Question about the $\textit{sup} B_t = \infty$ of a Brownian motion

Question

It has been answered here before using reflection principle, martingale and other methods. I have a basic technique, using scale invariance I believe, in my notes in which I don't understand a couple of steps.

Let $M = \text{sup}_t B_t$ and let $$\begin{align*} \widetilde{M} &= \text{sup}_t cB_{t/c^2} \\ &= \text{sup}_t cB_{t} \\ &= c\text{sup}_t B_{t} \\ &= cM \\ \implies \widetilde{M} &= M \text{ in distribution} \\ \implies cM &= M \text{ in distribution} \\ \iff M &\in\{0,\infty\} \end{align*}$$

May I know why

1) the second step is true i.e. $\text{sup}_t cB_{t/c^2} = \text{sup}_t cB_{t}$

2) $\widetilde{M}=cM\implies \widetilde{M}=M$ in distribution?

If the above are true I have the following to show that $M=\infty$ a.s. (about which I have a question):

$$ \begin{align*} p = P(M=0)\leq P(B_1\leq 0,\text{max}_{s\geq 1}B_s \leq 0) &\leq P(B_1\leq 0, \text{sup}_{s\geq 0}B_{1+s}-B_1 = 0) \\ &= P(B_1\leq 0)P(\text{sup}_{s\geq 0}B_{1+s}-B_1 = 0) \\ \implies p&= \frac{1}{2}p \\ \implies p&=0 \end{align*} $$

3) May I know why the second inequality is true i.e. $P(B_1\leq 0,\text{max}_{s\geq 1}B_s \leq 0) \leq P(B_1\leq 0, \text{sup}_{s\geq 0}B_{1+s}-B_1 = 0)$ or in other words why does $\text{max}_{s\geq 1}B_s \leq 0 \implies \text{sup}_{s\geq 0}B_{1+s}-B_1 = 0$? Thanks.

Answer 1

1. Because the set $\{t/c^2 : t > 0\}$ is the same as the set $\{t : t > 0\}$, i.e. $(0,\infty)$. So both are just taking the supremum of the same Brownian motion over all times.

2. $M$ and $\widetilde{M}$ are equal in distribution because both of them are the supremum of a standard Brownian motion (though not the same one) over all times. This is not a consequence of the fact that $\widetilde{M} = cM$, which you proved separately; but using these two facts together, you conclude that $cM = M$ in distribution.

3. The inequality is not correct for a general process; indeed, in general it goes the other way, since if $B_1 < 0$ then $B_{1+s} - B_1$ could be strictly positive for some $s$ and still have $\max_{s \ge 1} B_s \le 0$. So you need to go back to the drawing board on this step. (Of course, it happens to be true for Brownian motion, since both sides equal zero, but you can't assume that here or your proof will be circular.)