Assume I have a probability space $(\Omega, \mathcal{F}, P)$ that is mapped by a measurable function $X$ into $(E,\mathcal{E})$, moreover $P(X \in U)=P(-X \in U)$, now $Y$ maps this measurable space into $(G, \mathcal{G})$.
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=\inf\{t: 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:
This property gives you the answer of your two questions.
First of all, if $\mathbb P(X\in U)=\mathbb P(-X\in U)$ for all $U\in\mathcal E$, 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 $V\in\mathcal G$.
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 $B:\Omega\to C(\mathbb R_+,\mathbb R)$, which has the same distribution as $-B$. Let $F:C(\mathbb R_+,\mathbb R)\to\mathbb R$ be defined for all $b\in C(\mathbb R_+,\mathbb R)$ by $$ F(b)=\inf\{t\in\mathbb R_+\mid b(t)\notin(-1,1)\} $$
Clearly, for all $b\in C(\mathbb R_+,\mathbb R)$, we have $F(b)=F(-b)$. Let $H:C(\mathbb R_+,\mathbb R)\to\mathbb R$ be defined for all $b\in C(\mathbb R_+,\mathbb R)$ by $$ H(b)=b\left(F(b)\right) $$
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, $B_\tau\in\{-1,1\}$ by continuity of the sample paths. Since $B_\tau$ and $-B_\tau$ have the same distribution we have $\mathbb P(B_\tau=1)=\mathbb P(B_\tau=-1)=\frac12$.