6
$\begingroup$

Let $\left(B_t\right)_{t \in \left[0,\infty\right)}$ be a standard Brownian motion over the probability space $\left(\Omega, \mathcal{A}, P\right)$. For each $x \in \left(0, \infty\right)$, give an upper bound $q\left(x\right)$ on $$ P\left(\max_{t \in \left[0,1\right]} B\left(t\right) > x\right) $$

so that the following requirement is satisfied:

$$ \forall \epsilon \in \left(0, \infty\right),\ \Sigma_{n = 1}^\infty q\left(n\varepsilon\right) < \infty $$


Attempted solution #1

Consider Paul Lévy's construction of the Brownian motion on the interval $\left[0,1\right]$ as the limit of piecewise linear functions $F_n$ with knots at the dyadic fractions $\frac{m}{2^n} \in \left[0,1\right]$. Let $\omega \in \Omega$ be such that the path $t \in \left[0, \infty\right) \mapsto B_t\left(\omega\right)$ is continuous. Then if $B_d > x$ for some dyadic fraction, then clearly $\max_{t \in \left[0,1\right]} B_t > x$; whereas if $B_d > x$ for no dyadic fraction, then due to the density of the dyadics on the real line, $\max_{t \in \left[0,1\right]} B_t \leq x$. Therefore, denoting the set of dyadic fractions in the unit interval by $\mathcal{D}$, we have

$$ \left\{\max_{t \in \left[0,1\right]} B_t > x\right\} = \bigcup_{d \in \mathcal{D}}\left\{B_d > x\right\} $$

The only way I know how to bound the probability of a union from above is using the subadditivity property of the probability measure:

$$ P\left(\bigcup_{d \in \mathcal{D}}\left\{B_d > x\right\}\right) \leq \sum_{d \in \mathcal{D}}P\left(B_d > x\right) $$

Now, $B_d \sim N\left(0, d\right)$ and the tightest upper bound on the normal distribution that I'm familiar with is

$$ P\left(N\left(0, d\right) > x\right) = P\left(N\left(0,1\right) > \frac{x}{\sqrt{d}}\right) \leq \frac{1}{\sqrt{2\pi}}\frac{1}{x/\sqrt{d}}\exp\left(-\frac{\left(x/\sqrt{d}\right)^2}{2}\right) $$

The problem is, for any given dyadic point there are an infinity of other dyadic points that lie very closely, and hence yield values that lie very near the expression on the right. Therefore, the infinite sum of these expressions will go to infinity.


Attempted solution #2

According to Wikipedia, $E\left(M\right) = \sqrt{\frac{2}{\pi}}$, where $M$ denotes $\max_{t \in \left[0,1\right]} B_t$. Now, since $0$ is almost surely not a maximum point, $M \geq 0$ a.s., whence, by the Markov inequality

$$ P\left(M > x\right) \leq \frac{\sqrt{2/\pi}}{x} := q\left(x\right) $$

There are two problems with this solution:

  1. It does not satisfy the requirement $\sum q\left(n\varepsilon\right) < \infty$

  2. I haven't learned yet that $E\left(M\right) = \sqrt{\frac{2}{\pi}}$. Ideally, the solution should be based on Lévy's construction of the Brownian motion and on the most basic of properties that can be derived from it.

$\endgroup$
4
$\begingroup$

We have$$\begin{align*} \left\{\max_{t \in [0,1]} B_t> x \right\} &= \bigcup_{n \in \mathbb{N}} \bigcup_{k=1}^{2^n} \{B_{k/2^n}>x\} = \bigcup_{n \in \mathbb{N}} \bigcup_{k=1}^{2^n} \{B_{k/2^n}>x+c_n\} \\ &= \bigcup_{n \in \mathbb{N}} \bigcup_{k=1}^{2^n-1} \{B_{k/2^n}>x+c_n, B_{(k+1)/2^n}-B_{k/2^n} \leq -c_n\} \cup \{B_1>x\}\end{align*}$$

for any sequence $c_n \downarrow 0$. For the last equality note that $$\{B_{k/2^n}>x+c_n\} \subseteq \{B_{k/2^n}>x+c_n,B_{(k+1)/2^n}-B_{k/2^n} \leq -c_n\}\cup \{B_{(k+1)/2^n}>x\}.$$

Hence, by the independence of the increments,

$$\begin{align*} \mathbb{P}\left(\max_{t \in [0,1]} B_t>x \right) &\leq \sum_{n \in \mathbb{N}} \sum_{k=1}^{2^n-1} \mathbb{P}(B_{k/2^n}>x+c_n, B_{(k+1)/2^n}-B_{k/2^n} \leq -c_n) + \mathbb{P}(B_1>x) \\ &\leq \sum_{n \in \mathbb{N}} \sum_{k=1}^{2^n-1} \mathbb{P}(B_{k/2^n}>x) \cdot \mathbb{P}(B_{(k+1)/2^n}-B_{k/2^n} \leq -c_n)+ \mathbb{P}(B_1>x). \tag{1} \end{align*}$$

If we choose $c_n = 2^{-n/4}$, then we see from the stationarity of the increments and the scaling property that

$$\begin{align*} \mathbb{P}(B_{k/2^n}>x) \cdot \mathbb{P}(B_{(k+1)/2^n}-B_{k/2^n} \leq -c_n) &=\mathbb{P}(B_{k/2^n}>x) \cdot \mathbb{P}(2^{-n/2}B_1 \leq -c_n) \\ &= \mathbb{P}(B_{k/2^n}>x) \cdot \mathbb{P}(B_1 \geq 2^{n/4}) \\ &\leq \frac{1}{x^2} \mathbb{E}(B_{k/2^n}^2) \cdot \frac{1}{2^{2n}} \mathbb{E}(B_1^8) \\ &\leq \frac{C}{x^2 \cdot 2^{2n}} \end{align*}$$

for $C:= \mathbb{E}(B_1^8)<\infty$ where we used Markov's inequality in the preliminary inequality. Similarly, $$\mathbb{P}(B_1>x) \leq \frac{1}{x^2} \mathbb{E}(B_1^2) = \frac{1}{x^2}.$$ Plugging these estimates into $(1)$ yields

$$\mathbb{P}\left( \max_{t \in [0,1]} B_t > x \right) \leq \frac{C'}{x^2}.$$

$\endgroup$
  • $\begingroup$ Very clever of you. Thanks. $\endgroup$ – Evan Aad Mar 3 '14 at 4:14
3
$\begingroup$

Recall that $(B_t)_{t \geq 0}$ is a (continuous) martingale with respect to the canonical filtration. Therefore, Doob's maximal inequality shows

$$\mathbb{E} \left( \sup_{t \in [0,1]} B_t^2 \right) \leq 4 \mathbb{E}(B_1^2)=4.$$

Hence, by Markov's inequality

$$\mathbb{P} \left( \sup_{t \in [0,1]} B_t > x \right) \leq \frac{1}{x^2} \mathbb{E} \left( \sup_{t \in [0,1]} B_t^2 \right) \leq \frac{4}{x^2}.$$

Remark Using the reflection principle, one can show that $\max_{t \in [0,1]} B_t \sim |B_1|$ (see e.g. Brownian motion - An Introduction to Stochastic Processes-René Schilling/Lothar Partzsch).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.