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I'm trying to figure out a statement from page 166 of the book "Brownian Motion: An introduction to stochastic processes" by Schilling, Partzsch (2012). The chapters illustrates how fast the brownian paths grow as $t \rightarrow \infty$.

What I don't understand is the following statement:

"A first estimate (of the growth of the brownian paths) follows from the strong law of large numbers. Write for $n \geq1$

$$B(nt)=\sum_{j=1}^n [B(jt)-B((j-1)t)]$$

and observe that the increments of a brownian motion are stationary and independent random variables (with B(t) being a one dimensional standard brownian motion). By the strong law of large numbers we get that

$$B(t,\omega) \leq \epsilon t$$

$\forall \epsilon>0$ and all $t \geq t_0(\epsilon , \omega)$"

How can I use SLLN to obtain such bound?

Thanks in advance.

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    $\begingroup$ Since the increments are centered, SLLN tells you that $B(nt)/n\to0$ almost surely, in particular, for every fixed $t$ and every positive $\epsilon$, $B(nt)\leqslant\epsilon t$ for every $n\ge n_0(t,\epsilon)$. Can you complete this remark to get your result? $\endgroup$ – Did Jul 9 '17 at 11:08
  • $\begingroup$ No, i don't understand why for every fixed $t$ and every positive $\epsilon$, $B(nt) \leq \epsilon t$. I mean, I understand that $B(nt)/n \rightarrow 0$, but how is that related to $B(nt) \leq \epsilon t$? $\endgroup$ – Random-newbie Jul 9 '17 at 11:53
  • $\begingroup$ Sorry, typo: please replace $B(nt)\leqslant\epsilon t$ by $B(nt)\leqslant\epsilon n$. $\endgroup$ – Did Jul 9 '17 at 16:58

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