Is the absolute value of a stationary series also stationary? I know that linear transformations of time series arising from (weakly) stationary processes are also stationary. Is this true, however, for a transformation of a series via taking the absolute value of each element as well? In other words, if $\{x_i,i\in\mathbb{N}\}$ is stationary, then is $\{|x_i|,i\in\mathbb{N}\}$ stationary as well? 
 A: In one particular case this is somewhat true:
If your time series is stationary with normally distributed error, then the absolute values of your original time series follow a stationary folded normal distribution. Since even weak stationarity means both the mean and variance are constant over time, the absolute values will also be stationary. For other distributions this means that the absolute values of the original time series are at least weakly stationary, as constant variance of the original values translates to a constant mean of the new values. 
However, if your original time series only has a constant mean, the variance may change over time, which will affect the mean of the absolute values. Hence, the absolute values will not be (weakly) stationary themselves.
A more general answer would require some study of the moment generating function of the absolute value of a random variable. Perhaps someone with more mathematical background can answer that.
A: Let $\{X_n\colon n \in \mathbb Z\}$ be a time series where $X_n$ is a discrete random variable taking on values $\cos(n), \sin(n), -\cos(n), -\sin(n)$ with equal probability $\frac 14$. It is easily verified that $E[X_n] = 0$ and 
\begin{align}E[X_mX_{m+n}] &= \frac 14\bigg[\cos(m)\cos(m+n)+\sin(m)\sin(m+n)\\
&= ~~~~~ + (-\cos(m))(-\cos(m+n))+(-\sin(m))(-\sin(m+n))\bigg]\\
&= \frac 12\bigg[\cos(m)\cos(m+n)+\sin(m)\sin(m+n)\bigg]\\
&= \frac 12\,\cos(n)\end{align}
and so the process is weakly stationary. It is also obviously not strictly stationary since $X_0$ and $X_n$, $n\neq 0$ take on different values and so the distributions of $X_n$ and $X_m$ are different instead of being the same as is needed (along with many other requirements) for strict stationarity.
For the weakly stationary process described above, the process $\{|X_n|\colon n \in \mathbb Z\}$ is not weakly stationary because 
$E[|X_n|] = \left.\left.\frac 12\right[\cos(n) + \sin(n)\right]$ is not a constant as is needed for weak stationarity (though it is true that the autocorrelation function $E[|X_m|\cdot|X_{m+n}|]$ is a function of $n$ alone).

On the other hand, as noted by @bananach in a comment on the main question, if stationarity is interpreted as strict stationarity, then strict stationarity of $\{X_n\colon n \in \mathbb Z\}$ implies that $\{|X_n|\colon n \in \mathbb Z\}$ is also a strictly stationary process. Strictly stationary processes with finite variance are also weakly stationary processes, and thus for this subclass, it is true that weak stationarity of $\{X_n\colon n \in \mathbb Z\}$ implies weak stationarity of $\{|X_n|\colon n \in \mathbb Z\}$. But, as described in the first part of this answer, one cannot always conclude that weak stationarity of $\{X_n\colon n \in \mathbb Z\}$ implies weak stationarity of $\{|X_n|\colon n \in \mathbb Z\}$.
A: The answer is no. This can be seen by considering a sequence of
 independent r.vs. $X_i$ with their marginal distribution taken in
 a parametric family depending on three parameters. To get a generic
 example, we can consider a distribution which can be re-parameterized
 by using the first two moments along with the absolute moment $\mathbb{E}[|X|]$. We can
 then keep the first two parameters constant while the third
 $\mathbb{E}[|X_i|]$ depends on $i$.
As a specific example we can take a discrete distribution with
 support $\{-2, \, -1, \, 1, \, 2\}$; the three moments
 $\mathbb{E}[X]$, $\mathbb{E}[X^2]$ and $\mathbb{E}[|X|]$ express as
 linear combinations of the four probabilities $p_k := \text{Pr}\{X =
 k\}$. Since the three linear combinations are linearly independent,
 we can use the three moments to re-parameterize as wanted.
A: As several others have shown, weak stationarity does not necessarily remain when you take the absolute value of the time-series.  The reason for this is that taking the absolute value of each element of the time-series can change the mean and variance in a non-uniform way, due to differences in the underlying distributions of the values.  Although weak stationarity does not transfer over in this way, it is worth nothing that strong stationarity does remain under the absolute-value transformation.
