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Let $L_T$ the local time of a Brownian motion. Roughly speaking, it's the time spend by a BM at $0$, i.e. $$m\{s\in [0,T]\mid B_s=0\}.$$

Now, I know that $$\mathbb P\{L_T\geq x\}=2\mathbb P\{B_T\geq x\}.$$

And since $$2\mathbb P\{B_T\geq T\}>0,$$ there is no chance to have $$\mathbb P\{L_T\leq T\}=1.$$

I think this is rather strange. What am I mistaken here ? How is it possible that $(B_t)_{t\in [0,T]}$ spend more time in $0$ than $T$ ?

Todd
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  • In short, because your "roughly speaking" is too rough. The local time does not actually equal the measure of the time spent at zero; the latter measure is actually identically zero a.s. – Nate Eldredge Jun 17 '20 at 17:23
  • For instance, you might show that if $B_t$ were replaced by a constant path that just sits at zero, the corresponding local time $L_T$ would not equal $T$ but rather $\infty$. – Nate Eldredge Jun 17 '20 at 17:26
  • @NateEldredge the local time of the deterministic path sitting at zero is zero, which is evident from the Tanaka formula. But one can easily cook up other examples where local time at zero is larger than $T$. Local time is a density which integrates to $T$, which does not preclude it from being arbitrarily large at zero. – nakajuice Aug 01 '23 at 13:42

1 Answers1

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Things might be clearer from a wider perspective. The Brownian motion has a local time $L^x_t$ at each level $x\in\Bbb R$ (not just $0$). The collection $(L^x_t)_{x\in\Bbb R, t\ge 0}$ can be chosen to be jointly continuous in $(x,t)$, a.s., and then serves as occupation density: $$ \int_0^T f(B_t)\,dt =\int_{\Bbb R} L^x_T \, f(x)\,dx, $$ for each bounded measurable $f$ and each $T>0$, a.s. In particular, you have $$ \int_{\Bbb R} L^x_T\, dx = T. $$

John Dawkins
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