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 $0\le s\le t$, $$X_t-X_s=_d X_{t-s}-X_0.$$

I have already shown that

  • $(X_t)$ 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!

  • $\begingroup$ 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 $L^s_0$? $\endgroup$ – Thomas Rippl May 7 '14 at 7:08
  • $\begingroup$ @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. $\endgroup$ – Fang Jing May 7 '14 at 14:03
  • $\begingroup$ 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/… $\endgroup$ – Thomas Rippl May 9 '14 at 7:32
  • $\begingroup$ @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. $\endgroup$ – Fang Jing May 9 '14 at 19:28
  • $\begingroup$ @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. $\endgroup$ – Fang Jing May 9 '14 at 19:37

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