Supremum of Brownian motion

Question

I am trying to understand the proof in "Roger and Williams" for the Lemma

Lemma: Let $B_t$ be a Brownian motion, then$$P(\sup_t B_t=+\infty,\inf_t B_t=-\infty)=1$$

Let $Z:=\sup_t B_t$, they started with observing that, for $c>0$ and by scaling property of Brownian motion, $$cZ \stackrel{d}{=} Z$$. Therefore, the law of $Z$ is concentrated on $0$ or $+\infty$.

Question:

1. Why is $cZ\stackrel{d}{=}Z$? I understand that, under scaling property, $cB_{t/c^2}$ is again a Brownian motion and the law for $B_t$ and $cB_{t/c^2}$ are the same. How can I see that its also hold for their supremum rigorously?
2. Why is the law of $Z$ concentrated on ${0,\infty}$? Why do they have to restrict $c>0$? What goes wrong with the argument if we use $c\neq 0$ instead?

Answer 1

For 1., consider the function $f$ that takes a continuous function on $[0, \infty)$ and gives its supremum. If you already convinced yourself that $B_t$ and $\tilde{B}_t = c B_{t/c^2}$ have the same law as processes then you have that $Z = f(B_\cdot)$ and $\tilde{Z} = f(\tilde{B}_\cdot)$ have the same law too. Then, for any fixed $c > 0$ we have $$ P[Z \in A] = P[\tilde{Z} \in A] = P[f(c B_{\cdot/c^2}) \in A] = P[c f(B_{\cdot/c^2}) \in A] = P[c Z \in A], $$ where the last equality follow from noticing that the supremum of a function with domain $[0, \infty)$ is not altered by a temporal rescaling.

For 2., just notice that this property holds for all $c > 0$. Thus, for any fixed $M > 0$ if you make $c \to \infty$ $$ P[Z \geq M] = P[Z \geq M/c] = P[Z > 0], $$ implying that $P[Z = \infty] = P[Z \geq M, M \in \mathbb{N}] = \lim_M P[Z \geq M] = P[Z > 0]$. Since $Z$ is supported on $[0, \infty]$ we have $P[Z = 0] + P[Z = \infty] = 1$. Finally, you need to argue why one of these probabilities is zero. In general you can have a random variable $Z$ on $[0, \infty]$ that assumes ie, $0$ or $\infty$ with probability $1/2$ and it would satisfy $Z = cZ$ in distribution for every $c > 0$. In your specific case you should use some property of the Brownian motion to justify this last step.

Finally, I think it should be clearer now why we consider $c > 0$. If we use $c < 0$ then we would get an infimum when we make the first step. This is valid, and amounts to the fact the infimum of a Brownian motion has the same distribution as the supremum.

Answer 2

Since the Brownian motion has continuous sample paths, it holds that $$Z = \sup_{t \geq 0} B_t > a \iff \exists t \in (0,\infty) \cap \mathbb{Q}: B_t>a.$$ If we denote by $(t_k)_{k \in \mathbb{N}}$ an enumeration of $(0,\infty) \cap \mathbb{Q}$, this gives $$\mathbb{P}(Z>a) = \lim_{n \to \infty}\mathbb{P} \left( \bigcup_{k=1}^n \{B_{t_k}>a\} \right)$$ which shows that the distribution of $Z$ depends only on the finite dimensional distributions of $(B_{t_1},\ldots,B_{t_n})$, $n \in \mathbb{N}$. Consequently, if $(W_t)_{t \geq 0}$ is another process with continuous sample paths which has the same finite-dimensional distributions as $(B_t)_{t \geq 0}$, then $$\sup_{t \geq 0} B_t \stackrel{d}{=} \sup_{t \geq 0} W_t.$$ Applying this for $W_t := c B_{t/c^2}$ with $c>0$ we get

$$c \sup_{s \geq 0} B_s = \sup_{t \geq 0} W_t \stackrel{d}{=} \sup_{t \geq 0} B_t$$

which proves $c Z \stackrel{d}{=} Z$. Hence,

\begin{align*} \mathbb{E}\left(1_{\{Z<\infty\}} (1- e^{-Z}) \right) = \mathbb{E} \left( (1-e^{-cZ}) 1_{\{cZ<\infty\}} \right) &= \mathbb{E} \left( (1-e^{-cZ}) 1_{\{Z<\infty\}} \right) \\ &\xrightarrow[]{c \downarrow 0} 0. \end{align*}

As $1-e^{-Z} \geq 0$ on $\{Z<\infty\}$ this implies $1-e^{-Z}=0$ almost surely on $\{Z<\infty\}$, i.e. $Z=0$ on $\{Z<\infty\}$. Consequently, we have shown that the law of $Z$ is concentrated on $0$ and $+\infty$.

Regarding the 2nd part of your second question: For $c=0$ the process $(W_t)_{t \geq 0}$ is clearly not well-defined; for $c<0$ we still have $cZ\stackrel{d}{=} Z$ (which is not surprising because of the symmetry of Brownian motion).