Show that $\sup\limits_tX_t-t^{p/2}$ and $\sup\limits_t\left(\frac{X_t}{1+t^{p/2}}\right)^q$ have the same law

Question

I have a question: Let $X=B^{+}$ or $X=|B|$ where $B$ is the standard Brownian motion.

Set

$$J_p=\sup_{t\geq 0}(X_t-t^{\frac{p}{2}})$$

where $p>1$ and $q$ its conjugate number($p^{-1}+q^{-1}=1$). Prove $J_p$ is a.s. strictly positive and has the same law as

$$\sup_{t\geq 0}\left(\frac{X_t}{1+t^{\frac{p}{2}}}\right)^q$$ Thanks a lot for your help!

Answer 1

In both cases, $(X_t)$ and $(Y_t)$ coincide in law where $Y_t=a^{-1}X_{a^2t}$ for some $a\gt0$. Fix $x\gt0$, then $$ [J_p\leqslant x]=[\forall t,X_t\leqslant x+t^{p/2}]=[\forall t,Y_{t}\leqslant a^{-1}x+a^{p-1}t^{p/2}]. $$ Choosing $a=x^{1/p}$, one gets $$ [J_p\leqslant x]=[\forall t,Y_t\leqslant x^{1/q}+x^{1/q}t^{p/2}]=[\forall t,(1+t^{p/2})^{-q}Y_t^q\leqslant x]=[K_p\leqslant x], $$ where $$ K_p=\sup\limits_{t\geqslant0}\,(1+t^{p/2})^{-q}Y_t^q. $$ This holds for every $x$ hence $J_p=K_p$ almost surely. Finally, $(X_t)$ and $(Y_t)$ coincide in distribution hence $J_p$ coincides in distribution with $\sup\limits_{t\geqslant0}\, Z_t^q$, where $Z_t=X_t/(1+t^{p/2})$.