2

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 :-)

idm
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  • Draw a PICTURE! (And are you really sure you do not understand (3)?) – Did Jun 01 '15 at 13:38
  • I did it, but I still don't see. Could you give me more information ? – idm Jun 01 '15 at 17:52
  • What do you fail to understand in (3)? And in (4)? – Did Jun 01 '15 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 02 '15 at 09:50
  • 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 02 '15 at 10:05
  • But look, if $t<\tau$ it's clear that ${\max_u B_u\geq a, B_t\geq a}=\emptyset={\max_u\tilde B_u,B_t\geq a}$. Now if $t>\tau$, suppose $\omega\in {\max_u B_u\geq a,B_t\geq a}$. Then $\max_u\tilde B_u(\omega)=\max_u{2a-B_u(\omega)}\leq 2a-\max_u(\omega)\leq 2a-a\leq a$. What's wrong here ? – idm Jun 08 '15 at 13:59
  • @idm Isn't this reflection principle? – Idonknow Dec 30 '19 at 13:43

1 Answers1

1

Partial answer: (2) follows from the fact that B tilde is again a Brownian process, so the probability is the same.

wiskundeliefhebber
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  • 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 ? – idm Jun 02 '15 at 09:42
  • If both B and B tilde are Brownian motions, then they have the same probability distributions. Therefore, whatever you calculate for B tilde has the same result if you replace B tilde with B. – wiskundeliefhebber Jun 02 '15 at 12:30