Use the usual completed right-continuous <Brownian filtration>. The strict passage process $(T_a)_{a\geq0}$ 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 shows that the passage times above this level depend on a new independent standard <Brownian motion>. Thus for $0\leq a<b$, $T_b-T_a$ 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,
$$
\mathbb E e^{-\lambda T_a}=e^{-a\sqrt{2\lambda}}.
$$
For $\lambda,\eta>0$, this also proves continuity in probability at zero, since
$$
\mathbb P(T_h>\eta)\leq\frac{1-e^{-h\sqrt{2\lambda}}}{1-e^{-\lambda\eta}}\longrightarrow0.
$$
Finally, with $M_t=\sup_{s\leq t}B_s$, 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 $T_{a_n}\leq t$. 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 $\Delta T_j$. Consequently
$$
\mathbb E\exp\!\left(i\sum_{j=1}^ku_j\Delta X_j\right)
=\mathbb E\exp\!\left(-\frac12\sum_{j=1}^ku_j^2\Delta T_j\right)
=\prod_{j=1}^k\exp\bigl(-(a_j-a_{j-1})|u_j|\bigr).
$$
Factorization and dependence only on interval lengths prove <independent increments> and <stationary increments> for $X$. For small $h$, $T_h\to0$ in probability, and <independence> and continuity of $W$ imply $W_{T_h}\to0$ in probability. For example, bound its deviation probability by $\mathbb P(T_h>\eta)+\mathbb P(\sup_{t\leq\eta}|W_t|>\varepsilon)$ and then let $h\downarrow0$ and $\eta\downarrow0$. <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 $X_0=0$. Thus
$$
\boxed{(X_a)_{a\geq0}\text{ is a Lévy process}.}
$$
This 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 / 6 / d / Solution

{title2=$G(a)=\inf\{t\geq0:f(t)>a\}$}

For an unbounded nondecreasing function $f:[0,\infty)\to\mathbb R$, its strict generalized inverse is $G(a)=\inf\{t\geq0:f(t)>a\}$. 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 $G(a_n)\leq t$; also $G(a_n)\geq G(a)$. 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>.

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