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Let $(B_t)_{0\leq t\leq 1}$ be a Brownian motion. I want to show that $$\sup_{0\leq t\leq 1} |B_t| \stackrel{d}{=} \frac{1}{\sqrt{\tau}}$$ where $\tau:=\inf\{t>0: |B_t|\geq 1 \}$.

My idea was to pick an arbitrary $q\in \Bbb{R}$ and show that $\Bbb{P}\left(\sup_{0\leq t\leq 1} |B_t|\leq q\right)=\Bbb{P}\left( \frac{1}{\sqrt{\tau}}\leq q\right)$. But $$\begin{align}\Bbb{P}\left( \frac{1}{\sqrt{\tau}}\leq q\right)&=\Bbb{P}\left( \frac{1}{\tau}\leq q^2\right)\\&=\Bbb{P}\left(\sup\left\{\frac{1}{t}: t\in A\right\}\leq q^2\right)\end{align}$$ where $A:=\{t>0: |B_t|\geq 1\}$. But now I don't get further, can someone help me?

Summerday
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    Hint. Use the scale invariance of Brownian motion: $$\left(B_{t/q^2}\right){0\le t\le1}\stackrel d=\left(\tfrac1qB_t\right){0\le t\le 1}.$$ – nejimban Apr 29 '23 at 18:24
  • @nejimban thanks for your hint. So you mean use it at the beginning or is it okey up to the point above? – Summerday Apr 29 '23 at 18:26
  • I would start like this: $$\Bbb P!\left(\frac1{\sqrt\tau}\le q\right)=\Bbb P!\left(\tau\ge\frac1{q^2}\right)=\Bbb P!\left(\sup_{0\le t\le\frac1{q^2}}|B_t|\le1\right)=\ldots$$ but you may surely get to it via another path. – nejimban Apr 29 '23 at 18:28
  • @nejimban okey I will try it. Can I ask you also something. Why do I know that $\left(B_{t/q^2}\right){0\le t\le1}\stackrel d=\left(\tfrac1qB_t\right){0\le t\le 1}$? Because we have called the scale invariance of BM to be the following lemma: Let $B$ be a Brownian motion then the scaling invariance property says that $\forall c>0$, $X:=(c^{-1/2} B_{ct})_t$ is again a brownian motion. But I don't get from this that they are equal in distribution – Summerday Apr 29 '23 at 18:31
  • @Summerday They have the same distribution as a Process. What do you mean that you don't get that they are equal in distribution ? – Mr. Gandalf Sauron Apr 29 '23 at 18:33
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    Take $c=\frac1{q^2}$. So both processes $(B_t){0\le t\le 1}$ and $(qB{t/q^2})_{0\le t\le 1}$ are Brownian motions started from $0$. – nejimban Apr 29 '23 at 18:33
  • @nejimban ah I See it now you above have already divided by $q$ I was confused about this sorry – Summerday Apr 29 '23 at 18:35
  • @Summerday I have edited in a way which essentially expands on your method and completes it. But still, you have to use Nejimban's hint – Mr. Gandalf Sauron Apr 29 '23 at 19:22

1 Answers1

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Use the excellent hint of Nejimban. Notice that as $\large(\frac{B_{t}}c)_{t\in [0,1]}$ has the same distribution as a Process with $\displaystyle\large (B_{\frac{t}{c^{2}}})_{t\in [0,1]}$ . This means that these two processes have same finite dimensional distributions. So essentially they can be thought of as the same process(as far as distributions and hence probabilities are concerned) . So $(\frac{|B_{t}|}{q})_{t\in[0,1]}$ as a process has same distribution as $\displaystyle(|B_{\frac{t}{q^{2}}}|)_{t\in[0,1]}$ .

$$P(\max_{t\in [0,1]}\frac{|B_{t}|}{q}\leq 1) = P(\max_{t\in[0,1]}|B_{\frac{t}{q^{2}}}|\leq 1)$$

Now $\max_{t\in[0,1]}|B_{\frac{t}{q^{2}}}|\leq 1$ means $\{\tau\geq \frac{1}{q^{2}}\}=\{\sqrt{\tau}\geq \frac{1}{q}\}=\{\frac{1}{\sqrt{\tau}}\leq q\} $

(The first equality just means that if the maximum , $\max_{t\in[0,1]}|B_{\frac{t}{q^{2}}}|\leq 1$ then we could not have escaped $(-1,1)$ by time $\frac{1}{q^{2}}$ and vice versa) .

