$\mathbb{P}\left \{\sup_{t\in [0,\delta]} B_{t} \geq \delta^{2/3}\right\} \geq 1-\delta^\alpha$ as $\delta$ tends to zero.

Question

$\mathbb{P}\left \{\sup_{t\in [0,\delta]} B_{t} \geq \delta^{2/3}\right\} \geq 1-\delta^\alpha$ as $\delta$ tends to zero.

Does this type of tail estimate exist in the literature?

By Blumenthal $0-1$ law, the limit would apprach to $1$ as $\delta$ tends to zero since

$$\mathbb{P}\left \{\sup_{t\in [0,\delta]} B_{t} \geq \delta^{2/3}\right\} \geq \mathbb{P}\left \{B_{\delta} \geq \delta^{2/3}\right\} = \mathbb{P}\left \{\frac{B_{\delta}}{\sqrt{\delta}} \geq \delta^{1/6}\right\} = 1-\Phi(\delta^{1/6}) \geq 1/10$$ and I was wondering if there is a precise tail estimate.

Answer 1

By the reflection principle, it holds that

$$\sup_{t \in [0,\delta]} B_t \sim |B_{\delta}|,$$

and therefore

$$\mathbb{P} \left( \sup_{t \in [0,\delta]} B_t \geq \delta^{2/3} \right) = \mathbb{P}(|B_{\delta}| \geq \delta^{2/3}).$$

As $|B_{\delta}| \sim \sqrt{\delta} |B_1|$ this gives

$$\mathbb{P} \left( \sup_{t \in [0,\delta]} B_t \geq \delta^{2/3} \right) =\mathbb{P}(|B_1|\geq \delta^{4/3}) = 1-\mathbb{P}(|B_1|< \delta^{4/3}).$$

Noting that

$$\mathbb{P}(|B_1| < \delta^{4/3}) = \frac{1}{\sqrt{2\pi}} \int_{-\delta^{4/3}}^{\delta^{4/3}} \exp \left(- \frac{y^2}{2} \right) \, dy$$

we find that

$$\mathbb{P}(|B_1| < \delta^{4/3}) \approx \sqrt{\frac{2}{\pi}} \exp \left(- \frac{\delta^{8/3}}{2} \right) \delta^{4/3} $$

for small $\delta>0$. This means that

$$\mathbb{P} \left( \sup_{t \in [0,\delta]} B_t \geq \delta^{2/3} \right) \approx 1- \sqrt{\frac{2}{\pi}} \underbrace{\exp \left(- \frac{\delta^{8/3}}{2} \right)}_{\approx 1} \delta^{4/3}$$

for small $\delta>0$.