<s>[INST] Number of intervals needed to cross, Brownian motion 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.


*

*What is $\lim_{n \to \infty} \mathbb{P}\{K_n = 0\}$?

*What is $\lim_{n \to \infty} 2^{-n} \mathbb{E}[K_n]$?

*Does there exist $\rho > 0$ such that for $\mathbb{P}\{K_n \ge \rho2^{n}\} \ge \rho$ for all $n$?

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