$\\alpha(n) \\le \\varphi(n)$: strong mixing is dominated by uniform mixing

For a strictly stationary or arbitrary sequence $(Y_i)_{i\ge0}$ of random variables on a probability space, the strong (α-) mixing coefficient is dominated by the uniform (φ-) mixing coefficient at every lag:

Recall the two definitions. Writing $\mathcal A_k = \sigma(Y_0,\dots,Y_k)$ and $\mathcal B_{k+n} = \sigma(Y_{k+n}, Y_{k+n+1},\dots)$,

The comparison is immediate from the identity $|P(A\cap B) - P(A)P(B)| = P(A)\,|P(B \mid A) - P(B)|$ together with $P(A) \le 1$; the case $P(A) = 0$, which is excluded from the definition of $\varphi$, contributes the value $0$ and is covered by the nonnegativity of $\varphi$ (its defining set contains $0$, taking $A = B = \Omega$). Since both coefficients are defined as suprema, the argument also uses that the set defining $\varphi$ is bounded above by $1$, so that each of its elements is at most $\varphi(n)$.

This is the elementary end of the standard chain of comparisons between mixing coefficients, $2\alpha \le \beta \le \varphi$, and it lets a uniform-mixing hypothesis be fed into any result stated for strongly mixing sequences.