It is a standard fact that subordination of a Lévy process yields another Lévy process. That is, if $\{X_t\}$ is an $\mathbb{R}$-valued rcll process with independent stationary increments such that $X_0=0$, and $\{Y_t\}$ is another such process which is additionally non-negative and non-decreasing, then the composition $Z_t=X(Y_t)$ is again rcll with independent stationary increments.
The proofs I have seen of this statement in the literature always show that the characteristic function of the composed process has the correct form, and thus $Z$ is again a Lévy process. I wanted to make sure that I understand the necessary conditions from more basic principles.
I mean this in the following way: does one need $X$ and $Y$ to have stationary independent increments in order for $Z$ to have stationary increments? I think not. If $X$ and $Y$ have stationary increments, then this should be sufficient for $Z$ to have stationary increments, since
$$ Z_t-Z_s = X(Y_t)-X(Y_s) \overset{d}{=} X(Y_t-Y_s) \overset{d}{=} X(Y_{t-s})=Z_{t-s}, $$
where one should be able to prove the first equivalence in distribution by conditioning on $Y$, for example.
In the same way, is it true that $X$ and $Y$ having independent increments is sufficient for $Z$ to have independent increments? I'm not sure about this, but I think if $X$ has independent stationary increments and $Y$ has independent increments, then $Z$ has independent increments as follows from
\begin{align*}
P(Z_t-Z_s\in A, Z_s\in B)
&= E[P(X(Y_t)-X(Y_s)\in A\mid \sigma(Y))P(X(Y_s)\in B\mid \sigma(Y))] \\
&= E[P(X(Y_t-Y_s)\in A\mid \sigma(Y))P(X(Y_s)\in B\mid \sigma(Y))] \\
&= E[F(A,Y_t-Y_s)F(B,Y_s)] \\
&= E[F(A,Y_t-Y_s)] \cdot E[F(B,Y_s)] \\
&= P(Z_t-Z_s\in A)\cdot P( Z_s\in B)
\end{align*}
where $F(A,u):=P(X_u\in A)$, where the first equality follows from the conditional independence of the increments of $Z$ conditioned on $Y$.
Does the above proofs make sense or have I made a mistake somewhere? Also, does $Z$ have independent increments without the assumed stationary increments of $X$?
Best Answer
Does one need $X$ and $Y$ to have stationary (independent) increments in order for $Z$ to have stationary increments? No. Let $(X^1_t)_{t \geq 0}$ a process with stationary increments and define $$X(t) = \begin{cases} X^1(t) & t \in [0,1] \\ X^1(1) & t \in [1,2] \\ X^1(t-1) & t \geq 2 \end{cases}$$ Clearly, $X$ does not have stationary increments. But by definining $$Y(t) := \begin{cases} t & t \in [0,1] \\ t+1 & t \geq 1 \end{cases}$$ we obtain $Z(t) = X(Y(t))= X^1(t)$ - and by assumption, $X_1$ has stationary increments. Similarily, one can define $X$, $Y$ such that neither $X$ nor $Y$ have stationary increments, but $Z$.
$X$ independent, stationary increments, $Y$ independent increments $\Rightarrow$ $Z$ independent increments? No. Consider for example a Brownian motion $(B_t)_{t \geq 0}$ and define a process $(Y_t)$ by $$Y_t := \begin{cases} 1-t & t \in [0,1] \\ t-1 & t \geq 1 \end{cases}$$ Then, for $Z(t) := B(Y(t))$, by definition $$Z_1- Z_0 = B_0 - B_1 \qquad \qquad Z_2-Z_1 = B_1-B_0$$ and so $Z$ does not have independent increments. Consequently, by the reversion of time, we lose the independent of the increments.