and $V$ are independent standard normal r.v.'s.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .