Brownian motion: Why $p\{-x\leq B_t\leq x \mid B_{t_n}=\pm x_n,...,B_{t_1}=\pm x_1\}=p\{-x\leq B_t\leq x\mid B_{t_n}=x_n,...,B_{t_1}=x_1\}$

Question

Let $(B_t)$ be a Brownian motion. Why: $$p\{-x\leq B_t\leq x \mid B_{t_n}=\pm x_n,...,B_{t_1}=\pm x_1\}=p\{-x\leq B_t\leq x\mid B_{t_n}=x_n,...,B_{t_1}=x_1\}\ \ \ ?$$

Answer 1

Assuming $t > t_n$ you can reduced your relationship to

$$p\{-x\leq B_t\leq x \mid B_{t_n}=\pm x_n\}=p\{-x\leq B_t\leq x\mid B_{t_n}=x_n\}$$

because $B_t$ is Markov.

I'll proceed formally: \begin{eqnarray*} p\{-x\leq B_t\leq x \mid B_{t_n}=\pm x_n\} &=& \frac{p\{-x\leq B_t\leq x \text{ and } B_{t_n}=\pm x_n\}}{p\{B_{t_n}=\pm x_n\}} \\ &=& \frac{2p\{-x\leq B_t\leq x \text{ and } B_{t_n}= x_n\}}{2p\{B_{t_n}= x_n\}} \\ &=& p\{-x\leq B_t\leq x\mid B_{t_n}=x_n\} \\ \end{eqnarray*}