# Independence of two Brownian motions

Let $\{B_t:t\geq0\}$ be a Brownian motion defined on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Let $\widehat{B}_0 = B_0$ and set \begin{align} \widehat{B}_t = B_t - \int_0^t \frac{B_s}{s} ds, \qquad \text{for }t>0. \end{align} Now, I want to use the fact that $\widehat{B}$ is a Brownian motion to show that for all $t>0$, one has that $B_t$ is independent of $\widehat{B}_s$ for all $0\leq s < t$.

In order to do so, I first want to show that the function $s \to \frac{B_s}{s}$ is Lebesgue integrable on $[0,t)$. Therefore, I have to show that \begin{align} \int_0^t \mathbb{E}\big[\frac{|B_s|}{s}\big]\ ds < \infty. \end{align} However, I face some difficulties with showing the above inequality and hence to conclude the upper independence statement.

By the scaling property, we have $B_s \stackrel{d}{=} \sqrt{s} B_1$ which implies
$$\mathbb{E}(|B_s|) = \sqrt{s} \mathbb{E}(|B_1|).$$
Note that $\mathbb{E}(|B_1|)$ is finite since $B_1$ is Gaussian. Consequently,
$$\mathbb{E} \left( \int_0^t \frac{|B_s|}{s} \, ds \right) = \int_0^t \frac{\mathbb{E}(|B_s|)}{s} \, ds = \mathbb{E}(|B_1|) \int_0^t \frac{1}{\sqrt{s}} \, ds < \infty.$$