Subordination of a Levy process when the "subordinator" is not nondecreasing If $X_t$ is an $\mathbb{R}$-valued Levy process and $Z_t$ is an $\mathbb{R}$-valued subordinator, we know that $X_{Z_t}$ is also Levy process. My question is, are there processes $Z_t$ which are not non-decreasing, for which $X_{Z_t}$ is a Levy process? In addition to this, which part of the definition of a Levy process fails if the non-decreasing assumption is not satisfied? 
Any help is greatly appreciated as I am very new to Levy processes. 
 A: Usually, Lévy processes $(X_t)_{t \geq 0}$ are considered, i.e. the time $t$ is non-negative. This means that we need the subordinator $(Z_t)_{t \geq 0}$ to satisfy $Z_t \geq 0$. Since Lévy processes have very nice properties, the non-negativity already implies the monotonicity.
The deeper reason why we need the monotonicity lies somewhere else: The monotonicity of $(Z_t)_{t \geq 0}$ is necessary in order to ensure that $(X_{Z_t})_{t \geq 0}$ has independent increments. This becomes more obvious, if we consider a deterministic subordination, i.e. $f:[0,\infty) \to [0,\infty)$ and the corresponding "subordinated" process $Y_t := X_{f(t)}$. If $f$ is not increasing, then $Y_t$ has in general no independent increments. (E.g. choose
$$f(t) := \begin{cases} t & t < 1 \\ 2-t & t \in [1,2] \\ 0 & t>2 \end{cases},$$
then $Y_1-Y_0$ is not independent from $Y_{2}-Y_1$.) Obviously, $f$ is not a Lévy process; nevertheless, this shows that something with the independence goes wrong if the subordinator is not increasing.
