# If Bt is a Brownian movement. Show that for any t and s, P (Bt> Bs) = 1/2. [closed]

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

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

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• Try reviewing the properties that define the Brownian motion. You can reduce to the case $s = 0$. – Lorenzo May 6 at 5:05

$$B_t > B_s$$ 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.