I would need to verify if this answer is correct. Thanks in advance for the help.
Given $B_t$, $t \geq 0$ a brownian motion, determine
$ \lim_{t\to\infty} \sqrt{t} P[\max_{s \in [0,t]} \{B_s\} \leq 1] $
Where P stands for probability.
This is what i did:
$ \lim_{t\to\infty} \sqrt{t} P[\max_{s \in [0,t]} \{B_s\} \leq 1] = \lim_{t\to\infty} \sqrt{t} [1-2P[B_t \geq 1]] $ by application of the reflection principle. Now
$ \lim_{t\to\infty} \sqrt{t} [1-2P[B_t \geq 1]] = \lim_{t\to\infty} \sqrt{t} [2P[B_t < 1] - 1] = \lim_{t\to\infty} \sqrt{t} [2\Phi(\frac{1}{\sqrt{t}}) - 1] $
where $\Phi$ stands for the standard normal probability distribution. This is a $0 \times \infty$ indetermination since $\Phi(0)=1/2$. Applying l'hopital rule,
$ \lim_{t\to\infty} \frac{[2\Phi(\frac{1}{\sqrt{t}})-1]}{\frac{1}{\sqrt{t}}} = \lim_{t\to\infty} \frac{\frac{d}{dt}\left[\frac{2}{\sqrt{2\Pi}}\int_{-\infty}^{\frac{1}{\sqrt{t}}} \exp(-\frac{x^2}{2t}) dx -1 \right]}{\frac{d}{dt}\left[\frac{1}{\sqrt{t}} \right]} = \lim_{t\to\infty} \frac{\sqrt{\frac{2}{\Pi}} \exp(-\frac{1}{2t^2}) {\frac{d}{dt}\left[\frac{1}{\sqrt{t}} \right]}}{\frac{d}{dt}\left[\frac{1}{\sqrt{t}} \right]} = \sqrt{\frac{2}{\Pi}} $
Is this correct? Many thanks.