Let $(B_t)_{0\leq t\leq 1}$ be a Brownian motion. I want to show that $$\sup_{0\leq t\leq 1} |B_t| \stackrel{d}{=} \frac{1}{\sqrt{\tau}}$$ where $\tau:=\inf\{t>0: |B_t|\geq 1 \}$.
My idea was to pick an arbitrary $q\in \Bbb{R}$ and show that $\Bbb{P}\left(\sup_{0\leq t\leq 1} |B_t|\leq q\right)=\Bbb{P}\left( \frac{1}{\sqrt{\tau}}\leq q\right)$. But $$\begin{align}\Bbb{P}\left( \frac{1}{\sqrt{\tau}}\leq q\right)&=\Bbb{P}\left( \frac{1}{\tau}\leq q^2\right)\\&=\Bbb{P}\left(\sup\left\{\frac{1}{t}: t\in A\right\}\leq q^2\right)\end{align}$$ where $A:=\{t>0: |B_t|\geq 1\}$. But now I don't get further, can someone help me?
Use the excellent hint of Nejimban. Notice that as $\large(\frac{B_{t}}c)_{t\in [0,1]}$ has the same distribution as a Process with $\displaystyle\large (B_{\frac{t}{c^{2}}})_{t\in [0,1]}$ . This means that these two processes have same finite dimensional distributions. So essentially they can be thought of as the same process(as far as distributions and hence probabilities are concerned) . So $(\frac{|B_{t}|}{q})_{t\in[0,1]}$ as a process has same distribution as $\displaystyle(|B_{\frac{t}{q^{2}}}|)_{t\in[0,1]}$ .
$$P(\max_{t\in [0,1]}\frac{|B_{t}|}{q}\leq 1) = P(\max_{t\in[0,1]}|B_{\frac{t}{q^{2}}}|\leq 1)$$
Now $\max_{t\in[0,1]}|B_{\frac{t}{q^{2}}}|\leq 1$ means $\{\tau\geq \frac{1}{q^{2}}\}=\{\sqrt{\tau}\geq \frac{1}{q}\}=\{\frac{1}{\sqrt{\tau}}\leq q\} $
(The first equality just means that if the maximum , $\max_{t\in[0,1]}|B_{\frac{t}{q^{2}}}|\leq 1$ then we could not have escaped $(-1,1)$ by time $\frac{1}{q^{2}}$ and vice versa) .
Alternatively , here's a way to prove it using your initial idea:-
$$P(\tau\geq\frac{1}{q^{2}})=P(\max_{t\in[0,\frac{1}{q^{2}}]}|B_{t}|\leq 1)$$
And $(B_{t})_{t\in [0,\frac{1}{q^{2}}]}=(B_{t/q^{2}})_{t\in[0,1]}$
Thus again , $P(\max_{t\in[0,\frac{1}{q^{2}}]}|B_{t}|\leq 1)=P(\max_{t\in [0,1]}|B_{\frac{t}{q^{2}}}|\leq 1)=P(\max_{t\in[0,1]}q|B_{\frac{t}{q^{2}}}|\leq q)$
But again due to time-scale invariance , $(qB_{\frac{t}{q^{2}}})_{t\in[0,1]}$ is a Standard Brownian Motion. And hence ,
$P(\max_{t\in[0,1]}q|B_{\frac{t}{q^{2}}}|\leq q)=P(\max_{t\in[0,1]}|B_{t}|\leq q)$ and that's basically it.