0
$\begingroup$

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.

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

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.

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

$\endgroup$
  • $\begingroup$ F is measurable? $\endgroup$ – lucmobz Mar 18 at 13:03
  • $\begingroup$ 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. $\endgroup$ – Will Mar 18 at 13:23
  • $\begingroup$ Thank you you've been very helpful and detailed. $\endgroup$ – lucmobz Mar 18 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.