# Image of a symmetric law

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.

• First part is trivial: $P(Y(X) \in V)=P(X \in Y^{-1}(V)=P(-X \in Y^{-1}(V)P(Y(-X) \in V)$. – Kavi Rama Murthy Mar 18 at 10:20

I recall the following:

Let $$X,X':\Omega\to E$$ be two random variables and $$f:E\to F$$ 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.

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 $$\mathbb P(B_\tau=1)=\mathbb P(B_\tau=-1)=\frac12$$.

• F is measurable? – lucmobz Mar 18 at 13:03
• You can easily check that $\{F(b)\le t\}=\bigcap_{k\ge1}\bigcup_{s\in[0,t]\cap\mathbb Q_+}\{\vert b(s)\vert\ge1-\frac 1k\}$, hence it is measurable. – Will Mar 18 at 13:23
• Thank you you've been very helpful and detailed. – lucmobz Mar 18 at 15:54