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Let $\{W_t\}_{t\geq 0}$ be a standard Brownian motion (starting at $0$). For $T$ large enough, I would like to prove that $P(\max_{t\in[0,T]} |W_t| \leq c T^{1/3})$ is bigger than a negative power of $T$ (like $a T^{-1/6}$ for example, for an appropriate constant $a > 0$) or at least bigger than $e^{-b T^{1/3}}$ if $b$ can be made arbitrarily close to $0$.

We know that without the absolute value, i.e. $P(\max_{t\in[0,T]} W_t \leq c T^{1/3})$, this property is true because $\max_{t\in [0,T]} W_t$ has the same law as $|W_T|$.

So the result should be true intuitively because keeping $|W_t|$ in a cylinder should not be much less probable than restricting a standard Brownian motion to be lower than a bound of the same order. I've tried to prove this result without success.

Thanks for any help.

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  • $\begingroup$ Why do you think that this holds for any $T>0$? For small $T$, we have $T^{-1/6} \gg 1$, so we cannot expect to bound $\mathbb{P}(\dots)$ from below by $T^{-1/6}$. $\endgroup$ – saz Feb 28 '15 at 17:11
  • $\begingroup$ Sorry, the problem is meant when $T$ is big. I corrected it. $\endgroup$ – Frédéric Ouimet Feb 28 '15 at 17:12
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We have a series expansion for the wanted quantity, namely, $$\mathbb P\left(\max_{t\in[0,T]} |W_t| \leqslant c T^{1/3}\right)=\frac 4{\pi }\sum_{n=0}^{+ \infty } \frac{(-1)^n}{2n+1}\exp\left(-\frac{(2n+1)^2\pi^2}{8c^2}T^{1/3} \right) $$ holds (see Feller's An introduction to probability theory (1970) page 342).

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