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Given a probability space $(\Omega,\mathcal{A},\mathbb{P})$. Let $\{B_t:t\ge0\}$ be a standard Brownian Motion. We know that: $$\{\limsup B_n/\sqrt{n}\ge c\}\supseteq \{B_n/\sqrt{n}> c\text{ i.o.}\}$$

Could it be shown that: $$\mathbb{P}(\{B_n/\sqrt{n}> c\text{ i.o.}\})\ge \mathbb{P}(\{B_1>c\})$$? If so, how? Any hint?

Strictly_increasing
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Let $\{A_n\}$ be a sequence of events. Since for each $n\ge 1$, $\mathsf{P}\!\left(\bigcup_{m\ge n}A_m\right)\ge \sup_{m\ge n}\mathsf{P}(A_m)$, \begin{align} \mathsf{P}(\limsup_{n\to\infty}A_n)&=\mathsf{P}\!\left(\bigcap_{n\ge 1}\bigcup_{m\ge n} A_m\right)=\lim_{n\to\infty}\mathsf{P}\!\left(\bigcup_{m\ge n}A_m\right) \\ &\ge \lim_{n\to\infty}\left(\sup_{m\ge n}\mathsf{P}(A_m)\right)=\limsup_{n\to\infty}\mathsf{P}(A_n), \end{align} where the second equality follows from the continuity from above. Applying this result to your case, one gets $$ \mathsf{P}(B_n/\sqrt{n}>c\text{ i.o.})\ge \limsup_{n\to\infty}\mathsf{P}(B_n/\sqrt{n}>c)=\mathsf{P}(B_1>c) $$ because $B_n/\sqrt{n}\overset{d}{=}B_1$.

  • So, second equality corresponds to: $$\begin{align}\mathsf{P}!\left(\bigcap_{n\ge 1}\bigcup_{m\ge n} A_m\right)=\mathsf{P}!\left(\lim_{n\to\infty}\left(\sup_{m\ge n}A_m\right)\right)\&\overbrace{=}^{\text{by continuity from above}}\lim_{n\to\infty}\mathsf{P}\left(\sup_{m\ge n}A_m\right)=\lim_{n\to\infty}\mathsf{P}!\left(\bigcup_{m\ge n}A_m\right)\end{align}$$ right? One last fact: I know that $$\mathsf{P}(\bigcup_{m\ge n}A_m)\le\sum_{m\ge n}\mathsf{P}\left(A_m\right)$$by subadditivity. Why $$\mathsf{P}!\left(\bigcup_{m\ge n}A_m\right)\ge \sup_{m\ge n}\mathsf{P}(A_m)$$? @d.k.o. – Strictly_increasing Oct 07 '20 at 08:34
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    While last fact is given by the fact that $\limsup_{n\to\infty}\mathsf{P}(B_n/\sqrt{n}>c)= \limsup_{n\to\infty}\mathsf{P}(B_1>c)=\mathsf{P}(B_1>c)$. Last equality follows from fact that the event $(B_1>c)$ does not depend on the index $n$ and the first equality follows from $B_n/\sqrt{n}\overset{d}{=}B_1$. Correct? @d.k.o. – Strictly_increasing Oct 07 '20 at 08:46
  • @Strictly_increasing What is $\sup_{m\ge n}A_m$? The second equality is exactly as it appears in the answer. –  Oct 07 '20 at 09:20
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    Regarding the first inequality, $\mathsf{P}!\left(\bigcup_{m\ge n}A_m\right)\ge \mathsf{P}(A_m)$ for each $m\ge n$. –  Oct 07 '20 at 09:22
  • $\sup_{m\ge n}A_m$ is the supremum of the sequence ${A_m; m\ge n}$. I know that, by def.: $$\inf_{m\ge n}A_m=\bigcap_{m\ge n} A_m\text{ and }\sup_{m\ge n}A_n=\bigcup_{m\ge n} A_n$$ Is that correct? @d.k.o. – Strictly_increasing Oct 08 '20 at 06:25
  • @Strictly_increasing I've never seen this notation. https://math.stackexchange.com/questions/222192/what-does-the-supremum-of-a-sequence-of-sets-represent –  Oct 08 '20 at 06:37
  • $\mathsf{P}!\left(\bigcup_{m\ge n}A_m\right)\ge \mathsf{P}(A_m)\text{ each }m\ge n$: so, $\mathsf{P}!\left(\bigcup_{m\ge n}A_m\right)$ is an upper bound for $\mathsf{P}(A_m)$. By definition of $\sup$ as the smallest upper bound, it must hold that $\mathsf{P}!\left(\bigcup_{m\ge n}A_m\right)\ge \sup_{m\ge n}\mathsf{P}(A_m)$ – Strictly_increasing Oct 08 '20 at 07:24
  • Finally, for brownian motion $(B_n)_n$ we are allowed to say that $B_n/\sqrt{n}\overset{d}{=}B_1$ by scaling invariance property of Brownian motion – Strictly_increasing Oct 22 '20 at 14:26