This is a question from Revuz and Yor (exercise 3.18) for which I seem to get a different answer.
Show that $\lim_{t \to \infty}\,t^{1/2}\,\mathbb{P}\{B_s\leq1\,\forall\, s\in[0,t]\}=\sqrt{2/\pi}$.
Where $B_t, t\geq0$ is a standard Brownian Motion.
Here's my (I assume wrong) attempt:
Taking complements & using the reflection principle, $\mathbb{P}\{B_s\leq1\,\forall\, s\in[0,t]\}=1-\mathbb{P}\{\exists\, s\in[0,t]: B_s>1\,\}=1-2\mathbb{P}\{B_t>1\,\}$ $=1-2\mathbb{P}\{B_1>t^{-1/2}\,\}=1-\frac {2}{\sqrt {2\pi}}\int_{t^{-0.5}}^{\infty} e^{-\frac 12 x^2}dx =\frac {2}{\sqrt {2\pi}}\int_{0}^{t^{-0.5}} e^{-\frac 12 x^2}dx$
The last equality follows from the fact that $\frac {2}{\sqrt {2\pi}}\int_{0}^{\infty} e^{-\frac 12 x^2}dx=1$.
So the question is just to find $\lim_{t \to \infty}\,t^{1/2}\frac {2}{\sqrt {2\pi}}\int_{0}^{t^{-0.5}} e^{-\frac 12 x^2}dx$
Set $f(t):=\int_{0}^{t^{-0.5}} e^{-\frac 12 x^2}dx$ for $t>0$. Then by L'Hôpital's rule (taking $t^{-0.5}$ as the denominator function), the limit in question is just: $\lim_{t \to \infty}\,\frac{f'(t)}{-0.5t^{-1.5}}$ **
But $f'(t)=-0.5t^{-1.5}\,e^{-\frac{1}{2t}}$ so clearly ** tends to zero.
Could you please tell me what's wrong with this? Thanks in advance!
You simply forgot to apply the chain rule: Note that
$$\int_0^{t^{-1/2}} \exp \left(- \frac{1}{2} x^2 \right) \, dx = (g \circ h)(t)$$
for
$$h(t) := \frac{1}{\sqrt{t}} \qquad g(t) := \int_0^t \exp \left(- \frac{x^2}{2} \right) \, dx.$$
Hence,
$$\frac{d}{dt} \int_0^{t^{-1/2}}\exp \left(- \frac{1}{2} x^2 \right) \, dx = g'(h(t)) \cdot h'(t)= -\exp \left(- \frac{1}{2t} \right) \cdot \frac{1}{2 t^{3/2}}.$$