Let $B_t$ be a standard Brownian motion. Let $E_{j, n}$ denote the event$$\left\{B_t = 0 \text{ for some }{{j-1}\over{2^n}} \le t \le {j\over{2^n}}\right\},$$and let$$K_n = \sum_{j = 2^n + 1}^{2^{2n}} 1_{E_{j,n}},$$where $1$ denotes indicator function. I have three questions.

  1. What is $\lim_{n \to \infty} \mathbb{P}\{K_n = 0\}$?
  2. What is $\lim_{n \to \infty} 2^{-n} \mathbb{E}[K_n]$?
  3. Does there exist $\rho > 0$ such that for $\mathbb{P}\{K_n \ge \rho2^{n}\} \ge \rho$ for all $n$?

I'll address the second question on the expected value of the sum $K_n$.

Let $\phi(x)$ and $\Phi(x)$ be the probability density function and cumulative distribution functions for a standard normal distribution.

Let $h(a,b)$ be the probability that a Brownian motion without drift returns to $0$ at some time in $[a,b]$. Let $h(b)=h(1,b)$. Then by rescaling, $h(a,b)=h(1,b/a)=h(b/a)$. We can calculate this exactly.

For $x \gt 0$ let $f(x,t)$ be the probability that a Brownian motion released at position $x$ will hit $0$ by time $t$. Let $f(x) = f(x,1)$. By rescaling, $f(x,t)= f(\frac{x}{\sqrt{t}},1) = f(\frac{x}{\sqrt{t}})$. By reflection, $f(x) = 2 \Phi(-x) = 2-2\Phi(x)$.

$$\begin{eqnarray}h(b) &=& 2 \int_0^\infty \phi(x) f(x,b-1)~dx \newline &=&2 \int_0^\infty \phi(x)\cdot 2 \Phi\left(\frac{-x}{\sqrt{b-1}}\right)~dx\end{eqnarray}$$

We can use differentiation under the integral sign. $h(1)=0$ and $h(b) = \int_1^b h'(t) dt$.

$$\begin{eqnarray} h'(b) &=& 4 \int_0^\infty \phi(x) \phi\left(\frac{x}{\sqrt{b-1}}\right) \left(\frac{1}{2} \frac{x}{(b-1)^{3/2}}\right) dx \newline &=&\frac{2}{(b-1)^{3/2}}\int_0^\infty x \phi(x) \phi\left( \frac{x}{\sqrt{b-1}}\right) dx \newline &=& \frac{2}{(b-1)^{3/2}} \int_0^\infty \frac{x}{2 \pi} e^{-x^2 \cdot \left(\frac{1}{2} + \frac{1}{2(b-1)}\right)}dx \newline &=& \frac{1}{\pi b \sqrt{b-1}}\end{eqnarray}$$

So, $h(b) = \int_1^b \frac{dy}{\pi y \sqrt{y-1}} = 1-\frac{2}{\pi} \arcsin \frac{1}{\sqrt{b}}$.

$\mathbb{P}(E_{j+1,n})$ is the probability that the Brownian motion returns to $0$ on $\left[\frac{j}{2^n},\frac{j+1}{2^n}\right]$ which is $h(\frac{j}{2^n},\frac{j+1}{2^n}) = h(1 + \frac{1}{j}) = 1-\frac{2}{\pi} \arcsin \frac{1}{\sqrt{1+1/j}}.$ That can be simplified to $1-\frac{2}{\pi}(\frac{\pi}{2} - \arctan \frac{1}{\sqrt{j}}) = \frac{2}{\pi}\arctan \frac{1}{\sqrt{j}}$. For large $j$, this is approximately $\frac{2}{\pi} \frac{1}{\sqrt{j}}$.

$$\mathbb{E}(K_n) \sim \sum_{j=2^n}^{2^{2n}-1} \frac{2}{\pi} \frac{1}{\sqrt{j}} \approx \frac{2}{\pi} \int_{2^n}^{2^{2n}} \frac{1}{\sqrt{x}} dx =\frac{4}{\pi}(2^n -2^{n/2}) \sim \frac{4}{\pi} 2^n.$$


Regarding question 1, the limiting probability of not crossing 0 during the time interval $[1,2^n]$ is $$ \lim_{n \to \infty} \mathbb{P}\{K_n = 0\} = 0 $$ since Brownian motion is recurrent (in dimensions 1 and 2).


Is this it for my 2nd question? We want to compute the probability of zero crossing in $[t, t + \Delta]$ for $\Delta = 1/2^n$. This is (conditioning on the value $z$ at time $t$ and using the reflection principle):$$A = 2 \int_0^\infty dz {{e^{-z^2/2t}}\over{\sqrt{2\pi t}}} \cdot 2 \int_z^\infty dy {{e^{-y^2/2\Delta}}\over{2\pi\Delta}}.$$By change of variables, this the same as$${4\over{2\pi\sqrt{t}}} \int_0^\infty dz e^{-z^2/2t} \int_{z/\sqrt{\Delta}}^\infty dxe^{-x^2/2}.$$the main contribution comes (except for $t \sim \Delta$ or a bit more, which eventually only will contribute a constant/logarithmic term) from $z \sim \sqrt{\Delta}$ and so we can replace the first exponential by $1$ and finally get $A = {{2\sqrt{\Delta}}\over{\pi\sqrt{t}}}$. Summing over $t = i\Delta$ gives the limit as ${4\over\pi}$. I'm not very sure of this solution though, so anyone could check it that would be nice...


For 2.

Claim: $\lim_{n\rightarrow\infty}2^{-n/2}\mathbb{E}[K_n]=0$

The idea is this: The union of the events $E_{jn}$ is the event \begin{equation} F_n=\left\{B_t=0 \text{ for some $t\in\left[\frac{2^n+1}{2^n},\frac{2^{2n}}{2^n}\right]$} \right\} \end{equation} with probability less or equal than 1. The intersection of events $E_{jn}$ and $E_{j+1,n}$ is \begin{equation} G_{jn}=\left\{B_{j/2^n}=0\right\} \end{equation} which has probability equal to zero for all $j,n$. Then \begin{eqnarray} \mathbb{E}[K_n] & = & \mathbb{E}\left[\sum_{j=2^n+1}^{2^{2n}}1_{E_{jn}}\right]\\ & = & \sum_{j=2^n+1}^{2^{2n}}\mathbb{E}[1_{E_{jn}}]\\ & = & \sum_{j=2^n+1}^{2^{2n}}\mathbb{P}[E_{jn}]\\ & = & \mathbb{P}[F_{n}]+\sum_{j=2^n+1}^{2^{2n}}\mathbb{P}[G_{jn}]\\ & \leq & 1 \end{eqnarray}

This implies $\lim_{n\rightarrow\infty}2^{-n/2}\mathbb{E}[K_n]=0$

  • 3
    $\begingroup$ The intersection of $E_{j,n}$ with $E_{j+1,n}$ means that the Brownian motion hits $0$ on both of these intervals, not that it hits $0$ at the intersection of the intervals. This argument is not correct. By the way, the upper limit in the question is $2^{2n}$ rather than $2^{2^n}$. $\endgroup$ – Douglas Zare Oct 17 '15 at 11:58
  • $\begingroup$ @DouglasZare You are right. Thanks for pointing this out. $\endgroup$ – Nate River Oct 17 '15 at 12:06

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