Let $X_t$ and $Z_t$ be independent, $\mathbb{R}$-valued Brownian motions. For each $t$, the process $X_{|Z_t|}$ defined as $$\omega\mapsto X_{|Z_t(\omega)|}(\omega)$$ is measurable (with respect to the projection $\sigma$-algebra), but what is its distribution? Further, is it a Levy process?
Any help is greatly appreciated.
Let $p_t$ and $q_t$ denote the densities of $X_t$ and $Z_t$ respectively, then, by independence of $(X_t)$ and $(Z_t)$, the density of $X_{|Z_t|}$ is the function $$r_t(\ )=\int_{-\infty}^\infty p_{|z|}(\ )q_t(z)\mathrm dz.$$ The displacements of $X_{|Z|}$ seem not to be independent since $|Z|$ is not nondecreasing. And there is no reason to suspect that $$r_{t+s}(x)=\iint_{\mathbb R^2}p_{|z+y|}(x)q_t(z)q_s(y)\mathrm dz\mathrm dy,$$ coincides with $$r_t\ast r_s(x)=\int\!\!\!\!\!\iint_{\mathbb R^3}p_{|z|}(u)p_{|y|}(x-u)q_t(z)q_s(y)\mathrm dz\mathrm dy\mathrm d u,$$ for every positive $s$ and $t$.