Relation of mixing sequences (Phi, Rho) - Is there some inequality?

Question

Imagine I have the two mixing coefficients defined the following:

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $\sigma_1$ and $\sigma_2$ two sub $\sigma$-algebras of $\mathcal{F}$. Then we can define the the mixing coefficients:

$\phi(\sigma_1,\sigma_2):= \sup\limits_{B \in \sigma_1, A \in \sigma_2, \mathbb{P}(B)>0}\bigg \vert \mathbb{P}(A \mid B)-\mathbb{P}(A)\bigg \vert$

and

$\rho(\sigma_1,\sigma_2):= \sup\limits_{B \in \sigma_1, A \in \sigma_2}\bigg \vert\frac{\text{Cov}(A,B)}{(\text{Var}(A)\text{Var}(B))^{0.5}} \bigg \vert$.

Now I'm curious if there is some inequality, e.g. $\rho(\sigma_1,\sigma_2)\leq\phi(\sigma_1,\sigma_2)$ - but I don't see a way proving it?

Anyone here to help?

Answer 1

In the book: P. Billingsley, Convergence of Probability Measures, 2ed.(1999), p.260, App. M23(46), you could find the following inequality $\rho(\sigma_1,\sigma_2)\le 2\sqrt{\phi(\sigma_1,\sigma_2)}$.