1
$\begingroup$

Let $Z=(X,Y) : \Omega\rightarrow\mathbb{R}^2$ be a Borel-measurable random vector and $U\subset\mathbb{R}$ be open. Suppose that $Z$ is absolutely continuous with Lebesgue density $\zeta$.

I was wondering about the following question: If we assume that $\zeta$ factorizes on the square $U\times U$, i.e. that the restriction

$$\tag{1}q:=\left.\zeta\right|_{U\times U} \quad\text{is such that}\quad q\equiv q(x,y)=q_1(x)q_2(y) \quad\text{for some } \ q_1, q_2 : U\rightarrow\mathbb{R},$$ does this imply that $X$ and $Y$ are independent on $U$ in the sense that

$$\tag{2}\mathbb{P}(X^{-1}(A)\cap Y^{-1}(B)) = \mathbb{P}_X(A)\cdot\mathbb{P}_Y(B)\quad\forall\, A, B\subseteq U \ (\mathrm{Borel})?$$

Clearly this holds for both the special cases $U=\mathbb{R}$ and $q_i = \zeta_i$ ($i=1,2$, with $\zeta_i$ the marginal densities of $\zeta$), however I couldn't show it in the above generality; do you have a counterexample?

$\endgroup$

2 Answers 2

1
$\begingroup$

The answer is no. E.g., let $U=(0,1)$ and $$\zeta=\frac12\times 1_{(0,1)^2}+\frac12\times 1_{(1,2)^2};$$ that is, the joint distribution of $(X,Y)$ is the half-and-half mixture of the uniform distributions on the the squares $(0,1)^2$ and $(1,2)^2$. Then $$P_{X,Y}(U\times U)=\frac12\ne\frac12\times\frac12=P_X(U)P_Y(U).$$

$\endgroup$
0
1
$\begingroup$

No. For example suppose $Z=(X,Y)$ distributes uniformly in the triangle $\{(x,y):x,y\geq 0, x+y\leq 1\}$, and $U=(0,\frac 1 2)$. Let $A=(0,\frac 1 2)$ and $B=(0,\frac 1 {10})$. Then $P(Z\in A\times B)=\frac 1 {10}$, and $P(X\in A)P(Y\in B)=\frac 3 4 \times \frac{19}{100}$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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