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I am trying to clarify a certain part of a proof on non monotonic property of the Brownian motion. Suppose we assume $B_t$ is monotonic in the interval $[a,b]$ and let $\{a=a_1,a_2,...,b=a_n\}$ be a partition. Then on every sub interval for every $i$, $B_{a_i}-B_{a_{i+1}}$ has the same sign. The proof goes on to say that the probability of this is $2/2^{n}.$ I am not quite sure why this is true. By independent increments we do have $$P(B_{a_1}-B_{a_2}>0,...,B_{a_n}-B_{a_{n+1}}>0)=P(B_{a_1}-B_{a_2}>0)P(B_{a_2}-B_{a_3}>0)\cdot \cdot \cdot P(B_{a_n}-B_{a_{n+1}}>0)$$

And I think the idea is $P(B_{a_i}-B_{a_{i+1}}>0)=1/2$ for all $i$ and we get $P(B_{a_1}-B_{a_2}>0,...,B_{a_n}-B_{a_{n+1}}>0)=2^{-n}$. If we do the same for $<0$ and add the two probabilities we end of with $2/2^n$. I am not sure if it is correct to assume $P(B_{a_i}-B_{a_{i+1}}>0)=1/2$. Any clarification is appreciated. Thank you!

Heisenberg
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1 Answers1

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Your argument is correct. It is true that $P(B_{a_i}-B_{a_{i+1}}>0)=1/2$, because the random variable $B_{a_i}-B_{a_{i+1}}$ is normally distributed with mean zero.

(We typically consider the increments in the form $B_{a_{i+1}}-B_{a_i}$, where we subtract the earlier value of $B$ from the later value of $B$, but the argument you've posed is still correct.)

grand_chat
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