Distributional equality Let $(W_t)_{t\geq0}$ be a standard Brownian motion. I have to show that the following equality holds in distribution. Does someone has a good hint to show this?
$\sup_{t \geq 0}( |W_t| -t) = \sup_{t \geq 0} \left( W_t \cdot \frac{1}{1 + t} \right)^2$
Is there something particular to use to deal with the square on the RHS?
Any help is appreciated. Thanks
More general result (point 1°) for p):

 A: First note that $x^qa^{-q} = q\cdot\sup_{\lambda>0}(\lambda x - p^{-1}\lambda^p a^p)$. Then
$$
\sup_{t\ge 0} \left(\frac{X_t}{1+t^{p/2}}\right)^q = q\cdot\sup_{t\ge 0}\sup_{\lambda>0}\left(\lambda X_t - p^{-1}\lambda^p (1+t^{p/2})^p\right)\\ = \sup_{\lambda>0}\sup_{t\ge 0}\left(q\lambda X_t - qp^{-1}\lambda^p (1+t^{p/2})^p\right)\\ \overset{(d)}= \sup_{\lambda>0}\sup_{t\ge 0}\left( X_{q\vphantom{\lambda}^2\lambda^2t} - qp^{-1}\lambda^p (1+t^{p/2})^p\right) = \sup_{\lambda>0}\sup_{s\ge 0}\left( X_{s} - qp^{-1}\lambda^p (1+s^{p/2}\lambda^{-p}q^{-p})^p\right)\\
= \sup_{s\ge 0}\sup_{\lambda>0}\left( X_{s} - qp^{-1}\lambda^p (1+s^{p/2}\lambda^{-p}q^{-p})^p\right) = \Big| \lambda^* = (p-1)^{1/p}s^{1/2}q^{-1}\Big|\\
= \sup_{s\ge 0} \Big(X_s -qp^{-1}(p-1)s^{p/2}\Big) = \Big| q= p(p-1)^{-1}\Big| = \sup_{s\ge 0} \Big(X_s - s^{p/2}\Big),
$$
as required.

There is, however, a problem with the $\overset{(d)}{=}$ sign. It may be correct, but it is not straightforward. It would be obvious if we had the following equality in distribution:
$$
\big\{q\lambda X_t,t\ge 0,\lambda > 0\big\} \overset{(d)}{=} \big\{ X_{q^2\lambda^2 t},t\ge 0,\lambda > 0\big\}.
$$
But this is certainly false. We only have that
$$
\big\{q\lambda X_t,t\ge 0\big\} \overset{(d)}{=} \big\{ X_{q^2\lambda^2 t},t\ge 0\big\},
$$
for any $\lambda > 0$, which is not enough.

Thanks to @saz, here is a correct solution: for any $x>0$
$$ \mathbb{P} \left( \sup_{t \geq 0} \left( \frac{X_t}{1+t^{p/2}} \right)^q \leq x \right) = \mathbb{P} \left( \sup_{t \geq 0} \left( x^{1/p} \frac{X_t}{1+t^{p/2}} \right)^q \leq x^q \right) \\ = \mathbb{P} \left( \sup_{t \geq 0} \left( \frac{X_{t x^{2/p}}}{1+t^{p/2}} \right)^q \leq x^q \right)  = \mathbb{P} \left( \sup_{u \geq 0}  \frac{X_{u}}{1+u^{p/2}x^{-1}}  \leq x \right) \\ = \mathbb{P} \left( \sup_{u \geq 0}  \frac{X_u}{x + u^{p/2}}  \leq 1 \right) = \mathbb{P} \left( \forall\ u \geq 0\quad X_u \le x+ u^{p/2}\right)\\ = \mathbb{P} \left( \sup_{u \geq 0} \big(X_u -u^{p/2} \big)\le x\right). $$
