Abstract

Let $F_{x,t} (z)$ be the distribution function at time t of Brownian motion starting at $x \in [0,1]$ with absorbing boundary points 0 and 1. For \[ t \in \left\{ {0,\frac{1}{{2n^2 }}, \cdots ,\frac{N}{{2n^2 }}} \right\}\qquad {\text{and}}\qquad x \in \left\{ {0,\frac{1}{n}, \cdots ,\frac{k}{n}, \cdots ,1} \right\},\] let $F_{x,t}^{(n)} (z)$ be the distribution function at time t of the process obtained by stopping the standard random walk process at the boundaries 0 and 1. (Standard random walk is symmetric and has variance ${1 /{n^2 }}$ at ${{t = 1} /{2n^2 }}$.) We prove that \[ \left| {F_{x,t}^{(n)} (z) - F_{x,t} (z)} \right| \leq \frac{{3.6}}{{n\sqrt{t}}} = 3.6\sqrt 2 \frac{1} {{\sqrt N }},\] where $n \geq 10$, $t \geq \frac{1}{{40}}$ and N is even.The Berry and Esseen estimate can be used to give the rate of convergence of distribution functions of unconfined random walks to the distribution functions of Brownian motion. Our work extends this result.

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