If there are 3 vectors X, Y, Z of the same length, for any $x_i \in X,y_i \in Y,z_i \in Z$, we have $0<x_i<1,0<y_i<1,0<z_i<1$.
The correlation between Z, Y is greater than between X, Y. The inequality could be formulated as $\rho(Z, Y)>\rho(X, Y)$.
I want to ask if the inequality exists: $\rho(XZ, Y)>\rho(X, Y)$. If yes, please demonstrate it. If not, please give an example.
If you couldn't demonstrate it, please give me some clues or references.
The inequality does not always hold. Let $D \sim$ Bernoulli$(p)$, independent of $Y$. Let $Y'$ and $Y''$ be two independent copies of $Y$ (also independent of $Y$). Let $X=D Y+(1-D) Y'$ and $Z=DY'' + (1-D)Y$. Define also $m_k=E(Y^k)$ for $k=1,2$. Then it is easy to check that Cov$(X,Y) = p V(Y)$ and Cov$(Z,Y) = (1-p) V(Y)$. Because $X$ and $Z$ have the same distribution as $Y$, $$\rho(X,Y)=p < 1- p = \rho(Z,Y)$$ provided that $p<1/2$. Moreover, $XZ=Y Y'''$ where $Y'''$ is an independent copy of $Y$. Thus, Cov$(XZ,Y) = m_1 V(Y)$ and $V(XZ)=m_2^2 - m_1^4$. Hence, $$\rho(XZ,Y) = m_1 \sqrt{\frac{m_2 - m_1^2}{m_2^2 - m_1^4}} = \frac{m_1}{\sqrt{m_2 + m_1^2}}.$$ Consider a distribution of $Y$ such that $m_2> 3 m_1^2$. Such a distribution exists if $m_1 < 1/\sqrt{3}$ since there exist distributions with support included in $(0,1)$ such that $m_1 > m_2 \geq m_1^2$. For such a distribution, $\rho(XZ,Y)<1/2$ and any $p \in \left(m_1/\sqrt{m_2 + m_1^2}, 1/2\right)$ satisfies $\rho(X,Y) < \rho(Z,Y)$ and $\rho(XZ,Y) < \rho(X,Y)$.
