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The standard symmetric Cauchy process is the Lévy process with Lévy characteristic exponent $\Psi (u)=|u|$. Its time-$t$ distribution for $t>0$ is the Cauchy distribution of location zero and scale $t$. It can be constructed by subordination of a Lévy process: evaluate an independent standard Brownian motion at the Brownian first-passage subordinator, whose Laplace exponent is $\sqrt{2\lambda }$.

The Brownian first-passage subordinator has a countable dense set of jump levels almost surely, and is continuous at every other level. Its paths are finite nondecreasing functions, so on a bounded level interval there can be only finitely many jumps larger than any fixed positive size; a countable union proves countability. For density, suppose its path were continuous throughout an open level interval. It is strictly increasing, since the original continuous Brownian path takes different values at different passage times. Its continuous inverse on that interval would parametrize a positive-length time interval on which the Brownian path increases monotonically. This contradicts nowhere monotonicity of Brownian motion. Thus every level interval contains a jump. In particular dense discontinuities do not mean that the path is nowhere continuous: it is continuous at every level outside a countable random set.

Under the path convention used here, a real Lévy process $(L_a{)}_{a\ge 0}$ starts at zero, has independent increments and stationary increments, is stochastically continuous, and has càdlàg paths almost surely. We verify all four requirements for $S$.

The immediate positive excursions proved in part (b) give $S_0=0$. For a deterministic $a\ge 0$, $S_a$ is an almost-surely finite stopping time, and continuity gives $X_{S_a}=a$. The Strong Markov property implies that ${\tilde{X}}_t=X_{S_a+t}-a$ is a fresh Brownian motion independent of the information up to $S_a$. For $b\ge 0$,No crossing of $a+b$ could have occurred before $S_a$. The increment thus has the same law as $S_b$ and is independent of the past. Apply this successively at a finite increasing sequence of deterministic levels to prove stationary independent increments.

For $h>0$ and $\epsilon >0$, part (b) and reflection giveHence $S_h\to 0$ in probability. Stationarity of the increments gives stochastic continuity at every level, from either side where defined.

For path regularity, on a single probability-one event the Brownian running maximum $M_t={\max }_{s\le t}{X}_s$ is continuous, nondecreasing and unbounded, andThis inverse is nondecreasing. If $a_j\downarrow a$, then $S_{a_j}\downarrow L\ge S_a$. For any $t>S_a$, some $u<t$ has $M_u>a$; for all sufficiently large $j$, $M_u>a_j$, so $S_{a_j}\le u<t$. Thus $L=S_a$, proving right continuity at every level simultaneously. On each bounded level interval it is bounded by the finite value at a larger integer level. Monotonicity then gives finite left limits. ThereforeThe Brownian first-passage subordinator is the level-indexed increasing process, rather than a process indexed by the original Brownian time.

For $H$, the inverse relations for a continuous running maximum areFor example, if $b<a$ then $S_b\le H_a$; while for $t<H_a$, $M_t<a$ and any $b\in (M_t,a)$ has $S_b>t$. Taking limits gives $S_{a-}=H_a$. Similarly $H_b\ge S_a$ for $b>a$, and a crossing of any level strictly above $a$ before a given $t>S_a$ bounds $H_b$ from above for $b$ sufficiently close to $a$. This proves $H_{a+}=S_a$. These are the non-strict inverse of a continuous nondecreasing function identities.

At the random level $A$ in part (b), $H_A<S_A=H_{A+}$, so $H$ fails right continuity almost surely. Thus it is not a Lévy process under the stated càdlàg-path convention. It does have the same finite-dimensional distributions and stochastic continuity as $S$, so definitions of a Lévy process that require only the increment and stochastic-continuity properties would call $H$ a Lévy process with a càdlàg modification. The reason for excluding the particular version $H$ here is its path regularity.

Use the usual completed right-continuous Brownian filtration. The strict passage process $(T_a{)}_{a\ge 0}$ is finite at every level on one probability-one event: finiteness at all integer levels from (a) suffices by monotonicity. Also $T_0=0$ almost surely by (b).

For each fixed $a$, $T_a$ is a stopping time with $B_{T_a}=a$. The Strong Markov property shows that the passage times above this level depend on a new independent standard Brownian motion. Thus for $0\le a<b$, $T_b-T_a$ is independent of the past at $T_a$ and has the same law as $T_{b-a}$. Iteration gives independent increments and stationary increments in the level parameter. From (a) and fixed-level equality,For $\lambda ,\eta >0$, this also proves continuity in probability at zero, sinceFinally, with $M_t={\sup }_{s\le t}{B}_s$, we have $T_a=\inf \{t:M_t>a\}$. This strict generalized inverse of a nondecreasing function is right-continuous: if $a_n\downarrow a$, then for any $t>T_a$ we have $M_t>a$, and eventually $M_t>a_n$, which forces $T_{a_n}\le t$. Monotonicity gives the reverse bound. Monotonicity and local finiteness also give finite left limits. Hence $T$ is the Brownian first-passage subordinator, with càdlàg paths.

Now condition on this clock, which is independent of $W$. For a deterministic partition $0=a_0<a_1<\cdots <a_k$, write $\Delta T_j=T_{a_j}-T_{a_{j-1}}$ and $\Delta X_j=W_{T_{a_j}}-W_{T_{a_{j-1}}}$. Conditional on the clock these are independent centered Gaussian increments, with respective variances $\Delta T_j$. ConsequentlyFactorization and dependence only on interval lengths prove independent increments and stationary increments for $X$. For small $h$, $T_h\to 0$ in probability, and independence and continuity of $W$ imply $W_{T_h}\to 0$ in probability. For example, bound its deviation probability by $\mathbb{P}(T_h>\eta )+\mathbb{P}({\sup }_{t\le \eta }|W_t|>\epsilon )$ and then let $h\downarrow 0$ and $\eta \downarrow 0$. Stationary increments give stochastic continuity at every deterministic level. Composition of the continuous path of $W$ with the nondecreasing càdlàg clock gives càdlàg paths for $X$, and $X_0=0$. ThusThis is subordination of a Lévy process. Using strict passage times ensures the required right-continuous path choice, despite their fixed-level equality with the non-strict times.

For an unbounded nondecreasing function $f:[0,\infty )\to \mathbb{R}$, its strict generalized inverse is $G(a)=\inf \{t\ge 0:f(t)>a\}$. It is nondecreasing and right-continuous in $a$. If $a_n\downarrow a$, then for any $t>G(a)$ monotonicity gives $f(t)>a$, so eventually $f(t)>a_n$ and $G(a_n)\le t$; also $G(a_n)\ge G(a)$. Local finiteness gives finite left limits. Flat parts of $f$ create jumps of $G$. For the continuous running maximum of Brownian motion, this gives the càdlàg Brownian first-passage subordinator.