If we define that $X_t$ is Brownian motion over space $(\Omega,\mathcal F ,\mathcal F_t;P) $,
then why is it true that the fact that $X_t$ is Brownian motion implies that $P\{X_t=-X_t \}=1$ is impossible?
I would be grateful if anyone could help me.
best , Educ
Answered by Henrik in the comments: $P(X_t=-X_t)=P(X_t)=0$ because $X_t$ has a continuous distribution (namely, normal).
That said, the Brownian motion started at $0$ has the following symmetry property: for every measurable set $A\subset \mathbb R$ and every $t\ge 0$, $$P(X_t\in A) = P(-X_t\in A)$$ Which is a consequence of $X_t$ and $-X_t$ having the same distribution.