Comparison of mixing coefficients: $\alpha(n)\le\rho(n)/4$

For a measurable sequence $Y=(Y_i)_{i\ge 0}$ of random elements on a probability space, the strong mixing coefficient is dominated by a quarter of the maximal-correlation coefficient at the same lag:

Here $\alpha(n)$ is the supremum of $|P(A\cap B)-P(A)P(B)|$ over events $A$ in the past $\sigma(Y_0,\dots,Y_k)$ and $B$ in the future $\sigma(Y_{k+n},Y_{k+n+1},\dots)$, and $\rho(n)$ is the supremum of $|\mathrm{corr}(U,V)|$ over square-integrable $U,V$ measurable with respect to those same two $\sigma$-algebras.

The inequality is one of the standard comparisons between the classical mixing coefficients, and is what allows a $\rho$-mixing hypothesis to be fed into results stated for strongly mixing sequences: summability, or mere convergence to zero, of $\rho(n)$ implies the same for $\alpha(n)$.

The proof tests the supremum defining $\rho(n)$ against the indicator variables $U=\mathbf 1_A$, $V=\mathbf 1_B$. Their covariance is exactly $P(A\cap B)-P(A)P(B)$, and their variances are $P(A)(1-P(A))\le\tfrac14$ and $P(B)(1-P(B))\le\tfrac14$, so the correlation quotient contributes the factor $\tfrac14$. If one of the two variances vanishes, then the corresponding event has probability $0$ or $1$ and the covariance vanishes outright.

Formalization Note The measurability hypothesis on $Y$ is what makes events of the past and future $\sigma$-algebras measurable in the ambient space, which is needed for the indicator variables to be square-integrable test functions.