Let $B$ be the standard brownian motion: Why does $\lim\limits_{ s \to 0} sB_{\frac{1}{s}}=0$ a.s.?

I am thinking of applying the martingale convergence theorem here by some enumerated sequence $(t_{n})_{n \in \mathbb N}$ on $Y_{n}:=B_{t_{n}}$ such that

$t_{n} \xrightarrow{n \to \infty} \infty$. I am not sure whether the property $\sup\limits_{n \in \mathbb N}E(Y_{n})_{+}<\infty$ is satisfied, the best I could do was

$E(Y_{n})_{+}\leq E[\lvert Y_{n} \rvert ]\leq \sqrt{E(Y_{n})^{2}}=\sqrt{t_{n}}$ which clearly does not help me.

Assuming that it satisfies the necessary conditions for the martingale convergence theorem, then $\lim Y_{n} = Y_{\infty}$ almost surely where $Y_{\infty}$ is real-valued such that for a null set $N$

$B_{t_{n}}(\omega)\xrightarrow{n \to \infty}Y_{\infty}(\omega)$ for any $\omega \in N^{c}$

and hence $\frac{1}{t_{n}}B_{t_{n}}\xrightarrow{ n \to \infty} 0$ a.s.

Am I on the right track? If so, I need help to prove that $\sup\limits_{n \in \mathbb N}E(Y_{n})_{+}<\infty$

  • 1
    $\begingroup$ $Y_n$ has same distribution as $\sqrt {t_n} X$ where $X$ has standard normal distribution. So $sup EY_n^{+}=\infty$ and your approach doesn't work. But you can find a proof of this result in many books. $\endgroup$ Aug 5, 2020 at 10:45

1 Answer 1


Note that this is equivalent to proving that $\lim_{s\to \infty} \frac{1}{s} B_s=0$ almost surely and by symmetry, it suffices to prove that $\limsup_{s\to \infty} \frac{1}{s}B_s\leq 0$ almost surely.

It's probably easiest to prove this via the Browian Reflection Principle. So that for any $C>0$ and any $n $, we have

$$ \mathbb{P}(\sup_{0\leq t\leq n} B_t\geq Cn)=2\mathbb{P}(B_n\geq Cn)=2\mathbb{P}(\frac{B_n}{\sqrt{n}}\geq C\sqrt{n}) $$ Since $B_n/\sqrt{n}$ is standard Gaussian, we can apply Markov's inequality to get that

$$ \mathbb{P}(\frac{B_n}{\sqrt{n}}\geq C\sqrt{n})= \mathbb{P}(\frac{B_n^4}{n^2}\geq C^4n^2)\leq \frac{3}{C^4n^2} $$

Hence, we see that $$ \sum_{n=1}^{\infty} \mathbb{P}(\sup_{0\leq t\leq n} B_t\geq C n)<\infty $$ and hence, by Borel-Cantelli's lemma, $\limsup_{t\to\infty} \frac{B_t}{t}\leq C$ almost surely. Letting $C_k=\frac{1}{k},$ we see that

$$ \{\limsup_{t\to\infty} \frac{B_t}{t}\leq 0\}=\bigcap_{k=1}^{\infty}\{\limsup_{t\to\infty} \frac{B_t}{t}\leq C_k\} $$ is also almost sure and thus, we get the result.


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