Evaluating the distribution involving a Brownian motion. I'm trying to solve one exercise closely related to this question. Since I don't have an answer yet, I thought to post a new question with my thoughts about the problem. I hope this does not break any rule of this forum.
The problem is the following: we let $(X_u)_{u \ge 0}:=(u+1)b_{\frac{u}{u+1}},$ where $(b_u)_{0\le u \le 1}$ is a Brownian Bridge. One can show that $(X_u)_{u\ge 0}$ is a Brownian Motion. Then we want to show that

$$ \sup_{u \geq 0} \left( |X_u| - u \right) \overset{(d)}{=} \sup_{0 \leq u \leq 1 } b_u^2.$$

What we did (me and some friends) was this: first we noticed that 
$$\sup\limits_{u\ge 0}\left(\frac{X_u}{u+1}\right)^2=\sup\limits_{u\ge 0}b_{\frac{u}{u+1}}^2 =\sup\limits_{0 \le u \le 1}b_u^2.$$
Then to show that $\sup\limits_{0 \le u \le 1}b_u^2$ and $\sup\limits_{u\ge 0}\left(\frac{X_u}{u+1}\right)^2$ are equally distributed, we tried to manipulate the distribution function of the two random variables using time inversion and scaling of the Brownian Motion. So we tried to bring the LHS of the next expression to the RHS.
$$\Bbb{P}\left( \sup\limits_{u\ge 0}\left(\frac{X_u}{u+1}\right)^2\le x\right)= \Bbb{P}\left(  \sup_{u \geq 0} \left( |X_u| - u \right)\le x\right).$$
The main problem with this is the square factor on the LHS. So, to overcome the problem, we thought it would have been great if, instead of $x$, we had a $1$. Because then,
$$\Bbb{P}\left( \sup\limits_{u\ge 0}\left(\frac{|X_u|}{u+1}\right)^2\le 1\right) = \Bbb{P}\left( \sup\limits_{u\ge 0}\left(\frac{|X_u|}{u+1}\right)\le 1\right) =\Bbb{P}\left( \sup\limits_{u\ge 0}|X_u| - u \le 1\right), $$
which is the expression we have to prove. Also, I guess we proved correctly that 
$$\Bbb{P}\left( \sup\limits_{u\ge 0}\left(\frac{X_u}{u+1}\right)^2\le 1\right) = \Bbb{P}\left( \sup\limits_{u\ge 0}\left(\frac{|X_u|}{u+1}\right)^2\le 1\right).$$

Now we don't know how to explain the fact that we can let $x=1$, do you know first, whether this is true and secondly how to prove it? 

This idea came us also, because we saw, by a scaling argument, that for a Brownian motion $(B_t)_{t \ge 0}$ and for $M > 0$,
$$\Bbb{P}\left( \sup_{s\ge 0} B_s > 1\right) = \Bbb{P}\left( \sup_{s\ge 0}B_s> M\right).$$
 A: Fix $x > 0$. By the scaling property, we know that
$$\tilde{X}_t := \sqrt{x} X_{t/x}$$
is a Brownian motion. Consequently, we have
$$\begin{align*} \mathbb{P} \left( \sup_{u \geq 0} \left( \frac{|X_u|}{u+1} \right)^2 \leq x \right) &= \mathbb{P} \left( \sup_{u \geq 0} \left( \sqrt{x} \frac{|X_u|}{u+1} \right)^2 \leq x^2 \right) \\ &= \mathbb{P} \left( \sup_{u \geq 0} \left( \frac{|\tilde{X}_{ux}|}{u+1} \right)^2 \leq x^2 \right) \\ &= \mathbb{P} \left( \sup_{u \geq 0} \left( \frac{|X_{ux}|}{u+1} \right)^2 \leq x^2 \right) \\ &= \mathbb{P} \left( \sup_{u \geq 0}  \frac{|X_u|}{\frac{u}{x}+1}  \leq x \right) \\ &= \mathbb{P} \left( \sup_{u \geq 0} \frac{|X_u|}{x+u} \leq 1 \right) \end{align*}$$
Since
$$ \frac{y}{x+u} \leq 1 \iff y \leq x+u \iff y-u \leq x$$
it is not difficult to see that this implies
$$ \mathbb{P} \left( \sup_{u \geq 0} \left( \frac{|X_u|}{u+1} \right)^2 \leq x \right) = \mathbb{P} \left( \sup_{u \geq 0} (|X_u|-u) \leq x \right).$$
This proves the desired identity for $x > 0$. For $x \leq 0$ the identity is trivial since $\sup_u |X_u|^2/(u+1)^2$ is a non-negative random variable.
