Is binary mapping of simple stationary series still stationary Suppose I have a weakly stationary series with a support $\{0, 1, 2, 3\}$. If I were to map all values of this series into a binary series with support $\{0,1\}$ using the rule $\{0,1\}\rightarrow\{0\}$, $\{2,3\}\rightarrow\{1\}$, would the resulting binary series still be (weakly) stationary? Intuitively the answer would seem to be yes since this is just a "coarsening" of the series and not an alteration of its statistical properties, but how do I show this?
 A: To provide a counterexample, I will emulate an example described by Yves in responding to a similar inquiry.
Consider this family of distributions $p$ on $\{0,1,2,3\}$ parameterized by $p\in[0,1/4]:$
$$\left\{\eqalign{p(0) &= p \\ p(1) &= \frac{3}{4}-3p \\ p(2)&=3p \\ p(3)&=\frac{1}{4}-p.}\right.$$
These are all the distributions with mean $3/2$ and variance $1/2.$  (We knew a priori that such a family ought to exist because such distributions are determined by four probabilities but are subject to just three linear constraints, including the sum-to-unity constraint required of all probability distributions.)
For $p\approx 0$ most of the probability is concentrated on the value $1,$ with some placed near $3$ to maintain the mean and variance, while for $p\approx 1/4$ the distribution has been inverted and now concentrates on the value $2,$ with most of the rest of the probability placed near the value $0.$  This is evident in the first figure below, where $p$ grows from $0$ to $1/4$ as time goes on.
Thus, any sequence of parameters $p_i$ determines a sequence $X_i$ of independent random variables with corresponding distributions $p_i.$ This sequence is weakly second-order stationary because the mean and variance are constant while independence assures the covariances are constantly zero.

This figure shows one realization of $(X_i)$ corresponding to $p_i$ increasing linearly from $0$ to $1/4.$  The heights have been randomly jittered to resolve overlaps. The curve is a Loess smooth tracing out a local mean; as intended, it is essentially flat, indicating first-order weak stationarity.
Under the mapping in the question ($f(0)=f(1)=0;\ f(2)=f(3)=1$) the new distribution assigns probability $p + 3/4-3p = 3/4-2p$ to $0$ and the remaining probability $1/4+2p$ to $1.$  The mean of this distribution is $1/4+2p,$ whence the means of the $f(X_i)$ form a sequence $(1/4 + 2p_i).$  If the $p_i$ are not constant, then $f(X_i)$ is not even weakly first order stationary.

The same series, as transformed by $f.$  The Loess smooth demonstrates the lack of stationarity.
Appendix: The Code
This is the R code used to generate the figures.  It shows how to create realizations of $X_t$ and $f(X_t).$
#
# Sample from the distributional family.
#
r <- function(p, n=1) {
  p <- min(1/4, max(0, p)) 
  sample(0:3, n, replace=TRUE, prob=c(p, 3/4-3*p, 3*p, 1/4-p))
}
#
# Transform the values.
#
f <- function(x) floor(x/2)
#
# Simulate a process from an array of parameters (p).
#
n <- 300
# p <- seq(0, 1/4, length.out=n)
p <- rep(0:1/4, each=floor(n/2)) # An extreme example
x <- sapply(p, r)
#
# Plot (X) and (f(X)).
#
library(ggplot2)
plot.it <- function(var="x", title="Series X") {
  ggplot(X, aes_string("t", var)) + 
    geom_smooth(method="loess") + 
    geom_jitter(aes(fill=X), shape=21, alpha=0.5, width=0, height=0.05) +
    ggtitle(title)
}
X <- data.frame(t=1:n, p=p, x=x, y=f(x), X=factor(x))
plot.it()
plot.it("y", "Series f(X)")

