# Brownian motion: Why $p\{\max_{0\leq u\leq t} B_u\geq a\}=2p\{B_t\geq a\}$?

Let $(B_t)$ a standard Brownian motion (i.e. $B_t\sim\mathcal N(0,t)$). Let $a\geq 0$. Prove that $$p\left\{\max_{0\leq u\leq t} B_u\geq a\right\}=2p\{B_t\geq a\}.$$

The proof goes like this : Set $$\tau=\begin{cases}\inf\{u\geq 0\mid b_u=a\}&\text{if}\ \{u\geq 0\mid b_u=a\}\neq\emptyset\\ +\infty &\text{if}\ \{u\geq 0\mid b_u=a\}=\emptyset\end{cases}.$$ Let $$\tilde B_t=\begin{cases}B_t&\text{if }t<\tau\\ a-(B_t-a)&\text{if }t\geq\tau\end{cases}.$$

We have that \begin{align*} p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\geq a\right\}&\underset{(1)}{=}p\left\{\max_{0\leq u\leq t}\tilde B_u\geq a,\tilde B_t\leq a\right\}\\ &\underset{(2)}{=}p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\leq a\right\}. \end{align*}

Therefore \begin{align*} p\left\{\max_{0\leq u\leq t}B_u\geq a\right\}&\underset{(3)}{=} p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\geq a\right\}+p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\leq a\right\}\\ &=2p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\geq a\right\}\\ &\underset{(4)}{=}2p\{B_t\geq a\}. \end{align*}

I'm sincerely sorry, but I don't understand $(1)$, $(2)$, $(3)$ and $(4)$. Any explanation is welcome.

Thank you :-)

• Draw a PICTURE! (And are you really sure you do not understand (3)?)
– Did
Jun 1, 2015 at 13:38
• I did it, but I still don't see. Could you give me more information ?
– idm
Jun 1, 2015 at 17:52
• What do you fail to understand in (3)? And in (4)?
– Did
Jun 1, 2015 at 18:05
• for $(3)$ I thing this come from the fact that $([a,+\infty [\times ]-\infty ,a])\cap([a,+\infty [\times ]a+\infty [)=\emptyset$ and since $B_t$ is continuous, $p\{..., B_t>a\}=p\{...,B_t\geq a\}$, so $(3)$ is fine. For $(4)$ I think that $p\{\max B_u\geq a,B_t\geq a\}=p\{B_t\geq a\}\underbrace{p\{\max B_u\geq a\mid B_t\geq a\}}_{=1}$ so it's fine too. But I'm totally stuck with $(1)$ and $(2)$. Could you explain please ? (I know that I have an answer of wiskundeliefhebber for $(2)$ but I don't see why the fact that $(B_t)$ and $(\tilde B_t)$ are Brownien motion imply that $(2)$ is true).
– idm
Jun 2, 2015 at 9:50
• Since (3) and (4) were in fact clear from the start, let us have a look at (1) now: sure you cannot show that the two events involved, coincide? (The simplest way to show their probabilities are equal...) This uses only the pathwise definition of $\tilde B$.
– Did
Jun 2, 2015 at 10:05

• I'm sorry but I don't understand why the fact that there are Brownien motion $(2)$ is valid. Could you give me more explanation ?