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Let $W$ be a Brownian motion. In a calculation, I have to compute $$\mathbb P \left(\sup_{0\leq s\leq t}|W_s|>a\right).$$ My idea would be to use the reflexion principle that says that $$\mathbb P\left(\sup_{0\leq s\leq t}W_s>a \right)=2\mathbb P(W_t>a).$$ $\sup_{0\leq s\leq t}|W_s|$ instead of $\sup_{0\leq s\leq t}W_s$, I'm not sure how to pursue. Can I do as follow :

$$\mathbb P\left(\sup_{0\leq s\leq t}|W_s|>a\right)=\mathbb P\left(\sup_{0\leq s\leq t}W_s>a\right)+\mathbb P\left(\sup_{0\leq s\leq t}W_s<-a\right)$$ $$=2\mathbb P(W_t>a)+1-\mathbb P(W_t>-a)\tag{*}$$ where I used $$\mathbb P\left(\sup_{0\leq s\leq t}W_s<x\right)=1-\mathbb P\left(\sup_{0\leq s\leq t}W_s\geq x \right)=1-2\mathbb P(W_t\geq x).\tag{**}$$

My first doubt is on the correctness of the first equality in $(*)$. Also, for $(**)$ I'm not sure because in the reflexion principle inequality are strict, whereas I used non strict inequality.

saz
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user657324
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  • (**) looks correct for the first equality definitely. Based on this being true, the first equality is also true. Non strict inequality appears to hold according to wikipedia as well – fGDu94 Jun 19 '19 at 20:36
  • @GeorgeDewhirst: since $(W_t)$ is continuous, don't we have that $\sup_{0\leq s\leq t}W_s$ also continuous ? – user657324 Jun 19 '19 at 20:38
  • yes this is definitely true – fGDu94 Jun 19 '19 at 20:38

1 Answers1

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The first "=" in (*) fails to be true. Note that

$$\left\{ \sup_{0 \leq s \leq t} |W_s|> a \right\} \neq \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \sup_{0 \leq s \leq t} W_s < -a \right\}. \tag{1}$$

Say, for instance, we have a sample path with $\sup_{s \leq t} W_s(\omega)<a$ but $\inf_{s \leq t} W_s(\omega)<-a$, then $\omega \in \{\sup_{s \leq t} |W_s|>a\}$ but $\omega \notin \{ \sup_{0 \leq s \leq t} W_s > a\} \cup \left\{ \sup_{0 \leq s \leq t} W_s < -a \right\}$.

It does hold true that

\begin{align*} \left\{ \sup_{0 \leq s \leq t} |W_s|> a \right\} &= \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \color{red}{\inf_{0 \leq s \leq t}} W_s < -a \right\} \\ &= \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \sup_{0 \leq s \leq t} (-W_s) > a \right\}. \tag{2} \end{align*}

However, the events $$A:=\left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \qquad B := \left\{ \sup_{0 \leq s \leq t} (-W_s) > a \right\}$$ are not disjoint, and therefore $$\mathbb{P}(A \cup B) \neq \mathbb{P}(A) + \mathbb{P}(B),$$ i.e. $$ \mathbb{P} \left( \sup_{0 \leq s \leq t} |W_s|> a \right) \neq \mathbb{P} \left( \sup_{0 \leq s \leq t} W_s > a \right) + \underbrace{\mathbb{P} \left( \sup_{0 \leq s \leq t} (-W_s) > a \right)}_{=\mathbb{P} \left( \sup_{0 \leq s \leq t} W_s > a \right)}.$$ Remark: The distribution of $\sup_{0 \leq s \leq t} |W_s|$ has been discussed e.g. here and here.

saz
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  • I don't get why $A$ and $B$ are not disjoint. Set $W_t^=\sup_{0\leq s\leq t}W_t$. If $\omega \in A$, then $W_t^(\omega )>a$, so how can $\omega \in B$ ? We would have $W_t^(\omega )>a$ and $W_t^(\omega )<-a$ which is not possible (otherwise $W_t^*$ wouldn't be well defined as a function). – user657324 Jun 20 '19 at 08:13
  • @user657324 Sorry, my mistake; please see the edited answer. – saz Jun 20 '19 at 08:36
  • Sorry to ask you my question here, but I remember that once you answered the question on exit problem of solution of an equation of the form $dX_t=\alpha X_tdt+\varepsilon dW_t$ using martingal problem (i.e. take a solution of $xf'(x)+\frac{\varepsilon }{2}f=0$ and use Dynkin's formula). I tried to see in your post, but there are to many post. Do you think you could find this answer in your post or it's to complicated to find old answer ? – user657324 Aug 19 '19 at 13:23
  • @user657324 Do you mean this one...? – saz Aug 19 '19 at 14:32
  • Yes exactly. Thank you very much. Is there a technic to find quickly previous post or you remember where to search ? – user657324 Aug 19 '19 at 14:52
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    @user657324 Not really. You can try to specify tags and/or the user who wrote the question/answer but unfortunately it doesn't always work. I searched for "user:36150 [stopping-times] [sde]" (...which worked out) – saz Aug 19 '19 at 14:56
  • I understand now you often correct tags on questions. It makes sense now. Many thanks for your help :) – user657324 Aug 19 '19 at 16:27