This is a follow-up to [an earlier MO question](http://mathoverflow.net/questions/38451/is-the-nearest-walk-to-brownian-motion-uniform).


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 approximately the same distribution? Specifically, is it true that 
$$ \lim_{n\rightarrow\infty}\ \max_{f\in F_n}\ \left|\mathbb P(\phi_n=f)-\mathbb P(\psi_n=f)\right|\cdot 2^n=0\ ?
$$