As I am studying about Brownian motion, as a preparation for my exam, I came across the following problem that I would like some help for: Let $\{B(t), t ≥ 0\}$ be a standard Brownian motion and set $Y_n = \sum_{i=1}^{2^n} |B(\frac{i}{2^n}) - B(\frac{i-1}{2^n}) |$. Show that $E[Y_n] = 2^{n/2}\sqrt{2/\pi}$ and $\sum^∞_{k=1} P(Y_k < k) < \infty$.
Any help would be appreciated.
Let $Z_1, Z_2, \dots$ be the sequence of i.i.d. standard normal variables. Then
$$ Y_n \stackrel{d}= \frac{1}{2^{n/2}} \sum_{k=1}^{2^n} \left| Z_k \right|. $$
So by the Chebyshev's inequality,
$$ \mathbb{P}(\left| Y_n - \mathbb{E}[Y_n]\right| > \lambda) \leq \frac{\operatorname{Var}(Y_n)}{\lambda^2} = \frac{\operatorname{Var}(\left|Z_1\right|)}{\lambda^2}. $$
Using this, we estimate that
$$ \mathbb{P}(Y_n < n) = \mathbb{P}(Y_n - \mathbb{E}[Y_n] < n - \mathbb{E}[Y_n]) \leq \frac{\operatorname{Var}(\left|Z_1\right|)}{(\mathbb{E}[Y_n] - n)^2} \sim \frac{C}{2^n} $$
as $n\to\infty$, where $C = \frac{\pi}{2}\operatorname{Var}(\left|Z_1\right|)$ is an absolute constant. So $\sum_{n=1}^{\infty} \mathbb{P}(Y_n < n)$ converges by the limit comparison test.