Calculate the limit $$\lim_{t\to \infty} \mathbb E(1-e^{-e^{B_t}}),$$ where $B_t$ - standard Brownian motion.
$\lim_{t\to \infty} \mathbb E(1-e^{-e^{B_t}})=1-\lim_{t\to \infty} \mathbb E(e^{-e^{B_t}})$.
I think that next we should use the property of Brownian motion. Perhaps it is useful to know that $B_t \sim \mathcal N(0,t)$ or that $B_t \sim \sqrt t W_0$. However I have a problem to find this limit.
As $B_t \sim \sqrt{t}\cdot X$ where $X$ follows the standard normal distribution $\mathcal{N}(0,1)$, then $$I(t):= \mathbb{E}\left(e^{-e^{B_t}} \right)= \mathbb{E}\left(e^{-e^{\sqrt{t}\cdot X}} \right) = \int_{-\infty}^{+\infty} \underbrace{e^{-e^{x\sqrt t}} \varphi(x)}_{=:f(x,t)}dx $$ where $ \varphi(x)$ the density function of $\mathcal{N}(0,1)$.
According to the dominated convergence theorem, as the function $f(x,t)$ is bounded by $g(x) = \varphi(x)$( $\forall x, t$ we have $0 \le e^{-e^{x\sqrt t}} \varphi(x) \le \varphi(x)$) which is integrable and $$e^{-e^{x\sqrt t}} = \mathbf{1}_{\{x\le 0\}}\cdot e^{-e^{x\sqrt t}} + \mathbf{1}_{\{x \ge 0\}} \cdot e^{-e^{x\sqrt t}}\xrightarrow{t\to +\infty}\mathbf{1}_{\{x\le 0\}}\cdot e^0 + \mathbf{1}_{\{x \ge 0\}} \cdot 0 = \mathbf{1}_{\{x\le 0\}}$$
then, $$\begin{align} \lim_{t\to +\infty}I(t)&=\lim_{t\to +\infty}\int_{-\infty}^{+\infty} f(x,t)dx \\ &= \int_{-\infty}^{+\infty} \lim_{t\to +\infty}f(x,t)dx \\ &= \int_{-\infty}^{+\infty} \mathbf{1}_{\{x\le 0\}} \cdot \varphi(x)dx \\ &= \int_{-\infty}^0 \varphi(x)dx\\ &= \frac{1}{2} \end{align}$$
Then the result of the initial limit expectation is equal to $1/2$.