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!