How can I show that $\sup_{t \geq 0}(B^*_t-\mu t^{r/2})=_{law} \sup_{v \geq 0}(\lambda B^*_{v}-\mu \lambda^rv^{r/2})$ I am trying to show that
$\sup_{t \geq 0}(B^*_t-\mu t^{r/2})=_{law} \sup_{v \geq 0}(\lambda B^*_{v}-\mu \lambda^rv^{r/2})$ 
i.e the above two random variables have the same distribution where $(B_t)_{t \in \mathbb{R_+}}$ is  a brownian motion and $B^*_t=\sup_{s \leq t} |B_s|$
I am stuck. How should I proceed? I was thinking of using the scaling property of brownian motion but I cannot write down a precise answer. Can someone help me out
 A: You're right that you use the scaling property, ie $ \tilde B_t = c B_{t/c^2} $ ($c>0$) is a Brownian motion. Try this with some generic $c$, and then you should be able to guess what $c$ should be (in terms of $\lambda$). Recall also that "$\sup_{t\ge0} \iff \sup_{t/c^2\ge0}$", as $c^2 > 0$.
If you're still stuck, I'll show you in the morning. (I'm in the UK, so it's late at night here!)

Putting in as I suggested, the RHS becomes
$$ \sup_{v\ge0} (\lambda c B^*_{v/c^2} - \mu \lambda^r v^{r/2})
= \sup_{V\ge0} (\lambda c B^*_V - \mu \lambda^r (c^2 V)^{r/2} )
= \sup_{V\ge0} (\lambda c B^*_V - \mu (\lambda c)^r V^{r/2}), $$
where $V = v/c^2$. So choose $c \lambda = 1$, ie $c = 1/\lambda$, and since $v$ and $V$ are just dummy variables, this gives you your solution.
I leave it as a (hopefully!) very straightforward exercise to show that the change of variables for $B$ also holds for $B^*$. If you just plug in the change of variables I've done in the first line, it should drop out very quickly. :)
