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My teacher said it was because of the symmetry but I don't really understand ): somebody can help me?

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closed as off-topic by Lee David Chung Lin, Shailesh, José Carlos Santos, YuiTo Cheng, zhoraster May 6 at 11:54

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  • $\begingroup$ Try reviewing the properties that define the Brownian motion. You can reduce to the case $s = 0$. $\endgroup$ – Lorenzo May 6 at 5:05
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$B_t > B_s$ if and only if $B_t - B_s >0 $. Hence, $\mathbb{P}(B_t - B_s >0) = \mathbb{P}(B_t > B_s)$. Now, $B_t - B_s$ has distribution $\mathcal{N}(0,t-s)$, (by definition of Brownian motion) which is symmetric around the origin.

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