Growth speed of Brownian motion

Given a standard Brownian motion B, we have known that almost surely, $$\limsup_{n\to\infty}\frac{B(n)}{\sqrt{n}}=+\infty.$$

For any positive real number a and integer n, let $$E_n=\left\{\frac{B(n)}{\sqrt{n}}>a\right\}.$$ By CLT, we know that the measure of $E_n$ converges.

My question: Is there any good way to estimate the following term $$\mathbb{P}\left\{\bigcup_{k=n}^m E_k\right\}.$$

For simplify, if we take m=2n and fix a=1( or any number you want), I hope to know how to control the above measure in terms of n.

Thanks for any comments and suggestions.

• Look at any proof of the law of iterated logarithm. An important step in the proof are estimates of the type you ask about. Very roughly, you will get something like $1-c^{\log (m/n)}$. – ofer zeitouni Jun 12 '15 at 11:00
• Thanks for your comment. I have known this estimates, but it is too weak for me. In fact, using a method in the proof of LIL, we can get the measure is larger than 1-$c^{\log\log m/n}$ which is terrible for me. If you can get the measure is larger than the form you provided, please tell me, because it is perfect for me. – Bing Jun 12 '15 at 23:41
• Your question is not clear. If you take $m=2n$ as you asked, the probability is just a constant, bounded away from 0 and 1. You only need an upper bound (since $P(E_n)=c(a)>0$). To see the upper bound, it is convenient to lower bound the probability of the complement: look at the event that $B_{n}<-\sqrt{n}$ and simply require the BM to stay below 0 up to time $m$. This has positive probability. – ofer zeitouni Jun 13 '15 at 5:39
• So if you want $m>>n$, you need to wait for decorrelation, and this is precisely what you do for the LIL. In particular, you need the logaritmic scale in order to induce decorrelation. – ofer zeitouni Jun 13 '15 at 5:45
• Thanks for your help. In fact, I need to get lower bounds of $$\mathbb{P}\left\{\bigcup_{k=n}^m E_k\right\},$$ when $n$ large and $m=n^2$. In fact, I need to know how fast it convergence to 1, when $m$ and $n$ are large. (The more fast, the better for me.) If the way we used here is similar as the way used in the proof of LIL, the estimate of the speed on convergence is too weak for me. Thanks for your patient. – Bing Jun 13 '15 at 18:32

$X_t = e^{-t}B(e^{2t})$ is an Ornstein-Uhlenbek process, and it is above a fixed level when the Brownian motion is above a square root boundary. The probability you want is the probability that it is ever above the level 1 over some time interval. I don't know how explicitly you can get an answer but the elements are: conditional on being at $x < 1$ at time $t_0$ the probability that you cross the boundary 1 before time $t_1$