Second moment method, Brownian motion?

Let Bt be a standard Brownian motion. Let Ej,n denote the event{Bt=0 for some j12ntj2n},and letKn=22nj=2n+11Ej,n,where 1 denotes indicator function. Does there exist ρ>0 such that for P{Knρ2n}ρ 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 2n+1 terms).

Denote pn(ρ)=P(Kn>ρ2n)=P(Kn/2n>ρ)
Observe that pn(ρ1)>pn(ρ2) if ρ1<ρ2

First point: if you ask whether such ρ exists for all n, you need to show that for some δ the limit is positive lim
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

Source : Link , Question Author : Student , Answer Author : krzmip

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