What can be deduced about the covariance between f(x) and f(x + K) for constant K? The question concerns transformations of a random variable $X$ by a function $f$. I’d like to understand whether there are conditions that could make:
$$\operatorname{covar}(f(X), f(X + K) ) \geq \operatorname{covar}(f(X),f(X)) = \operatorname{var}(f(X)).$$
In other words, is it possible that the covariance of a function of $X$ with a shifted version of itself can be greater than the covariance of the function with itself? If so, when exactly?
Assumptions
$X$ is a real random variable with unknown distribution
$f \colon \Re \rightarrow \Re$ is a continuous, bounded, differentiable function (perhaps also positive, monotonically increasing if that simplifies things)
$K \neq 0$ is a real constant
I suspect there are particular distributions of $X$, properties of $f$, and/or values of $K$ that might make the above inequality true, but I'm not sure how to start.
I’ve tried applying various Gruss-type inequalities but it’s not clear to me how they might help. The only starting point I'm aware of is that if $\operatorname{covar}(f(X), f(X + K) ) \geq \operatorname{covar}(f(X),f(X))$, then $\operatorname{var}(f(X+K)) \geq \operatorname{var}(f(X))$.
I'd appreciate any suggestions on how to approach this problem.
 A: Because $f$ may have very different behavior on values of the form $X+K$ than on values of $X,$ there is no general relationship of this kind that will hold, nor is there any simple alternative way to characterize it.
The following simple example uncovers the essence of the situation.
For any real number $\lambda \gt 1$ define a smooth, invertible transformation $f:\mathbb{R}\to\mathbb{R}$ by setting $f(x)=x$ for $x\le 1,$ $f(x) = \lambda x$ for $x\ge 2,$ and connecting these two partial functions smoothly on the segment $(1, 2).$
Let $X$ be a random variable with a uniform distribution on $[0,1].$
Since $X\le 1$ almost surely, $f(X)$ and $X$ have identical distributions.  However, when $K\ge 2,$ $X+K\ge 2$ almost surely, whence $f(X+K) = \lambda(X+K)$ almost surely.
Compute
$$\begin{aligned}
\operatorname{Cov}(f(X), f(X+K)) &= \int_0^1 f(x)f(x+K)\mathrm{d}x - E[f(X)]E[f(X+K)]\\
&= \int_0^1 x\,\lambda(x+K)\mathrm{d}x - E[X] E[\lambda(X+K)]\\
&= \lambda\left(E[X^2] + KE[X]\right) - E[X]\left(\lambda (E[X] + K\right)) \\
&= \lambda\,\operatorname{Var}(X)\\
& \gt \operatorname{Var}(X) = \operatorname{Var}(f(X)).
\end{aligned}$$
The point is that the effect of $f$ on the values $X+K$ may change the scale, thereby rescaling the covariance of $f(X+K)$ (up or down) relative to any other random variable.  You cannot generally guarantee any relationship between that covariance and any of the variances of $X$, $f(X),$ or $f(X+K).$

Characterizing this situation (where the covariance exceeds the variance) in any other way would be difficult: it has to depend on the distribution of $X,$ the details of $f,$ and the value of $K.$
