# Brownian motion maximum inequality

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.

• en.wikipedia.org/wiki/Reflection_principle_(Wiener_process) – d.k.o. Jan 23 at 2:14
• Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions? – John Doe Jan 23 at 16:55
• What is $|B(t)|$ for higher dimensions? – d.k.o. Jan 23 at 23:06
• 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 ...? – John Doe Jan 24 at 0:08
• Is it $\ell_2$ norm? – 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)}|
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|