Assume μ>0 .Using the climax properties of the Brownian Motion prove that these 2 given random values $Z=\underset{t\geq0}{sup}(\left|W_{t}\right|-\mu t)$ and $Y=\frac{1}{μ}\underset{t\geq0}{sup}\left(\frac{W_{t}}{1+t}\right)^{2}$ have the same distribution.
Attemp: $u=\frac{t}{1-t}$ so $\frac{W_{t}}{1+t}$ becomes $(1-u)W\left(\frac{u}{1-u}\right)$ I also used some properties from the brownian bridge but I can't figure out how exactly can this be solved.
$\mathbb{P}\left[\frac{1}{μ}\underset{t\geq0}{sup}\left(\frac{W_{t}}{1+t}\right)^{2}\leq a\right]=\mathbb{P}\left[\frac{1}{μ}\underset{t\geq0}{sup}\left[\left(\frac{W_{t}}{1+t}\right)^{2}-μa\right]\leq0\right]=\mathbb{P}\left[\underset{t\geq0}{sup}\left[\left(\frac{W_{t}}{1+t}\right)^{2}-μa\right]\leq0\right]$
$=\mathbb{P}\left[\underset{0\leq t<1}{sup}\left[\left((1-t)W_{\frac{t}{1-t}}\right)^{2}-μa\right]\leq0\right]$
For an l>0
$\mathbb{P}\left[Z\leq l\right]=\mathbb{P}\left[\underset{t\geq0}{sup}\left\{ \left|W_{t}\right|-\mu t\right\} \leq l\right]\overset{\forall a>0}{=}\mathbb{P}\left[\underset{s\geq0}{sup}\left\{ \frac{\left|W_{as}\right|}{\sqrt{a}}-\mu\sqrt{a}s\right\} \leq\frac{l}{\sqrt{a}}\right]=$
$=\mathbb{P}\left[\underset{t\geq0}{sup}\left\{ \left|W_{s}^{(a)}\right|-\mu\sqrt{a}s\right\} \leq\frac{l}{\sqrt{a}}\right]$we choose $\alpha_{0}=\frac{l}{\mu}$
$\mathbb{P}\left[\underset{s\geq0}{sup}\left\{ \left|W_{s}^{(a_{0})}\right|-\mu\sqrt{\frac{l}{\mu}}s\right\} \leq\frac{l}{\sqrt{\frac{l}{\mu}}}\right]=\mathbb{P}\left[\underset{s\geq0}{sup}\left\{ \left|W_{s}^{(a_{0})}\right|-\sqrt{\mu}\sqrt{l}s\right\} \leq\sqrt{\mu}\sqrt{l}\right]=$
$=\mathbb{P}\left[\underset{s\geq0}{sup}\left\{ \left|W_{s}^{(a_{0})}\right|-\sqrt{\mu}\sqrt{l}(1+s)\right\} \leq0\right]=\mathbb{P}\left[\forall s\geq0,\left|W_{s}^{(a_{0})}\right|-\sqrt{\mu}\sqrt{l}(1+s)\leq0\right] $
and we divide with$ \left\{ 1+s\right\} \geq1,\forall s\geq0$
$\mathbb{P}\left[\forall s\geq0,\left\{ \frac{{\left|W_{s}^{(a_{0})}\right|}}{(1+s)}-\sqrt{\mu}\sqrt{l}\right\} \leq0\right]=\mathbb{P}\left[\underset{s\geq0}{sup}\left\{ \frac{{\left|W_{s}^{(a_{0})}\right|}}{(1+s)}-\sqrt{\mu}\sqrt{l}\right\} \leq0\right]=$ $\mathbb{P}\left[\underset{s\geq0}{sup}\left\{ \frac{\left|W_{s}^{(a_{0})}\right|}{1+s}\right\} \leq\sqrt{\mu}\sqrt{l}\right]$
$\mathbb{P}\left[\underset{s\geq0}{sup}\left(\frac{{\left|W_{s}^{(a_{0})}\right|}}{1+s}\right)\leq\sqrt{\mu}\sqrt{l}\right]=\mathbb{P}\left[\underset{s\geq0}{sup}\left(\frac{{\left|W_{s}^{(a_{0})}\right|}}{1+s}\right)^{2}\leq\mu l\right]=\mathbb{P}\left[\frac{1}{\mu}\underset{s\geq0}{sup}\left(\frac{{\left|W_{s}^{(a_{0})}\right|}}{1+s}\right)^{2}\leq l\right]=$ $=\mathbb{P}\left[\frac{1}{μ}\underset{t\geq0}{sup}\left(\frac{\left|W_{t}\right|}{1+t}\right)^{2}\leq l\right]=\mathbb{P}\left[Y\leq l\right]$