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

idm
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1 Answers1

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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*}

muaddib
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  • I just don't understand why: $$\frac{p{-x\leq B_t\leq x \text{ and } B_{t_n}=\pm x_n}}{p{B_{t_n}=\pm x_n}} =\frac{2p{-x\leq B_t\leq x \text{ and } B_{t_n}= x_n}}{2p{B_{t_n}= x_n}}$$ Thanks – idm Jun 09 '15 at 12:13
  • @idm Those both follow from Brownian Motion being symmetric about the origin. If $B_t(\omega)$ is a path then so is $-B_t(\omega)$. Alternatively, $-B_t$ is a Brownian Motion. – muaddib Jun 09 '15 at 12:16