When is the autocorrelation function of a stationary process decreasing/nonincreasing? Markovian? When is the autocorrelation function of a stationary process strictly decreasing or nonincreasing?
Can being Markovian make it true?
When is the autocorrelation function of a stationary process (strictly) increasing?
Thanks!
 A: The autocorrelation function $R_X(n)$ of a stationary process 
$\{X(t)\colon t \in \mathbb Z\}$, whether Markovian or not, is
given by $$R_X(n) = E[X(m)X(m+n)]$$ where, because of stationarity,
the choice of $m$ does not matter: $(X(m), X(m+n))$ has the same
joint distribution as $(X(m^\prime), X(m^\prime + n))$ and so
$E[X(m)X(m+n)]= E[X(m^\prime)X(m^\prime+n)]$. Hence,
using an argument which is essentially a version of the
argument that allows us to assert that the Pearson correlation
coefficient $\rho \in [-1,1]$, we get that
$$R_X(0) \geq |R_X(n)|.$$
It is also easy to verify that $R_X(n) = R_X(-n)$ and so, if $R_X(n)$
is an increasing function of $n$ for $n < 0$ (as $R_X(n)$ rises towards its
peak at $0$, say), then $R_X(n)$ must be a decreasing function of $n$ for
$n > 0$. In short, $R_X(n)$ cannot be a strictly increasing function of $n$ for
all $n$. What goes up must come down.
The only other possibilities are 


*

*$R_X(n)=R_X(0)$ for all $n$, that is, the autocorrelation function
is a constant. An example of such a process is in @whuber's example in the comments on the main question: $X(n) = X$ for all $n$ (Aksakal's suggestion 
of $X$ being a constant is just a special (degenerate) case of this when
the random variable $X$ equals a constant almost surely).

*$R_X(n) = (-1)^n R_X(0)$ for all $n$, that is, the autocorrelation
function is periodic with period $2$ as in whuber's example in the
comments where $X(n) = (-1)^nX$ with $X$ being a zero-mean random
variable with symmetric distribution, that is, $X$ and $-X$ are identically
distributed and so all the $X(n)$'s have the same distribution (important
for stationarity).
A: *

*The autocorrelation function of the stationary process does not depend on the time $m$, i.e.


$$R_X(m,n)=E[X(m)X(m+n)] = R_X(n)$$
Then we can make a shift $m\rightarrow m-n$
$$R_X(n) = E[X(m-n)X(m-n+n)] = E[X(m-n)X(m)] = E[X(m)X(m-n)] = R_X(-n)$$
Thus we have the symmetry property
$$R_X(n) = R_X(-n)$$


*The proof of the non-increasing property is borrowed from Kun Il Park, Fundamentals of Probability and Stochastic Processes with Applications to Communications:


$$E[(X(m) - X(m+n))^2] = E[X^2(m) + X^2(m+n) - 2X(m)(m+n)] = 2 E[X^2(m)] + 2E[X(m)(m+n)] \geq 0$$
from which we get
$$E[X^2(m)] \geq E[X(m)(m+n)]$$
or
$$R(0) \geq R(n)$$
