show that $\mathbb{E}[T^{\alpha}_1] < \infty$ for $\alpha < 1/2$

Question

let $T_1 = \inf \{t \geq 0 \,|\, B_t = 1 \}$ where $B = (B_t)_{t \geq 0}$ is a standard brownian motion

Question : show that $\mathbb{E}[T^{\alpha}_1] < \infty$ for $\alpha < 1/2$

My Failed attempt :

since $T^{\alpha}_1 > 0$ then :

\begin{align} \mathbb{E}[T^{\alpha}_1] & = \int_0^{\infty}P(T^{\alpha}_1 > t) dt = \int_0^{\infty}P(T_1 > t^{1/\alpha}) dt \end{align}

$u = t^{1/\alpha} \implies du = \frac{1}{\alpha}t^{{1-\alpha}/\alpha}dt \implies \alpha du = u^{1 - \alpha} dt \implies dt = \alpha u^{\alpha - 1} du $

\begin{align} \mathbb{E}[T^{\alpha}_1] & = \alpha \int_0^{\infty}u^{\alpha - 1}P(T_1 > u) du \leq \alpha \int_0^{\infty}\frac{1}{u^{1-\alpha}} du \end{align}

but $ 1 - \alpha > \frac12$

something that would solve this problem is if $P(T_1 > u) \sim u^{-1/2}$

what I know is $P(T_1 > u) = P(\max_{s \leq u} B_s \leq 1)$

but the lack of independence makes it hard for me to compute it

Answer 1

The distribution of $\max_{0\leq s\leq u}B_s$ is well-known. See https://en.wikipedia.org/wiki/Reflection_principle_(Wiener_process).

So $$ \mathbb P(T_1>u) = \mathbb P(\max_{0\leq s\leq u}B_s \leq 1) = 1-\mathbb P(\max_{0\leq s\leq u}B_s > 1) = 1-2\mathbb P(B_u>1). $$ Since $B_u$ have normal $\mathcal N(0,u)$ distribution, $Y=B_u/\sqrt{u}$ is standard normal and for $u\to\infty$ $$ 1-2\mathbb P(B_u>1) = 1-2\mathbb P(Y>1/\sqrt{u}) = 1-2\Phi(-1/\sqrt{u}) \sim 2 \frac{\phi(1/\sqrt{u})}{\sqrt{u}} \sim \frac{2}{\sqrt{2\pi u}}. $$ This asymptotic equivalence takes place by L'Hôpital's rule $$ \lim\limits_{x\to 0}\frac{1-2\Phi(-x)}{x} = \lim\limits_{x\to 0}\frac{2\phi(-x)}{1} = \frac{2}{\sqrt{2\pi}}. $$ So, as you supposed, $\mathbb P(T_1>u)\sim cu^{-1/2}$.