I am supposed to show this inequality for a 2 or 3 dimensional Brownian motion:

$\textbf{P}^0\{\sup\limits_{t\leq k}{|B(t)|}\geq\frac{1}{2}\}\leq 2\textbf{P}^0\{{|B(k)|\geq\frac{1}{2}}\}$ (where $\textbf{P}^0$ means the BM is started in zero).

This looks a lot like Doob's maximum inequality to me, but on the RHS there is a probabilty instead of an expectation and also $|B|$ is not a martingale for d=2,3. I think you can fix the last part by just taking $\ln|B|$ and $|B|^{-1}$ instead, but I don't see how to replace the expectation with a probability.

Thanks in advance!

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Reflection_principle_(Wiener_process) $\endgroup$ – d.k.o. Jan 23 at 2:14
  • $\begingroup$ Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions? $\endgroup$ – John Doe Jan 23 at 16:55
  • $\begingroup$ What is $|B(t)|$ for higher dimensions? $\endgroup$ – d.k.o. Jan 23 at 23:06
  • $\begingroup$ I am not sure what you mean ... it's of course length of the random vector with independent Brownian motions in it's entries. But I don't think $|B(t)|$ is a Brownian motion again, is it ...? $\endgroup$ – John Doe Jan 24 at 0:08
  • $\begingroup$ Is it $\ell_2$ norm? $\endgroup$ – d.k.o. Jan 24 at 7:52

For $a>0$ let $\tau_a:=\inf\{t\ge 0:|B_t|=a\}$. Then $$ \mathsf{P}^0(\tau_a\le t)=\mathsf{P}^0(\tau_a\le t,|B_t|\ge a)+\mathsf{P}^0(\tau_a\le t,|B_t|< a).\tag{1} $$

Using the strong Markov property (e.g. Theorem 8.3.7 on page 314 here), \begin{align} \mathsf{P}^0(\tau_a\le t,|B_t|< a)&=\mathsf{E}^0[1\{\tau_a\le t\}\mathsf{P}^0(|B_{\tau_a+(t-\tau_a)}|<a\mid \mathcal{F}_{\tau_a})] \\ &=\mathsf{E}^0[1\{\tau_a\le t\}\mathsf{P}^{B_{\tau_a}}(|B_{t-\tau_a}|<a)]\le \frac{1}{2}\mathsf{P}^0(\tau_a\le t).\tag{2} \end{align}

Combining $(1)$ and $(2)$, we get $$ \mathsf{P}^0\left(\sup_{s\le t}|B_t|\ge a\right)=\mathsf{P}^0(\tau_a\le t)\le 2\mathsf{P}^0(|B_t|\ge a). $$

It remains to verify the last inequality in (2), i.e. for any $x$ with $|x|=a$ and $s\ge 0$, $$ \mathsf{P}^x(|B_s|<a)\le 1/2. $$


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