I am studying Brownian local time processes and several references mentioned the scaling invariance of local time. For example, page 10 of this reference (https://hal.archives-ouvertes.fr/hal-00091335/document) says "It is well known that the Brownian motion and its local time have the fol-lowing scaling property: $(B_u,L_u)_{u \geq 0} \sim (\sqrt{t}B_{\frac{u}{t}},\sqrt{t}L_{\frac{u}{t}})_{u \geq 0} $," where $B$ is a standard Brownian motion and $L$ is its local time. The definition of local time that I am familiar with is $L(t,a)=\lim_{\epsilon \rightarrow 0}\frac{1}{2\epsilon}\mu(\{s \in [0,t]: |B_t-a| \leq \epsilon\})=\lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon} \int_{0}^{t}1_{Bs \in [a-\epsilon,a+\epsilon]}ds,$ where $\mu$ is the Lebesgue measure. I tried to prove $L(u,a) = \sqrt{t}L(\frac{u}{t},\frac{a}{\sqrt{t}})$ in distribution using the scale invariance of Brownian motion and the definition: $$\int_{0}^{u/t}1_{B_s \in [\frac{a}{\sqrt{t}}-\epsilon,\frac{a}{\sqrt{t}}+\epsilon]}ds=\int_{0}^{u/t}1_{\frac{B_{ts}}{\sqrt{t}} \in [\frac{a}{\sqrt{t}}-\epsilon,\frac{a}{\sqrt{t}}+\epsilon]}ds=\int_{0}^{u/t}1_{B_{ts} \in [a-\epsilon \sqrt{t},a+\epsilon \sqrt{t}]}ds\\=\frac{1}{t}\int_{0}^{u}1_{B_x \in [a-\epsilon \sqrt{t},a+\epsilon \sqrt{t}]}dx,$$ but I'm not sure what to do next.

  • $\begingroup$ So far so good. Now divide by $2\epsilon$ and let $\epsilon\to 0$. $\endgroup$ – zhoraster May 6 at 18:06
  • $\begingroup$ But how does that give me $\frac{1}{\sqrt{t}} L(u,a)$? $\endgroup$ – user0617 May 6 at 21:46
  • $\begingroup$ Since in the rhs you would need to have $\frac1{2\epsilon \sqrt{t}} $ in order to get the definition of local time. And then you would end up with $\sqrt t$. $\endgroup$ – zhoraster May 7 at 5:43
  • $\begingroup$ So did you succeed? $\endgroup$ – zhoraster May 8 at 15:47
  • $\begingroup$ @zhoraster I did! Thanks so much!!! $\endgroup$ – user0617 May 9 at 3:06

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