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For standard Brownian motion started at zero, the strict first-passage times form a subordinator in the level parameter $a$. The Strong Markov property gives independent increments and stationary increments; the strict inverse of the continuous running maximum has càdlàg paths. The Brownian first-passage Laplace transform gives Laplace exponent $\Phi (\lambda )=\sqrt{2\lambda }$, so the process is strictly stable of index $1/2$. Choosing the non-strict hitting times preserves each fixed-level law but generally loses right continuity at random levels, as in the fixed-level versus simultaneous Brownian passage-time equality.

For standard Brownian motion, set $H_a=\inf \{t:B_t\ge a\}$ and $T_a=\inf \{t:B_t>a\}$. The Strong Markov property and immediate positive excursions give $T_a=H_a$ almost surely for each fixed $a\ge 0$. However, these level-indexed processes are not indistinguishable stochastic processes. Almost surely the random level $A={\sup }_{t\le 1}{B}_t$ is positive and exceeds $B_1$. It is first attained before time one but is not exceeded until after time one, so $H_A<1<T_A$. The simultaneous-equality event therefore has probability zero. The strict passage process $T_a$ is the right-continuous choice needed for a subordinator.

If $X$ is a Lévy process and $T$ an independent subordinator, then $Y_t=X_{T_t}$ is a Lévy process. Conditional on the clock, increments of $X$ over disjoint clock intervals are independent; averaging over the independent stationary increments of the clock gives the same properties for $Y$. If $\mathbb{E}{e}^{iu{X}_s}=e^{-s\Psi (u)}$, its characteristic function is $\mathbb{E}{e}^{-T_t\Psi (u)}$. For standard Brownian motion, $\Psi (u)=u^2/2$, so this is the clock's Laplace transform at $u^2/2$.