Suppose we have a continuous-time stochastic process $s(t)$ that's mixing. What properties would a function $f$ have to have so that $f(s(t))$ would also be mixing?
I'm sure this is a well-known thing, I'm just having trouble locating a term for such functions or any relevant literature.
$\newcommand{\R}{\mathbb R}$ $\newcommand{\P}{\mathsf P}$ $\newcommand{\F}{\mathscr F}$ $\newcommand{\S}{\mathscr S}$ $\newcommand{\T}{\mathscr T}$ $\newcommand{\Si}{\Sigma}$ Let $X:=(X_t)_{t\in\R}$ be a stochastic process in a state space $S$, endowed by a sigma-algebra $\S$, so that, for each $t\in\R$, $X_t$ is a random variable (r.v.) on a probability space $(\Omega,\F,\P)$ with values in $S$. The strong mixing coefficient for $X$ is $$a_X(s):=\sup\{|\P(A\cap B)-\P(A)\P(B)|\colon t\in\R,A\in X_{-\infty}^t,B\in X_{t+s}^\infty\}, $$ where $s>0$ and $X_u^v$ is the sigma-algebra generated by the set $\{X_t\colon t\in\R, u\le t\le v\}$ of r.v.'s. Assume that the process $X$ is strongly mixing, i.e., $a_X(s)\to0$ as $s\to\infty$.
Let $(T,\T)$ be any measurable space. For each $t\in\R$, let $Y_t:=f(X_t):=f\circ X_t$, where $f\colon S\to T$ is a measurable function.
Then $Y:=(Y_t)_{t\in\R}$ is a strongly mixing stochastic process in the state space $T$. This follows immediately, because $Y_u^v\subseteq X_u^v$ for all $u$ and $v$ in $[-\infty,\infty]$ and hence $0\le a_Y(s)\le a_X(s)$ for all real $s>0$.
Thus, the strong mixing is preserved under any measurable transformation $f$.
