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:
- 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?
- 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?
Best Answer
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.