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This is a followup to my question here.

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. Does there exist $\rho > 0$ such that for $\mathbb{P}\{K_n \ge \rho2^{n}\} \ge \rho$ for all $n$? I have tried messing around with the second moment method to show existence, but not to much avail. Would the second moment method be a viable approach? Or is this doomed to fail?

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Here is some notation I used in the related problem with some results. Let the probability hat a Brownian motion starting at $0$ returns to home on $[a,b]$ be $h(a,b) = h(b/a)$. By the calculation in my answer there, $h(b) = 1- \frac{2}{\pi} \arcsin \frac{1}{\sqrt{b}} = \frac{2}{\pi} \arctan \sqrt{b-1} $. So, $\mathbb{P}(E_{j+1,n}) = h(\frac{j}{2^n},\frac{j+1}{2^n}) = h(1+\frac{1}{j}) = \frac{2}{\pi} \arctan \frac{1}{\sqrt{j}} \le \frac{2}{\pi \sqrt{j}}.$

$\mathbb{E}(K_n) \sim \frac{4}{\pi}2^n.$

To see if the second moment method can work, let's estimate the second moment of $K_n$.

$$\begin{eqnarray}\mathbb{E} [K_n^2] &=& \sum_{i=2^n}^{2^{2n}-1} \sum_{j=2^n}^{2^{2n}-1} \mathbb{P}(E_{i+1,n} \cap E_{j+1,n}) \newline &=& \sum_{i=2^n}^{2^{2n}-1} \left[ \mathbb{P}(E_{i+1,n}) \sum_{j=2^n}^{2^{2n}-1} \mathbb{P}(E_{j+1,n} | E_{i+1,n}) \right] \newline &\le& 2 \sum_{i=2^n}^{2^{2n}-1} \left[\mathbb{P}(E_{i+1,n}) \sum_{k=0}^{2^{2n}} \mathbb{P}(E_{i+k+1}|E_{i+1,n}) \right] \newline &\le& 2 \sum_{i=2^n}^{2^{2n}-1} \mathbb{P}(E_{i+1,n}) \left(2 + \sum_{k=1}^{2^{2n} } h(k,k+1)\right)\newline &\le& 2 \mathbb{E}(K_n)(c+ \frac{2}{\pi}2^n) \newline &\sim& \frac{16}{\pi^2}2^{2n}\end{eqnarray}.$$

I think the constant factor is right. Once we have this upper estimate on the second moment, approximately the square of the expected value, the Paley-Zygmund inequality gives us bounds on the probability that $K_n$ is small.

$$\mathbb{P}(K_n \gt \rho \mathbb{E}[K_n]) \ge (1-\rho)^2 \frac{\mathbb{E}[K_n]^2}{\mathbb{E}[K_n^2]}. $$

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