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I am trying to convince myself of continuity at zero of time inverted Brownian motion.

Let $X(t) = \begin{cases} 0 \quad \quad \quad t=0 \\ tB(1/t) \quad t > 0 \\ \end{cases}$

I am happy with the fact that X(t) has the same FDDs as Brownian motion and is continuous for t>0.

https://people.bath.ac.uk/maspm/book.pdf

On page 13 of this book by Peres, they prove this and state that clearly when t tends to 0 through the rationals, $\lim_{t \rightarrow 0, t \in Q} X(t) = 0$. I don't see why this is obvious, and to me it seems so close to what we are trying to prove it is almost cheating. Can anyone help me with why it is obvious?

  • You can show that $X$ satisfies all properties of standard Brownian motion.i.e. the increments are independent and $X(t)-X(s)$ follows Gaussian distribution $N(0,t-s)$ and $X(t)$ is continue a.e.. – Ben Mar 14 '20 at 21:31
  • Thanks, I've already shown the distribution properties of X(t). I've also shown that X(t) is continuous a.e. for t>0. I just need to show it's continuous at 0. –  Mar 14 '20 at 21:36
  • I think that's fine. You just want X to be continuous a.e. It doesn't matter if it is continuous at 0. – Ben Mar 14 '20 at 21:42
  • No I need it to be continuous everywhere, for almost all $\omega \in \Omega$, i.e. almost surely. –  Mar 14 '20 at 21:48
  • Oh right. It is way harder than I thought. – Ben Mar 14 '20 at 21:59
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    I found a note proving it: https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec6.pdf – Ben Mar 14 '20 at 22:15
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    Take a look at this question – saz Mar 15 '20 at 06:25
  • There's always the Law of the Iterated Logarithm, if you're not averse to overkill. – John Dawkins Mar 16 '20 at 16:14
  • @JohnDawkins haha thanks, for this kind of simple result I was hoping for a proof that doesn't rely on other theorems I don't know how to prove. I was also hoping to get the SLLN for Brownian motion as a consequence. I guess the question saz pointed to works. –  Mar 18 '20 at 17:09

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There's also a martingale proof (taking off from Doob's martingale proof of the SLLN). Namely, $B(t)/t$ is a "backward martingale" with respect to the backward filtration $\mathcal G_t:=\sigma(B(s):s\ge t)$, in the sense that for $0<t<s$, $\Bbb E[B(t)/t\mid\mathcal G_s]=B(s)/s$. As such, $\lim_{t\to\infty}B(t)/t$ exists a.s. and in $L^1$. Because $\Bbb E[|B(t)|] = c\cdot\sqrt{t}$, the $L^1$ limit is $0$, hence so is the a.s. limit.

John Dawkins
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  • Great ! Such a nice application of Doob's proof of the SLLN. Also it avoids the standard exponential moments estimates (see Sangchul's Lee answer in the question pointed by Saz). – Olivier Mar 18 '20 at 21:26