Brownian Motion-Independence of Increments

Question

Consider a Brownian Motion $B(t)$ with $B(0)=0$. Suppose $s<t$. I read in a book that while $B(t)-B(s)$ is independent of the past, $2B(t)-B(s)$ or $B(t) - 2B(s)$ is not.

Why is this the case?

Thank your for your helps.

Answer 1

Let $(\Omega\,,\mathbb{P},\mathcal{F_t})$ be probability space and $\mathcal{F_t}=\sigma(B_t)$. Set $s<t$ we have \begin{align} & {{E}^{\mathbb{P}}}[2{{B}_{t}}-{{B}_{s}}|\,\mathcal{F_s}]={{E}^{\mathbb{P}}}[2({{B}_{t}}-{{B}_{s}})|\,{\mathcal{F_s}}]+{{E}^{\mathbb{P}}}[{{B}_{s}}|\,\mathcal{F_s}] \\ \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\underbrace{2{{E}^{\mathbb{P}}}[({{B}_{t}}-{{B}_{s}})]}_{0}+{{B}_{s}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={{B}_{s}} \\ \end{align} and \begin{align} & {{E}^{\mathbb{P}}}[{{B}_{t}}-2{{B}_{s}}|\,{\mathcal{F_s}}]={{E}^{\mathbb{P}}}[{{B}_{t}}-{{B}_{s}}|\,{\mathcal{F_s}}]-{{E}^{\mathbb{P}}}[{{B}_{s}}|\,{\mathcal{F_s}}] \\ \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\underbrace{{{E}^{\mathbb{P}}}[{{B}_{t}}-{{B}_{s}}]}_{0}{-{B}_{s}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-{{B}_{s}} \\ \end{align}

Answer 2

Because the exact statement is that increments $B(t)-B(s)$ are independent from $B(r)$ for $r\leq s\leq t$. If you do something like $B(t)-2B(s)=[B(t)-B(s)]-B(s)$, the bracketed term is independent of $B(s)$ but clearly the second $B(s)$ term is not.