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I know that by the reflection principle, $$ P\left[\sup_{0 < s < t} B_s > a \right] = 2P[B_t> a] $$ where $B_t$ is a Brownian Motion. But what is $P\left[\sup_{0 < s < t} |B_s|> a \right]$?

Tohiko
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    Hint: $B_t$ has the same distribution as $-B_t$ and $P(\sup_{0 < s < t} |B_s| > a) = P(\sup_{0 < s < t}B_t > a$ or $\inf_{0 < s < t} B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly. – Chris Janjigian Jul 30 '18 at 16:52

1 Answers1

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Let $\Phi_t(x)$ be the cdf for $B_t$. Then $$ \mathbb P\left(\sup_{0< s< t} |B_s|\le a\right)=\sum_{k=-\infty}^\infty(-1)^k\Big(\Phi_t\big(a(2k+1)\big)-\Phi_t\big(a(2k-1)\big)\Big). $$

Mike Earnest
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  • Nice! How do you prove this? Also, do you know a way to lower bound this for large $t$? – Thomas Ahle Nov 23 '20 at 12:50
  • @ThomasAhle Here's my proof. Since the terms decrease in absolute value as $k\to \pm\infty$, you get a lower bound by taking only the $k=-1,0,1$ terms. That is, $P(\sup |B_s|\le a)\ge -\Phi(3a)+2\Phi(a)-2\Phi(-a)+\Phi(-3a)$. – Mike Earnest Nov 23 '20 at 17:09
  • Thanks! I'm interested in the case $a$ constant and $t$ large, (same as $a\to 0$ and $t=1$). It seems I need at least $t$ terms of the sum for the lower bound to still be meaningful ($> 0$). Ideally, I'm hoping for a bound like $\Pr(\sup_{0<s<t}|B_s|\le a) \ge e^{-(t/a)^2/2}$. – Thomas Ahle Nov 23 '20 at 19:43
  • I feel like this should be the same probability as https://math.stackexchange.com/a/2332298/7072 (the first time $W_t\not\in(-a,a)$. However, I get quite different probabilities. E.g. for $a=1$ and $t=2$ your answer gives $P\approx 0.0087$, while that answer is $\approx 0.108$. Am I missing a distinction here? – Thomas Ahle Nov 23 '20 at 20:38
  • @ThomasAhle Using Mathematica, I also get $\approx 0.108$ with my formula. Here is an online Mathematica compiler which verifies this. – Mike Earnest Nov 24 '20 at 16:03
  • I thought NormalDistribution[] took the variance and not the standard deviation :D. That explains it. Thanks! – Thomas Ahle Nov 24 '20 at 16:18