# A right-inverse of Brownian motion local time at zero has stationary independent increments

Let $L_0^t$ be the local time for a standard Brownian motion at $0$ and define $$X_t=\sup\{s\ge0:L_0^s\le t\}, t\ge0.$$ I would like to show that $(X_t)$ has stationary independent increments. That is, for $0\le t_1<\cdots<t_k$, the increments $X_{t_2}-X_{t_1},\ldots, X_{t_k}-X_{t-{k-1}}$ are independent, and for $0\le s\le t$, $$X_t-X_s=_d X_{t-s}-X_0.$$

• $(X_t)$ is right-continuous and nondecreasing;
• For $t\ge0$, $X_t=_dt^2X_1$;
• $X_t$ has Laplace transform $$\mathbb{E} e^{-sX_t}=\int_0^1\frac{2t\sqrt{s}}{\sqrt{-2\pi\log x}}\int_0^\infty e^{\frac{st^2y^2}{2\log x}}\text{d} y\text{d} x, \text{ where } s\ge0.$$
I am not sure if these results are useful for the proof. I think my main obstacle is that the joint distribution of $X_{t_2}-X_{t_1},\ldots, X_{t_k}-X_{t-{k-1}}$ is unknown. On the other hand, since $X_t$ isn't a stopping time, I cannot use the strong Markov property (in the style of this proof) either. Any help is appreciated!
• Isn't it true that $\sup \{s\geq 0: L^s_0 \leq t\} = \inf \{s\geq 0: L^s_0 =t \}$ almost surely by the almost sure continuity of $s\mapsto$L^s_0$? – Thomas Rippl May 7 '14 at 7:08 • @thomas I don't think so. By a consequence of Tanaka's formula,$L_0^s$is constant on each component of the complement of the set$Z:=\{t:B_t=0\}$. (Since$Z$is nowhere dense and closed,$Z^c$is open and dense.) Given a nonempty open interval$(a,b)\subset Z^c$,$L_0^s$is constant on$(a,b)$, say at$t$, then $$\sup\{s:L_0^s\le t\}\ge b$$ and $$\inf\{s:L_0^s=t\}\le a$$ which are not equal. – Fang Jing May 7 '14 at 14:03 • you are right. However, it should be true that$\sup\{s\geq 0: L_0^s \leq t\} = \inf \{s\geq 0: L_0^s >t\}$for a non-decreasing and continuous function$s\mapsto L_0^s$. The definition with$>$can be found in some sources, e.g. books.google.de/… – Thomas Rippl May 9 '14 at 7:32 • @thomas Yes I agree that$\sup\{s\ge 0: L_0^s \leq t\} = \inf \{s\geq 0: L_0^s >t\}$. But I'm not quite sure how to go from here. – Fang Jing May 9 '14 at 19:28 • @thomas Since$L_0^s$is right-continuous and$(t,\infty)$is open, I think this makes each$X_t$an optional time. Then if the underlying filtration is right-continuous, each$X_t\$ is a stopping time. I think these might be some helpful implications but I'm not a-hundred-percent positive. – Fang Jing May 9 '14 at 19:37