showing $\mathbb{P}(\sup_{t}B_t>0) =1$ where $B_t$ is a brownian motion

Question

Showing $\mathbb{P}(\sup_{t}B_t>0) =1$ where $B_t$ is a brownian motion

What I have done so far: $$\mathbb{P}(\sup_{t}B_t>0) =1- \mathbb{P}(\sup_{t}B_t\leq 0) $$ so it suffices to show $\mathbb{P}(\sup_{t}B_t\leq 0) = 0$ $$\mathbb{P}(\sup_{t}B_t\leq 0) = \mathbb{P}(\forall t \geq 0,\; B_t \leq 0)$$ let $\{t_n\}_{n\in \mathbb{N}}$ be a strictly increasing sequence $$\mathbb{P}(\bigcap_{n\geq 1} \{B(t_n-t_{n-1})\leq 0\} ) \stackrel{(1)}{=}\mathbb{P}(\bigcap_{n\geq 1} \{B(t_{n-1}+ t_n-t_{n-1})-B(t_{n-1})\leq 0\} ) \stackrel{indep.\; increment}{=} \prod_{n\geq 1}\mathbb{P}(B(t_n)-B(t_{n-1})\leq 0 ) \stackrel{(2)}{=} \prod_{n\geq 1}\mathbb{P}(B(t_n-t_{n-1})\leq 0 ) = \prod_{n\geq 1} \frac{1}{2}=0$$ To conclude $$\mathbb{P}(\forall t \geq 0,\; B_t \leq 0) \leq \mathbb{P}(\bigcap_{n\geq 1} \{B(t_n-t_{n-1})\leq 0\} ) =0 $$ My main question: Are equalities (1) and (2) correct? My reasoning would be $$B_{t_{n-1}}(t) = B(t_{n-1}+t)-B(t_{n-1})$$ is a brownian motion hence $$B(t_n -t_{n-1}) \sim B_{t_{n-1}}(t_n -t_{n-1}) =B(t_{n-1}+ t_n-t_{n-1})-B(t_{n-1})$$ from this observation, equality (1) and (2) follow.

Finally was there a better technique to show $\mathbb{P}(\sup_{t}B_t>0) =1$

Answer 1

Just consider the Brownian motion restricted to $[0,1]$ . By Desire-Andre Reflection Principle or by just using Strong Markov Property, it can be shown that $\sup_{t\in[0,1]}B_{t}$ has same distribution as $|B_{1}|$ which has same distribution $|X|$ where $X\sim N(0,1)$.

Now $P(\sup_{t\in[0,1]} B_{t}>0)\leq P(\sup_{t}B_{t}>0)$

But $P(\sup_{t\in[0,1]} B_{t}>0)=P(|X|>0)=1-P(|X|=0)=1-0=1$ as $|X|$ has continuous distribution.

Thus $P(\sup_{t}B_{t}>0)=1$

To prove that $\sup_{t\in [0,s]}B_{t}\sim |B_{s}|$, consider $\tau_{x}$ to be the hitting time of $x$ such that $x>0$ and let $M_{s}=\sup_{t\in [0,s]}B_{t}$

Thus $P(M_{s}\leq x)=P(\tau_{x}\geq s)$

Now, $P(B_{s}>x)=P(B_{s}>x,\tau_{x}>s)+P(B_{s}>x,\tau_{x}\leq s)=P(B_{s}>x,\tau_{x}\leq s)$ as the first probability is $0$ by intermediate value property of the continuous brownian paths.

Now $P(B_{s}>x,\tau_{x}\leq s)=P(B_{\tau_{x}}+(s-\tau_{x})>x,\tau_{x}\leq s)=P(B_{\tau_{x}+(s-\tau_{x})}-B_{\tau_{x}}>0)P(\tau_{x}\leq s)$.

But $P(B_{\tau_{x}+(s-\tau_{x})}-B_{\tau_{x}}>0)=\frac{1}{2}$ as by Strong Markov Property, $B_{\tau_{x}+t}-B_{\tau_{x}}$ is a standard brownian motion.

Thus $P(B_{s}>x)=P(B_{s}>x,\tau_{x}\leq s)=\frac{1}{2}P(\tau_{x}\leq s)$

Thus $P(M_{s}\geq x)=P(\tau_{s}\leq x)=2\cdot P(B_{s}>x)=P(|B_{s}|\geq x)$

Answer 2

I will address your last question. By reflection principle we have for $a>0$ $$P(\sup_{s\leq t}B_s\geq a)=2P(B_t\geq a)=2(1-\Phi(a/\sqrt{t}))$$ Since $\lim_{a\downarrow 0}P(\sup_{s\leq t}B_s\geq a)=P(\sup_{s\leq t}B_s>0)$ and $\Phi$ (the Gaussian cdf) is continuous we have $$P(\sup_{s\leq t}B_s>0)= \lim_{a\downarrow 0}2(1-\Phi(a/\sqrt{t}))=2(1-\Phi(0))=2\frac{1}{2}=1$$ This holds for any $t$, and $\sup_sB_s\geq \sup_{s\leq t}B_s$ for any $t$. We conclude.