# Second moment method, Brownian motion?

Let $B_t$ be a standard Brownian motion. Let $E_{j, n}$ denote the eventand letwhere $1$ 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?

Not the answer, but possibly useful reformulation

I assume that comment made above is right (that is sum has $2^{n+1}$ terms).

Denote
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
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 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