Dyadic Brownian bridge increment tree

Assume the Axiom of Choice as in the finite-dimensional Gaussian and Brownian interfaces. Let $m\ge 1$ be a finite integer. A node represents an interval $[u,u+h]$, where $h>0$, and has an already sampled parent increment $\mathrm{\Delta }W\in {\mathbb{R}}^m$. Draw a fresh vector $Z\sim N_m(0,(h\mathrm{/}4)I_m)$ independently of the parent and of every innovation used at another node. Store the two child increments $\mathrm{\Delta }{W}_L=\frac{1}{2}\mathrm{\Delta }W+Z,\mathrm{\Delta }{W}_R=\frac{1}{2}\mathrm{\Delta }W-Z\mathrm{.}$

For a finite ordered list of $q\in {\mathbb{N}}_0$ split requests, each naming an existing leaf, apply the rule once to each requested leaf and retain every parent. Process a rejected trial's left child before its right child; if a child is later refined, draw a new independent innovation for it. Each request replaces one leaf by two children, so the result is a finite rooted tree with the sampled increments, interval lengths, and left-to-right order. Never discard a sampled parent and redraw a replacement for the same interval. The procedure assigns increments at dyadic tree nodes; it does not by itself define values at nondyadic times or prove that the node-selection rule is predictable. In an adaptive run, a finite work budget supplies a finite list of requests.

Conditional on a fixed parent, the child sum is fixed and their conditional cross-covariance is $-(h\mathrm{/}4)I_m$, so they are not conditionally independent. The exact unconditional Brownian increment laws at fixed dyadic depth are proved separately on the A page; this procedure makes no law claim for a data-dependent set of requested leaves.