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Assume I have a probability space $(Omega, mathcalF, P)$ that is mapped by a measurable function $X$ into $(E,mathcalE)$, moreover $P(X in U)=P(-X in U)$, now $Y$ maps this measurable space into $(G, mathcalG)$.

Is it true that: $P(Y(X) in V)=P(Y(-X) in V)$ or not?

Under which conditions it is.

If we have a continuos Brownian motion $B$ and a stopping time $tau=inft: B_t notin(-1,1) $, is it true (and why) that:
$B_tau$ has the same law as $-B_tau$ ?
And how do you compute those laws.

Thanks.

I recall the following:

Let $X,X':Omegato E$ be two random variables and $f:Eto F$ be a
measurable map (of course with the same $sigma$-algebra for $E$).

If $X$ and $X'$ have the same distribution, then $f(X)$ and $f(X')$
have the same distribution as well.

This property gives you the answer of your two questions.

First of all, if $mathbb P(Xin U)=mathbb P(-Xin U)$ for all $Uinmathcal E$, then $X$ and $-X$ have the same distribution, so $Y(X)$ and $Y(-X)$ have the same distribution and $mathbb P(Y(X)in V)=mathbb P(Y(-X)in V)$ for all $Vinmathcal G$.

Second, let $C(mathbb R_+,mathbb R)$ denote the set of continuous functions from $mathbb R_+$ to $mathbb R$. Then your brownian motion $B$ is a random variable $B:Omegato C(mathbb R_+,mathbb R)$, which has the same distribution as $-B$. Let $F:C(mathbb R_+,mathbb R)tomathbb R$ be defined for all $bin C(mathbb R_+,mathbb R)$ by
$$
F(b)=inftinmathbb R_+mid b(t)notin(-1,1)
$$

Clearly, for all $bin C(mathbb R_+,mathbb R)$, we have $F(b)=F(-b)$. Let $H:C(mathbb R_+,mathbb R)tomathbb R$ be defined for all $bin C(mathbb R_+,mathbb R)$ by
$$
H(b)=bleft(F(b)right)
$$

Since $B$ and $-B$ have the same distribution, $H(B)=B_F(B)$ and $H(-B)=-B_F(-B)=-B_F(B)$ have the same distribution. Yet $F(B)=tau$. So $B_tau$ and $-B_tau$ have the same distribution.

Moreover, $B_tauin-1,1$ by continuity of the sample paths. Since $B_tau$ and $-B_tau$ have the same distribution we have $mathbb P(B_tau=1)=mathbb P(B_tau=-1)=frac12$.