1

Let $(B(t))_{t∈[0,\infty)}$ be a Brownian Motion. We want to show that $$ \frac{B(t)}{t}\to^{a.s.} 0$$ when $t\to \infty$.

For that we want to use the fact that the Brownian motion $\tilde B(t):=t B(t)$ is continuous at $0$.

The proof then says that for almost all $w \in \Omega $: $$\lim_{t \to \infty} \frac{B(t)(w)}{t}=\lim_{t \to \infty} \tilde B(\frac{1}{t})(w)=0$$

But I feel that something is wrong here. Isn't $\tilde B(\frac{1}{t})=B(\frac{1}{t})/t$ rather than $B(t)/t$?

Kilkik
  • 2,172
  • '$\tilde B(t):=t B(t)$ is continuous at $0$': I think you are misquoting this result. This statement is too trivial to be of any use. – Kavi Rama Murthy Jul 07 '23 at 08:34
  • 1
    For a different proof see: https://math.stackexchange.com/questions/1099655/law-of-large-numbers-for-brownian-motion-direct-proof-using-l2-convergence – Kavi Rama Murthy Jul 07 '23 at 08:38
  • You should have written `the Brownian motion defined $\tilde{B} ():=(1/)$ for $t>0$ and $\tilde{B}(0)=0$ is almost surely continuous at 0'. – Christophe Leuridan Jul 07 '23 at 08:53
  • @ChristopheLeuridan I also think it is the right thing to do – Kilkik Jul 07 '23 at 08:54
  • 1
    See my answer here .It was almost exactly the same question. Perhaps both posters are using a common text with a common printing mistake. – Mr. Gandalf Sauron Jul 07 '23 at 08:58
  • @geetha290krm $tB_{t}$ is not a Brownian Motion ($Var, (tB_{t})=t^{3}$). The correct statement is $tB_{1/t}$ is a Brownian Motion called the projective reflection at $\infty$. They have the same law as processes but the continuity at $0$ (which is equivalent to the question above) is the main step in the proof. – Mr. Gandalf Sauron Jul 07 '23 at 09:13
  • 1
    @Mr.GandalfSauron I was fully aware of that. I was just pointing out why OP was getting into trouble by quoting a wrong result. – Kavi Rama Murthy Jul 07 '23 at 09:22
  • Perhaps think of $B(t)/t \to 0$ a.s. as a "law of large numbers". – GEdgar Jul 07 '23 at 10:14
  • @geetha290krm Ofcourse an user who is viewing the question with intention of helping the op will be aware of the correct results. Your first comment should also include the "correct" result so as to not confuse the op. The statement your first comment is referring to is btw no "misquotation" and is a correct statement. I wrote my comment so that the op additionally sees the flaw that $tB_{t}$ is not a brownian motion and that $tB_{1/t}$ is a brownian motion. I personally think that one should state the correct version in addition to pointing out the flaws in a statement. – Mr. Gandalf Sauron Jul 07 '23 at 13:16
  • @Mr.GandalfSauron I strongly feel that your second comment above should not have been addressed to me, if it was meant for OP. – Kavi Rama Murthy Jul 07 '23 at 23:29
  • @geetha290krm Comments are made in reply to another comment. It is very unfortunante if you consider an elaboration/addition of more context and more results as an assumption that "you don't know what you are talking about" which you clearly do since you feel so "strongly" about it. m – Mr. Gandalf Sauron Jul 08 '23 at 07:08

1 Answers1

1

Showing continuity of the reflected(at infinity) brownian motion is equivalent to showing that $\frac{B_{t}}{t}\to 0$.

That is, $X_{t}=tB_{\frac{1}{t}}$ is a process with same law as that of Brownian motion. But if you manage to show that $\lim_{t\to 0}X_{t}=0$ , then you have that $X_{t}$ is a Standard Brownian Motion called the projective reflection at infinity.

Continuity away from $0$ is trivial to show but, showning continuity at $0$ is the main step.

If you do that, you'll get that $\lim_{t\to 0}\frac{B_{1/t}}{1/t}\to 0$ which is same as $\lim_{t\to\infty}\frac{B_{t}}{t}\to 0$.

And conversely if you manage to show $\lim_{t\to\infty}\frac{B_{t}}{t}\to 0$ then you have that $X_{t}$ is a Brownian Motion.

See my answer here for a proof which imitates the one given in Rene Schilling.

One can use Doob's Maximal Inequality to give a direct proof of $\lim_{t\to\infty}\frac{B_{t}}{t}\to 0$ by restricting to intervals of the form $[n,n+1]$. See here or here for a proof.

Mr. Gandalf Sauron
  • 17,114
  • This result has been proved long back on MSE in various posts. You have yourself answered it before. Please avoid duplicates. – Kavi Rama Murthy Jul 07 '23 at 09:24
  • @geetha290krm The answers should be based on context of the problem and mere existence of a proof of a theorem/problem in this site does not mean that one can treat all questions about it/similar to it as duplicates. The confusion of the op lies in the fact that they are using $tB_{t}$ which is not a brownian motion. Also, they are using continuity without understanding it's significance. When the op states something like $tB_{t}$ is a brownian motion, then it is up to the users to correct it. If you think that this qualifies as a duplicate, then I suggest you vote as such. – Mr. Gandalf Sauron Jul 07 '23 at 13:12