Alternatively , here's a way to prove it using your initial idea:-

$$P(\tau\geq\frac{1}{q^{2}})=P(\max_{t\in[0,\frac{1}{q^{2}}]}|B_{t}|\leq 1)$$

And $(B_{t})_{t\in [0,\frac{1}{q^{2}}]}=(B_{t/q^{2}})_{t\in[0,1]}$

Thus again , $P(\max_{t\in[0,\frac{1}{q^{2}}]}|B_{t}|\leq 1)=P(\max_{t\in [0,1]}|B_{\frac{t}{q^{2}}}|\leq 1)=P(\max_{t\in[0,1]}q|B_{\frac{t}{q^{2}}}|\leq q)$

But again due to time-scale invariance , $(qB_{\frac{t}{q^{2}}})_{t\in[0,1]}$ is a Standard Brownian Motion. And hence ,

$P(\max_{t\in[0,1]}q|B_{\frac{t}{q^{2}}}|\leq q)=P(\max_{t\in[0,1]}|B_{t}|\leq q)$ and that's basically it.

Mr. Gandalf Sauron
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  • Perfect thanks a lot for the second approach. Can I ask you for interest, is it possible/easy to compute $\Bbb{E}(1/\sqrt{\tau})$? – Summerday Apr 29 '23 at 19:19
  • Aha @Summerday , good question . I suggest you look at this . It is not so very easy to handle . But let me think if there's an easy way out . – Mr. Gandalf Sauron Apr 29 '23 at 19:22
  • In class he said something that the expectation should give something like $\sqrt{\frac{\pi}{2}}$ therefore I thought maybe it is not really hard to get it. – Summerday Apr 29 '23 at 19:24
  • You can try work with this . Maybe you can get an approximation for this. – Mr. Gandalf Sauron Apr 29 '23 at 19:26
  • Okey thanks a lot. I also thought about the following. In the last lecture we did a proof exactly before he told us this remark above, in this proof he showed the following claim : $$X\stackrel{d}{=}\sup_{0\leq t\leq 1} |B_t|$$where $X:=\sup_{0\leq s\leq 1}\left{\sup_{0\leq t\leq s} {B_t-B_s}\right}$ therefore I thought that maybe this is also useful since we now would know with this exercise that $X\stackrel{d}{=} \sqrt{\frac{1}{\tau}}$. But this is only an idea – Summerday Apr 29 '23 at 19:31
  • Yeah that's right. But still, that won't make the job of finding the expectation any easier though. – Mr. Gandalf Sauron Apr 29 '23 at 19:37
  • Okey perfect thanks for your help I will try a bit more. – Summerday Apr 29 '23 at 19:38
  • You'll end up with this ugly integral $\sum_{n=0}^{\infty}\int_{0}^{\infty}(-1)^{n}\exp(-\frac{\pi^{2}(2n+1)^{2}}{8x^{2}})\frac{(2n+1)\pi}{8x^{4}},dx$ . – Mr. Gandalf Sauron Apr 29 '23 at 19:41
  • Yes I also got something like this with your post. – Summerday Apr 29 '23 at 19:42
  • I'm afraid I don't see a way out other than numerical approximation – Mr. Gandalf Sauron Apr 29 '23 at 19:44
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    No problem. Thanks anyways! – Summerday Apr 29 '23 at 19:45
  • I have asked a new question and don't know how to tag your name, therefore I wanted to ask here if you maybe also could help me with my new question? – Summerday Apr 30 '23 at 15:23
  • Sorry, I have not done Ito Calculus in a long while. But I'll take a look at it. – Mr. Gandalf Sauron Apr 30 '23 at 15:26
  • Okey no worries, but it is not a deep question it's one for beginners since I have started with it really new – Summerday Apr 30 '23 at 15:27