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 suspect the answer is yes; I've tried messing around with the second moment method, but not to much avail. Can this be shown with the second moment method? Or should I be trying something else?
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$\begingroup$ First, should your sum not be: $$ K_n = \sum_{j=2^n+1}^{2^{n+1}} $$ as your event hints that the rate of growth of $K_n$ is $2^n$ so one would expect your sum to have $2^{n+1}$ terms, no? $\endgroup$– Grant IzmirlianDec 22, 2015 at 21:32
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Not the answer, but possibly useful reformulation
I assume that comment made above is right (that is sum has $2^{n+1}$ terms).
Denote $$p_n(\rho)=P(K_n>\rho 2^n)=P(K_n/2^n>\rho )$$ Observe that $p_n(\rho_1)>p_n(\rho_2)$ if $\rho_1 < \rho_2$
First point: if you ask whether such $\rho$ exists for all n, you need to show that for some $\delta $ the limit is positive $$\lim_{n\rightarrow \infty} p_n(\delta)>0$$ then, if $p_n(\delta)$ has positive limit and all values are positive, it must be separated from zero, let's say $p_n(\delta)>\varepsilon$. Then $$p_n(\min(\varepsilon,\delta)) \geq p_n(\delta)>\varepsilon \geq \min(\varepsilon,\delta)$$ so you have desired property for $\rho=\min(\varepsilon, \delta)$.
So you just need to show the limit of $p_n$ to be positive.
I would then investigate the variable $K_n/2^n$ and its expected value
