# Supremum of Brownian motion

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?

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.

• Why can you say $P(Z=\infty)=P(Z \geq 1/M,M \in \Bbb N)$? – quallenjäger Jan 21 at 15:48
• Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = \infty] = P[Z \geq M, M \in \mathbb{N}]$ and then you use that these events are nested. – Daniel Jan 21 at 17:13

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 $$W_t := c W_{t/c^2}$$ 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).

• Thank you for your answer! Why can I infer from $cZ \stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+\infty}$ – quallenjäger Jan 21 at 14:53
• @quallenjäger $cZ \stackrel{d}{=} Z$ implies $1_{\{cZ<\infty\}} cZ \stackrel{d}{=} Z 1_{\{Z<\infty\}} =: \tilde{Z}$. If you consider the characteristic function of $\tilde{Z}$ , you will thus find that $\phi(c) = \phi(1)$ for all $c$, i.e. $\tilde{Z}=0$. – saz Jan 21 at 14:58
• sorry, I accepted the other answer. Your answer is great as well as always. Thank you – quallenjäger Jan 21 at 15:34
• @quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+\infty$. – saz Jan 21 at 17:14