Let $W:[0,1]\rightarrow\mathbb R$ be standard Brownian motion with $W(0)=0$.


Let $F\_n$ denote the collection of all the $2^n$ many piecewise linear continuous functions $f:[0,1]\rightarrow\mathbb R$ such that $f(0)=0$ and $f$ is linear with slope $\pm \sqrt{n}$ on the intervals $[\frac in,\frac{i+1}n]$ for $0\le i<n$. 


Let $\psi\_n$ denote a uniformly randomly chosen element of $F\_n$, i.e., $\mathbb P(\psi\_n=f)=2^{-n}$ for each $f\in F\_n$.


Let $\phi\_n$ denote a uniformly randomly chosen element of
$$
\text{arg min}\_{f\in F\_n}\left(\sup\_{0\le x\le 1}|W(x)-f(x)|\right).
$$
In other words, $\phi\_n$ is an element of $F\_n$ that minimizes the sup-norm distance to $W$. More simply, we can say that $\phi\_n$ is a *nearest walk* to Brownian motion.


*Question:* Do $\phi\_n$ and $\psi\_n$ have the same distribution? In other words, is $\mathbb P(\phi\_n=f)=2^{-n}$ for all $f\in F\_n$? I am mostly interested in the case $n\rightarrow\infty$, but will accept a rigorous answer for $n=2$.


*Motivation*: Donsker's Theorem says that $\psi\_n$ converges weakly to $W$, whereas it is clear that almost surely, $\phi\_n$ converges to $W$ uniformly. 


EDIT: Here is a [follow-up question](https://mathoverflow.net/questions/38481/is-the-nearest-walk-to-brownian-motion-approximately-uniform).



Is the nearest walk to Brownian motion uniform?
My understanding of the question is this:


There is a function $V\_n$ that maps the $L^\infty([0,1])$ to the set $F\_n$ of $2^n$ piecewise linear functions as defined. $V\_n$ gives the Voronoi subdivision, taking each point of the big set to the nearest neighbor of the smaller set.


I believe you're asking whether $V\_n$ pushes Brownian measure on $L^\infty$ to uniform measure(?)


This doees not seem plausible, starting with $F\_2$, because unlike random walks and Brownian motion, this process has a memory effect. 


Think of the provisional grouping of paths into an upper group and a lower group, based on behavior in the interval $[0,1/2]$. The two distributions of positions at time 1/2 are skewed, because of how Gaussian tails decay rapidly. The provisionally upper group has mean $\sqrt 2/2$, but more of them are less than $\sqrt 2/2$ than greater than $\sqrt 2/2$. This means that those that start provisionally up have less than probability .5 to continue upward, more than probability .5 to go down or to switch to down-up, and low probability of switching to down-down. In other words, I believe the Voronoi measure is biased toward paths that end at 0, rather than at $\pm \sqrt 2$. 


It should be possible for someone to compute the exact distribution in this case.


